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ee22d8a1fba41847184f4be14a53f4b59f06e7b07bc89a5758d090c78d8b61cd
Argentina
Cono Sur Mathematical Olympiad
2,023
null
CS.5.
Let $ABC$ be an acute triangle. Denote by $D, E, F$ the midpoints of sides $BC, CA, AB$ respectively. The circle with diameter $AB$ intersects lines $AB$ and $AC$ again at $P$ and $Q$ respectively. The line through $P$ parallel to $BC$ meets line $DE$ at $R$, the line through $Q$ parallel to $BC$ meets line $DF$ at $S$...
[ "Since $D$ and $E$ are midpoints we know that $DE \\parallel AB$, and by definition $PR \\parallel BC$, hence $PRDB$ is a parallelogram. Analogously, $QSDC$ is a parallelogram. Thus $PR = BD = DC = SQ$.\n\n![](images/ARG_Other_2024_p31_data_6f1de77da3.png)\n\nSince $AD$ is a diameter, we know that $AB \\perp PD$ an...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Distance chasing" ] ]
[ "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Miscellaneous > Distance chasing" ]
English
ARG_2024
Argentina
true
1
In an acute triangle, certain points are constructed using midpoints, a circle based on a diameter, and lines parallel to the base. Two circumcircles are then drawn from these constructions, and their second intersection is shown to lie on the line joining the other intersection points while also bisecting that segment...
[ "Use mid-segment parallels to form parallelograms and deduce equal lengths", "Apply Thales-type right-angle properties from diameters to infer perpendiculars", "Identify diameters of circumcircles via right angles in cyclic quadrilaterals", "Leverage equal circle diameters to obtain an isosceles configuration...
null
proof only
0.79
Let $ABC$ be an acute triangle. Denote by $D, E, F$ the midpoints of sides $BC, CA, AB$ respectively. The circle with diameter $AB$ intersects lines $AB$ and $AC$ again at $P$ and $Q$ respectively. The line through $P$ parallel to $BC$ meets line $DE$ at $R$, the line through $Q$ parallel to $BC$ meets line $DF$ at $S$...
45574682689791b27dd43b1c4775f990929979f9455ed715de0fd2dad6a2896e
Asia Pacific Mathematics Olympiad (APMO)
APMO
2,024
null
1
Let $ABC$ be an acute triangle. Let $D$ be a point on side $AB$ and $E$ be a point on side $AC$ such that lines $BC$ and $DE$ are parallel. Let $X$ be an interior point of $BCED$. Suppose rays $DX$ and $EX$ meet side $BC$ at points $P$ and $Q$, respectively, such that both $P$ and $Q$ lie between $B$ and $C$. Suppose t...
[ "![](https://cdn.mathpix.com/cropped/2025_11_20_0a2d7e826a69e1356aa9g-01.jpg?height=700&width=992&top_left_y=691&top_left_x=502)\nLet $\\ell$ be the radical axis of circles $BQX$ and $CPX$. Since $X$ and $Y$ are on $\\ell$, it is sufficient to show that $A$ is on $\\ell$. Let line $AX$ intersect segments $BC$ and $...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Circles", "Radical axis theorem" ], [ "Topics", "Geometry", "Plane Geometry", "Transformations", "Homothety" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ ...
[ "Geometry > Plane Geometry > Circles > Radical axis theorem", "Geometry > Plane Geometry > Transformations > Homothety", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
English
apmo2024_sol
Asia Pacific Mathematics Olympiad (APMO)
false
0
Given a triangle with a segment inside parallel to one side and an interior point, two rays from that point meet the base side. The circumcircles through these ray intersections and the point meet again at another point. Show that the vertex opposite the base, the interior point, and this second intersection point all ...
[ "Use the radical axis of the two circumcircles to reduce collinearity to a power-of-a-point equality", "Exploit parallel lines to get segment ratios and equal products that place a constructed point on the radical axis", "Apply power of a point via intersections of the circles with the sides to show the vertex ...
null
proof only
0.93
Let $ABC$ be an acute triangle. Let $D$ be a point on side $AB$ and $E$ be a point on side $AC$ such that lines $BC$ and $DE$ are parallel. Let $X$ be an interior point of $BCED$. Suppose rays $DX$ and $EX$ meet side $BC$ at points $P$ and $Q$, respectively, such that both $P$ and $Q$ lie between $B$ and $C$. Suppose t...
736afae63f32703ce33ebebc7dd53c974738f87002a0531ed3666c61c9d7a8c3
Austria
Austrian Mathematical Olympiad
2,024
Junior Regional Competition
2
Let $ABCD$ be a trapezoid with parallel sides $AB$ and $CD$, with $\angle BAD = 90^\circ$ and with $AB + CD = BC$. Furthermore, let $M$ be the mid-point of $AD$. Prove that $\angle CMB = 90^\circ$.
[ "We reflect the points $B$ and $C$ in $M$ and obtain the points $E$ and $F$, respectively. We clearly have $EC = BF = AB + AF = AB + CD = BC = EF$, therefore, the quadrilateral $BCEF$ is a rhombus. Since the diagonals in a rhombus are orthogonal, we get $BE \\perp CF$ and we obtain $\\angle BMC = 90^\\circ$ as desi...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Quadrilaterals with perpendicular diagonals" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Constructions and loc...
[ "Geometry > Plane Geometry > Quadrilaterals", "Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals", "Geometry > Plane Geometry > Miscellaneous > Constructions and loci" ]
English
AUT_2024
Austria
true
1
In a trapezoid with one right angle and where the sum of the parallel sides equals the length of the non-parallel side, show that the angle at the midpoint of one leg formed by connecting this midpoint to the two non-parallel vertices is a right angle.
[ "Reflect points across the midpoint of a segment to create symmetric counterparts", "Use the given length condition to show a constructed quadrilateral is a rhombus", "Exploit the property that diagonals of a rhombus are perpendicular and bisect each other", "Identify the midpoint as the intersection point of...
null
proof only
0.83
Let $ABCD$ be a trapezoid with parallel sides $AB$ and $CD$, with $\angle BAD = 90^\circ$ and with $AB + CD = BC$. Furthermore, let $M$ be the mid-point of $AD$. Prove that $\angle CMB = 90^\circ$.
8bfc7dde1274ce124a0760afa80ea51833e1be1dc0677f30bf672c15eec42cbf
Balkan Mathematical Olympiad
Balkan Mathematical Olympiad Shortlisted Problems
2,025
Combinatorics
C2
A graph is *good* if its edges can be colored with 2 colors so that no cycle has two consecutive edges of the same color. What is the maximum number of edges in a *good* graph with 1000 vertices?
[ "We will prove the answer to be $4 \\cdot 333 = 1332$.\nFirst we prove that a *good* graph with $n$ vertices has at most $\\frac{4(n-1)}{3}$ edges.\n\n*Claim 1.* If we have 3 paths going from vertex $A$ to vertex $B$, then 2 of them have a common vertex different from $A$, $B$.\n*Proof.* Suppose not. By the Pigeonh...
[]
[ [ "Topics", "Discrete Mathematics", "Graph Theory" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Coloring schemes, extremal arguments" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Pigeonhole principle" ], [ "Topics", "Discrete Ma...
[ "Discrete Mathematics > Graph Theory", "Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments", "Discrete Mathematics > Combinatorics > Pigeonhole principle", "Discrete Mathematics > Combinatorics > Induction / smoothing" ]
English
BMO_2025
Balkan Mathematical Olympiad
true
1
Determine the largest number of edges possible in a graph with one thousand vertices such that its edges can be colored with two colors in a way that around every cycle no two consecutive edges have the same color.
[ "Edge 2-coloring forces cycles to alternate colors, hence all cycles are even and of length at least four", "Pigeonhole principle on the first edge color of multiple paths to show two paths must intersect internally", "Deduce no edge can belong to more than one cycle; cycles are edge-disjoint", "Use a spannin...
1332
proof and answer
0.93
A graph is *good* if its edges can be colored with 2 colors so that no cycle has two consecutive edges of the same color. What is the maximum number of edges in a *good* graph with 1000 vertices?
a9bf638e76aef4d97ddeb7aa76a1d6907e1b35c052ff8d4fe0676a66dc42a72e
Baltic Way
Baltic Way 2023 Shortlist
2,023
Combinatorics
C 1
Let $n$ be a positive integer. Each cell of an $n \times n$ table is coloured in one of $k$ colours where every colour is used at least once. Two colours $A$ and $B$ are said to touch each other, if there exists a cell coloured in $A$ sharing a side with a cell coloured in $B$. The table is coloured in such a way that ...
[ "$k = 2n-1$ when $n \\neq 2$ and $k = 4$ when $n = 2$.\n$k = 2n - 1$ is possible by colouring diagonally as shown in the figure below and when $n = 2$, $k = 4$ is possible by colouring each cell in a unique colour.\n![](images/BW23_Shortlist_2023-11-01_p37_data_39b900ef38.png)\n\nWe consider the graph, where each n...
[]
[ [ "Topics", "Discrete Mathematics", "Combinatorics", "Coloring schemes, extremal arguments" ], [ "Topics", "Discrete Mathematics", "Graph Theory" ] ]
[ "Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments", "Discrete Mathematics > Graph Theory" ]
English
BWS_2023
Baltic Way
true
2
Color the cells of a square grid so that each color type touches at most two other color types, and determine the largest number of distinct colors that can be used.
[ "Model colors as vertices of a graph with edges for touching colors; degree at most two implies the graph is a path or a cycle", "Use connectivity and graph diameter bounds derived from grid distances from the center (or central block) to limit the number of vertices", "Handle odd grid sizes via a single centra...
k = 2n − 1 for n ≠ 2, and k = 4 for n = 2
proof and answer
0.86
Let $n$ be a positive integer. Each cell of an $n \times n$ table is coloured in one of $k$ colours where every colour is used at least once. Two colours $A$ and $B$ are said to touch each other, if there exists a cell coloured in $A$ sharing a side with a cell coloured in $B$. The table is coloured in such a way that ...
f14af4d8643741f68196430cf27ab2692a4dec47cbb2a83a3eea993f37ae71ff
Belarus
SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO
2,024
Fourth Selection Test
15
Inside an isosceles triangle $ABC$ ($AB = BC$), a point $D$ is chosen so that $\angle ADC = 150^\circ$. On the segment $CD$, a point $E$ is chosen so that $AE = AB$. Prove that if $\angle BAE + \angle CBE = 60^\circ$, then $\angle BDC + \angle EAC = 90^\circ$.
[ "Let us rotate $\\triangle AED$ around the point $A$ so that $E \\to B$, $D \\to T$, and reflect $\\triangle ABT$ symmetrically with respect to $BT$ ($A \\to F$). Since $\\angle ADE = 150^\\circ$, then the triangle $ATF$ is equilateral. Let us prove that $FD = DC$. Denote $\\angle EAB = \\alpha$ and $\\angle AED = ...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles" ], [ "Topics", "Geometry", "Plane Geometry", "Transformations", "Rotation" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ], [ "Topics", "Geometry", "P...
[ "Geometry > Plane Geometry > Triangles", "Geometry > Plane Geometry > Transformations > Rotation", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Miscellaneous > Constructions and loci" ]
English
BLR_2024
Belarus
true
1
In an isosceles triangle, points are selected inside with one angle condition. Show that if the sum of two specified angles equals sixty degrees, then another sum of angles equals ninety degrees.
[ "Rotation around a point to map one chosen point to another and track corresponding images", "Reflection across a line to construct an equilateral triangle configuration", "Angle chasing to establish equal angles and isosceles relationships", "Using perpendicular bisector properties via equal distances to ded...
null
proof only
0.87
Inside an isosceles triangle $ABC$ ($AB = BC$), a point $D$ is chosen so that $\angle ADC = 150^\circ$. On the segment $CD$, a point $E$ is chosen so that $AE = AB$. Prove that if $\angle BAE + \angle CBE = 60^\circ$, then $\angle BDC + \angle EAC = 90^\circ$.
be509de9cf64fff50092bc5d5a46fd8af60e5d701e5fd7909ced068597c13ab2
Benelux Mathematical Olympiad
17th Benelux Mathematical Olympiad
2,025
null
Problem 3
Problem: Let $ABC$ be a triangle with incentre $I$ and circumcircle $\Omega$. Let $D, E, F$ be the midpoints of the arcs $\overline{BC}, \overline{CA}, \overline{AB}$ of $\Omega$ not containing $A, B, C$, respectively. Let $D'$ be the point of $\Omega$ diametrically opposite to $D$. Show that $I, D'$, and the midpoint ...
[ "Solution:\n![](attached_image_solution_1.png)\nBy definition of $D$, $E$ and $F$, we know that $ID, IE$ and $IF$ are the angle bisectors of $ABC$. Using angles in $\\Omega$, we can compute \n$$\n\\overline{EFI} = \\overline{EFC} = \\overline{EBC} = \\frac{\\overline{ABC}}{2} = 90^{\\circ} - \\frac{\\overline{ACB}...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Transformations", "Homothety" ], [ "Topics", "Geometry", "Plane Geometr...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Transformations > Homothety", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
null
Benelux_MO__md__Benelux_en-olympiad_en-bxmo-problems-2025-zz
Benelux Mathematical Olympiad
false
0
In a triangle, consider the incenter and the circumcircle. Take the midpoints of the arcs opposite the vertices, then the point on the circle opposite one of these midpoints and the midpoint of the segment connecting the other two arc midpoints. Show that these three points lie on a single straight line.
[ "Use properties of arc midpoints and angle bisectors to relate angles on the circumcircle", "Establish parallelism (D'E parallel FI and D'F parallel EI) to form a parallelogram ED'FI", "Diagonals of a parallelogram bisect each other, implying the midpoint of EF lies on the line through I and D'", "Interpret t...
null
proof only
0.9
Problem: Let $ABC$ be a triangle with incentre $I$ and circumcircle $\Omega$. Let $D, E, F$ be the midpoints of the arcs $\overline{BC}, \overline{CA}, \overline{AB}$ of $\Omega$ not containing $A, B, C$, respectively. Let $D'$ be the point of $\Omega$ diametrically opposite to $D$. Show that $I, D'$, and the midpoint ...
d1cd64c1995dd73742e35b0cc65f8dc1c136fa7224b65a47d213fc585ecffb95
Brazil
Al doilea baraj de selecție pentru OBMJ
2,022
null
Problema 3.
Problem: Se consideră o rețea formată din 49 de puncte, ce reprezintă vârfurile a 36 de pătrate de latură 1 în care este descompus un pătrat de latură 6. Spunem că un pătrat cu vârfurile în punctele rețelei este bun, dacă laturile și diagonalele sale nu sunt pe laturile pătratelor rețelei. a) Aflați numărul de pătrate ...
[ "Solution:\na) Spunem că un pătrat este normal, dacă are vârfurile în punctele rețelei și laturile sale se află pe drepte ale rețelei paralele cu laturile pătratului $6 \\times 6$, sau pe laturile pătratului $6 \\times 6$. Orice pătrat bun are vârfurile pe laturile unui pătrat normal, a cărui lungime a laturii poat...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Inscribed/circumscribed quadrilaterals" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Enumeration with symmetry" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Dist...
[ "Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals", "Discrete Mathematics > Combinatorics > Enumeration with symmetry", "Geometry > Plane Geometry > Miscellaneous > Distance chasing", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
null
Romania_Olympiad__md__ro-19-Olimpiada Nationala de Matematica 2022 Al doilea baraj de selectie juniori pentru OBMJ-baraj2_juniori
Brazil
false
0
On a lattice of points formed by subdividing a large square into unit squares, count all tilted squares whose sides and diagonals do not lie on the grid lines, and show there exist two such squares, disjoint and of different sizes, whose closest points are separated by the square root of five divided by five.
[ "Reduce to axis-aligned host squares (normal squares) within the grid; any good square has vertices on the sides of such a host square with side 3, 4, 5, or 6", "Parametrize good squares by integer offsets along adjacent sides of the host square: two positive integers summing to the host side length and unequal, ...
a) 70; b) sqrt(5)/5
proof and answer
0.86
Problem: Se consideră o rețea formată din 49 de puncte, ce reprezintă vârfurile a 36 de pătrate de latură 1 în care este descompus un pătrat de latură 6. Spunem că un pătrat cu vârfurile în punctele rețelei este bun, dacă laturile și diagonalele sale nu sunt pe laturile pătratelor rețelei. a) Aflați numărul de pătrate ...
56ee6e609bf5236044bed55d57c11370429ac244278871e4af719d888425e87f
Bulgaria
TST for BMO
2,024
Day 2, problem 4
6.6
Let $n$ be a natural number. King Arthur has invited $2^n - 1$ knights to an audience in Camelot. Merlin the Magician arranged the knights in a list numbered from $1$ to $2^n - 1$. It turned out that any two knights with numbers $a, b, a < b$ are friends if and only if $0 \le b - 2a \le 1$. The king chose a natural num...
[ "Let us construct a graph $T$ with vertex-set which is the set of all knights, numbered as in the first list. Two vertices are adjacent if the corresponding knights are friends. It can be seen that $T$ is a fully balanced binary tree - see fig. 1. We label each vertex with the knight's number in the first list. Let...
[]
[ [ "Topics", "Discrete Mathematics", "Graph Theory" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Coloring schemes, extremal arguments" ], [ "Topics", "Discrete Mathematics", "Algorithms" ] ]
[ "Discrete Mathematics > Graph Theory", "Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments", "Discrete Mathematics > Algorithms" ]
English
BGR_2024
Bulgaria
true
1
You must reorder a list of knights so that for each knight, all of their later friends appear not too far ahead in the list, and prove that the smallest such allowed forward distance grows on the order of an exponential divided by a linear factor, with explicit absolute constant bounds.
[ "Model the friendship relation as edges of a graph and observe it forms a complete balanced binary tree on the knights", "Use the tree’s diameter (longest path length) to derive a lower bound on the window size by summing gaps between consecutive vertices along a path from first to last in the ordering", "Desig...
null
proof only
0.86
Let $n$ be a natural number. King Arthur has invited $2^n - 1$ knights to an audience in Camelot. Merlin the Magician arranged the knights in a list numbered from $1$ to $2^n - 1$. It turned out that any two knights with numbers $a, b, a < b$ are friends if and only if $0 \le b - 2a \le 1$. The king chose a natural num...
d2d9fbfe2e56df19a24f1a27f1b7eafdf8b83b38eb4801054c3293206afe9073
Canada
Canadian Mathematical Olympiad
2,025
null
P5.
Problem: A rectangle $R$ is divided into a set $S$ of finitely many smaller rectangles with sides parallel to the sides of $R$ such that no three rectangles in $S$ share a common corner. An ant is initially located at the bottom-left corner of $R$. In one operation, we can choose a rectangle $r \in S$ such that the ant...
[ "Solution:\nConsider the following version of the problem:\nA rectangle $R$ is divided into a set $S$ of finitely many smaller rectangles such that no three rectangles in $S$ share a common corner. For each $r \\in S$, draw two non-intersecting arcs inside $r$, connecting the pairs of adjacent corners of $r$ (there...
[]
[ [ "Topics", "Discrete Mathematics", "Combinatorics", "Invariants / monovariants" ], [ "Topics", "Geometry", "Plane Geometry", "Combinatorial Geometry" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Constructions and loci" ] ]
[ "Discrete Mathematics > Combinatorics > Invariants / monovariants", "Geometry > Plane Geometry > Combinatorial Geometry", "Geometry > Plane Geometry > Miscellaneous > Constructions and loci" ]
null
CANADA_MO__md__en-CMO2025-solutions
Canada
false
0
In a rectangle subdivided into smaller rectangles sharing corners pairwise only, an ant moves along corners of chosen rectangles, starting at the bottom left and aiming for the top right. Show that reaching the target requires selecting some rectangle more than once.
[ "Model the configuration with an undirected graph whose vertices are rectangle corners and edges are arcs connecting adjacent corners within each small rectangle; interior vertices have degree two and the outer corners have degree one, so the graph decomposes into disjoint paths and cycles.", "Planarity argument:...
null
proof only
0.84
Problem: A rectangle $R$ is divided into a set $S$ of finitely many smaller rectangles with sides parallel to the sides of $R$ such that no three rectangles in $S$ share a common corner. An ant is initially located at the bottom-left corner of $R$. In one operation, we can choose a rectangle $r \in S$ such that the ant...
05f740d3f5fb9ac9f739db3a02b3cbbf22023196b78172855cf8f7746bda9404
China
China-TST-2025A
2,025
Test 1 · Day 1
Problem 1
Prove that the real-coefficient polynomial in $x, y, z$, $$ x^{4}(x - y)(x - z) + y^{4}(y - z)(y - x) + z^{4}(z - x)(z - y), $$ cannot be expressed as a finite sum of squares of real-coefficient polynomials in $x, y, z$.
[ "Denote the given polynomial by $F(x, y, z)$. By contradiction, assume there exist real-coefficient polynomials $f_1(x, y, z)$, $f_2(x, y, z)$, ..., $f_m(x, y, z)$ satisfying\n$$\nF(x, y, z) = \\sum_{i=1}^{m} f_i(x, y, z)^2. \\qquad (1)\n$$\nFirst, all $f_i$ must have degree at most 3. If some $f_i$ had degree $d >...
[]
[ [ "Topics", "Algebra", "Algebraic Expressions", "Polynomials", "Polynomial operations" ] ]
[ "Algebra > Algebraic Expressions > Polynomials > Polynomial operations" ]
English
CHN_2025
China
true
1
Show that a specific sixth degree polynomial in three real variables cannot be represented as a finite sum of squares of real polynomials.
[ "Argue by contradiction via a sum of squares decomposition and reduce to homogeneous components", "Degree bound: only cubic forms can appear in the summands since the target polynomial has degree six", "Specialization by setting variables equal to force divisibility by linear factors like the difference of two ...
null
proof only
0.84
Prove that the real-coefficient polynomial in $x, y, z$, $$ x^{4}(x - y)(x - z) + y^{4}(y - z)(y - x) + z^{4}(z - x)(z - y), $$ cannot be expressed as a finite sum of squares of real-coefficient polynomials in $x, y, z$.
a36998fc2e8f18c168a6dbffeb225f57b5e8685a62b9c409a2d4b31e5be35b01
Croatia
Croatian Mathematical Olympiad
2,019
Day 2
G2
On the side $\overline{AB}$ of the cyclic quadrilateral $ABCD$ there is a point $X$ such that the diagonals $\overline{BD}$ and $\overline{AC}$ bisect the segments $\overline{CX}$ and $\overline{DX}$, respectively. Find the smallest possible value of $|AB| : |CD|$. (Belarus)
[ "Let $M$ and $N$ be the midpoints of the segments $\\overline{CX}$ and $\\overline{DX}$, respectively. Also, denote $\\alpha = \\angle BAC = \\angle BDC$ and $\\beta = \\angle CAD = \\angle CBD$.\n\n![](images/HRV_ABooklet_2019_p35_data_677614dbdf.png)\n\nSince the triangles $AXN$ and $ADN$ have equal areas, we hav...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle trigonometry" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ],...
[ "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Triangles > Triangle trigonometry", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean" ]
English
HRV_2019
Croatia
true
1
In a cyclic quadrilateral, a point is chosen on one side so that each diagonal passes through the midpoint of a segment from that point to a vertex. Determine the smallest possible ratio of the length of that side to the opposite side.
[ "Use that a segment from a vertex to the midpoint of the opposite side bisects the area of a triangle (median area property)", "Express triangle areas via two sides and the sine of the included angle to derive length ratios", "Relate sines of angles in a cyclic quadrilateral to corresponding chord lengths (sin ...
2
proof and answer
0.83
On the side $\overline{AB}$ of the cyclic quadrilateral $ABCD$ there is a point $X$ such that the diagonals $\overline{BD}$ and $\overline{AC}$ bisect the segments $\overline{CX}$ and $\overline{DX}$, respectively. Find the smallest possible value of $|AB| : |CD|$. (Belarus)
8b3ce64b0961e6fb283da137e4c8be8dce875ee15caea79d01493fc6cd169c84
Czech Republic
First Round of the 73rd Czech and Slovak Mathematical Olympiad (take-home part)
2,024
null
1
Ten boys and ten girls met at a party. Assume that every girl likes exactly $k$ boys and every boy likes exactly $k$ girls. Is it always possible to find a couple where both the partners like each other? Solve the problem for: a) $k = 5$, b) $k = 6$.
[ "a) For $k=5$, it may happen that there are no such couples, with one counterexample given as follows. Split the boys into two disjoint quintuples $A, B$ and the girls into two disjoint quintuples $C, D$. Consider the configuration where every boy from $A$ likes all the girls in $C$, every boy in $B$ likes all the ...
[]
[ [ "Topics", "Discrete Mathematics", "Combinatorics", "Pigeonhole principle" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Counting two ways" ] ]
[ "Discrete Mathematics > Combinatorics > Pigeonhole principle", "Discrete Mathematics > Combinatorics > Counting two ways" ]
English
CZE_2024
Czech Republic
true
1
There are equal numbers of boys and girls at a party, and each person likes exactly a fixed number of people of the opposite group. For two given values of this number, determine whether one can always guarantee at least one mutually liking pair, by either building a counterexample or proving a general overlap.
[ "Explicit bipartite construction to avoid reciprocal edges (partition each side into two equal groups with cross-like directions)", "Double counting the number of likes from degrees: total likes equals number of people times k", "Pigeonhole/inclusion–exclusion argument: two subsets of size sixty within one hund...
a) No. b) Yes.
proof and answer
0.93
Ten boys and ten girls met at a party. Assume that every girl likes exactly $k$ boys and every boy likes exactly $k$ girls. Is it always possible to find a couple where both the partners like each other? Solve the problem for: a) $k = 5$, b) $k = 6$.
b769cb0193756f0aa3eeea4fb5901a29513b96a25d776eb05e9f75fb7b07814c
Czech-Polish-Slovak Mathematical Match
CAPS Match 2025
2,025
Second day – 18 June 2025
4
The plane was divided by vertical and horizontal lines into unit squares. Determine whether it is possible to write integers into cells of this infinite grid so that: (i) every cell contains exactly one integer (ii) every integer appears exactly once (iii) for every two cells $A$ and $B$ sharing exactly one vertex, if ...
[ "Yes, this is possible. Consider the spiral depicted below and write consecutive integers along the spiral:\n![](images/2025CZPS_p3_data_8c68622425.png)\n\nWe claim that this works. Consider any two cells $A$ and $B$ sharing exactly one vertex. Consider the $2 \\times 2$ square containing $A$ and $B$. If the $2 \\t...
[]
[ [ "Topics", "Discrete Mathematics", "Combinatorics", "Recursion, bijection" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Coloring schemes, extremal arguments" ] ]
[ "Discrete Mathematics > Combinatorics > Recursion, bijection", "Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments" ]
null
CZA_2025
Czech-Polish-Slovak Mathematical Match
true
2
Decide whether you can place all whole numbers exactly once on an infinite square grid so that for any pair of diagonally adjacent cells, one of the two cells that touch both of them by a side contains a number lying between the two.
[ "Construct a bijection between the integers and grid cells by listing numbers along a square spiral", "Local case analysis on each 2 by 2 block, distinguishing whether the spiral turns at a corner or passes straight", "Use monotonicity of consecutive integers along the spiral to guarantee an intermediate value ...
Yes, it is possible.
proof and answer
0.86
The plane was divided by vertical and horizontal lines into unit squares. Determine whether it is possible to write integers into cells of this infinite grid so that: (i) every cell contains exactly one integer (ii) every integer appears exactly once (iii) for every two cells $A$ and $B$ sharing exactly one vertex, if ...
d099a086fc815c6d1c9277805bf26c12933ecbae297b9b5f36b1e814ef75083d
Estonia
Estonian Mathematical Olympiad
2,025
Selected Problems from Open Contests
O2
Let $ABCD$ be a rectangle. The bisector of the angle $CAD$ meets the side $CD$ at point $L$. Let $M$ be the midpoint of the line segment $AL$. The line $DM$ meets lines $AC$ and $AB$ at points $E$ and $F$, respectively. Given that line segments $AE$ and $AF$ are equal, prove that $ABCD$ is a square.
[ "Let $\\angle DAL = \\angle LAC = \\alpha$. Then $\\angle FAE = 90^\\circ - 2\\alpha$ (Fig. 1).\n\nAs $AE = AF$, we obtain $\\angle AFE = \\frac{180^\\circ - (90^\\circ - 2\\alpha)}{2} = 45^\\circ + \\alpha$.\n\nBut as $M$ bisects the hypotenuse $AL$ of the right triangle $ALD$, it follows that $M$ is the circumcen...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ], [ "Topics", "Geometry", ...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Quadrilaterals" ]
English
EST_2025
Estonia
true
1
In a rectangle, draw from one corner the bisector of the angle formed by the diagonal and a side, meeting the opposite side. Take the midpoint of the segment from that corner to the meeting point, and join it to the opposite corner; this line intersects the diagonal and an adjacent side. If the two distances from the o...
[ "Midpoint of hypotenuse is the circumcenter in a right triangle, giving equal distances to the endpoints and key angle equalities", "Isosceles triangle angle relationships from the equality of two segments", "Properties of rectangles: diagonals are equal and bisect each other", "Angle chasing to deduce a fort...
null
proof only
0.86
Let $ABCD$ be a rectangle. The bisector of the angle $CAD$ meets the side $CD$ at point $L$. Let $M$ be the midpoint of the line segment $AL$. The line $DM$ meets lines $AC$ and $AB$ at points $E$ and $F$, respectively. Given that line segments $AE$ and $AF$ are equal, prove that $ABCD$ is a square.
b50ee206d17ec64c0439afccb2855802b8868bebd8c63c564e6a6868778abca6
European Girls' Mathematical Olympiad (EGMO)
EGMO
2,025
Day 2
P5.
Problem: Let $n > 1$ be an integer. In a configuration of an $n \times n$ board, each of the $n^{2}$ cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one sq...
[ "Solution:\n\nWe will show that the maximum number of good cells over all possible starting configurations is\n$$\n\\frac{n^{2}}{4} \\quad \\text{if } n \\text{ is even and}\n$$\n$$\n0 \\quad \\text{if } n \\text{ is odd.}\n$$\n\n## Odd $n$\n\nFirst, we will prove that there are no good cells if $n$ is an odd numbe...
[]
[ [ "Topics", "Discrete Mathematics", "Combinatorics", "Coloring schemes, extremal arguments" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Invariants / monovariants" ] ]
[ "Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments", "Discrete Mathematics > Combinatorics > Invariants / monovariants" ]
null
EGMO__md__en-2025-solutions
European Girls' Mathematical Olympiad (EGMO)
false
0
On a square grid, each cell has an arrow and after every step all arrows rotate a quarter turn counterclockwise. A snail moves by following the arrow in its current cell. A starting cell is called good if the snail visits every cell exactly once without leaving the board and returns to the start. Find, as a function of...
[ "Chessboard parity coloring to rule out a Hamiltonian return tour on odd-sized boards", "Explicit Hamiltonian snake-like cycle covering all cells exactly once", "Pre-rotating arrows by an index to synchronize with global quarter-turn rotations so the snail follows the designed cycle", "Periodicity modulo four...
n^2/4 if n is even; 0 if n is odd
proof and answer
0.94
Problem: Let $n > 1$ be an integer. In a configuration of an $n \times n$ board, each of the $n^{2}$ cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one sq...
b41d0e5397c2706740f2845d58424a40610a5926f24a80c75e57a17cb28b0b9e
France
PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri
2,024
Juniors
Exercice 1.
Problem: Soit $ABC$ un triangle et $\Omega$ son cercle circonscrit. On note $A'$ le point diamétralement opposé à $A$ dans le cercle $\Omega$. Soit $I$ le centre du cercle inscrit au triangle $ABC$, $E$ et $F$ les points de contact du cercle inscrit avec les côtés $AC$ et $AB$ respectivement. Le cercle circonscrit au ...
[ "Solution:\n\n![](attached_image_solution_1.png)\n\nPuisque $F$ est le point de contact du cercle inscrit avec le côté $[AC]$, l'angle $\\widehat{IFA}$ est droit et le segment $[IA]$ est un diamètre du cercle circonscrit au triangle $AEF$. On en déduit que\n$$\n\\widehat{AXI} = 90^\\circ = \\widehat{AXA'}\n$$\noù o...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Circles", "Tangents" ], [ "Topics", "Geometry", "Plane Geo...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
null
French__md__envois__fr-Corrige-envoi-5-2023-2024
France
false
0
In a triangle with its incircle touching two sides, consider the circle through the vertex and the two touchpoints, and its second intersection with the triangle’s circumcircle. Prove that the antipode of the vertex on the circumcircle, the incenter, and this intersection point lie on a single straight line.
[ "Radius to a point of tangency is perpendicular to the tangent side (incircle tangency property)", "Thales' theorem: an angle subtended by a diameter is a right angle in a circle", "Using equal right angles at a common vertex to deduce that two lines coincide, yielding collinearity", "Intersecting circumcircl...
null
proof only
0.92
Problem: Soit $ABC$ un triangle et $\Omega$ son cercle circonscrit. On note $A'$ le point diamétralement opposé à $A$ dans le cercle $\Omega$. Soit $I$ le centre du cercle inscrit au triangle $ABC$, $E$ et $F$ les points de contact du cercle inscrit avec les côtés $AC$ et $AB$ respectivement. Le cercle circonscrit au ...
9c37f9dfd679abc1943504f6a6364f11865cd29ede0d0bce95400f77410c1a08
Germany
Auswahlwettbewerb zur Internationalen Mathematik-Olympiade 2022
2,022
1. Auswahlklausur
Aufgabe 1
Problem: Der größte gemeinsame Teiler zweier positiver ganzer Zahlen $m$ und $n$ sei mit $\operatorname{ggT}(m, n)$ bezeichnet. Es sei eine unendliche Menge $S$ positiver ganzer Zahlen gegeben, sodass es vier paarweise verschiedene Zahlen $v, w, x, y \in S$ gibt, für die $\operatorname{ggT}(v, w) \neq \operatorname{gg...
[ "Solution:\n\nIm folgenden nennen wir eine dreielementige Teilmenge $\\{s, t, u\\} \\subset S$ ein ausgewogenes Dreieck, falls die Menge $\\{\\mathrm{ggT}(s, t), \\operatorname{ggT}(s, u), \\operatorname{ggT}(t, u)\\}$ genau zwei verschiedene Elemente hat. Es ist zu zeigen, dass es ein ausgewogenes Dreieck gibt.\n\...
[]
[ [ "Topics", "Number Theory", "Divisibility / Factorization", "Greatest common divisors (gcd)" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Pigeonhole principle" ] ]
[ "Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)", "Discrete Mathematics > Combinatorics > Pigeonhole principle" ]
null
Germany_TST__md__de-2022-2022_IMO_Auswahlklausuren_Lsg_HP
Germany
false
0
From an infinite set of positive integers in which there exist two pairs with different greatest common divisors, show that there are three distinct numbers such that one of them has the same greatest common divisor with each of the other two, and this value is different from the greatest common divisor of the other tw...
[ "Define a structural target: a triple where two pairwise greatest common divisors coincide and differ from the third", "For a fixed element, the set of possible greatest common divisors with other elements is finite since they must divide the fixed element", "Apply the pigeonhole principle to extract an infinit...
null
proof only
0.91
Problem: Der größte gemeinsame Teiler zweier positiver ganzer Zahlen $m$ und $n$ sei mit $\operatorname{ggT}(m, n)$ bezeichnet. Es sei eine unendliche Menge $S$ positiver ganzer Zahlen gegeben, sodass es vier paarweise verschiedene Zahlen $v, w, x, y \in S$ gibt, für die $\operatorname{ggT}(v, w) \neq \operatorname{gg...
16e818de0cd3ba1f353a9a9fab03ad0d677eae00222adcb2a602f3d0c389cfb6
Greece
Hellenic Mathematical Olympiad
2,024
A. Juniors
Problem 2.
Let $A B \Gamma$ be an acute angled triangle with circumcircle $\omega$. A circle $\gamma$ with center $A$ intersects the arc $AB$ of the circle $\omega$, not containing $\Gamma$, at point $\Delta$ and the arc $A\Gamma$, not containing $B$, at point $E$. We suppose that the point of intersection $K$ of the lines $BE$ a...
[ "From the relationship of a central angle and an inscribed angle that go on the same arc $\\Delta K$ of the circle $\\gamma$, we have:\n$$\n\\angle A\\Delta K = 2 \\cdot \\angle E\\Delta K \\quad (1)\n$$\nAlso we have the equality of inscribed angles\n$$\n\\angle E\\Delta K = \\angle AB\\Delta \\quad (2)\n$$\nFrom ...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ] ]
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
English
GRC_2024
Greece
true
1
From a triangle, a circle centered at one vertex meets specific arcs of the circumcircle at two points. The intersection of the lines through those points lies on the same circle. Prove that the line from the vertex to this intersection is perpendicular to the side opposite that vertex.
[ "Relate central and inscribed angles subtending the same arc in a circle", "Use equality of inscribed angles on a circumcircle that subtend the same arc", "Exploit isosceles triangle properties arising from equal radii of a circle centered at a vertex", "Identify the orthocenter by showing two altitudes are p...
null
proof only
0.86
Let $A B \Gamma$ be an acute angled triangle with circumcircle $\omega$. A circle $\gamma$ with center $A$ intersects the arc $AB$ of the circle $\omega$, not containing $\Gamma$, at point $\Delta$ and the arc $A\Gamma$, not containing $B$, at point $E$. We suppose that the point of intersection $K$ of the lines $BE$ a...
596919c118fe74f4cb81759366f91ac78aea5875b332b5efada59732d4580848
Hong Kong
IMO HK TST
2,023
Test 1
3
A point $P$ lies inside an equilateral triangle $ABC$ such that $AP = 15$ and $BP = 8$. Find the maximum possible value of the sum of areas of triangles $ABP$ and $BCP$.
[ "Rotate $\\triangle PBC$ about point $B$ by $60^\\circ$ in the anticlockwise direction, so that $BA$ is the image of $BC$ after rotation. Let $Q$ be the image of $P$ after rotation. Then\n$$\n[ABP] + [BCP] = [PAQB] = [PQB] + [QPA].\n$$\n\n![](images/1.2_1997-2023_IMO_HK_TST_Solutions_p204_data_51114dfb22.png)\n\nNo...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Transformations", "Rotation" ], [ "Topics", "Geometry", "Plane Geometry", "Geometric Inequalities", "Optimization in geometry" ], [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle trigonomet...
[ "Geometry > Plane Geometry > Transformations > Rotation", "Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry", "Geometry > Plane Geometry > Triangles > Triangle trigonometry", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
null
HKG_TST_2023
Hong Kong
true
1
Inside an equilateral triangle, a point has two given distances to two vertices. Find how to place the point to maximize the combined areas of two triangles formed with this point and adjacent sides.
[ "Rotate the configuration by sixty degrees about a vertex to align sides and relate areas", "Express the sum of areas as the area of a composed quadrilateral formed by the rotation", "Identify an equilateral triangle arising from the rotation to compute a fixed area", "Use the triangle area formula with sine ...
60 + 16√3
proof and answer
0.92
A point $P$ lies inside an equilateral triangle $ABC$ such that $AP = 15$ and $BP = 8$. Find the maximum possible value of the sum of areas of triangles $ABP$ and $BCP$.
7579e8da5e03edb7dbb297b1abbb8c8c05ce2aaf284c8292ba60ded6a226c989
IMO
IMO2024 Shortlisted Problems
2,024
Combinatorics
C4
On a board with $2024$ rows and $2023$ columns, Turbo the snail tries to move from the first row to the last row. On each attempt, he chooses to start on any cell in the first row, then moves one step at a time to an adjacent cell sharing a common side. He wins if he reaches any cell in the last row. However, there are...
[ "First we demonstrate that there is no winning strategy if Turbo has $2$ attempts.\nSuppose that $(2, i)$ is the first cell in the second row that Turbo reaches on his first attempt. There can be a monster in this cell, in which case Turbo must return to the first row immediately, and he cannot have reached any oth...
[]
[ [ "Topics", "Discrete Mathematics", "Combinatorics", "Games / greedy algorithms" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Coloring schemes, extremal arguments" ] ]
[ "Discrete Mathematics > Combinatorics > Games / greedy algorithms", "Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments" ]
English
IMO2024SL
IMO
false
0
A snail repeatedly tries to move from the top to the bottom of a rectangular grid, restarting whenever it steps on a hidden monster. There is one monster in each intermediate row and all monsters are in different columns. Find the minimum number of tries needed to guarantee reaching the bottom and describe a strategy t...
[ "Adversarial lower bound by placing monsters at the first reached cells in consecutive rows to defeat any two-attempt plan", "Sweep the entire second row on the first attempt to learn the monster’s column in that row", "Design complementary descent paths via neighboring columns so that at least one avoids the u...
3
proof and answer
0.87
On a board with $2024$ rows and $2023$ columns, Turbo the snail tries to move from the first row to the last row. On each attempt, he chooses to start on any cell in the first row, then moves one step at a time to an adjacent cell sharing a common side. He wins if he reaches any cell in the last row. However, there are...
733d391ec58ed70a8191c273278764eda1bc465558099c00ac3778cdefd006b7
Ibero-American Mathematical Olympiad
Iberoamerican Mathematical Olympiad
2,003
18th Iberoamerican
B2
Problem: $\mathrm{ABCD}$ is a square. $\mathrm{P}, \mathrm{Q}$ are points on the sides $\mathrm{BC}, \mathrm{CD}$ respectively, distinct from the endpoints such that $\mathrm{BP}=\mathrm{CQ}$. $\mathrm{X}, \mathrm{Y}$ are points on $\mathrm{AP}, \mathrm{AQ}$ respectively. Show that there is a triangle with side length...
[ "Solution:\n\n![](attached_image_solution_1.png)\n\nWe have $DY < BY \\leq BX + XY$ (this is almost obvious, but to prove formally use the cosine formula for $BAY$ and $DAY$ and notice that $\\angle BAY > \\angle DAY$). Similarly, $BX < DX \\leq DY + YX$. So it remains to show that $XY < BX + DY$.\n\nTake $Q'$ on t...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle trigonometry" ], [ "Topics", "Geometry", "Plane Geometry", "Geometric Inequalities", "Triangle inequalities" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasi...
[ "Geometry > Plane Geometry > Triangles > Triangle trigonometry", "Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Miscellaneous > Distance chasing" ]
null
IberoAmerican_MO__md__en-1985-2003-IberoamericanMO
Ibero-American Mathematical Olympiad
false
0
Given a square with chosen points on two adjacent sides and points along two segments from a corner, prove that the distances from one corner to the first point, between the two points, and from the opposite corner to the second point can be the sides of a triangle.
[ "Use triangle inequality conditions to ensure three given lengths can form a triangle", "Compare side sums by applying the law of cosines to relate side lengths to angles", "Construct auxiliary points on extensions to create comparable segments with controlled lengths", "Employ angle comparison via a circle w...
null
proof only
0.86
Problem: $\mathrm{ABCD}$ is a square. $\mathrm{P}, \mathrm{Q}$ are points on the sides $\mathrm{BC}, \mathrm{CD}$ respectively, distinct from the endpoints such that $\mathrm{BP}=\mathrm{CQ}$. $\mathrm{X}, \mathrm{Y}$ are points on $\mathrm{AP}, \mathrm{AQ}$ respectively. Show that there is a triangle with side length...
1b53f98637200a95e870019db7d6d16d04eb5ede7571b18cb59e5a7610d56070
India
IMO TST
2,024
IMO TST 2024 Day 4
3
Let $ABC$ be an acute-angled triangle with $AB < AC$, and let $O, H$ be its circumcentre and orthocentre respectively. Points $Z, Y$ lie on segments $AB, AC$ respectively, such that $$ \angle ZOB = \angle YOC = 90^\circ. $$ The perpendicular line from $H$ to line $YZ$ meets lines $BO$ and $CO$ at $Q, R$ respectively. L...
[ "Define $K$ to be the point on $YZ$ such that $HK \\perp YZ$. Let $A'$ be a point on the circumcircle of $\\triangle ABC$ such that $AA' \\parallel BC$.\n\n**Lemma 1.** $YZ$ is the perpendicular bisector of $HA'$.\n*Proof.* Let $H_B$ denote the reflection of $H$ in $AC$. Then note that $A'H_B \\parallel CO$. This i...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Advanced Configurations", "Miquel point" ], [ "Topics", "Geometry", "Plane Geometry", "Advanced Configurations", "Isogonal/isotomic conjugates, barycentric coordinates" ], [ "Topics", "Geometry", "Plane Geometry",...
[ "Geometry > Plane Geometry > Advanced Configurations > Miquel point", "Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates", "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geomet...
null
IND_2024
India
true
1
In an acute triangle, pick points on two sides so that lines from the circumcenter to these points are perpendicular to the lines from the circumcenter to the opposite vertices. From the orthocenter, drop a perpendicular to the line through the two chosen points and extend this perpendicular to meet the lines from the ...
[ "Use reflections of the orthocenter and perpendicular bisector arguments to show the key symmetry YZ is the perpendicular bisector of a constructed chord through A", "Establish that the foot from H to YZ and the circumcenter define a line perpendicular to the base, yielding cyclic configurations with O", "Prove...
null
proof only
0.74
Let $ABC$ be an acute-angled triangle with $AB < AC$, and let $O, H$ be its circumcentre and orthocentre respectively. Points $Z, Y$ lie on segments $AB, AC$ respectively, such that $$ \angle ZOB = \angle YOC = 90^\circ. $$ The perpendicular line from $H$ to line $YZ$ meets lines $BO$ and $CO$ at $Q, R$ respectively. L...
b1efe77fad266fde61ca0042f86173f9e47a2102a18336081f5698239b9d3525
Iran
Iranian Mathematical Olympiad
2,025
Second Round
4
In the triangle $ABC$ the point $M$ is the midpoint of $AB$, and the point $B'$ is the foot of the altitude from $B$ to $AC$. The circle ($CB'M$) intersects $BC$ again at $D$. The circles ($ABD$) and ($CB'M$) intersect again at $K$. The line parallel to $AB$ passing through $C$ intersects circle ($CB'M$) again at $L$. ...
[ "Since $CL \\parallel AB$ and $A$, $B$, $D$, $K$ are concyclic, we have\n$$\n180^\\circ - \\angle AKD = \\angle ABD = \\angle BCL = \\angle DKL\n$$\nThus, $\\angle AKD + \\angle DKL = 180^\\circ$. This implies that $A$, $K$, $L$ are collinear.\n\n![](images/IRN_booklet2025_p23_data_bdbdc6f267.png)\n\nNotice that $B...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ] ]
[ "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
null
IRN_2025
Iran
true
1
In a triangle, pick the midpoint of one side and the foot of the altitude from another vertex. Form a circle through certain points and define new intersection points with lines and circles. Show that the line through two of these constructed points passes through the midpoint of the segment joining the vertex and the ...
[ "Angle chasing with parallel lines and cyclic quadrilaterals to show three points are collinear", "Use the right-triangle fact that the midpoint of the hypotenuse is equidistant from the vertices to equate base angles", "Exploit the cyclicity of four points to transfer an angle to establish pairs of parallel si...
null
proof only
0.86
In the triangle $ABC$ the point $M$ is the midpoint of $AB$, and the point $B'$ is the foot of the altitude from $B$ to $AC$. The circle ($CB'M$) intersects $BC$ again at $D$. The circles ($ABD$) and ($CB'M$) intersect again at $K$. The line parallel to $AB$ passing through $C$ intersects circle ($CB'M$) again at $L$. ...
aeca9641b28e331526e864cafaf1328ceee572040c9b6b168eb1bd34eb14ab16
Ireland
IRL_ABooklet_2025
2,025
Problems
0
A Sudoku grid is a $9 \times 9$ table in which each row and each column contains each of the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ in some order. In addition, the nine $3 \times 3$ subgrids contain each of the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ exactly once. Suppose the product of all nine numbers on one diagonal is $M$...
[ "Because $2025 = 3^4 \\cdot 5^2$, we have $2025^3 = 3^{12} \\cdot 5^6$ and $2025^4 = 3^{16} \\cdot 5^8$. Each diagonal can have at most three $5$s. Therefore, $MN$ can have at most six factors $5$, and this only happens when the number in the centre of the grid is $5$. Hence $MN$ might be divisible by $2025^3$ but ...
[]
[ [ "Topics", "Number Theory", "Divisibility / Factorization", "Factorization techniques" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Coloring schemes, extremal arguments" ] ]
[ "Number Theory > Divisibility / Factorization > Factorization techniques", "Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments" ]
null
IRL_2025
Ireland
true
2
Show that in a standard Sudoku, the product of the numbers on both diagonals can be arranged to be divisible by the cube of a specific composite number but cannot reach divisibility by its fourth power, using structural limits from the subgrid layout and a concrete construction.
[ "Prime factorization of the target number to reduce divisibility to exponents of primes", "Bounding the number of occurrences of a specific digit on a diagonal using the 3×3 subgrid structure (at most one per subgrid, three subgrids per diagonal)", "Recognizing the center cell lies on both diagonals, maximizing...
null
proof only
0.86
A Sudoku grid is a $9 \times 9$ table in which each row and each column contains each of the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ in some order. In addition, the nine $3 \times 3$ subgrids contain each of the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ exactly once. Suppose the product of all nine numbers on one diagonal is $M$...
e75c1a9122493562daeabe1d7c288f93895abd8f386e42efed2ba85df0aa7174
Italy
Olimpiadi di Matematica
2,024
Problemi a risposta multipla
4.
Problem: Sia $ABC$ un triangolo rettangolo in $C$ di lati $BC=3$ e $AB=12$. Siano $M$ il punto medio di $AB$, e $D$ l'intersezione tra $AC$ e la circonferenza circoscritta a $BCM$. Sia infine $P$ il punto di intersezione tra $BC$ e $MD$. Quanto misura il segmento $PA$? (A) $\frac{28}{5} \sqrt{15}$ (B) $6 \sqrt{15}$ ...
[ "Solution:\n\nLa risposta è (C). Poiché $BCDM$ è ciclico, si ha $\\angle BMD = \\angle BCD = 90^{\\circ}$. I triangoli $ABC$ e $BMP$ sono dunque simili in quanto sono rettangoli e condividono l'angolo in $B$. Si ha quindi che $BM : BC = BP : BA$, cioè $6 : 3 = BP : 12$, da cui $BP = 24$ e $CP = 21$. Infine, anche i...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Distance chasing" ], ...
[ "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Miscellaneous > Distance chasing", "Geometry > Plane Geometry > Miscellaneous > Constructions and loci" ]
null
Italy__md__it-febbraio__it-soluzioni2024
Italy
false
0
In a right triangle with hypotenuse twelve and one leg three, take the midpoint of the hypotenuse and the circle through that midpoint and the two vertices around the right angle. Intersect this circle with the side along the right angle, connect this intersection to the midpoint, and find where this line meets the sho...
[ "Use cyclic quadrilateral angle properties to deduce a right angle at the midpoint-line intersection", "Establish similarity between two right triangles sharing an acute angle at the same vertex", "Apply the perpendicular bisector property through a midpoint to deduce an isosceles triangle", "Compute lengths ...
C
MCQ
0.96
Problem: Sia $ABC$ un triangolo rettangolo in $C$ di lati $BC=3$ e $AB=12$. Siano $M$ il punto medio di $AB$, e $D$ l'intersezione tra $AC$ e la circonferenza circoscritta a $BCM$. Sia infine $P$ il punto di intersezione tra $BC$ e $MD$. Quanto misura il segmento $PA$? (A) $\frac{28}{5} \sqrt{15}$ (B) $6 \sqrt{15}$ ...
825c5394243ec4b11c031495dff39a61fda63034bea2ed5b21f7a62565a0536e
JBMO
Balkan Mathematical Olympiad
2,023
null
Problem 4.
Problem: Let $ABC$ be an acute triangle with circumcenter $O$. Let $D$ be the foot of the altitude from $A$ to $BC$ and let $M$ be the midpoint of $OD$. The points $O_{b}$ and $O_{c}$ are the circumcenters of triangles $AOC$ and $AOB$, respectively. If $AO = AD$, prove that the points $A$, $O_{b}$, $M$ and $O_{c}$ are ...
[ "Solution:\n![](attached_image_solution_1.png)\nNote that $AB = AC$ cannot hold since $AO = AD$ would imply that $O$ is the midpoint of $BC$, which is not possible for an acute triangle. So we may assume without loss of generality that $AB < AC$.\nLet $M_{b}$ and $M_{c}$ be the midpoints of $AC$ and $AB$, respectiv...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Circles", "Radical axis theorem" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ], ...
[ "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Circles > Radical axis theorem", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line,...
null
JBMO__md__en-official__en-jbmo_2023_final_paper_-_with_solutions_1
JBMO
false
0
In an acute triangle, under the condition that the distance from a vertex to the circumcenter equals the distance from that vertex to the foot of its altitude, show that the vertex, the circumcenters of two auxiliary triangles, and the midpoint of a segment related to the circumcenter and altitude foot all lie on a sin...
[ "Use the equality of distances from the vertex to the circumcenter and to the altitude foot to conclude the vertex lies on the perpendicular bisector of the segment joining the circumcenter and the altitude foot; hence the line from the vertex to the midpoint of that segment is perpendicular to it, giving a right a...
null
proof only
0.86
Problem: Let $ABC$ be an acute triangle with circumcenter $O$. Let $D$ be the foot of the altitude from $A$ to $BC$ and let $M$ be the midpoint of $OD$. The points $O_{b}$ and $O_{c}$ are the circumcenters of triangles $AOC$ and $AOB$, respectively. If $AO = AD$, prove that the points $A$, $O_{b}$, $M$ and $O_{c}$ are ...
fef72c333c923280a8362a13dbeedf395abc328c2a55bd206f545c902303c727
Japan
The 35th Japanese Mathematical Olympiad
2,025
FIRST ROUND
1
As shown in the figure, seven regular hexagonal cells form a hexagonal pattern. We write one integer from $1$ to $7$ in each cell without repetition. For any two cells that share an edge, the sum of the integers written in those cells must be at most $10$. How many ways are there to write the integers under these condi...
[ "$72$\n\nSince the sum of the integers in any two adjacent cells must be at most $10$, the only possible integers that can appear in the neighbors of the cell containing $7$ are $1$, $2$, or $3$. Therefore, the cell labeled $7$ cannot be the central cell, and there are exactly $6$ possible cells in which to place t...
[]
[ [ "Topics", "Discrete Mathematics", "Combinatorics", "Coloring schemes, extremal arguments" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Counting two ways" ] ]
[ "Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments", "Discrete Mathematics > Combinatorics > Counting two ways" ]
English
JPN_JMO35
Japan
true
3
Place the numbers one through seven in a cluster of seven hexagonal cells forming a larger hexagon so that any two neighboring cells have a total of at most ten. Determine how many distinct placements there are, with different orientations considered different.
[ "Use the adjacency sum constraint to deduce that seven cannot be placed in the central cell and that its neighbors must be from the three smallest numbers", "Exploit symmetry to fix the position of seven on a rim cell and multiply by the number of equivalent rim positions", "Partition the remaining cells by adj...
72
proof and answer
0.86
As shown in the figure, seven regular hexagonal cells form a hexagonal pattern. We write one integer from $1$ to $7$ in each cell without repetition. For any two cells that share an edge, the sum of the integers written in those cells must be at most $10$. How many ways are there to write the integers under these condi...
83df18ec4d0617278da8e912b060af3a781c4e7cbf59c5f036065aadb83327d7
Mexico
LVI Olimpiada Matemática Española (Concurso Final)
2,020
Enunciados y Soluciones
5
En un triángulo acutángulo $ABC$, sea $M$ el punto medio del lado $AB$ y $P$ el pie de la altura sobre el lado $BC$. Prueba que si $AC + BC = \sqrt{2}AB$, entonces la circunferencia circunscrita del triángulo $BMP$ es tangente al lado $AC$.
[ "Sea $S$ el punto de $AC$, al mismo lado de $A$ que $C$, tal que $AS = \\sqrt{2}AB/2$. Este punto cumple que $AS^2 = \\frac{AB^2}{2} = AB \\cdot AM$, que es la potencia de $A$ con respecto a la circunferencia circunscrita de $BMP$; luego si esta circunferencia circunscrita pasa por $S$ entonces es tangente a $AB$. ...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Circles", "Tangents" ], [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geo...
[ "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
Spanish
MEX_OME56
Mexico
true
1
In an acute triangle, with a midpoint on one side and the foot of the altitude to another side, show that if the sum of two sides equals a constant multiple of the third side, then the circumcircle of the triangle formed by the midpoint and the foot is tangent to one side of the original triangle.
[ "Construct a point on a side so that its distance from a vertex matches the tangent length implied by the power of that vertex with respect to the target circumcircle", "Use the midpoint of a side to translate the given side-length condition into an isosceles configuration, yielding key angle equalities", "Appl...
null
proof only
0.92
En un triángulo acutángulo $ABC$, sea $M$ el punto medio del lado $AB$ y $P$ el pie de la altura sobre el lado $BC$. Prueba que si $AC + BC = \sqrt{2}AB$, entonces la circunferencia circunscrita del triángulo $BMP$ es tangente al lado $AC$.
b258958f933843b9ee93ad9713cf8848dd9ce31aeef01c18e166ef749a705f86
Middle European Mathematical Olympiad (MEMO)
MEMO Szeged
2,024
Team
T-5
Problem: Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $D$ be a point on the line $AC$ such that $AB = AD$ and $A$ lies between $C$ and $D$. Suppose that there are two points $E \neq F$ on the circumcircle of the triangle $DBC$ such that $AE = AF = BC$. Prove that the line $EF$ passes through the circumc...
[ "Solution:\n\nLet $N$ be the midpoint of arc $BAC$. Then triangle $NBC$ is equilateral as $\\angle BNC = 60^{\\circ}$ and $N$ lies on the perpendicular bisector of $BC$. Moreover, $N$ lies on the angle bisector of the angle $DAB$, which is the perpendicular bisector of segment $BD$ considering the isosceles triangl...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Circles", "Radical axis theorem" ], [ "Topics", "Geometry", "Plane Geom...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Circles > Radical axis theorem", "Geometry > Plane Geometry > Triangles > Triangle trigonometry", "Geometry > Plane Geometry > Miscellaneous > Angle chasing...
null
MEMO__md__en-2024-SolutionBooklet
Middle European Mathematical Olympiad (MEMO)
false
0
In a triangle with one angle of sixty degrees, a point is chosen on a side so that two segments from the vertex are equal. If there are two points on the circumcircle of the triangle formed with that point and the other vertices whose distances to the original vertex equal the opposite side length, show that the line j...
[ "Use the midpoint of arc construction to identify a special point that serves as the circumcenter of the auxiliary triangle and forms an equilateral triangle with a side", "Exploit equal distances to form a rhombus, implying a perpendicular bisector line that passes through the circumcenter via chord properties",...
null
proof only
0.9
Problem: Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $D$ be a point on the line $AC$ such that $AB = AD$ and $A$ lies between $C$ and $D$. Suppose that there are two points $E \neq F$ on the circumcircle of the triangle $DBC$ such that $AE = AF = BC$. Prove that the line $EF$ passes through the circumc...
ba034e2671a7ea98dbf8c2e7e7ebd14934d7141933708e988c00612e1466a0c7
Moldova
Olimpiada Republicană la Matematică
2,023
Prima zi, Clasa X-a
10.2.
Problem: Fie $ABC$ un triunghi ascuțitunghic, iar $H$ un punct din interiorul triunghiului, astfel încât $AB^{2} + CH^{2} = BC^{2} + AH^{2} = AC^{2} + BH^{2}$. Demonstrați că $H$ este ortocentrul triunghiului $ABC$.
[ "Solution:\n\n1. Fie $CC_{1} \\perp AB$, $C_{1} \\in AB$ și $HH_{1} \\perp AB$, $H_{1} \\in AB$.\n\n2. $AC^{2} - AC_{1}^{2} = BC^{2} - BC_{1}^{2}$,\n\n(1),\n$$\nAH^{2} - AH_{1}^{2} = BH^{2} - BH_{1}^{2}\n$$\n\n3. Din ipoteză obținem $AC^{2} - BC^{2} = AH^{2} - BH^{2}$, din relația (1) obținem $AC^{2} - BC^{2} = AC_...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Distance chasing" ] ]
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Miscellaneous > Distance chasing" ]
null
Moldova__md__ro-national__ro-omrm_matem_10_solutii_ziua_1_2023
Moldova
false
0
In an acute triangle, an interior point is such that for each side the sum of the square of that side and the square of the distance from the point to the opposite vertex is the same. Prove that this point is the orthocenter.
[ "Introduce feet of perpendiculars from a vertex and from the point onto a side to compare distances", "Apply the Pythagorean theorem to relate differences of squared side lengths to squared altitudes (Pythagorean differences)", "Use the given equal-sum conditions to equate these differences and transfer them to...
null
proof only
0.9
Problem: Fie $ABC$ un triunghi ascuțitunghic, iar $H$ un punct din interiorul triunghiului, astfel încât $AB^{2} + CH^{2} = BC^{2} + AH^{2} = AC^{2} + BH^{2}$. Demonstrați că $H$ este ortocentrul triunghiului $ABC$.
6fb600278a32fc57cadaeaf3a4cd56a7c5aac07d99a2f759ebdab7bd38f7d8b0
Mongolia
MMO2025 Round 4
2,025
Category T (Secondary School Teacher)
T5
Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, and $CF$. Let $\omega$ be the circle with diameter $BC$, and suppose it intersects the segment $AD$ at point $K$ inside triangle $ABC$. On ray $KD$, let $L$ be a point such that $KA = KL$. Let lines $BL$ and $CL$ intersect the circle $\omega$ again at points $P$...
[ "![](images/MNG_ABooklet_2025_p35_data_0af9375826.png)\n\nSince $\\angle BEC = \\angle BFC = 90^\\circ$, points $E$ and $F$ lie on circle $\\omega$.\n\nWe observe that:\n$$\n\\angle LAF = \\angle DAB = \\angle FCB = \\angle FPL,\n$$\nso quadrilateral $AFLP$ is cyclic; denote its circumcircle by $\\omega_1$. Similar...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Circles", "Radical axis theorem" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ] ]
[ "Geometry > Plane Geometry > Circles > Radical axis theorem", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
English
MNG_2025
Mongolia
true
1
In an acute triangle, a circle with the base as diameter meets one altitude. A point is chosen on the extension of that altitude so it is equally distant from the vertex and the intersection point. Lines from this new point to the other two vertices meet the circle again, and the task is to show that three specific lin...
[ "Feet of the altitudes lie on the circle with diameter through the base endpoints, giving right angles on that circle", "Angle chasing to prove certain quadruples of points are concyclic", "Recognizing that lines through pairs of intersection points of two circles are their radical axes", "Applying the Radica...
null
proof only
0.93
Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, and $CF$. Let $\omega$ be the circle with diameter $BC$, and suppose it intersects the segment $AD$ at point $K$ inside triangle $ABC$. On ray $KD$, let $L$ be a point such that $KA = KL$. Let lines $BL$ and $CL$ intersect the circle $\omega$ again at points $P$...
3ad9522b7a3044b22da8f7494711770b50d9066dabe182984ab1698a546a191b
Netherlands
BxMO/EGMO Team Selection Test
2,025
null
3
A group of 4050 friends is playing a video game tournament. There are 2025 computers labelled $a_1, \dots, a_{2025}$ in one room and 2025 computers labelled $b_1, \dots, b_{2025}$ in another room at the tournament. The player on computer $a_i$ always plays against the players $b_i, b_{i+2}, b_{i+3}$ and $b_{i+4}$ (in p...
[ "For the opponent computers of $a_i$, we look at the $a_j$ they are playing against, see the following table.\n\n$b_i:$\n| $a_{i-4}$ | $a_{i-3}$ | $a_{i-2}$ | $a_i$ |\n|---|---|---|---|\n\n$b_{i+2}:$\n| $a_{i-2}$ | $a_{i-1}$ | $a_i$ | $a_{i+2}$ |\n|---|---|---|---|\n\n$b_{i+3}:$\n| $a_{i-1}$ | $a_i$ | $a_{i+1}$ | $...
[]
[ [ "Topics", "Discrete Mathematics", "Combinatorics", "Induction / smoothing" ], [ "Topics", "Number Theory", "Other" ] ]
[ "Discrete Mathematics > Combinatorics > Induction / smoothing", "Number Theory > Other" ]
null
NLD_2025
Netherlands
true
1
Two equal rooms have labeled computers and each player on a computer faces four specific opponents determined by the labels. After a reshuffle within each room, every player ends up with exactly the same set of opponents as before. Show that if any one player did not move, then in fact no one moved.
[ "Model the setup as a fixed bipartite circulant graph between two cycles with edges defined by fixed shifts", "Use common-opponent structure to identify uniquely determined seats: two specific neighbors share the same unique common opponent with a given vertex", "If one vertex is fixed, deduce its matched oppon...
null
proof only
0.79
A group of 4050 friends is playing a video game tournament. There are 2025 computers labelled $a_1, \dots, a_{2025}$ in one room and 2025 computers labelled $b_1, \dots, b_{2025}$ in another room at the tournament. The player on computer $a_i$ always plays against the players $b_i, b_{i+2}, b_{i+3}$ and $b_{i+4}$ (in p...
cae17420ffb96d8e6098343cb41ced053dca1f3e8980e2cdab70acb751d3c847
New Zealand
NZMO Round One
2,025
null
2.
Problem: Let $ABC$ be a right-angled triangle with $\angle BAC = 90^{\circ}$, $\angle ABC = 70^{\circ}$, and $AB = 1$. Let $M$ be the midpoint of $BC$. Let $D$ be the point on the extension of $AM$ beyond $M$ such that $\angle CDA = 110^{\circ}$. Find the length of $CD$.
[ "Solution:\n\nConstruct point $E$ so that $ABEC$ is a rectangle. The diagonals of any rectangle bisect each other, that is, they meet at each other's midpoints. Hence $AE$ and $BC$ meet at $M$, i.e. $E$ lies on line $AM$.\n\n![](attached_image_solution_1.png)\n\nBy symmetry in rectangle $ABEC$, we have\n$$\n\\angle...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Constructions and loci" ] ]
[ "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Miscellaneous > Constructions and loci" ]
null
NewZealand_MO__md__en-nzmo1_2025_solutions
New Zealand
false
0
In a right triangle with specified angles and a given side length, a point is placed on the line through the midpoint so that a particular angle is fixed, and the task is to determine the distance from this point to one vertex.
[ "Construct a rectangle using the perpendicular legs of the right triangle to introduce a helpful auxiliary point", "Use the property that diagonals of a rectangle (parallelogram) bisect each other to place the constructed point on the midpoint line", "Angle chasing: relate angles at the constructed point and th...
1
proof and answer
0.91
Problem: Let $ABC$ be a right-angled triangle with $\angle BAC = 90^{\circ}$, $\angle ABC = 70^{\circ}$, and $AB = 1$. Let $M$ be the midpoint of $BC$. Let $D$ be the point on the extension of $AM$ beyond $M$ such that $\angle CDA = 110^{\circ}$. Find the length of $CD$.
502d9c4e0bc649f08cc3ed52aee938e7328e0a80aed2f00c1af1e482c5503cae
Nordic Mathematical Olympiad
Nordic Mathematical Contest
2,022
null
Problem 4
Problem: Let $ABC$ be an acute-angled triangle with circumscribed circle $k$ and centre of the circumscribed circle $O$. A line through $O$ intersects the sides $AB$ and $AC$ at $D$ and $E$. Denote by $B'$ and $C'$ the reflections of $B$ and $C$ over $O$, respectively. Prove that the circumscribed circles of $ODC'$ an...
[ "Solution:\n\nLet $P$ be the intersection of the circles $k$ and the circumscribed circle of triangle $ADE^{1}$. Let $C_{1}$ be the second intersection of the circumscribed circle of $\\triangle DOP$ with $k$. We will prove that $C_{1}=C'$, i.e. the reflection of $C$ over $O$. We know that $|OC_{1}|=|OP|$, and henc...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Transformations", "Spiral similarity" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ...
[ "Geometry > Plane Geometry > Transformations > Spiral similarity", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler ...
null
Nordic_MO__md__en-2022-sol
Nordic Mathematical Olympiad
false
0
In an acute triangle, draw a line through the circumcenter meeting two sides. Reflect the two corresponding vertices across the circumcenter. Show that the circles through the circumcenter with each reflection and the corresponding side intersection share a common point on the original circumcircle.
[ "Introduce the intersection point of the triangle's circumcircle with the circumcircle through the two side-intersection points, serving as the spiral similarity center mapping one segment to another", "Use similarity of triangles induced by the spiral similarity to equate key angles", "Apply inscribed angle re...
null
proof only
0.86
Problem: Let $ABC$ be an acute-angled triangle with circumscribed circle $k$ and centre of the circumscribed circle $O$. A line through $O$ intersects the sides $AB$ and $AC$ at $D$ and $E$. Denote by $B'$ and $C'$ the reflections of $B$ and $C$ over $O$, respectively. Prove that the circumscribed circles of $ODC'$ an...
6708e88824f30bb2bdd1b67f0e59426592ca1a6ee84d97695a58a5184afdc633
North Macedonia
Macedonian Mathematical Olympiad
2,018
XXV Macedonian Mathematical Olympiad
5
Let $\triangle ABC$ be an acute triangle with orthocenter $H$. The point $H'$ is symmetric with $H$ with respect to the line $AB$. Let $N$ be the intersection point of $HH'$ and $AB$. The circle passing through the points $A$, $N$ and $H'$ intersects again the line $AC$ at $M$, and the circle passing through the points...
[ "First, we will show the following lemma.\n\n**Lemma.** Let $H$ be an orthocenter in $\\triangle ABC$, and $H'$ be the symmetric point of $H$ with respect to $AB$. Then $H'$ lies on the circle around the triangle $\\triangle ABC$.\n\n**Proof.**\n\n**First proof of the lemma.** Let $N$ be the intersection point of $...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Advanced Configurations", "Simson line" ], [ "Topics", "Geometry", "Plane Geometry", "Concurrency and Collinearity", "Menelaus' theorem" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cycl...
[ "Geometry > Plane Geometry > Advanced Configurations > Simson line", "Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter...
English
MKD_2018
North Macedonia
true
3
In an acute triangle, reflect the orthocenter across one side and define a point where the reflection line meets that side. Construct two circles through this reflected point and one vertex each, meeting the adjacent sides again at two new points. Show that these two new points and the intersection point on the side al...
[ "Reflection of the orthocenter across a side lies on the circumcircle of the triangle", "Simson line: for a point on the circumcircle, the perpendicular feet to the triangle’s sides are collinear", "Using diameters in auxiliary circles to deduce right angles and identify perpendicular feet", "Power of a point...
null
proof only
0.86
Let $\triangle ABC$ be an acute triangle with orthocenter $H$. The point $H'$ is symmetric with $H$ with respect to the line $AB$. Let $N$ be the intersection point of $HH'$ and $AB$. The circle passing through the points $A$, $N$ and $H'$ intersects again the line $AC$ at $M$, and the circle passing through the points...
5722c6540c33f1365b9bda8e8613c964d628379cc60303449051a5d87ac55187
Philippines
25th Philippine Mathematical Olympiad Area Stage
2,023
PART II
2.
Problem: Let $ABC$ be an acute scalene triangle with orthocenter $H$. Let $M$ be the midpoint of $BC$, and suppose that the line through $H$ perpendicular to $AM$ intersects $AB$ and $AC$ at points $E$ and $F$ respectively. Denote by $O$ the circumcenter of triangle $AEF$, and $D$ the foot of the perpendicular from $H...
[ "Solution:\n\nWLOG assume $AB < AC$. Let $AM$ intersect the circumcircle of $ABC$ again at $Y \\neq A$. We first need to prove a claim.\n\n![](attached_image_solution_1.png)\n\nClaim: Quadrilateral $B H D C$ is cyclic.\n\nProof of Claim: Consider the reflection with respect to $M$. This maps $B$ and $C$ to each oth...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geome...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Transformations > Rotation", "Geometry > Plane Geometry > Miscellaneous >...
null
Philippines__md__en-pmo__en-PMO-25-Area-Stage
Philippines
false
0
In an acute triangle, construct points from the orthocenter and the midpoint of one side, including a line through the orthocenter perpendicular to the median and the circumcenter of the resulting triangle through the vertex. Show that the line from the vertex to this circumcenter meets the line through the constructed...
[ "Use a half-turn about the midpoint of a side to map the orthocenter to the antipode of the opposite vertex on the circumcircle", "Establish that the foot from the orthocenter to the median is the reflection of the median’s second intersection with the circumcircle, implying a cyclic quadrilateral", "Angle chas...
null
proof only
0.86
Problem: Let $ABC$ be an acute scalene triangle with orthocenter $H$. Let $M$ be the midpoint of $BC$, and suppose that the line through $H$ perpendicular to $AM$ intersects $AB$ and $AC$ at points $E$ and $F$ respectively. Denote by $O$ the circumcenter of triangle $AEF$, and $D$ the foot of the perpendicular from $H...
cd0c50e45f78af1f7d3b9b5b6e6190a399ba4e928c1fed7aa1f8eac2544f0451
Romania
75th Romanian Mathematical Olympiad
2,025
Final Round - 8th GRADE
Problem 4
From a point $O$ inside the square $ABCD$ the perpendicular line $OS$ is raised to the plane of the square. Let $M, N, P, Q$ be projections of point $O$ onto the planes $(SAB), (SBC), (SCD)$, respectively $(SDA)$. Prove that the points $M, N, P, Q$ are coplanar if and only if $O$ lies on one of the diagonals of the squ...
[ "We assume that $O$ lies, for example, on the diagonal $AC$. Let $OE \\perp AB$, $E \\in AB$ and $OF \\perp AD$, $F \\in AD$. Then we have successively $OE = OF$, $\\triangle SOE \\equiv \\triangle SOF$ (C.C.), $SE = SF$. Then $M \\in SE$ and $OM \\perp SF$, $Q \\in SF$ and $OQ \\perp SF$, $\\triangle SOM \\equiv \...
[]
[ [ "Topics", "Geometry", "Solid Geometry", "Other 3D problems" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Analytic / Coordinate Methods", "Trigonometry" ], [ ...
[ "Geometry > Solid Geometry > Other 3D problems", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry", "Geometry > Plane Geometry > Triangles > Triangle trigonometry" ]
English
ROU_2025
Romania
true
1
In a square with a point inside and a point directly above it, project the interior point onto the four triangular faces formed with adjacent sides. Show that these four projected points lie in one plane exactly when the interior point lies on a diagonal of the square.
[ "Exploit symmetry when the interior point lies on a diagonal to get equal distances to adjacent sides, leading to congruent right triangles with the apex above the square and parallelism of segments implying coplanarity", "Use properties of orthogonal projections onto side planes: the feet lie on lines through th...
null
proof only
0.78
From a point $O$ inside the square $ABCD$ the perpendicular line $OS$ is raised to the plane of the square. Let $M, N, P, Q$ be projections of point $O$ onto the planes $(SAB), (SBC), (SCD)$, respectively $(SDA)$. Prove that the points $M, N, P, Q$ are coplanar if and only if $O$ lies on one of the diagonals of the squ...
b2930ff1b8cad4740727da4a5b6ef11a34d1edb47d922f51ac98cee362638e0e
Romanian Master of Mathematics (RMM)
Romanian Master of Mathematics Competition
2,021
Day 2
Problem 5
Problem: Let $n$ be a positive integer. The kingdom of Zoomtopia is a convex polygon with integer sides, perimeter $6 n$, and $60^{\circ}$ rotational symmetry (that is, there is a point $O$ such that a $60^{\circ}$ rotation about $O$ maps the polygon to itself). In light of the pandemic, the government of Zoomtopia wou...
[ "Solution:\nLet $P$ denote the given polygon, i.e., the kingdom of Zoomtopia. Throughout the solution, we interpret polygons with integer sides and perimeter $6 k$ as $6 k$-gons with unit sides (some of their angles may equal $180^{\\circ}$). The argument hinges on the claim below:\n\nClaim. Let $P$ be a convex pol...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Transformations", "Rotation" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Constructions and loci" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Distance chasing" ] ]
[ "Geometry > Plane Geometry > Transformations > Rotation", "Geometry > Plane Geometry > Miscellaneous > Constructions and loci", "Geometry > Plane Geometry > Miscellaneous > Distance chasing" ]
null
RMM__md__en-2021-RMM2021-Day2-English_Solutions
Romanian Master of Mathematics (RMM)
false
0
Given a convex polygon with sixty degree rotational symmetry, integer side lengths, and perimeter equal to six times a positive whole number, show that one can place exactly three times the square of that number plus three times that number plus one points inside or on the boundary so that every two points are at least...
[ "Exploit sixty-degree rotational symmetry to perform symmetric attachments along six corresponding edges", "Inductive construction reducing perimeter by six via attaching six unit equilateral triangles and unit lozenges, then removing them", "Verification that attached lozenges have all angles at least sixty de...
null
proof only
0.88
Problem: Let $n$ be a positive integer. The kingdom of Zoomtopia is a convex polygon with integer sides, perimeter $6 n$, and $60^{\circ}$ rotational symmetry (that is, there is a point $O$ such that a $60^{\circ}$ rotation about $O$ maps the polygon to itself). In light of the pandemic, the government of Zoomtopia wou...
be50024c3a1d4107dda5d2c4834fbf3bcb2985902323baa0e8de328846345b7b
Russia
LI Всероссийская математическая олимпиада школьников
2,025
9 класс
9.2
Diagonals of a convex quadrilateral *ABCD* intersect at *E*. The four points of tangency of the circles (*ABE*) and (*CDE*) with their external common tangents lie on a circle $\omega$. Analogously, the four points of tangency of the circles (*ADE*) and (*BCE*) with their external common tangents lie on a circle $\gamm...
[ "Let us denote the centers of the circumscribed circles of triangles *ABE*, *BCE*, *CDE*, *ADE* by $O_{AB}$, $O_{BC}$, $O_{CD}$, $O_{AD}$ respectively. Let $T_1, T_2$ be the points of tangency of one of the common tangents with the circumscribed circles of triangles *ABE* and *CDE*, respectively; denote by *O* and ...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Circles", "Tangents" ], [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ] ]
[ "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ]
Russian
RUS_2025
Russia
true
2
Given a convex quadrilateral with diagonals intersecting, take the circumcircles of two opposite triangles formed with the intersection point. For each pair, the four points where their external common tangents touch lie on a circle. Prove that the circles arising from the two opposite pairs have the same center.
[ "Exploit geometry of external common tangents to two circles: radii to tangency points are perpendicular to the tangent, forming a right trapezoid with the centers and tangency points", "Use the midline in a right trapezoid to identify the perpendicular bisector of the segment between tangency points and thus loc...
null
proof only
0.79
Diagonals of a convex quadrilateral *ABCD* intersect at *E*. The four points of tangency of the circles (*ABE*) and (*CDE*) with their external common tangents lie on a circle $\omega$. Analogously, the four points of tangency of the circles (*ADE*) and (*BCE*) with their external common tangents lie on a circle $\gamm...
a2b7a1aedcc46e0f077c90899d35fc53bff76b4addeed64aedfd090f9cf85e16
Saudi Arabia
Saudi Booklet
2,025
Preselection tests - Test 1
4
Let $ABC$ be a triangle inscribed in circle $(O)$ with $\angle A = 45^\circ$. Two rays $BO$, $CO$ intersect $AC$, $AB$ at $E$, $F$ respectively. The circumcircles of triangles $BOC$ and $EOF$ intersect at $K$. Let $J$ be circumcenter of triangle $AEF$. Prove that $JK$ passes through the orthocenter of triangle $ABC$.
[ "We have\n$$\n\\angle BOC + \\angle EOF = 90^\\circ + 90^\\circ = 180^\\circ\n$$\nso according to the familiar property of isogonal conjugates in quadrilaterals, we see that there exists a point $O'$ which is the isogonal conjugate of $O$ in $BFEC$. On the other hand, $BH$, $BO$ and $CH$, $CO$ are isogonal pairs in...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Advanced Configurations", "Isogonal/isotomic conjugates, barycentric coord...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilate...
null
SAU_2025
Saudi Arabia
true
1
Given a triangle where one angle is forty five degrees, define points where lines from two vertices toward the circle center meet the opposite sides. Consider the circumcircles through those points and another circle through two vertices and the center that meet at a point. Show that the line from the center of the cir...
[ "Use central angle facts: angle at the center over a given arc equals twice the inscribed angle, giving a right angle at the center since the given vertex angle is forty-five", "Recognize that the lines from vertices to the circumcenter yield a right angle at the center for triangle formed by intersection points ...
null
proof only
0.77
Let $ABC$ be a triangle inscribed in circle $(O)$ with $\angle A = 45^\circ$. Two rays $BO$, $CO$ intersect $AC$, $AB$ at $E$, $F$ respectively. The circumcircles of triangles $BOC$ and $EOF$ intersect at $K$. Let $J$ be circumcenter of triangle $AEF$. Prove that $JK$ passes through the orthocenter of triangle $ABC$.
1afb4f87d778339fcd1a62d59a93080865aa4e44dff06997253a0cc33da187f9
Serbia
14. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА
2,020
Други дан
4.
Problem: У трапезу $ABCD$ чији унутрашњи углови нису прави, дијагонале $AC$ и $BD$ секу се у тачки $E$. Нека су $P$ и $Q$ редом подножја нормала из темена $A$ и $B$ на праве $BC$ и $AD$. Описане кружнице троуглова $CEQ$ и $DEP$ секу се у тачки $F \neq E$. Доказати да се праве $AP$, $BQ$ и $EF$ секу у једној тачки или ...
[ "Solution:\n\nСлучај када је $AD \\parallel BC$ је једноставан. Наиме, тада је $E$ средиште дијагонале $AC$, те је $EA = EC = EP$, а слично важи и $EB = ED = EQ$. Одатле следи да су кругови $CEQ$ и $DEP$ симетрични у односу на симетралу дужи $CP$ и $DQ$, те је $EF \\perp CP$, тј. $EF \\parallel AP \\parallel BQ$.\n...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Circles", "Radical axis theorem" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Concurrency and Collinearity", "Pappus ...
[ "Geometry > Plane Geometry > Circles > Radical axis theorem", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem" ]
null
Serbia_MO__md__sr-2020_smo_resenja
Serbia
false
0
In a trapezoid, from two vertices drop perpendiculars to the opposite sides and form two circles using the intersection of the diagonals and these points. Show that the lines from the vertices to the foot points and the line through the intersection of the two circles all meet at one point or are parallel.
[ "Handle the special case of parallel sides by symmetry of the two circumcircles, yielding a parallel common line", "Use similarity of triangles formed with the intersection of the two non-parallel sides to derive a product relation implying four points are concyclic", "Apply power of a point to show a construct...
null
proof only
0.84
Problem: У трапезу $ABCD$ чији унутрашњи углови нису прави, дијагонале $AC$ и $BD$ секу се у тачки $E$. Нека су $P$ и $Q$ редом подножја нормала из темена $A$ и $B$ на праве $BC$ и $AD$. Описане кружнице троуглова $CEQ$ и $DEP$ секу се у тачки $F \neq E$. Доказати да се праве $AP$, $BQ$ и $EF$ секу у једној тачки или ...
5ff0903552c59c969bfc65cddc7e62a5f565fb0944d6de733d759fe4e266fa2d
Silk Road Mathematics Competition
SILK ROAD MATHEMATICAL COMPETITION
2,025
null
2
Altitudes of an acute scalene triangle $ABC$ meet at point $H$. Points $M$ and $N$ are the midpoints of the segments $AB$ and $CH$, respectively. $R$ is the foot of the perpendicular from $H$ to line $CM$. $T$ is the second intersection of the line $MN$ with the circumcircle of triangle $CNR$. Let $P$ be the circumcent...
[ "We will make several uses of the fact that the lines connecting a vertex of a triangle with the circumcentre and the orthocentre are symmetrical with respect to the angle bisector at this vertex.\nLet $AC > BC$, $AA_1$ and $BB_1$ be the altitudes of the triangle $ABC$, $O$ its circumcentre. The points $A_1$, $B_1$...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Advanced Configurations", "Brocard point, symmedians" ], [ "Topics...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane G...
English
SRM_2025
Silk Road Mathematics Competition
true
2
In an acute scalene triangle, construct the orthocenter, certain midpoints, and the foot of a perpendicular from the orthocenter to a median. Intersect a specific midline with a circle through two of these points to define another point. Also define a center associated with three given lines. Prove that the line throug...
[ "Use isogonal conjugacy: the lines from a vertex to the circumcenter and orthocenter are symmetric about the internal angle bisector", "Exploit similarity between the original triangle and the triangle formed by feet of altitudes to relate lengths via cosine and compare circumradii", "Show COMN is a parallelogr...
null
proof only
0.79
Altitudes of an acute scalene triangle $ABC$ meet at point $H$. Points $M$ and $N$ are the midpoints of the segments $AB$ and $CH$, respectively. $R$ is the foot of the perpendicular from $H$ to line $CM$. $T$ is the second intersection of the line $MN$ with the circumcircle of triangle $CNR$. Let $P$ be the circumcent...
ce3fe1aa595b92d78a4af6b7734558c2f949e089d7aa006c6d410ea0082fe2ac
Singapore
Singapore Mathematical Olympiad (SMO)
2,025
Junior Section, Round 2
1
Two rhombi *ABCD* and *AXYZ* are external to each other with a common vertex *A*. The vertices of the rhombi are labelled clockwise. If $\angle DAX = \angle BAZ$, prove that the centres of the rhombi and the midpoint of the segment *BZ* form an isosceles triangle.
[ "Let the centres be $P$, $Q$ and the midpoint of the segment $BZ$ be $R$ as shown in the figure. Since $\\angle DAX = \\angle BAZ$, we have $DZ = BX$. Since $P$ and $R$ are the midpoints of $XZ$ and $BZ$, respectively, $PR = BX/2$. Similarly $QR = DZ/2$. Therefore $QR = PR$ and it follows that $\\triangle PQR$ is i...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Distance chasing" ] ]
[ "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Miscellaneous > Distance chasing" ]
null
SIN_2025
Singapore
true
1
Two rhombi share a common corner and lie outside each other. Given that two specific angles at the common corner are equal, show that the line segment connecting the centers of the two rhombi and the midpoint of a certain segment between their vertices forms a triangle with two equal sides.
[ "Identify rhombus centers as the midpoints of both diagonals", "Apply the triangle midpoint (mid-segment) theorem to relate segments between midpoints to half of the third side", "Use the equality of adjacent side lengths in rhombi together with the given angle condition to deduce equality of certain cross dist...
null
proof only
0.73
Two rhombi *ABCD* and *AXYZ* are external to each other with a common vertex *A*. The vertices of the rhombi are labelled clockwise. If $\angle DAX = \angle BAZ$, prove that the centres of the rhombi and the midpoint of the segment *BZ* form an isosceles triangle.
fd0871709f459e3a6ef0f841dcf907710760e50bd6187fdb18fb7970796bb363
Slovenia
67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje
2,023
1. letnik
A2.
Problem: Dan je pravokotni trikotnik $ABC$ s pravim kotom pri $A$. Na stranicah $AB$, $BC$ in $CA$ zaporedoma ležijo točke $D$, $E$ in $F$, tako da velja $|BD| = |BE|$ in $|CF| = |CE|$ (glej sliko). Koliko stopinj je velikost kota $\angle FED$? (A) 30 (B) 37,5 (C) 45 (D) 52,5 ![A right triangle ABC with right angle a...
[ "Solution:\n\nOznačimo kote trikotnika $ABC$ kot običajno z $\\alpha$, $\\beta$ in $\\gamma$. Tedaj je $\\alpha = 90^\\circ$ in zato je $\\beta + \\gamma = 90^\\circ$.\n\nKer sta trikotnika $DBE$ in $ECF$ enakokraka z vroma pri $B$ in $C$, je\n$$\n\\angle DEB = \\frac{180^\\circ - \\beta}{2} = 90^\\circ - \\frac{\\...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ], [ "Topics", "Geometry", "Plane Geometry", "Triangles" ] ]
[ "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Triangles" ]
null
Slovenia__md__sl-massa__sl-MaSSA_Drzavno_2023
Slovenia
true
1
In a right triangle, points are placed on its sides so that two segments from one vertex are equal and two segments from the other vertex are equal. Determine the measure of the angle formed at the point between the two constructed segments.
[ "Use that the two acute angles in a right triangle sum to ninety degrees", "Recognize isosceles triangles formed by equal segments from a vertex to points on adjacent sides", "Compute base angles of the isosceles triangles to express adjacent angles at a common point", "Sum of angles around the configuration ...
C
MCQ
0.97
Problem: Dan je pravokotni trikotnik $ABC$ s pravim kotom pri $A$. Na stranicah $AB$, $BC$ in $CA$ zaporedoma ležijo točke $D$, $E$ in $F$, tako da velja $|BD| = |BE|$ in $|CF| = |CE|$ (glej sliko). Koliko stopinj je velikost kota $\angle FED$? (A) 30 (B) 37,5 (C) 45 (D) 52,5 ![](attached_image_problem_1.png) (E) Nem...
b1b7184d9c28c73fdf5ba345ae1febb3b50b56735c9a796d045742b6d0d5ba20
South Africa
The South African Mathematical Olympiad Third Round
2,024
Senior Division (Grades 10 to 12)
5
Consider three circles $\Gamma_1, \Gamma_2$ and $\Gamma_3$, with centres $O_1, O_2$ and $O_3$, respectively, such that each pair of circles is externally tangent. Suppose we have another circle $\Gamma$ with centre $O$ on the line segment $O_1O_3$ such that $\Gamma_1, \Gamma_2$ and $\Gamma_3$ are each internally tangen...
[ "Let the points of tangency of $\\Gamma$ with $\\Gamma_1, \\Gamma_2$ and $\\Gamma_3$ be $A, B$ and $C$, respectively.\n\n![](images/s3s2024_p3_data_815ea3bdb8.png)\n\nThen $AOB$ is a straight line, and since it is a diameter, $\\angle ACB = 90^\\circ$. Let $E$ be the other intersection point of $AC$ with $\\Gamma_3...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Circles", "Tangents" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Distance chasing" ] ]
[ "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Miscellaneous > Distance chasing" ]
null
ZAF_2024
South Africa
true
1
Given three circles that are pairwise externally tangent and all internally tangent to a larger circle whose center lies on the line through two of the small centers, prove that the angle at the center of the middle small circle formed with the other two centers is less than a right angle.
[ "Use the fact that an angle subtended by a diameter of a circle is a right angle", "Exploit collinearity of centers and tangency points, and equality of radii to infer parallel lines", "Construct auxiliary points on the diameter to compare and bound the central angle", "Relate radii of internally tangent and ...
null
proof only
0.86
Consider three circles $\Gamma_1, \Gamma_2$ and $\Gamma_3$, with centres $O_1, O_2$ and $O_3$, respectively, such that each pair of circles is externally tangent. Suppose we have another circle $\Gamma$ with centre $O$ on the line segment $O_1O_3$ such that $\Gamma_1, \Gamma_2$ and $\Gamma_3$ are each internally tangen...
87b2baed4bf16dfe3815d0e8d60a4f421b81ceeff40b7416c8da3147b371fc22
South Korea
20th Korean Mathematical Olympiad Final Round
2,007
First Day
2
Consider the sixteen tiles fixed on a wall as shown below. How many ways are there to write either $0$ or $1$ on each tile so that the product of the two numbers written on every neighboring pair of tiles (sharing a common side) is always $0$? ![](images/Juzna_Koreja_2007_p2_data_d1e857f48a.png)
[ "When $0$ or $1$ is written on each tile in such a way that the product of the two numbers written on every neighboring pair of tiles is always $0$, we'll call the status a *z-pattern*. First, we prove a couple of lemmas.\n\n**Lemma 1.** Let $a_n$ be the number of z-patterns for $n$ tiles laid in a row. Then\n$$\na...
[]
[ [ "Topics", "Discrete Mathematics", "Combinatorics", "Recursion, bijection" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Induction / smoothing" ] ]
[ "Discrete Mathematics > Combinatorics > Recursion, bijection", "Discrete Mathematics > Combinatorics > Induction / smoothing" ]
null
KOR_2007
South Korea
true
1
Count the ways to fill a specific sixteen-tile arrangement with zeros and ones so that no two tiles sharing a side both show one.
[ "Encode the adjacency constraint as forbidding neighboring ones and derive a linear recurrence for sequences on a line (Fibonacci numbers).", "Extend the recurrence to a cyclic ring by conditioning on a chosen tile to obtain counts in terms of the linear case.", "Perform case analysis on the central block to re...
1234
proof and answer
0.93
Consider the sixteen tiles fixed on a wall as shown below. How many ways are there to write either $0$ or $1$ on each tile so that the product of the two numbers written on every neighboring pair of tiles (sharing a common side) is always $0$? ![](images/Juzna_Koreja_2007_p2_data_d1e857f48a.png)
68f84bfea82cd7d30a0824cb5d10c49e411d9921af6e4ff1f3893f5f285fd002
Soviet Union
1st CIS
1,992
null
Problem 2
Problem: $E$ is a point on the diagonal $BD$ of the square $ABCD$. Show that the points $A$, $E$ and the circumcenters of $ABE$ and $ADE$ form a square.
[ "Solution:\n\n![](attached_image_solution_1.png)\n\nLet $O$, $O'$ be the circumcenters of $ABE$, $ADE$ respectively. Then $OA = OE$ and $\\angle AOB = 2 \\angle ABE = 90^\\circ$. Similarly, $O'A = O'E$ and $\\angle AOE = 2\\angle ADE = 90^\\circ$. Hence $AOEO'$ is a square." ]
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ] ]
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
null
SovietUnion__md__en-ASU-1961-1991_chunk_1981-1991
Soviet Union
false
0
In a square, pick any point on one diagonal. Consider the centers of the circles passing through the chosen point and two adjacent vertices. Show that these two centers together with the chosen point and one fixed vertex become the four corners of a square.
[ "Use properties of circumcenters: equal distances from triangle vertices imply equal adjacent sides in the constructed quadrilateral", "Apply the inscribed angle theorem: central angle equals twice the inscribed angle", "Exploit the fact that a diagonal of a square makes a forty-five degree angle with its sides...
null
proof only
0.86
Problem: $E$ is a point on the diagonal $BD$ of the square $ABCD$. Show that the points $A$, $E$ and the circumcenters of $ABE$ and $ADE$ form a square.
c706295410cc34445e17b0be9326e1acc352cf549c7176db7ca0a2a191491c2c
Spain
LIX Olimpiada Matemática Española
2,023
null
2
Sea $ABC$ un triángulo acutángulo y escaleno con incentro $I$ y ortocentro $H$. Sea $M$ el punto medio de $AB$. Sobre la recta $AH$ se consideran puntos $D$ y $E$ tales que la recta $MD$ es paralela a $CI$ y $ME$ es perpendicular a $CI$. Prueba que $AE = DH$.
[ "Demostraremos que los segmentos $AH$ y $DE$ tienen el mismo punto medio, lo cual probará que $AE = DH$. Sea $F = AH \\cap BC$. Sea $N$ el punto\n\n![](images/ome59-2023_probs_sols_p3_data_f3b80d3020.png)\n\nEsquema para resolver el problema 2.\nmedio de $DE$, que es el circuncentro del triángulo rectángulo $DEM$. ...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ], [ "Topics", "Geometry", ...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Miscellaneous > Constructions and loci" ]
Spanish
ESP_2023
Spain
true
1
In an acute scalene triangle, pick the midpoint of one side. On the line through a vertex and the orthocenter, choose two points so that one segment from the midpoint is parallel to the line from the incenter and another is perpendicular to it. Show that the distance from the vertex to one point equals the distance fro...
[ "Use properties of the orthocenter to relate perpendicular directions: the altitude from a vertex is perpendicular to the opposite side", "Construct triangle DEM right at M since one segment is parallel to the incenter line and the other is perpendicular to it, implying EM ⟂ MD", "Identify the circumcenter of t...
null
proof only
0.78
Sea $ABC$ un triángulo acutángulo y escaleno con incentro $I$ y ortocentro $H$. Sea $M$ el punto medio de $AB$. Sobre la recta $AH$ se consideran puntos $D$ y $E$ tales que la recta $MD$ es paralela a $CI$ y $ME$ es perpendicular a $CI$. Prueba que $AE = DH$.
100eebfea73cbe7863725615810d78b69d8b130849675d2f35054c5481423357
Switzerland
Swiss Mathematical Olympiad
2,023
First Exam
1.
Problem: Let $ABC$ be an acute triangle with incentre $I$. On its circumcircle, let $M_{A}$, $M_{B}$ and $M_{C}$ be the midpoints of minor arcs $BC$, $CA$ and $AB$ respectively. Prove that the reflection of $M_{A}$ over the line $IM_{B}$ lies on the circumcircle of the triangle $IM_{B}M_{C}$.
[ "Solution:\n\n![](attached_image_solution_1.png)\n\nLet $X$ be the reflection of $M_{A}$ over the line $IM_{B}$. We wish to prove that $X$, $M_{C}$, $I$, $M_{B}$ lie on a circle. By WUM, observe that $A$, $I$, $M_{A}$, $B$, $I$, $M_{B}$ and $C$, $I$, $M_{C}$ are collinear. Therefore, by symmetry, we get\n$$\n\\angl...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geome...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
null
Switzerland__md__mix-olympiad__mix-finalRoundSolution2023
Switzerland
false
0
In an acute triangle, take its incenter and the midpoints of the minor arcs on the circumcircle. Reflect the arc midpoint opposite one side across the line from the incenter to the arc midpoint opposite another side. Show that this reflected point lies on the circle passing through the incenter and the other two arc mi...
[ "Use that each internal angle bisector passes through the midpoint of the opposite minor arc on the circumcircle", "Reflection across a line preserves angles with that line", "Inscribed angle theorem and its converse to establish concyclicity by matching equal angles subtending the same chord", "Angle chasing...
null
proof only
0.9
Problem: Let $ABC$ be an acute triangle with incentre $I$. On its circumcircle, let $M_{A}$, $M_{B}$ and $M_{C}$ be the midpoints of minor arcs $BC$, $CA$ and $AB$ respectively. Prove that the reflection of $M_{A}$ over the line $IM_{B}$ lies on the circumcircle of the triangle $IM_{B}M_{C}$.
4f33e9ee52b770219450c6e9ee9136c979014b34798dc4322418e8b505e78ebb
Taiwan
IMO 1J, Mock Exam 2
2,024
null
5
令 $m$ 與 $n$ 為大於 1 的正整數。在 $m \times n$ 方格紙上的每一格都有一枚背面向上的硬幣。每一步,我們依次進行以下動作: (1) 選擇一個 $2 \times 2$ 的區域; (2) 將該區域左上角與右下角的硬幣翻面; (3) 將該區域左下角與右上角的硬幣擇一翻面。 試求所有 $(m, n)$,使得我們能透過有限步將硬幣全部翻成正面。 Let $m$ and $n$ be positive integers greater than 1. In each square of a $m \times n$ grid lies a coin with its tail-side up. A *move* c...
[ "答案為滿足 $3 \\mid mn$ 的所有 $(m, n)$。\n\n構造:不失一般性假設 $3 \\mid m$。當 $2 \\mid n$ 時,我們可用左圖方式將所有硬幣翻面:\n![](images/2024-TWN_p44_data_bb9cfb2211.png)\n\n當 $2 \\nmid n$ 時,先用上述的方法將左邊的 $m \\times (n-1)$ 全部翻成正面,剩下最右邊一排是背面。接著,令 $L(i, j)$ 為將 $(i, j)$,$(i+1, j)$ 和 $(i, j+1)$ 位置翻面的 $L$ 型操作,而 $R(i, j)$ 為將 $(i, j)$,$(i, j-1)$ 和 $(i-1, ...
[]
[ [ "Topics", "Discrete Mathematics", "Combinatorics", "Invariants / monovariants" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Coloring schemes, extremal arguments" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Games / greedy algorithms" ...
[ "Discrete Mathematics > Combinatorics > Invariants / monovariants", "Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments", "Discrete Mathematics > Combinatorics > Games / greedy algorithms" ]
Chinese; English
TWN_2024
Taiwan
true
1
Find exactly which rectangular grid sizes allow a sequence of permitted local flips to turn all coins from tails to heads.
[ "Three-color the grid by assigning repeating labels to cells and track counts of heads in each class", "Establish parity invariants: the parity of differences between class head counts is preserved by each move", "Compute the distribution of labels in the grid based on the area modulo three to obtain a necessar...
All pairs (m, n) such that 3 divides mn
proof and answer
0.9
令 $m$ 與 $n$ 為大於 1 的正整數。在 $m \times n$ 方格紙上的每一格都有一枚背面向上的硬幣。每一步,我們依次進行以下動作: (1) 選擇一個 $2 \times 2$ 的區域; (2) 將該區域左上角與右下角的硬幣翻面; (3) 將該區域左下角與右上角的硬幣擇一翻面。 試求所有 $(m, n)$,使得我們能透過有限步將硬幣全部翻成正面。 Let $m$ and $n$ be positive integers greater than 1. In each square of a $m \times n$ grid lies a coin with its tail-side up. A *move* c...
08da7d779d92e270a71614d4045a05a886820d04cd378f28096462639e2f1af0
Thailand
The 13th Thailand Mathematical Olympiad
2,016
Mathematical Competitions in Thailand 2016
1
Let $ABC$ be a triangle with $AB \neq AC$. Let the angle bisector of $\angle BAC$ intersect $BC$ at $P$, and intersect the perpendicular bisector of segment $BC$ at $Q$. Prove that $\frac{PQ}{AQ} = \left(\frac{BC}{AB+AC}\right)^2$.
[ "We first show that $Q$ lies on the circumcircle of triangle $ABC$.\n\n![](images/THA_National_2016_p2_data_9162532cf3.png)\n\nLet the perpendicular bisector of $BC$ intersect the circumcircle of $ABC$ at $Q'$. Since $Q'$ bisects the arc $BC$ we have $\\angle BAQ' = \\angle CAQ'$. Thus $Q'$ lies on the angle bisect...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Circles" ], [ "Topics", "Geometry", "Plane Geometry", "Mis...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Circles", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
null
THA_2016
Thailand
true
1
In a nonisoceles triangle, the angle bisector from one vertex meets the opposite side and also meets the perpendicular bisector of that side. Prove that the ratio of the distance between these two intersection points to the distance from the vertex to the second intersection point equals the square of the ratio of the ...
[ "Show the intersection point lies on the circumcircle using perpendicular bisector properties and equal angles", "Angle Bisector Theorem to express segments on the base in terms of side lengths", "Similarity of triangles to relate a length ratio involving the intersection point", "Power of a point with respec...
null
proof only
0.86
Let $ABC$ be a triangle with $AB \neq AC$. Let the angle bisector of $\angle BAC$ intersect $BC$ at $P$, and intersect the perpendicular bisector of segment $BC$ at $Q$. Prove that $\frac{PQ}{AQ} = \left(\frac{BC}{AB+AC}\right)^2$.
13f74e8086c9d7eaab2620d5184be67ceb08f3ebb81507c78a0205e313008d6c
Turkey
Team Selection Test for IMO 2024
2,024
null
1
In scalene triangle $ABC$, the incenter is $I$ and the circumcenter is $O$. $AI$ intersects the circumcircle of $ABC$ a second time at $P$. The line passing through $I$ and perpendicular to $AI$ intersects $BC$ at $X$. The foot of the perpendicular from $X$ to $IO$ is $Y$. Show that the points $A$, $P$, $X$, $Y$ are co...
[ "*Claim 1.* $A$, $X$, $P$, $E$ are concyclic.\n\n*Proof.* $X$, $I$, $M$, $P$ are concyclic because $\\angle XIP = \\angle XMP = 90^\\circ$. It is well known that $S$ lies on the incircle and since $DI = SI$ and $DM = EM$, we get $IM \\parallel AE$. So, $\\angle EAP = \\angle PIM = \\angle PXE$, which means that $A$...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Circles", "Tangents" ], [ "Topics", "Geometry", "Plane Geo...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Miscellaneous > Distance chasing...
English
TUR_2024
Turkey
true
2
Given a scalene triangle with its incenter and circumcenter, construct points using the angle bisector, a perpendicular to this angle bisector through the incenter meeting the opposite side, and a foot from this point to the line joining the centers. Prove that these four constructed points lie on a single circle.
[ "Use angle bisector relations to get similarity of triangles involving the incenter, yielding XI squared equals XB times XC", "Apply power of a point with respect to the circumcircle to relate distances from points on lines perpendicular to the line IO", "Exploit right-angle configurations (perpendiculars to AI...
null
proof only
0.78
In scalene triangle $ABC$, the incenter is $I$ and the circumcenter is $O$. $AI$ intersects the circumcircle of $ABC$ a second time at $P$. The line passing through $I$ and perpendicular to $AI$ intersects $BC$ at $X$. The foot of the perpendicular from $X$ to $IO$ is $Y$. Show that the points $A$, $P$, $X$, $Y$ are co...
e86c8241c88fc404bae424aeaca29c8ca95b278c35b219f0f520b577f90467b1
Ukraine
62nd Ukrainian National Mathematical Olympiad
2,023
Third Round, February 2023
4
For a natural number $n \ge 2$, consider an $n \times n$ board. Let $n^2$ points denote the centers of each of the $1 \times 1$ squares on this board. What is the largest number of these points that can be marked in such a way that no three marked points form the vertices of a right triangle? *(Mykhailo Shtandenko)*
[ "We will show that there is an example where the marked number of points satisfies the conditions of the problem. For this, we will denote the centers of all the cells in the first row and the first column, except for the center of the cell at the intersection of the first row and the first column (see Fig. 3). By ...
[]
[ [ "Topics", "Discrete Mathematics", "Combinatorics", "Pigeonhole principle" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Coloring schemes, extremal arguments" ] ]
[ "Discrete Mathematics > Combinatorics > Pigeonhole principle", "Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments" ]
English
UKR_2022
Ukraine
true
1
On a square grid of unit cells, you consider the centers of all cells. Determine the maximum number of these centers you can choose so that no three chosen centers are the corners of a right triangle.
[ "Constructive example: choose all centers in one fixed row and one fixed column, omitting their intersection, to avoid any right triangle.", "Observation that any axis-aligned right triangle would require the omitted intersection point as the right-angle vertex in that construction.", "Upper bound via counting ...
2n-2
proof and answer
0.93
For a natural number $n \ge 2$, consider an $n \times n$ board. Let $n^2$ points denote the centers of each of the $1 \times 1$ squares on this board. What is the largest number of these points that can be marked in such a way that no three marked points form the vertices of a right triangle? *(Mykhailo Shtandenko)*
1bac32aa131c082483f5d043ead34aaadcb83b5a40bd70f9fbcc74484582633b
United States
AIME II
2,025
null
1
Six points $A$, $B$, $C$, $D$, $E$, and $F$ lie in a straight line in that order. Suppose that $G$ is a point not on the line and that $AC = 26$, $BD = 22$, $CE = 31$, $DF = 33$, $AF = 73$, $CG = 40$, and $DG = 30$. Find the area of $\triangle BGE$.
[ "Because $CD = AF - AC - DF = 14$, the side lengths of $\\triangle CDG$ are $14$, $30$, and $40$. By Heron's Formula,\n$$\n\\text{Area}(\\triangle CDG) = \\sqrt{42(42 - 14)(42 - 30)(42 - 40)} = 168,\n$$\nimplying that the distance from $G$ to line $CD$ is $\\frac{2\\cdot168}{CD} = 24$. Then because $BE = BD + CE - ...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Distance chasing" ] ]
[ "Geometry > Plane Geometry > Miscellaneous > Distance chasing" ]
null
USA_2025a
United States
true
1
Six points lie on a straight line in order and there is another point off the line. Several distances along the line and from the off-line point to two of the collinear points are given. Determine the area of the triangle formed by the off-line point and the second and fifth points on the line.
[ "Use segment additivity along collinear points to express unknown lengths (compute CD and BE from given totals).", "Apply Heron's formula to a triangle with known side lengths to find its area.", "Convert area to altitude to obtain the perpendicular distance from the off-line point to the baseline.", "Exploit...
468
final answer only
0.93
Six points $A$, $B$, $C$, $D$, $E$, and $F$ lie in a straight line in that order. Suppose that $G$ is a point not on the line and that $AC = 26$, $BD = 22$, $CE = 31$, $DF = 33$, $AF = 73$, $CG = 40$, and $DG = 30$. Find the area of $\triangle BGE$.
c664026a894bf999cf42a3d12e6d591c5c82ba4a4fdca29a21667f381acc1385
Vietnam
Vietnamese MO
2,024
Day 1
3
Let $ABC$ be an acute triangle with circumcenter $O$. Let $A'$ be the center of the circle passing through $C$ and tangent to $AB$ at $A$, let $B'$ be the center of the circle passing through $A$ and tangent to $BC$ at $B$, let $C'$ be the center of the circle passing through $B$ and tangent to $CA$ at $C$. a) Prove t...
[ "a) Let $(A'), (B'), (C')$ respectively represent the circle passing through point $C$ and touching the line $AB$ at point $A$, the circle passing through point $B$ and touching the line $BC$ at point $B$, and the circle passing through point $C$ and touching the line $CA$ at point $C$.\n\nLet $K$ be the second int...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Circles", "Tangents" ], [ "Topics", "Geometry", "Plane Geo...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Transformations > Homothety", "Geometry > Plane Geometry > Transformations > Rotation", "Ge...
English
VNM_2024
Vietnam
true
2
Given an acute triangle, define three circle centers, each center belonging to a circle tangent to one side at a vertex and passing through the opposite vertex. First, prove that the triangle formed by these centers has area at least that of the original triangle. Then, project the circumcenter of the original triangle...
[ "Construct the common second intersection point of the three tangent circles and use equal tangent-angle relations to show it lies on all three", "Apply Erdős-type inequality relating sums of distances from a point to vertices and to sides to bound an angle and control similarity ratio", "Use a rotational homot...
null
proof only
0.68
Let $ABC$ be an acute triangle with circumcenter $O$. Let $A'$ be the center of the circle passing through $C$ and tangent to $AB$ at $A$, let $B'$ be the center of the circle passing through $A$ and tangent to $BC$ at $B$, let $C'$ be the center of the circle passing through $B$ and tangent to $CA$ at $C$. a) Prove t...
68f19c96f8652d343e56ba1cc97c3f124438823de9878756e1c5174d83958c16
Zhautykov Olympiad
Zhautykov Olympiad
2,021
null
№2
Problem: In a convex cyclic hexagon $A B C D E F$, $B C = E F$ and $C D = A F$. Diagonals $A C$ and $B F$ intersect at point $Q$, and diagonals $E C$ and $D F$ intersect at point $P$. Points $R$ and $S$ are marked on the segments $D F$ and $B F$ respectively so that $F R = P D$ and $B Q = F S$. The segments $R Q$ and ...
[ "Solution:\n\nIt follows obviously that $B F \\parallel C E$ and $A C \\parallel D F$. We denote the circumcircles of $\\triangle A B Q$ and $\\triangle D E P$ by $\\omega_{1}$ and $\\omega_{2}$, respectively. Note that the lines $A D$ and $B E$ are internal common tangents to $\\omega_{1}$ and $\\omega_{2}$. Indee...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Circles", "Tangents" ], [ "Topics", "Geometry", "Plane Geometry", "Concurrency and Collinearity", "Menelaus' theorem" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ] ]
[ "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
null
IZho__md__en-2021_zhautykov_resenja_e
Zhautykov Olympiad
false
0
In a cyclic six sided figure with two pairs of equal sides, certain intersections of diagonals are used to define points on two edges at matching distances. Lines through these points meet, and the task is to prove that the line from this intersection to a fixed vertex cuts the opposite diagonal into two equal parts.
[ "Equal chords in a circle imply parallel chords and diagonals, yielding multiple parallelograms in the configuration", "Power of a point with tangent lengths: equating products along a secant with the square of a tangent length to deduce equal segment lengths on a baseline", "Using parallels from the constructe...
null
proof only
0.83
Problem: In a convex cyclic hexagon $A B C D E F$, $B C = E F$ and $C D = A F$. Diagonals $A C$ and $B F$ intersect at point $Q$, and diagonals $E C$ and $D F$ intersect at point $P$. Points $R$ and $S$ are marked on the segments $D F$ and $B F$ respectively so that $F R = P D$ and $B Q = F S$. The segments $R Q$ and ...
e6f1e232262009f6b6fb1a332fb2df5222662262ed16b2a93c17b5b1d5442fcf
Argentina
Rioplatense Mathematical Olympiad
2,023
Level A
A.2.
Ana placed the numbers from $1$ to $9$ in the squares of the figure, one in each square, without repeating numbers. It turned out that, for each of the four arrows indicated, the sum of the three numbers in that direction is equal to the number of Ana's cats. How many cats does Ana have? Find all possibilities. ![](ima...
[ "Denote the numbers in the squares as shown in the figure, and let $x$ be the number of cats Ana has. If we add up both vertical arrows plus the top horizontal arrow we find that each number appears exactly once on this sum, except for $a$ which is added twice, and $e$ which does not appear on the sum. Hence, since...
[]
[ [ "Topics", "Discrete Mathematics", "Combinatorics", "Counting two ways" ], [ "Topics", "Algebra", "Equations and Inequalities", "Linear and quadratic inequalities" ], [ "Topics", "Algebra", "Prealgebra / Basic Algebra", "Integers" ] ]
[ "Discrete Mathematics > Combinatorics > Counting two ways", "Algebra > Equations and Inequalities > Linear and quadratic inequalities", "Algebra > Prealgebra / Basic Algebra > Integers" ]
English
ARG_2024
Argentina
true
2
Arrange the numbers from one to nine in a small grid so that the total along each indicated arrow is the same, and determine what that common total can be, listing all possible values.
[ "Count the sums in two ways by aggregating selected arrow sums to relate the common sum to the total of all entries", "Use the known total of the numbers from one to nine to derive an equation involving the common sum and two specific positions", "Bound a difference of two distinct entries to produce an interva...
13, 14, 16, 17
proof and answer
0.93
Ana placed the numbers from $1$ to $9$ in the squares of the figure, one in each square, without repeating numbers. It turned out that, for each of the four arrows indicated, the sum of the three numbers in that direction is equal to the number of Ana's cats. How many cats does Ana have? Find all possibilities. ![](ima...
5b1ccf41cc80e3ca454bbc3ada356c4ba401bfaa601f95e99f9d9f936d6d24f0
Asia Pacific Mathematics Olympiad (APMO)
APMO
2,024
null
5
Line $\ell$ intersects sides $B C$ and $A D$ of cyclic quadrilateral $A B C D$ in its interior points $R$ and $S$ respectively, and intersects ray $D C$ beyond point $C$ at $Q$, and ray $B A$ beyond point $A$ at $P$. Circumcircles of the triangles $Q C R$ and $Q D S$ intersect at $N \neq Q$, while circumcircles of the ...
[ "We start with the following lemma.\nLemma 1. Points $M, N, P, Q$ are concyclic.\nPoint $M$ is the Miquel point of lines $A P=A B, P S=\\ell, A S=A D$, and $B R=B C$, and point $N$ is the Miquel point of lines $C Q=C D, R C=B C, Q R=\\ell$, and $D S=A D$. Both points $M$ and $N$ are on the circumcircle of the trian...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Circles", "Radical axis theorem" ], [ "Topics", "Geometry", "Plane Geometry", "Advanced Configurations", "Miquel point...
[ "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Circles > Radical axis theorem", "Geometry > Plane Geometry > Advanced Configurations > Miquel point", "Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates...
English
apmo2024_sol
Asia Pacific Mathematics Olympiad (APMO)
false
0
In a cyclic quadrilateral, a line meets two sides and the extensions of the other two sides. Two pairs of circumcircles define special intersection points, and the claim is that the intersection of the lines through these points and the external points lies on the line joining the intersections of the opposite sides of...
[ "Use Miquel points associated to complete quadrilaterals formed by the given lines", "Angle chasing with directed angles to prove concyclicity of four key points", "Exploit the property that the Miquel point of the quadrilateral lies on the line joining intersections of opposite sides", "Consider the circumci...
null
proof only
0.89
Line $\ell$ intersects sides $B C$ and $A D$ of cyclic quadrilateral $A B C D$ in its interior points $R$ and $S$ respectively, and intersects ray $D C$ beyond point $C$ at $Q$, and ray $B A$ beyond point $A$ at $P$. Circumcircles of the triangles $Q C R$ and $Q D S$ intersect at $N \neq Q$, while circumcircles of the ...
fb61e31fd2a556b6e140fa18eabdfde0e75526393bbd63ee24deda7433f8ee91
Austria
Austrian Mathematical Olympiad
2,024
Regional Competition
6
Let $\mathcal{ABC}$ be an acute triangle with orthocenter $H$. The circumcircle of the triangle $\mathcal{BHC}$ intersects $\mathcal{AC}$ a second time in point $P$ and $\mathcal{AB}$ a second time in point $Q$. Prove that $H$ is the circumcenter of the triangle $\mathcal{APQ}$.
[ "![](images/AUT_ABooklet_2024_p9_data_9ab510c722.png)\nFigure 2: Problem 6\n\nLet $H_a$ be the foot of the altitude on $BC$. With the angle sum in triangle $AH_aC$, we get\n$$\n\\angle HAC = 90^\\circ - \\angle BCA.\n$$\nLet $H_b$ be the foot of the altitude on $AC$. With the angle sum in triangle $CH_bB$, we get\n...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ] ]
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
English
AUT_2024
Austria
true
1
In an acute triangle with its orthocenter, take the circle through the orthocenter and two vertices and mark where this circle meets the two sides again. Show that the orthocenter is the center of the circle passing through the original vertex and those two new points.
[ "Use angle sums in right triangles formed by altitudes to relate key angles at the orthocenter and feet of altitudes", "Apply the inscribed angle theorem on the circumcircle through two vertices and the orthocenter", "Establish isosceles triangles to show equal distances from the orthocenter to the relevant poi...
null
proof only
0.9
Let $\mathcal{ABC}$ be an acute triangle with orthocenter $H$. The circumcircle of the triangle $\mathcal{BHC}$ intersects $\mathcal{AC}$ a second time in point $P$ and $\mathcal{AB}$ a second time in point $Q$. Prove that $H$ is the circumcenter of the triangle $\mathcal{APQ}$.
91a2a7b1da65e719783b7c3ff5d8da4413423e3f1b450b38598567224d72b7ff
Balkan Mathematical Olympiad
Balkan Mathematical Olympiad Shortlisted Problems
2,025
Combinatorics
C3
A rabbit is at some point $(x, y)$ in the Euclidean plane. There are some (possibly infinitely many) landmines, which are circles with any radius that do not intersect except possibly at one point (tangent). Every move, the rabbit can hop a distance of exactly $1$, but cannot land in the interior of a landmine (but may...
[ "We claim the answer is $\\frac{1}{\\sqrt{2}}$.\n\nFirstly, we prove that there is a configuration of landmines such that the rabbit cannot be closer than $\\frac{1}{\\sqrt{2}}$ to the origin.\nConsider a square packing of circles, with centers $\\left(\\frac{2m+1}{\\sqrt{2}}, \\frac{2n+1}{\\sqrt{2}}\\right)$ for a...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Circles", "Tangents" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Distance chasing" ], [ "Topics", "Geometry", "Plane Geometry", "Geometric Inequalities", "Optimization in geometry" ] ]
[ "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Miscellaneous > Distance chasing", "Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry" ]
English
BMO_2025
Balkan Mathematical Olympiad
true
4
A rabbit moves in the plane by fixed step lengths and must avoid the interiors of circular hazards that do not overlap. Determine the smallest guaranteed proximity to the origin that the rabbit can always achieve, no matter how the hazards are placed.
[ "Construct a square packing of equal tangent circles to give a lower bound that traps the rabbit on tangency points", "Use intersection arcs of the unit circle around the rabbit with a circle centered at the origin to find a safe landing point", "Perpendicular bisector region argument to show no single landmine...
1/sqrt(2)
proof and answer
0.86
A rabbit is at some point $(x, y)$ in the Euclidean plane. There are some (possibly infinitely many) landmines, which are circles with any radius that do not intersect except possibly at one point (tangent). Every move, the rabbit can hop a distance of exactly $1$, but cannot land in the interior of a landmine (but may...
bff74d2b8dc9adfc07435727d64f64e4613e62c1cd1c8f3e8e2955127287600a
Baltic Way
Baltic Way 2023 Shortlist
2,023
Combinatorics
C 7
Determine if there exists a triangle that can be cut into 101 congruent triangles.
[ "Answer: Yes, there is.\nChoose an arbitrary positive integer $m$ and draw a height in the right triangle with ratio of legs $1 : m$. This height cuts the triangle in two similar triangles with similarity coefficient $m$. The largest of them can further be cut into $m^2$ smaller equal triangles by splitting all sid...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles" ], [ "Topics", "Geometry", "Plane Geometry", "Transformations", "Homothety" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Constructions and loci" ] ]
[ "Geometry > Plane Geometry > Triangles", "Geometry > Plane Geometry > Transformations > Homothety", "Geometry > Plane Geometry > Miscellaneous > Constructions and loci" ]
English
BWS_2023
Baltic Way
true
1
Decide whether it is possible to split a single triangle into one hundred and one identical smaller triangles, and if so, describe how to construct such a division.
[ "Use altitude in a right triangle to split it into two similar sub-triangles with a prescribed integer similarity ratio", "Choose legs in ratio one to m so the two sub-triangles differ by a linear scale factor of m", "Partition the larger sub-triangle into m^2 congruent copies by dividing sides into m equal par...
Yes
proof and answer
0.9
Determine if there exists a triangle that can be cut into 101 congruent triangles.
32812a21b75780bf25d0131b939194e9ea14fc6f181df9e07fd3084396f108ec
Belarus
SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO
2,024
Fourth Selection Test
16
Given positive integers $n$ and $k \le n$. Consider an equilateral triangular board with side $n$, which consists of circles: in the first (top) row there is one circle, in the second row there are two circles, ..., in the bottom row there are $n$ circles (see the figure below). Let us place checkers on this board so t...
[ "a. Let us prove that in any such placement of checkers on a triangular board with side $n$, the number of checkers $q$ satisfies the inequality\n$$\nq \\le \\frac{2n + 1}{3} \\cdot k.\n$$\nThe main idea of the proof is double counting. Note that wherever a checker stands, if you count all the cells of the three li...
[]
[ [ "Topics", "Discrete Mathematics", "Combinatorics", "Counting two ways" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Coloring schemes, extremal arguments" ], [ "Topics", "Algebra", "Equations and Inequalities", "Combinatorial optimization" ] ]
[ "Discrete Mathematics > Combinatorics > Counting two ways", "Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments", "Algebra > Equations and Inequalities > Combinatorial optimization" ]
English
BLR_2024
Belarus
true
5
On a triangular grid of points arranged in rows, place as many checkers as possible so that no line parallel to a side contains more than a given limit. Prove a universal upper bound and determine the exact maximum when the limit equals one and when it equals two.
[ "Double counting by summing contributions of three families of parallel lines through each checker", "Weighted sum over lines by their lengths to bound the total, with a greedy/majorization-type argument to maximize weight under per-line caps", "Decompose the total number of checkers as multiples of the cap plu...
Upper bound: T(k, n) ≤ floor(k(2n + 1)/3). Exact values: T(1, n) = floor((2n + 1)/3) and T(2, n) = floor((4n + 2)/3).
proof and answer
0.82
Given positive integers $n$ and $k \le n$. Consider an equilateral triangular board with side $n$, which consists of circles: in the first (top) row there is one circle, in the second row there are two circles, ..., in the bottom row there are $n$ circles (see the figure below). Let us place checkers on this board so t...
12b450047e32f9c67bc26886565f7f85753205025f84519e596fcd539798dfee
Benelux Mathematical Olympiad
16th Benelux Mathematical Olympiad
2,024
null
Problem 3
Problem: Let $ABC$ be a triangle with incentre $I$ and circumcircle $\Omega$ such that $|AC| \neq |BC|$. The internal angle bisector of $\angle CAB$ intersects side $[BC]$ in $D$, and the external angle bisectors of $\angle ABC$ and $\angle BCA$ intersect $\Omega$ again in $E$ and $F$, respectively. Let $G$ be the int...
[ "Solution:\n\nWe first notice the general fact that $EF \\perp AI$. This can be proved using the following argument. Denote $S$ for the intersection of $EF$ and $AI$. Then $\\angle BIS = (\\angle IBA + \\angle IAB) = \\frac{1}{2}(\\angle ABC + \\angle BAC) = \\frac{1}{2}(180^{\\circ} - \\angle BCA) = \\angle BCF = ...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geome...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Miscellaneou...
null
Benelux_MO__md__Benelux_en-olympiad_en-bxmo-problems-2024-zz
Benelux Mathematical Olympiad
true
1
In a triangle, certain points are constructed using angle bisectors and the circumcircle. The task is to prove that a specific point lies on a circle through three fixed points if and only if another constructed point lies on the same circle.
[ "Establish that the line through the external bisector points is perpendicular to the internal bisector line via cyclic angle chasing", "Use cyclic quadrilateral equivalences and right-angle conditions to link positions of points on a common circumcircle", "Employ the excenter as the concurrency point of angle ...
null
proof only
0.9
Problem: Let $ABC$ be a triangle with incentre $I$ and circumcircle $\Omega$ such that $|AC| \neq |BC|$. The internal angle bisector of $\angle CAB$ intersects side $[BC]$ in $D$, and the external angle bisectors of $\angle ABC$ and $\angle BCA$ intersect $\Omega$ again in $E$ and $F$, respectively. Let $G$ be the int...
0f25312086932d12d6d234e81ae46e217c2a2706c50c49202065d84df483fb1f
Brazil
Brazilian Mathematical Olympiad
2,020
Nível 2
4
Problem: Seja $ABCD$ um quadrilátero tal que $AC = BC + CD$. Se $\angle BCD = 120^{\circ}$, $\overline{CA}$ é bissetriz e $AB = x$, qual o valor de $BD$, em função de $x$?
[ "Solution:\n\nVamos analisar a figura:\n![](attached_image_solution_1.png)\n\nAplicando a Lei dos Cossenos no triângulo $ABC$, temos:\n$$\n\\begin{aligned}\n& x^{2} = b^{2} + (b + c)^{2} - 2 \\cdot b \\cdot (b + c) \\cdot \\cos 60^{\\circ} \\\\\n& x^{2} = b^{2} + b^{2} + 2 b c + c^{2} - 2 b^{2} \\frac{1}{2} - 2 b c...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle trigonometry" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ] ]
[ "Geometry > Plane Geometry > Triangles > Triangle trigonometry", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
null
Brazilian_MO__md__pt-bq2020_N2
Brazil
false
0
Given a quadrilateral where one diagonal equals the sum of two adjacent sides, the angle at a vertex is one hundred twenty degrees, and that diagonal bisects this angle, determine the length of the other diagonal in terms of a given side length.
[ "Use the angle bisector information to deduce a sixty-degree angle at the shared vertex", "Apply the Law of Cosines in the triangle formed by the diagonal and one side, using the given sum relation for a side", "Apply the Law of Cosines in the adjacent triangle with an angle of one hundred twenty degrees", "E...
x
proof and answer
0.93
Problem: Seja $ABCD$ um quadrilátero tal que $AC = BC + CD$. Se $\angle BCD = 120^{\circ}$, $\overline{CA}$ é bissetriz e $AB = x$, qual o valor de $BD$, em função de $x$?
57ee64670cc54740bf11bebf3091ca6f790a0f3fc764310ffac1dea8763c9f59
Bulgaria
Bulgarian Winter Tournament
2,024
9.4
2.8
There are 11 points equally spaced on a circle. Some of the segments having endpoints among these vertices are drawn and colored in two colors, so that each segment meets at an internal point at most one other segment from the same color. What is the greatest number of segments that could be drawn? (Mladen Vylkov)
[ "![](images/BGR_ABooklet_2024_p17_data_7e352642f3.png)\n\nPut $n$ instead of 11 and let the points be $A_1, A_2, \\dots, A_n$. We may assume that the points are vertices of a regular $n$-gon. Let us first calculate the maximum number of diagonals that we can draw so that any diagonal meets no more than one other di...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Combinatorial Geometry" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Coloring schemes, extremal arguments" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Induction / smoothing" ] ]
[ "Geometry > Plane Geometry > Combinatorial Geometry", "Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments", "Discrete Mathematics > Combinatorics > Induction / smoothing" ]
English
BGR_2024
Bulgaria
true
2
Among eleven equally spaced points on a circle, draw as many colored segments between points as possible so that within each color any segment intersects at most one other segment in its interior. Determine the maximum number of segments that can be drawn.
[ "Reduce to counting diagonals per color since polygon sides create no interior crossings and can be added freely at the end.", "Define R(n) as the maximal number of diagonals with at most one interior crossing per diagonal and prove by induction that R(n) = ceil(3(n−3)/2).", "In an extremal configuration pick t...
35
proof and answer
0.79
There are 11 points equally spaced on a circle. Some of the segments having endpoints among these vertices are drawn and colored in two colors, so that each segment meets at an internal point at most one other segment from the same color. What is the greatest number of segments that could be drawn? (Mladen Vylkov)
07b0e238d532ce500667ab3e5075ad14a03bcc91f484b42afce0e8413f9098d0
Canada
CMO 2023
2,023
Official Solutions
3
Problem: An acute triangle is a triangle that has all angles less than $90^{\circ}$ ($90^{\circ}$ is a Right Angle). Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, and $CF$ meeting at $H$. The circle passing through points $D$, $E$, and $F$ meets $AD$, $BE$, and $CF$ again at $X$, $Y$, and $Z$ respectively....
[ "Solution:\n\nLet the circumcircle of $ABC$ meet the altitudes $AD$, $BE$, and $CF$ again at $I$, $J$, and $K$ respectively.\n\n![](attached_image_solution_1.png)\n\nLemma (9-point circle). $I$, $J$, $K$ are the reflections of $H$ across $BC$, $CA$, $AB$. Moreover, $D$, $E$, $F$, $X$, $Y$, $Z$ are the midpoints of ...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geome...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Transformations > Homothety", "Algebra > Equations and Inequalities > QM-...
null
CANADA_MO__md__en-cmo2023-solutions-en
Canada
false
0
Given an acute triangle with its altitudes and the circle through the feet of those altitudes, prove that the sum of three ratios comparing distances from the orthocenter to certain segments along the altitudes is at least three.
[ "Identify the circle through the altitude feet as the nine-point circle; use properties that it also contains the midpoints of segments from the orthocenter to the vertices", "Use reflections of the orthocenter across the sides via the circumcircle intersections of the altitudes to relate key points", "Apply a ...
null
proof only
0.86
Problem: An acute triangle is a triangle that has all angles less than $90^{\circ}$ ($90^{\circ}$ is a Right Angle). Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, and $CF$ meeting at $H$. The circle passing through points $D$, $E$, and $F$ meets $AD$, $BE$, and $CF$ again at $X$, $Y$, and $Z$ respectively....
e11d174d1b2b7b1a1f241e29d973af4c80a34b2855220e9895c04b39791f7089
China
China-TST-2025A
2,025
Test 1 · Day 1
Problem 2
Let point $P$ lie on the nine-point circle of triangle $ABC$. A line through $P$ perpendicular to $AP$ intersects $BC$ at $Q$. A line through $A$ perpendicular to $AQ$ intersects $PQ$ at $X$. Let $H$ be the orthocenter of triangle $ABC$, and let $D$ and $M$ be the midpoints of segments $BC$ and $AQ$, respectively. Prov...
[ "![](images/China-TST-2025A_p9_data_eb7ee2987d.png)\n\n**Proof:** Let $N$ be the midpoint of $AH$. By the properties of the nine-point circle, $DN$ is its diameter, so $DP \\perp PN$. Since $AH \\perp BC$ and $AP \\perp PQ$, we have $\\triangle DPQ \\sim \\triangle NPA$. Therefore, $\\frac{DQ}{NA} = \\frac{PQ}{PA}$...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geome...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
English
CHN_2025
China
true
1
From a triangle, construct points using perpendicular lines through a point on the nine-point circle and through a vertex. Show that the line from the constructed intersection to the orthocenter is perpendicular to the line joining the midpoints of a side and a constructed segment.
[ "Use the nine-point circle property that the midpoints of the altitude and the opposite side form a diameter, implying a right angle subtended at the point on the circle", "Establish chains of similar triangles induced by perpendicular constructions to relate key length ratios", "Leverage midpoint relationships...
null
proof only
0.9
Let point $P$ lie on the nine-point circle of triangle $ABC$. A line through $P$ perpendicular to $AP$ intersects $BC$ at $Q$. A line through $A$ perpendicular to $AQ$ intersects $PQ$ at $X$. Let $H$ be the orthocenter of triangle $ABC$, and let $D$ and $M$ be the midpoints of segments $BC$ and $AQ$, respectively. Prov...
6f98270b09de2f00caf29a4099536ca189b7fda836736e5bdf82a4a1c353af39
Croatia
Croatian Mathematical Olympiad
2,019
Final test for IMO team selection
G4
A point $T$ is chosen inside the triangle $ABC$. Let $A_1, B_1$ and $C_1$ be the reflections of $T$ across the lines $BC, CA$ and $AB$, respectively. The lines $A_1T, B_1T$ and $C_1T$ intersect the circle $k$ circumscribed to the triangle $A_1B_1C_1$ again at $A_2, B_2$ and $C_2$, respectively. Prove that the lines $AA...
[ "Let $K$ be the intersection of $CC_2$ and $k$.\n\n![](images/HRV_ABooklet_2019_p40_data_317fb561be.png)\n\nSince $CB$ and $CA$ are the bisectors of $\\overline{TA_1}$ and $\\overline{TB_1}$, respectively, the point $C$ is the circumcentre of the triangle $A_1TB_1$. Hence,\n$$\n\\triangle(CA_1, CB) = \\triangle(CB,...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ] ]
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
English
HRV_2019
Croatia
true
1
Pick a point inside a triangle and reflect it across the sides. Draw the circle through the three reflected points. From the point connect to each reflection and extend to meet the circle again. Show that the lines from the triangle’s vertices to these new points all meet at a single point on that circle.
[ "Reflection across a side implies that the side is the perpendicular bisector of the segment joining the original and reflected point", "Identify a vertex as the circumcenter of a triangle formed by the original point and two of its reflections", "Use inscribed angle equalities and chord-angle relationships on ...
null
proof only
0.86
A point $T$ is chosen inside the triangle $ABC$. Let $A_1, B_1$ and $C_1$ be the reflections of $T$ across the lines $BC, CA$ and $AB$, respectively. The lines $A_1T, B_1T$ and $C_1T$ intersect the circle $k$ circumscribed to the triangle $A_1B_1C_1$ again at $A_2, B_2$ and $C_2$, respectively. Prove that the lines $AA...
28e876620721f67a678a8ae499dc9f5e99e38150f6933bd6cc8f75e051cc943c
Czech Republic
First Round of the 73rd Czech and Slovak Mathematical Olympiad (take-home part)
2,024
null
3
Let $T$ be the centroid of a triangle $ABC$. Consider two isosceles right-angled triangles $BTK$ and $CTL$ so that $K$ lies in the half-plane $BTC$ and $L$ lies in the half-plane $CTA$. Finally, denote the centre of the side $BC$ as $D$ and the centre of $KL$ as $E$. Determine all the possible values of the ratio $\fra...
[ "We shall prove that the ratio has to be equal to $2\\sqrt{2}$.\n![](images/CZE_ABooklet_2024_p1_data_db4f1a6942.png)\nFirst, we shall observe that the triangles $BTC$ and $KTL$ (coloured turquoise and yellow in the diagram) are similar, since\n$$\n\\angle BTC = \\angle BTK + \\angle KTC = 45^\\circ + \\angle KTC =...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ], [ "Topics", "Geometry", ...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Miscellaneous > Distance chasing" ]
English
CZE_2024
Czech Republic
true
1
In a triangle, build two right isosceles triangles at the centroid along two sides, then take the midpoints of one original side and of the segment joining the new apexes. Find the constant value of the ratio between the distance from a vertex to the centroid and the distance between these two midpoints.
[ "Establish similarity between the triangles formed by the centroid with two vertices and by the apexes of the constructed right isosceles triangles, using angle chasing and the isosceles right condition", "Use midpoints corresponding under similarity to infer similarity of auxiliary triangles involving the midpoi...
2√2
proof and answer
0.9
Let $T$ be the centroid of a triangle $ABC$. Consider two isosceles right-angled triangles $BTK$ and $CTL$ so that $K$ lies in the half-plane $BTC$ and $L$ lies in the half-plane $CTA$. Finally, denote the centre of the side $BC$ as $D$ and the centre of $KL$ as $E$. Determine all the possible values of the ratio $\fra...
1c69b9f511429c3d9114e6ce0b205400c206915bc62dc3126bba1c0aaec7cf09
Czech-Polish-Slovak Mathematical Match
CAPS Match 2025
2,025
Second day – 18 June 2025
5
We are given an acute triangle $ABC$. Point $D$ lies in the halfplane $AB$ containing $C$ and satisfies $DB \perp AB$ and $\angle ADB = 45^\circ + \frac{1}{2}\angle ACB$. Similarly, $E$ lies in the halfplane $AC$ containing $B$ and satisfies $AC \perp EC$ and $\angle AEC = 45^\circ + \frac{1}{2}\angle ABC$. Let $F$ be ...
[ "Denote $\\angle ABC = \\beta$ and $\\angle ACB = \\gamma$. The conditions translate as $\\angle BAD = 45^\\circ - \\gamma$ and $\\angle EAC = 45^\\circ - \\beta$. Denote by $G$ the intersection point of $BD$ and $CE$. Clearly $\\angle BAG = 90^\\circ - \\gamma = 2\\angle BAD$, and so $AD$ is the angle bisector of ...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Transformations", "Homothety" ], [ "Topics", "Geometry", "Plane Geometry", "Transformations", "Rotation" ], [ "Topics", "Geometry", "Plane Geometry", "Transformations", "Spiral similarity" ], [ "To...
[ "Geometry > Plane Geometry > Transformations > Homothety", "Geometry > Plane Geometry > Transformations > Rotation", "Geometry > Plane Geometry > Transformations > Spiral similarity", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Circles" ]
null
CZA_2025
Czech-Polish-Slovak Mathematical Match
true
2
In an acute triangle, construct two points by dropping perpendiculars from one side of each vertex and fixing certain angle measures. Reflect one vertex across the midpoint of the arc opposite that vertex on the circumcircle. Prove that these four constructed points all lie on the same circle.
[ "Introduce midpoints of the given segments and use a homothety centered at the vertex to relate the desired circle to an auxiliary circle", "Angle chasing to identify angle bisectors and key equal angles in the configuration", "Use congruent triangles derived from equal angles and equal radii to obtain length e...
null
proof only
0.79
We are given an acute triangle $ABC$. Point $D$ lies in the halfplane $AB$ containing $C$ and satisfies $DB \perp AB$ and $\angle ADB = 45^\circ + \frac{1}{2}\angle ACB$. Similarly, $E$ lies in the halfplane $AC$ containing $B$ and satisfies $AC \perp EC$ and $\angle AEC = 45^\circ + \frac{1}{2}\angle ABC$. Let $F$ be ...
8e09bb98bbd9248a0edb7eb983badf752dfc951e85403d2a2e2ce035b90b3841
Estonia
Estonian Mathematical Olympiad
2,025
Selected Problems from Open Contests
O6
In an isosceles triangle $ABC$ with $AB = AC$, the bisector of the angle $BAC$ intersects $BC$ at $D$. The bisector of the angle $ABC$ intersects the perpendicular bisector of $AD$ at $E$. Prove that the bisector of the angle $ACB$ is perpendicular to $DE$.
[ "As the $A$-bisector is also the altitude, we have $BD \\perp AD$. Let $I$ be the incenter of $ABC$ and $E' \\neq B$ the intersection of $BI$ and the circumcircle of $ABD$ (Fig. 2). As the angles $\\angle ABE' = \\angle DBE'$ correspond to the arcs $E'A$ and $E'D$ of this circle, we have $E'A = E'D$. Thus $E'$ lies...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geome...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Miscellaneous > Angle c...
English
EST_2025
Estonia
true
3
In an isosceles triangle, a point is defined by intersecting one angle bisector with a perpendicular bisector. Show that the other base angle bisector is perpendicular to the line through this point and a specific point on the base.
[ "Use the fact that in an isosceles triangle the vertex-angle bisector is also the altitude, giving perpendicularity to the base", "Identify the incenter and incircle touchpoints to relate equal tangency segments and form an isosceles subtriangle", "Construct E as the intersection of the B-angle bisector with th...
null
proof only
0.88
In an isosceles triangle $ABC$ with $AB = AC$, the bisector of the angle $BAC$ intersects $BC$ at $D$. The bisector of the angle $ABC$ intersects the perpendicular bisector of $AD$ at $E$. Prove that the bisector of the angle $ACB$ is perpendicular to $DE$.
43d1dbf778e086b837f87170c1ae84fee60879839fce1c85f7c16eec0910bd43
European Girls' Mathematical Olympiad (EGMO)
EGMO
2,025
Day 2
P4.
Problem: Let $ABC$ be an acute triangle with incentre $I$ and $AB \neq AC$. Let lines $BI$ and $CI$ intersect the circumcircle of $ABC$ at $P \neq B$ and $Q \neq C$, respectively. Consider points $R$ and $S$ such that $AQRB$ and $ACSP$ are parallelograms (with $AQ \parallel RB$, $AB \parallel QR$, $AC \parallel SP$, a...
[ "Solution:\n\nWe will prove that $\\triangle BIR \\sim \\triangle CIS$, since the statement then follows from $\\angle TRI = \\angle BRI = \\angle CSI = \\angle TSI$.\n\n![](attached_image_solution_1.png)\n\nStep 1. Let us prove $\\angle RBI = \\angle SCI$. We will use directed angles:\n\n$$(BR,BI) = (BR,AB) + (AB,...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geome...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Transformations > Spiral similarity", "Geometry > Plane Geometry > Transf...
null
EGMO__md__en-2025-solutions
European Girls' Mathematical Olympiad (EGMO)
false
0
In a triangle, construct points by extending lines from the incentre to meet the circumcircle and forming two parallelograms. Show that the incentre and three of these constructed points lie on a single circle.
[ "Establish similarity between triangles formed with the incentre via angle chasing and length ratios", "Use properties of the parallelograms to equate segments (e.g., opposite sides equal) and transfer relations", "Prove similarity of two key triangles built from the chord intersections and the incentre", "Ap...
null
proof only
0.86
Problem: Let $ABC$ be an acute triangle with incentre $I$ and $AB \neq AC$. Let lines $BI$ and $CI$ intersect the circumcircle of $ABC$ at $P \neq B$ and $Q \neq C$, respectively. Consider points $R$ and $S$ such that $AQRB$ and $ACSP$ are parallelograms (with $AQ \parallel RB$, $AB \parallel QR$, $AC \parallel SP$, a...
169b65ff5ef3b97c11450580756ee14dbd61929f84e8579b54ad3bbea9ec4e4f
France
PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri
2,024
Seniors
Exercice 10.
Problem: Soit $ABC$ un triangle et $\Omega$ son cercle circonscrit. On note $A'$ le point diamétralement opposé à $A$ dans le cercle $\Omega$. Soit $I$ le centre du cercle inscrit au triangle $ABC$, $E$ et $F$ les points de contact du cercle inscrit avec les côtés $AC$ et $AB$ respectivement. Le cercle circonscrit au ...
[ "Solution:\n\n![](attached_image_solution_1.png)\n\nPuisque $F$ est le point de contact du cercle inscrit avec le côté $[AC]$, l'angle $\\widehat{IFA}$ est droit et le segment $[IA]$ est un diamètre du cercle circonscrit au triangle $AEF$. On en déduit que\n$$\n\\widehat{AXI} = 90^\\circ = \\widehat{AXA'}\n$$\nou o...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Circles", "Tangents" ], [ "Topics", "Geometry", "Plane Geo...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
null
French__md__envois__fr-Corrige-envoi-5-2023-2024
France
false
0
In a triangle, consider the incenter and the points where the incircle touches two sides through a vertex. Take the circle through that vertex and the two touchpoints, and intersect it with the circumcircle of the triangle. Show that the incenter, the intersection point, and the point opposite the vertex on the circumc...
[ "Radius to a tangency point is perpendicular to the tangent side, giving a right angle at the touchpoint", "In a triangle, a right angle implies the opposite side is a diameter of the circumcircle (inscribed angle theorem)", "Using the antipode property on the circumcircle to obtain a right angle with the given...
null
proof only
0.94
Problem: Soit $ABC$ un triangle et $\Omega$ son cercle circonscrit. On note $A'$ le point diamétralement opposé à $A$ dans le cercle $\Omega$. Soit $I$ le centre du cercle inscrit au triangle $ABC$, $E$ et $F$ les points de contact du cercle inscrit avec les côtés $AC$ et $AB$ respectivement. Le cercle circonscrit au ...
fc9ef6291be5347b96ad52ec3cdc3ce002370d9ab991a0f18ccdc5c61f181ca1
Germany
Auswahlwettbewerb zur Internationalen Mathematik-Olympiade 2022
2,022
1. Auswahlklausur
Aufgabe 2
Problem: Es sei $ABCD$ ein Parallelogramm mit $|AC| = |BC|$. Ein Punkt $P$ sei auf dem Strahl $AB$ gewählt, sodass $B$ zwischen $A$ und $P$ liegt. Der Umkreis des Dreiecks $ACD$ und die Strecke $PD$ haben außer dem Punkt $D$ noch den Punkt $Q$ gemeinsam. Der Umkreis des Dreiecks $APQ$ und die Strecke $PC$ haben außer ...
[ "Solution:\n\nWir arbeiteten mit gerichteten Winkeln modulo $180$ Grad. Zunächst erkennen wir, dass aus der Voraussetzung $|AC| = |BC|$ unmittelbar folgt, dass die Winkel $\\angle BAC$, $\\angle CBA$, $\\angle DCA$ und $\\angle ADC$ allesamt gleich groß sind.\n\n![](attached_image_solution_1.png)\n\nEs genügt zu ze...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Circles", "Tangents" ], [ "Topics", "Geometry", "Plane Geometry", "Circles", "Radical axis theorem" ], [ "Topi...
[ "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Circles > Radical axis theorem", "Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem", "Geometry > Plane Geometry > Miscellaneous > Angl...
null
Germany_TST__md__de-2022-2022_IMO_Auswahlklausuren_Lsg_HP
Germany
false
0
In a parallelogram where one diagonal equals a side, choose a point on the extension of a side beyond a vertex. Define two new points by intersecting certain circles of triangles with specific segments. Show that three natural lines associated to the parallelogram and these constructed points all meet at a single point...
[ "Directed angles to streamline angle equalities and cyclicity arguments", "Establishing multiple cyclic quadrilaterals via angle chasing", "Tangent–chord angle theorem and secant–tangent (power of a point) to identify a midpoint on a diagonal", "Radical axis theorem: concurrency of pairwise radical axes of th...
null
proof only
0.86
Problem: Es sei $ABCD$ ein Parallelogramm mit $|AC| = |BC|$. Ein Punkt $P$ sei auf dem Strahl $AB$ gewählt, sodass $B$ zwischen $A$ und $P$ liegt. Der Umkreis des Dreiecks $ACD$ und die Strecke $PD$ haben außer dem Punkt $D$ noch den Punkt $Q$ gemeinsam. Der Umkreis des Dreiecks $APQ$ und die Strecke $PC$ haben außer ...
36735e570bfb01e460aee84d6030ea061064e0329cf02a3ad1b81e79727e4208
Greece
Hellenic Mathematical Olympiad
2,024
B. Seniors
Problem 2
Let $ABC$ be a triangle with $AB < AC < BC$ and circumcircle $\Gamma_1$ of center $O$. We consider the circle $\Gamma_2$ with center $D$ lying on the circle $\Gamma_1$, and tangent to the line $BC$ at the point $E$ and tangent to the extension of the side $AB$ at point $F$. The circles $\Gamma_1$ and $\Gamma_2$ interse...
[ "We have $DE \\perp BC$ and $DF \\perp AB$ (because $\\Gamma_2$ is tangent to the side $BC$ at $E$ and to the line $AB$ at $F$).\n\n![](images/GreekMO2024finalbook_p9_data_7723d00d75.png)\nFigure 4\n\n![](images/GreekMO2024finalbook_p9_data_cb1e958d9b.png)\nFigure 5\n\nHence the quadrilateral $BEDF$ is cyclic and l...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Circles", "Tangents" ], [ "Topics", "Geometry", "Plane Geometry", "Circles", "Radical axis theorem" ], [ "Topi...
[ "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Circles > Radical axis theorem", "Geometry > Plane Geometry > Transformations > Inversion", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
English
GRC_2024
Greece
true
2
In a scalene triangle, a second circle centered on the circumcircle is tangent to one side and to the extension of another side. The two circles intersect at two points. The line through these intersections meets a tangent line and a line through the second circle’s center at two points. Show that these two points toge...
[ "Use tangency to get perpendicular radii, implying a right-angle cyclic quadrilateral through equal angles", "Apply the radical axis theorem: intersection of two radical axes lies on the third, yielding a key collinearity through the common chords", "Establish congruence of two right triangles to deduce equal d...
null
proof only
0.86
Let $ABC$ be a triangle with $AB < AC < BC$ and circumcircle $\Gamma_1$ of center $O$. We consider the circle $\Gamma_2$ with center $D$ lying on the circle $\Gamma_1$, and tangent to the line $BC$ at the point $E$ and tangent to the extension of the side $AB$ at point $F$. The circles $\Gamma_1$ and $\Gamma_2$ interse...
affabe4ada8a25068323f7ee88e7dd7d98b1233add7a35c422c26d404d32a86e
Hong Kong
IMO HK TST
2,023
Test 1
5
Let $ABCD$ be a convex quadrilateral with $AB = 5$, $AD = 17$, and $CD = 6$. If the angle bisectors of $\angle BAD$ and $\angle ADC$ intersect at the midpoint of $BC$, find the area of $ABCD$.
[ "Let $M$ be the midpoint of $BC$. Let $B'$ and $C'$ be points on $AD$ such that $AB' = AB = 5$ and $DC' = DC = 6$. Then $B'C' = 17 - 5 - 6 = 6$. Note that $\\triangle ABM \\cong \\triangle AB'M$ and $\\triangle DCM \\cong \\triangle DC'M$. Note also that $\\triangle MB'C'$ is isosceles as $MB' = MB = MC = MC'$. Let...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ], [ "Topics", "...
[ "Geometry > Plane Geometry > Quadrilaterals", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Miscellaneous > Constructions and loci" ]
null
HKG_TST_2023
Hong Kong
true
1
You are given a convex quadrilateral with three side lengths and the property that the bisectors of two opposite angles meet at the midpoint of the other side. Determine the area of the quadrilateral.
[ "Construct points on a side so that two given sides are replicated and use the angle bisectors to create congruent triangles by reflection symmetry", "Use the midpoint condition to equate distances from the midpoint to endpoints and to the constructed points, yielding an isosceles configuration", "Establish a c...
14√21
proof and answer
0.74
Let $ABCD$ be a convex quadrilateral with $AB = 5$, $AD = 17$, and $CD = 6$. If the angle bisectors of $\angle BAD$ and $\angle ADC$ intersect at the midpoint of $BC$, find the area of $ABCD$.
34fef6d0047c31a8efbd3e4f2e0bb673d708384737134de96b2b25e9e2514a9e
IMO
IMO2024 Shortlisted Problems
2,024
Combinatorics
C8
Let $n$ be a positive integer. Given an $n \times n$ board, the unit cell in the top left corner is initially coloured black, and the other cells are coloured white. We then apply a series of colouring operations to the board. In each operation, we choose a $2 \times 2$ square with exactly one cell coloured black and w...
[ "Now we prove that if such a colouring is possible for $n$ then $n$ must be a power of 2. Suppose it is possible to colour an $n \\times n$ board where $n>1$. Identify the top left corner of the board by $(0,0)$ and the bottom right corner by $(n, n)$. Whenever an operation takes place in a $2 \\times 2$ square cen...
[]
[ [ "Topics", "Discrete Mathematics", "Combinatorics", "Invariants / monovariants" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Coloring schemes, extremal arguments" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Recursion, bijection" ] ]
[ "Discrete Mathematics > Combinatorics > Invariants / monovariants", "Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments", "Discrete Mathematics > Combinatorics > Recursion, bijection" ]
English
IMO2024SL
IMO
false
0
Starting with a square board where only the top left cell is black, you repeatedly choose any two by two block that has exactly one black cell and turn all its cells black. Determine for which board sizes it is possible to eventually make every cell black.
[ "Model operations via a graph built from diagonally connected markers, proving the graph is a tree (acyclic and connected)", "Parity invariant on allowable positions for operations, forcing constraints on coordinates and board edge behavior", "Perimeter forcing shows the board size must be even; uniqueness of c...
n is a power of 2
proof and answer
0.95
Let $n$ be a positive integer. Given an $n \times n$ board, the unit cell in the top left corner is initially coloured black, and the other cells are coloured white. We then apply a series of colouring operations to the board. In each operation, we choose a $2 \times 2$ square with exactly one cell coloured black and w...
2960dbc512066eca65a01450cd6a268f9063ac92696d23a9f295eb1d903e910e
Ibero-American Mathematical Olympiad
Iberoamerican Mathematical Olympiad
2,003
18th Iberoamerican
A2
Problem: $C$ is a point on the semicircle with diameter $AB$. $D$ is a point on the arc $BC$. $M$, $P$, $N$ are the midpoints of $AC$, $CD$ and $BD$. The circumcenters of $ACP$ and $BDP$ are $O$, $O'$. Show that $MN$ and $OO'$ are parallel.
[ "![](attached_image_solution_1.png)\nLet the center of the circle be $X$ and the radius $r$. Let $\\angle AXM = \\theta$, $\\angle BXN = \\varphi$. Note that $O$ is the intersection of $XM$ and the perpendicular to $CD$ at $Q$, the midpoint of $CP$. We have $XM = r \\cos \\theta$. Let $CD$ and $XM$ meet at $Y$. The...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Transformations", "Homothety" ], [ "Topics", "Geometry", "...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Transformations > Homothety", "Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry", "Geometry > Plane Geometry > Miscella...
null
IberoAmerican_MO__md__en-1985-2003-IberoamericanMO
Ibero-American Mathematical Olympiad
false
0
In a semicircle with points chosen on the arc and midpoints of certain segments, prove that the line connecting two circumcenters is parallel to the line connecting two midpoints.
[ "Characterize circumcenters as intersections of perpendicular bisectors, using that the line through the circle’s center and a chord midpoint is the perpendicular bisector of the chord", "Relate distances from the circle’s center to chord midpoints and circumcenters via cosine and secant in right triangles", "A...
null
proof only
0.86
Problem: $C$ is a point on the semicircle with diameter $AB$. $D$ is a point on the arc $BC$. $M$, $P$, $N$ are the midpoints of $AC$, $CD$ and $BD$. The circumcenters of $ACP$ and $BDP$ are $O$, $O'$. Show that $MN$ and $OO'$ are parallel.
e517c15e6151b16fdf6456ae249506198eeac390582f35456fd078baa9766ad6
India
INMO
2,024
Indian National Mathematical Olympiad
1.
In triangle $ABC$ with $CA = CB$, point $E$ lies on the circumcircle of $ABC$ such that $\angle ECB = 90^\circ$. The line through $E$ parallel to $CB$ intersects $CA$ in $F$ and $AB$ in $G$. Prove that the centre of the circumcircle of triangle $EGB$ lies on the circumcircle of triangle $ECF$.
[ "![](images/Contests_Booklet_2023_2024_p2_data_dbbd814a54.png)\nWe have $FG = FA$ since $FG$ is parallel to $BC$. But also $\\triangle GAE$ is a right angle triangle. Thus, if $F'$ is the midpoint of $GE$, then $\\angle GAF = \\angle FGA = \\angle F'GA = \\angle GAF'$ which implies $F \\equiv F'$. Thus, $F$ is the ...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Transformations", "Homothety" ], [ "Topics", "Geometry", "...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Transformations > Homothety", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Quadrilaterals > Cyclic...
null
IND_2023_2024
India
true
1
In an isosceles triangle, choose a point on the circumcircle so that the angle at that point with one vertex is a right angle. Through this point draw a line parallel to one side, meeting the other two sides at two points. Show that the center of the circle through the chosen point and one vertex and one intersection p...
[ "Use parallel lines in an isosceles triangle to get equal segments and deduce that a constructed point is the midpoint of a chord", "Exploit right triangle properties: the midpoint of the hypotenuse is equidistant from the vertices", "Relate central and inscribed angles in a circle; use that the perpendicular b...
null
proof only
0.86
In triangle $ABC$ with $CA = CB$, point $E$ lies on the circumcircle of $ABC$ such that $\angle ECB = 90^\circ$. The line through $E$ parallel to $CB$ intersects $CA$ in $F$ and $AB$ in $G$. Prove that the centre of the circumcircle of triangle $EGB$ lies on the circumcircle of triangle $ECF$.
f97b272460f14e1919eef0cb9eb229718cebd9c9689c8a005111d5fe1f5ffc2c
Iran
Iranian Mathematical Olympiad
2,025
Third Round - Geometry
2
Point $M$ is the midpoint of side $BC$ of triangle $ABC$. The line perpendicular to $AM$ at point $A$ intersects the circumcircle of $ABC$ for the second time at $K$. The altitudes $BE$ and $CF$ of triangle $ABC$ intersect line $AK$ at points $P$ and $Q$, respectively. Prove that the radical axis of the circumcircles o...
[ "**Claim 1.** *If $\\Xi$ is the $\\Xi$-point corresponding to vertex $A$, then $K\\Xi \\perp BC$.*\n\n*Proof*. Let $\\Xi'$ and $X$ be the reflections of $\\Xi$ with respect to $BC$ and $M$ respectively. According to the properties of the $\\Xi$-point, we know that these points lie on the circumcircle of $ABC$ and $...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Circles", "Radical axis theorem" ], [ "Topics", "Geometry", "Plane Geometry", "Circles", "Coaxal circles" ], [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, ci...
[ "Geometry > Plane Geometry > Circles > Radical axis theorem", "Geometry > Plane Geometry > Circles > Coaxal circles", "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Miscellaneous > Angle cha...
null
IRN_2025
Iran
true
1
In a triangle, take the midpoint of one side. Draw the line through the opposite vertex that is perpendicular to the segment to this midpoint and let it meet the circumcircle again. The altitudes from the other two vertices meet this line at two points. Show that the line that is the radical axis of the two circles det...
[ "Identify a special point whose line through the constructed intersection on the circumcircle is perpendicular to the base, using reflection properties and circle geometry", "Use that the radical axis of two intersecting circles is the common chord line through their intersection points", "Angle chasing to show...
null
proof only
0.78
Point $M$ is the midpoint of side $BC$ of triangle $ABC$. The line perpendicular to $AM$ at point $A$ intersects the circumcircle of $ABC$ for the second time at $K$. The altitudes $BE$ and $CF$ of triangle $ABC$ intersect line $AK$ at points $P$ and $Q$, respectively. Prove that the radical axis of the circumcircles o...
ed0eaee316e2f546a24d8451c1f0c8b337433514f20bb140c061c2480e34bce4
Ireland
IRL_ABooklet_2025
2,025
Problems
6
How many crosswords can we make from the word “CROSSWORD”, assuming the following rules? 1. A crossword has two words, one horizontal and one vertical, intersecting at a single letter. 2. A 'word' is any ordered set of at least two letters. It does not need to be a real word in any language. 3. Letters should be used a...
[ "We consider first the sequence of the letters, and then the configuration in which they are arranged. There are 9 letters on CROSSWORD, consisting of 3 unique letters and three duplicate pairs. If all letters were distinct there would be $9! = 362,880$ permutations, but because of the three duplicates, this counts...
[]
[ [ "Topics", "Discrete Mathematics", "Combinatorics", "Recursion, bijection" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Counting two ways" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Inclusion-exclusion" ] ]
[ "Discrete Mathematics > Combinatorics > Recursion, bijection", "Discrete Mathematics > Combinatorics > Counting two ways", "Discrete Mathematics > Combinatorics > Inclusion-exclusion" ]
null
IRL_2025
Ireland
true
1
Count the number of ways to form a pair of letter sequences from the given word that cross at one shared letter, considering repeated letters, orientations, and all possible crossing positions.
[ "Multiset permutation counting: adjust total permutations for repeated letters by dividing by powers of two", "Casework on word lengths and intersection positions; summing products of available crossing locations", "Encoding configurations as strings with separators and using combinations to count valid placeme...
6667920
proof and answer
0.9
How many crosswords can we make from the word “CROSSWORD”, assuming the following rules? 1. A crossword has two words, one horizontal and one vertical, intersecting at a single letter. 2. A 'word' is any ordered set of at least two letters. It does not need to be a real word in any language. 3. Letters should be used a...
ebbd004c7dc4e0b4a8227606bfa8cba05b11aa7cc3035dcf393acaddf0e2f387
Italy
Italian Mathematical Olympiad - February Round
2,024
Problemi a risposta numerica
13.
Problem: Su un foglio di carta sono disegnati due esagoni regolari. Il più piccolo ha area $18$, e una diagonale minore dell'esagono più grande coincide con una diagonale maggiore dell'esagono più piccolo. Quanto misura l'area dell'unione dei due esagoni?
[ "Solution:\n\nLa risposta è $29$. La figura disegnata sul foglio si può rappresentare come segue:\n\n![](attached_image_solution_1.png)\n\nIl punto $O$ in figura è dato dal centro dell'esagono maggiore $A B C D E F$. Poiché l'esagono $A B C D E F$ è regolare, il triangolo $O C D$ è equilatero. Inoltre, essendo $D A...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Circles" ], [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle trigonometry" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ] ]
[ "Geometry > Plane Geometry > Circles", "Geometry > Plane Geometry > Triangles > Triangle trigonometry", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
null
Italy__md__it-febbraio__soluzioni2024
Italy
false
0
Two regular hexagons are drawn so that a small diagonal of the larger one coincides with a large diagonal of the smaller one. Given the area of the smaller hexagon, determine the area covered by the two hexagons together.
[ "Use properties of a regular hexagon: center-to-vertex triangles are equilateral and diagonals relate via the circumscribed circle", "Apply Thales' theorem to get a right triangle and Pythagoras to compute the ratio between a long diagonal and a short diagonal", "Scale areas by the square of linear ratios to fi...
29
proof and answer
0.9
Problem: Su un foglio di carta sono disegnati due esagoni regolari. Il più piccolo ha area $18$, e una diagonale minore dell'esagono più grande coincide con una diagonale maggiore dell'esagono più piccolo. Quanto misura l'area dell'unione dei due esagoni?
ff3ba994a1fd2eb65f6ca1c9bd7ca2aca46d0dde16d990c02f449586bd174e23
JBMO
JBMO
2,022
null
Problem 2
Problem: Let $ABC$ be an acute triangle such that $AH = HD$, where $H$ is the orthocenter of $ABC$ and $D \in BC$ is the foot of the altitude from the vertex $A$. Let $\ell$ denote the line through $H$ which is tangent to the circumcircle of the triangle $BHC$. Let $S$ and $T$ be the intersection points of $\ell$ with...
[ "Solution:\n\nIn order to prove that $SM$ and $TN$ are parallel, it suffices to prove that both of them are perpendicular to $ST$. Due to symmetry, we will provide a detailed proof of $SM \\perp ST$, whereas the proof of $TN \\perp ST$ is analogous. In this solution we will use the following notation: $\\angle BAC ...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Circles", "Tangents" ], [ "Topics", "Geometry", "Plane Geo...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle c...
null
JBMO__md__en-official__en-jbmo-2022-solutions
JBMO
false
0
In an acute triangle where the orthocenter is the midpoint of the altitude from one vertex, a line through the orthocenter tangent to a related circumcircle meets two sides. With midpoints on segments from the orthocenter to two vertices, show that the lines from these midpoints to the tangent points are parallel by pr...
[ "Use tangency at the orthocenter to relate angles on the circumcircle of the triangle formed by two vertices and the orthocenter", "Exploit the condition that the orthocenter is the midpoint of the altitude to infer midpoint relations and parallel lines", "Introduce midpoints on sides and use parallelism to est...
null
proof only
0.92
Problem: Let $ABC$ be an acute triangle such that $AH = HD$, where $H$ is the orthocenter of $ABC$ and $D \in BC$ is the foot of the altitude from the vertex $A$. Let $\ell$ denote the line through $H$ which is tangent to the circumcircle of the triangle $BHC$. Let $S$ and $T$ be the intersection points of $\ell$ with...
efabb828ec044a391437341c40a768fbcc83dd929ea731600cdf2d498d672f5f
Japan
The 35th Japanese Mathematical Olympiad
2,025
FIRST ROUND
11
The JMO Cluster initially consists of five stars $O$, $A$, $B$, $C$, and $D$. Each star is assigned a value called its **importance**. The importance of $O$ is $0$, and the importance of each of $A$, $B$, $C$, and $D$ is $1$. Moreover, there are one-way direct flights from $O$ to $A$ and $C$; from $A$ to $B$ and $D$; f...
[ "Let the initial configuration be called state $0$, and for each integer $n \\ge 1$, let state $n$ denote the configuration after the $n$-th operation.\n\n**Lemma 1.** There exists an assignment of every star into exactly one of groups $0$, $1$, or $2$ such that:\n(1) Star $O$ is assigned to group $0$.\n(2) For eve...
[]
[ [ "Topics", "Discrete Mathematics", "Graph Theory" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Recursion, bijection" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Induction / smoothing" ], [ "Topics", "Discrete Mathematics", ...
[ "Discrete Mathematics > Graph Theory", "Discrete Mathematics > Combinatorics > Recursion, bijection", "Discrete Mathematics > Combinatorics > Induction / smoothing", "Discrete Mathematics > Combinatorics > Invariants / monovariants", "Algebra > Algebraic Expressions > Sequences and Series > Recurrence relat...
English
JPN_JMO35
Japan
true
1
Starting from a directed network of five nodes with given importance values, repeatedly replace each edge by a new node whose importance is the sum of its endpoints and connect new nodes according to compatible edge endpoints. After performing this process one hundred times, determine the total importance of the nodes ...
[ "Partition the stars into three cyclic groups that track arrivals and departures, establishing an invariant structure across operations", "Use induction to prove the group assignment persists and to fix the outdegree of stars by group", "Set up and solve linear recurrences for group importances, including total...
2^68 * (2^102 - 1) / 3
proof and answer
0.83
The JMO Cluster initially consists of five stars $O$, $A$, $B$, $C$, and $D$. Each star is assigned a value called its **importance**. The importance of $O$ is $0$, and the importance of each of $A$, $B$, $C$, and $D$ is $1$. Moreover, there are one-way direct flights from $O$ to $A$ and $C$; from $A$ to $B$ and $D$; f...
e3be9e437d4c7344f84d09ea0df6a078709bcfa6aaf7ff1575fd4f958c9dff1b
Mexico
LVI Olimpiada Matemática Española (Concurso Final)
2,020
Enunciados y Soluciones
1
Decimos que un polinomio $p(x)$, con coeficientes reales, es *almeriense* si tiene la forma $$ p(x) = x^3 + ax^2 + bx + a $$ y sus tres raíces son números reales positivos en progresión aritmética. Halla todos los polinomios almerienses tales que $p(7/4) = 0$.
[ "Llamemos $\\alpha \\le \\beta \\le \\gamma$ a las raíces del polinomio. De la condición de estar en progresión aritmética tenemos que existe un número real no negativo $\\delta$ de manera que $\\alpha = \\beta - \\delta$ y $\\gamma = \\beta + \\delta$. Por su parte, utilizando las fórmulas de Cardano–Viète, result...
[]
[ [ "Topics", "Algebra", "Algebraic Expressions", "Polynomials", "Vieta's formulas" ] ]
[ "Algebra > Algebraic Expressions > Polynomials > Vieta's formulas" ]
Spanish
MEX_OME56
Mexico
false
0
Find all monic cubic polynomials whose constant term equals the coefficient of the squared term, whose three real positive zeros form an arithmetic progression, and which evaluate to zero at seven fourths.
[ "Parameterize roots in arithmetic progression as a central term plus or minus a nonnegative difference", "Apply Vieta’s formulas to equate the sum and product of the roots due to the special cubic form", "Derive the constraint that the product of the two outer roots equals three", "Casework on which term of t...
p(x) = x^3 - \frac{21}{4}x^2 + \frac{73}{8}x - \frac{21}{4} and p(x) = x^3 - \frac{291}{56}x^2 + \frac{14113}{1568}x - \frac{291}{56}.
proof and answer
0.86
Decimos que un polinomio $p(x)$, con coeficientes reales, es *almeriense* si tiene la forma $$ p(x) = x^3 + ax^2 + bx + a $$ y sus tres raíces son números reales positivos en progresión aritmética. Halla todos los polinomios almerienses tales que $p(7/4) = 0$.
788788f00f1b41589e97255e3c1320f121291214bc7afca0e842314f2cdfb419
Middle European Mathematical Olympiad (MEMO)
MEMO Szeged
2,024
Team
T-6
Problem: Let $ABC$ be an acute triangle. Let $M$ be the midpoint of the segment $BC$. Let $I, J, K$ be the incenters of triangles $ABC$, $ABM$, $ACM$, respectively. Let $P, Q$ be points on the lines $MK$, $MJ$, respectively, such that $\angle AJP = \angle ABC$ and $\angle AKQ = \angle BCA$. Let $R$ be the intersection ...
[ "Solution:\nNote that $MK \\perp MJ$. By simple angle chasing we get that\n$$\n\\begin{gathered}\n\\angle PJM = \\angle AJM - \\angle AJP = 90^\\circ + \\frac{1}{2} \\angle ABC - \\angle ABC = 90^\\circ - \\frac{1}{2} \\angle ABC, \\\\\n\\angle JPM = 90^\\circ - \\angle PJM = \\frac{1}{2} \\angle ABC .\n\\end{gathe...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ], [ "Topics", "Geometry", ...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem", "Geometry > Plane Geometry > Qu...
null
MEMO__md__en-2024-SolutionBooklet
Middle European Mathematical Olympiad (MEMO)
false
0
In an acute triangle, take the midpoint of one side and the incenters of the whole triangle and of the two smaller triangles formed with that midpoint. Choose points on lines through these incenters to satisfy given angle conditions, and define the intersection of two resulting lines. Prove that the line through the in...
[ "Use that the lines from the midpoint to the incenters of the two subtriangles are perpendicular", "Reflect constructed points across the midpoint to obtain excenters via the incenter–excenter lemma", "Identify the intersection of lines through excenters as the excenter of the original triangle and use midpoint...
null
proof only
0.86
Problem: Let $ABC$ be an acute triangle. Let $M$ be the midpoint of the segment $BC$. Let $I, J, K$ be the incenters of triangles $ABC$, $ABM$, $ACM$, respectively. Let $P, Q$ be points on the lines $MK$, $MJ$, respectively, such that $\angle AJP = \angle ABC$ and $\angle AKQ = \angle BCA$. Let $R$ be the intersection ...
5cb9851185e0ee322701e94a70237aaa4b36a2656d74a4ca870793b6d9ad7183
Moldova
Olimpiada Republicană la Matematică
2,023
Clasa X-a
10.6.
Problem: Fie $ABC$ un triunghi ascuțitunghic, în care $m(\angle BAC)=60^{\circ}$, iar $BC=a$. Fie $M$ mijlocul laturii $BC$, iar $[BB_{1}]$ și $[CC_{1}]$ două înălțimi ale triunghiului. Calculați suma distanțelor de la ortocentrul triunghiului $ABC$ până la laturile triunghiului $MB_{1}C_{1}$.
[ "Solution:\n\n1. În triunghiul dreptunghic $BCB_{1}$ segmentul $MB_{1}$ este mediană, atunci $MB_{1}=\\frac{BC}{2}=\\frac{a}{2}$. Idem $MC_{1}=\\frac{a}{2}$.\n\n2. Considerăm triunghiurile $AB_{1}C_{1}$ și $ABC$. Vom arăta că ele sunt asemenea.\nAceste triunghiuri au unghiul $A$ comun.\nVom arăta că laturile care f...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle trigonometry" ], [ "Topics", "Geometry",...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Triangles > Triangle trigonometry", "Geometry > Plane Geometry > Miscellaneous > Distance chasing" ]
null
Moldova__md__ro-national__ro-omrm_matem_10_solutii_ziua_2_2023
Moldova
false
0
Given an acute triangle with one angle equal to sixty degrees, along with the midpoint of the base and the feet of the two altitudes from the other vertices, find the sum of the perpendicular distances from the orthocenter of the original triangle to the three sides of the triangle formed by those three points.
[ "Median to the hypotenuse in a right triangle equals half the hypotenuse to get lengths from the midpoint to altitude feet", "Use cosine of sixty degrees and projections to establish similarity between two triangles and deduce a halving scale factor", "Recognize the constructed triangle is equilateral and deter...
a*sqrt(3)/4
proof and answer
0.9
Problem: Fie $ABC$ un triunghi ascuțitunghic, în care $m(\angle BAC)=60^{\circ}$, iar $BC=a$. Fie $M$ mijlocul laturii $BC$, iar $[BB_{1}]$ și $[CC_{1}]$ două înălțimi ale triunghiului. Calculați suma distanțelor de la ortocentrul triunghiului $ABC$ până la laturile triunghiului $MB_{1}C_{1}$.
313a1c891686968a3f3f8393e16327710be7bf068198db5290a06ccf667abcc2
Mongolia
Mongolian Mathematical Olympiad
2,024
E (Grades 9-10)
E5
Let $ABC$ be an isosceles triangle with $AB = BC$. Let $M$ and $N$ be midpoints of $AC$ and $BM$, respectively. $P$ is the foot of the altitude from $A$ to $AN$ of triangle $AMN$. Prove that triangles $APM$ and $CPB$ are similar. (Khulan Tumenbayar)
[ "Let us denote $\\angle PNM = \\alpha$. Then $\\angle PMC = \\angle PNB = 180^\\circ - \\alpha$ and $\\angle PAC = \\angle PMB$. Hence $\\triangle APM \\sim \\triangle MPN$. As we have $AM = MC$, $MN = NB$ it yields $\\triangle APC \\sim \\triangle MPB$. It implies $\\angle APC = \\angle MPB$ and $\\angle MPC = \\a...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ] ]
[ "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
English
MNG_2024
Mongolia
true
1
In an isosceles triangle, take the midpoints of two chosen segments and drop a perpendicular from one vertex onto a segment connecting these midpoints. Show that the triangle formed with this perpendicular is similar to a triangle built from the base vertex and the midpoint.
[ "Use the perpendicular from a vertex to establish right angles and trigger initial triangle similarity", "Exploit midpoint properties to obtain equal segment lengths and correspondences", "Derive cyclicity of a quadrilateral from angle equalities", "Leverage properties of cyclic quadrilaterals to infer additi...
null
proof only
0.77
Let $ABC$ be an isosceles triangle with $AB = BC$. Let $M$ and $N$ be midpoints of $AC$ and $BM$, respectively. $P$ is the foot of the altitude from $A$ to $AN$ of triangle $AMN$. Prove that triangles $APM$ and $CPB$ are similar. (Khulan Tumenbayar)
ad07046d3618fbbfdaf9bb1f9a4dc8e35365b25d2a7e0fb5f8f83d665ef1fbb9
Netherlands
IMO Team Selection Test 1
2,025
null
2
Consider a rectangular board of $m \times n$ cells with $m, n \ge 1$. The vertices of the cells form a $(m+1) \times (n+1)$-grid. We say a triangle whose vertices are points on the grid is *low* if there is at least one side of the triangle that is parallel to a side of the board and for which the height of the triangl...
[ "If $m, n \\ge 2$ and at least one of the two is even, the answer is 0. Otherwise (at least one of the two is 1, or they are both odd), the answer is 2.\n\nWe first draw an example for $n = 1$ and $m \\ge 1$ with two special triangles, an example for $n = 2$ and $m \\ge 3$ with zero special triangles, and the speci...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Constructions and loci" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Coloring schemes, extremal arguments" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Invariants / monovar...
[ "Geometry > Plane Geometry > Miscellaneous > Constructions and loci", "Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments", "Discrete Mathematics > Combinatorics > Invariants / monovariants" ]
null
NLD_2025
Netherlands
true
2
You have a rectangular grid of unit squares and must divide it into triangles whose altitude to a horizontal or vertical side is one unit. Some triangles have two sides parallel to the rectangle. Among all such divisions, find the smallest number of these special triangles.
[ "Constructive partitions using strips of width two to eliminate special triangles when an even dimension exists", "Midpoint connection method: place points at midpoints of slanted sides and connect them to form orthogonal paths and cycles", "Characterization of turn points: red paths can change direction only a...
0 if m,n ≥ 2 and at least one of m or n is even; otherwise 2
proof and answer
0.9
Consider a rectangular board of $m \times n$ cells with $m, n \ge 1$. The vertices of the cells form a $(m+1) \times (n+1)$-grid. We say a triangle whose vertices are points on the grid is *low* if there is at least one side of the triangle that is parallel to a side of the board and for which the height of the triangl...
7a69756413b698464bf5db06417031908be97f092105e10d58f20776d00104e6
New Zealand
NZMO Round Two
2,025
null
3.
Problem: Let $ABC$ be an acute scalene triangle with $AC > BC > AB$. Let the orthocentre be $H$ and circumcentre be $O$. Suppose that lines $BO$ and $CH$ intersect at a point $D$. Point $E$ (where $E \neq C$) lies on side $AC$ so that $OECD$ is cyclic. Point $F$ (where $F \neq C$) lies on side $BC$ such that $CE = FE$...
[ "Solution:\n\nLet $\\alpha = \\angle BAC$. Let $BB'$ and $CC'$ be altitudes in triangle $ABC$, as shown.\n\n![](attached_image_solution_1.png)\n\nClaim. Triangle $CDE$ is isosceles with $CE = DE$.\n\nProof.\n\n$$\n\\angle DEC = \\angle DOC \\\\\n\\qquad = \\angle BOC \\\\\n\\qquad = 2\\angle BAC \\\\\n\\qquad = 2\\...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Triangles", "Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geome...
[ "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
null
NewZealand_MO__md__en-nzmo2_2025_solutions
New Zealand
false
0
In an acute scalene triangle with its orthocenter and circumcenter, construct one point as the intersection of two notable lines, and choose two points on the sides to satisfy a cyclic condition and an equal length condition. Prove that four of these constructed points lie on a common circle.
[ "Use cyclicity of a constructed quadrilateral to equate angles at two points", "Apply central–inscribed angle relationships on the circumcircle to connect angles at the center with angles of the triangle", "Exploit altitude right-angle properties to express angles as complements of a vertex angle", "Show an i...
null
proof only
0.86
Problem: Let $ABC$ be an acute scalene triangle with $AC > BC > AB$. Let the orthocentre be $H$ and circumcentre be $O$. Suppose that lines $BO$ and $CH$ intersect at a point $D$. Point $E$ (where $E \neq C$) lies on side $AC$ so that $OECD$ is cyclic. Point $F$ (where $F \neq C$) lies on side $BC$ such that $CE = FE$...
b37e4006bc156d5ee4910d2977ded9bf814754b6162820cabab7684672ca33aa
Nordic Mathematical Olympiad
The 35th Nordic Mathematical Contest
2,021
null
Problem 4
Problem: Let $A$, $B$, $C$ and $D$ be points on the circle $\omega$ such that $ABCD$ is a convex quadrilateral. Suppose that $AB$ and $CD$ intersect at a point $E$ such that $A$ is between $B$ and $E$ and that $BD$ and $AC$ intersect at a point $F$. Let $X \neq D$ be the point on $\omega$ such that $DX$ and $EF$ are p...
[ "Solution:\n\nIt can be difficult to find out what to do, but the key is to show that $AYFE$ is cyclic. The motivation for this is that we want to show that $\\angle BAY = \\angle BAX$. But $\\angle BAX = \\angle BDX$ since $BXDA$ is cyclic, and $\\angle BDX = \\angle FDX = \\angle DFE$ since $DX$ and $EF$ are para...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Constructions and loci"...
[ "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Miscellaneous > Constructions and loci" ]
null
Nordic_MO__md__en-2021-sol
Nordic Mathematical Olympiad
false
0
Four points lie on a circle forming a convex shape. The pairs of opposite sides and diagonals intersect at two points. A fifth point on the circle is chosen so that a line through one vertex is parallel to the line through the intersection points, and that vertex is reflected across this line to obtain a new point insi...
[ "Exploit cyclicity of the original quadrilateral on the circle to relate angles", "Use parallelism between the chord through the fixed point and the line through the intersection points to transfer angles", "Apply properties of reflection across a line to equate corresponding angles", "Establish that a constr...
null
proof only
0.92
Problem: Let $A$, $B$, $C$ and $D$ be points on the circle $\omega$ such that $ABCD$ is a convex quadrilateral. Suppose that $AB$ and $CD$ intersect at a point $E$ such that $A$ is between $B$ and $E$ and that $BD$ and $AC$ intersect at a point $F$. Let $X \neq D$ be the point on $\omega$ such that $DX$ and $EF$ are p...
7b2c81311124d11e41081f2d9e7eac7be6a23d3c27347df655884dc2f1881136
North Macedonia
Junior Mathematical Olympiad
2,018
22nd Junior Mathematical Olympiad 2018
2
Let $k$ be a semicircle with center $O$ and diameter $AB$. Let $C$ be a point on $k$ such that $CO \perp AB$. The symmetrical of $\angle ABC$ intersects $k$ at the point $D$. Let $E$ be the point of $AB$ such that $DE \perp AB$ and let $F$ be the midpoint of $CB$. Prove that the quadrilateral $EFCD$ is cyclic.
[ "![](images/Macedonia2018Binder_kniga_MMO_2018_p3_data_e1084c720c.png)\n\nNote that the triangle $ABC$ is isosceles right triangle. Let $CD \\cap AB = \\{H\\}$. From\n$$\n\\angle AED = \\angle ADB = 90^\\circ \\text{ and } \\angle DAE = \\angle DAB\n$$\nfollows that $\\triangle ADE \\sim \\triangle ABD$. Since $ABC...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ], [ "Topics", "Geometry", "Plane Geometry", "Advanced Configurations", "Isogonal/isot...
[ "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates" ]
English
MKD_2018
North Macedonia
true
1
Given a semicircle with a right isosceles triangle on its diameter, a point is constructed by reflecting an angle at one vertex across its bisector to meet the semicircle. Dropping a perpendicular from that point to the diameter and taking the midpoint of a side, show that the resulting four points lie on a common circ...
[ "Use Thales’ theorem to recognize the right isosceles configuration in the semicircle with diameter", "Interpret the symmetric ray of the given angle as an isogonal reflection to control key angles", "Exploit the cyclicity of the initial quadrilateral on the semicircle to relate inscribed angles", "Show a per...
null
proof only
0.79
Let $k$ be a semicircle with center $O$ and diameter $AB$. Let $C$ be a point on $k$ such that $CO \perp AB$. The symmetrical of $\angle ABC$ intersects $k$ at the point $D$. Let $E$ be the point of $AB$ such that $DE \perp AB$ and let $F$ be the midpoint of $CB$. Prove that the quadrilateral $EFCD$ is cyclic.
0ba14623357407f3ad37cc40a484760b652037769a4f272ec03d25221889ee00
Philippines
25th Philippine Mathematical Olympiad
2,023
National Stage (Day 1)
3.
Problem: In $\triangle ABC$, $AB > AC$. Point $P$ is on line $BC$ such that $AP$ is tangent to its circumcircle. Let $M$ be the midpoint of $AB$, and suppose the circumcircle of $\triangle PMA$ meets line $AC$ again at $N$. Point $Q$ is the reflection of $P$ with respect to the midpoint of segment $BC$. The line throu...
[ "Solution:\n\n![](attached_image_solution_1.png)\n\nLet lines $BE$ and $AC$ meet at $R$. It suffices to show that the points $P$, $M$ and $R$ are collinear. Since $AP$ is tangent to the circumcircle of $ABC$ and $PNA M$ is cyclic, we have $\\angle PNM = \\angle PAB = \\angle ACB$ and $\\angle BAC = \\angle MPN$. Th...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Circles", "Tangents" ], [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Concurrency and Collinearity", "Menelaus' theorem" ...
[ "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
null
Philippines__md__en-pmo__en-PMO-25-National-Stage
Philippines
false
0
In a triangle where a tangent from a point on one side is drawn to the circumcircle, several points are defined using midpoints, reflection, and parallel lines. The goal is to prove that a line through one constructed point, a line through a vertex and another constructed point, and the side opposite the vertex all mee...
[ "Use the tangent–chord angle relation and a cyclic configuration to get triangle similarity between two constructed triangles and derive a key length ratio", "Exploit parallel lines to transfer angles, yielding cyclic quadrilaterals and further triangle similarities that produce additional length ratios", "Esta...
null
proof only
0.78
Problem: In $\triangle ABC$, $AB > AC$. Point $P$ is on line $BC$ such that $AP$ is tangent to its circumcircle. Let $M$ be the midpoint of $AB$, and suppose the circumcircle of $\triangle PMA$ meets line $AC$ again at $N$. Point $Q$ is the reflection of $P$ with respect to the midpoint of segment $BC$. The line throu...
f5511d244ba9ca22f76a0f05cd6dbcefeb0c167176d7c7ec4fc29784b507f061
Romania
75th NMO Selection Tests
2,025
Junior Balkan Mathematical Olympiad - FIRST SELECTION TEST
Problem 2.
Let $\triangle ABC$ be an isosceles triangle with $\angle BAC > 90^\circ$, and let $C$ be the circle centered at $A$ with radius $AB$. Let $M$ be the midpoint of side $AC$. The line $BM$ intersects the circle $C$ a second time at point $D$. Let $E$ be a point on the circle $C$ such that $BE \perp AC$, and suppose that ...
[ "From the condition $AC \\perp BE$, it follows that $AC$ is the perpendicular bisector of segment $BE$, so $MB = ME$. Since $AB = AE$, we conclude that $\\triangle MAB \\equiv \\triangle MAE$ (congruent triangles). It follows that $\\angle MAB = \\angle MAE$ (1), and $\\angle MBA = \\angle MEA$ (2).\nFrom $AB = AD$...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Angle chasing" ], [ "Topics", "Geometry", "Plane Geometry", "Miscellaneous", "Distance chasing" ] ]
[ "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Miscellaneous > Angle chasing", "Geometry > Plane Geometry > Miscellaneous > Distance chasing" ]
English
ROU_2025
Romania
true
1
In an isosceles triangle with an obtuse apex, draw the circle centered at the apex with radius equal to the equal side. From the midpoint of one equal side, draw the line to the other vertex and extend it to meet the circle again. Pick a point on the circle so that the segment to the other vertex is perpendicular to th...
[ "Use that the chosen point on the circle centered at the apex makes the line through the midpoint perpendicular to the base, and since the apex is equidistant from the two endpoints, this line is the perpendicular bisector, implying equal distances from the midpoint to those endpoints", "Apply SSS congruence to t...
null
proof only
0.83
Let $\triangle ABC$ be an isosceles triangle with $\angle BAC > 90^\circ$, and let $C$ be the circle centered at $A$ with radius $AB$. Let $M$ be the midpoint of side $AC$. The line $BM$ intersects the circle $C$ a second time at point $D$. Let $E$ be a point on the circle $C$ such that $BE \perp AC$, and suppose that ...
07c4beed7252e1fdead1ad439ba2801bdb231ffe1445dfe75c4a9e17bbd3971f
Romanian Master of Mathematics (RMM)
Romanian Master of Mathematics Competition
2,021
Day 1
Problem 1
Problem: Let $T_{1}$, $T_{2}$, $T_{3}$, $T_{4}$ be pairwise distinct collinear points such that $T_{2}$ lies between $T_{1}$ and $T_{3}$, and $T_{3}$ lies between $T_{2}$ and $T_{4}$. Let $\omega_{1}$ be a circle through $T_{1}$ and $T_{4}$; let $\omega_{2}$ be the circle through $T_{2}$ and internally tangent to $\om...
[ "Solution:\n\nLet $O_{i}$ be the centre of $\\omega_{i}$, $i=1,2,3,4$. Notice that the isosceles triangles $O_{i} T_{i} T_{i-1}$ are similar (indices are reduced modulo $4$), to infer that $\\omega_{4}$ is internally tangent to $\\omega_{1}$ at $T_{4}$, and $O_{1} O_{2} O_{3} O_{4}$ is a (possibly degenerate) paral...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Circles", "Tangents" ], [ "Topics", "Geometry", "Plane Geometry", "Transformations", "Homothety" ], [ "Topics", "Geometry", "Plane Geometry", "Analytic / Coordinate Methods", "Vectors" ], [ "Topics...
[ "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Transformations > Homothety", "Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors", "Geometry > Plane Geometry > Miscellaneous > Distance chasing" ]
null
RMM__md__en-2021-RMM2021-Day1-English_Solutions
Romanian Master of Mathematics (RMM)
false
0
Four points lie on a line in order, and four circles are constructed so that each passes through one of these points and is tangent to the previous circle in a specified way. A straight line cuts the circles in eight points in a fixed order. Prove that the sum of two specific adjacent chord segments on this line equals...
[ "Centers of tangent circles lie on the line through the tangency point, and successive tangencies imply the four centers form a parallelogram (via similar isosceles triangles or homothety composition).", "Project the centers orthogonally onto the transversal; these projections are midpoints of the intercepted cho...
null
proof only
0.86
Problem: Let $T_{1}$, $T_{2}$, $T_{3}$, $T_{4}$ be pairwise distinct collinear points such that $T_{2}$ lies between $T_{1}$ and $T_{3}$, and $T_{3}$ lies between $T_{2}$ and $T_{4}$. Let $\omega_{1}$ be a circle through $T_{1}$ and $T_{4}$; let $\omega_{2}$ be the circle through $T_{2}$ and internally tangent to $\om...
cd1fbbfbd7c11cd917ce187baa9f47d6412f1938573fa3276b2e6a08785da44f
Russia
LI Всероссийская математическая олимпиада школьников
2,025
9 класс
9.4
The chess King was placed on a cell of the board $8 \times 8$, and then he made $64$ moves so that he visited all the cells and returned back to the initial cell. At each moment, we calculate the distance from the center of the cell occupied by the King to the center of the board. We call the move *pleasant* if after t...
[ "Ответ. $44$ moves.\n\nLet us prove that there must have been at least $20$ unpleasant moves (and thus the number of pleasant moves cannot exceed $44$). Let's place numbers in the cells as shown in рис. 3; cells with the same numbers are equidistant from the center, and cells with smaller numbers are closer to the ...
[]
[ [ "Topics", "Discrete Mathematics", "Combinatorics", "Pigeonhole principle" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Coloring schemes, extremal arguments" ] ]
[ "Discrete Mathematics > Combinatorics > Pigeonhole principle", "Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments" ]
Russian
RUS_2025
Russia
true
2
A king makes a closed tour of all squares of a standard board. A move is called pleasant if it brings the king closer to the board center. Determine the maximum possible number of pleasant moves over such tours.
[ "Partition the board into level sets by equal distance from the board center and label squares accordingly", "Observe monotonicity constraints: from the closest level no move can reduce distance; from the next level only moves into the closest level can be pleasant", "Apply the pigeonhole principle to bound how...
44
proof and answer
0.88
The chess King was placed on a cell of the board $8 \times 8$, and then he made $64$ moves so that he visited all the cells and returned back to the initial cell. At each moment, we calculate the distance from the center of the cell occupied by the King to the center of the board. We call the move *pleasant* if after t...
13e00afaa5afafa0463170bd8e295f8a252d91c28bf9681d0428753670c50705
Saudi Arabia
Saudi Booklet
2,025
Preselection tests - Test 3
9
Let $ABC$ be a triangle. Point $D$ lies on side $BC$, such that the incircles of triangles $ABD$ and $ACD$ are congruent. Let $\Omega_B$ be the circle with diameter $AB$, and let $\Omega_C$ be the circle with diameter $AC$. Prove that line $AD$ is perpendicular to one of the common tangents to the circles $\Omega_B$ an...
[ "Denote $AH$ as the altitude of triangle $ABC$ then clearly $H \\in \\Omega_B, \\Omega_C$ so $AH$ is the common chord of the two circles. Let $EF$ be the common tangent near $A$ of the two circles with $E \\in \\Omega_B, F \\in \\Omega_C$. According to the familiar property, $AH$ passes through the midpoint $K$ of ...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Circles", "Tangents" ], [ "Topics", "Geometry", "Plane Geometry", "Circles", "Radical axis theorem" ], [ "Topics", "Geometry", "Plane Geometry", "Advanced Configurations", "Polar triangles, harmonic conjug...
[ "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Circles > Radical axis theorem", "Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates", "Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter...
null
SAU_2025
Saudi Arabia
true
1
In a triangle, choose a point on the base so that the two smaller triangles formed with the vertex have equal inscribed circles. Show that the line from the vertex to this point is perpendicular to one of the lines that touch both of the circles built on the two sides through the vertex as diameters.
[ "Use that the radical axis (common chord) of two intersecting circles passes through the midpoint of the segment joining the tangency points of any common tangent", "Identify the altitude foot as the second intersection of the two circles with diameters on two sides, so the altitude line is their radical axis", ...
null
proof only
0.73
Let $ABC$ be a triangle. Point $D$ lies on side $BC$, such that the incircles of triangles $ABD$ and $ACD$ are congruent. Let $\Omega_B$ be the circle with diameter $AB$, and let $\Omega_C$ be the circle with diameter $AC$. Prove that line $AD$ is perpendicular to one of the common tangents to the circles $\Omega_B$ an...
7bb1faccff84b0a8022a4cc4a750a5fb7bbeecb56048f9bb97310754bc0a2982
Serbia
14. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА
2,020
Први дан
2.
Problem: Дат је конвексан полиедар са бар 5 темена у чијем се сваком темену састају тачно по три ивице. Доказати да је могуће доделити сваком темену тог полиедра неки рационалан број тако да буду задовољени следећи услови: (i) бар један од додељених бројева је једнак 2020; (ii) за сваку страну полиедра, производ броје...
[ "Solution:\n\nОзначимо $c = 2020$. Прво размотримо случај када постоји страна $\\mathcal{F}$ са парним бројем темена. Тада је довољно доделити теменима стране $\\mathcal{F}$ наизменично бројеве $c$ и $\\frac{1}{c}$, а свим осталим теменима полиедра број $1$. Заиста, производ бројева на страни $\\mathcal{F}$ је једн...
[]
[ [ "Topics", "Geometry", "Solid Geometry", "3D Shapes" ], [ "Topics", "Geometry", "Solid Geometry", "Other 3D problems" ], [ "Topics", "Discrete Mathematics", "Combinatorics", "Coloring schemes, extremal arguments" ] ]
[ "Geometry > Solid Geometry > 3D Shapes", "Geometry > Solid Geometry > Other 3D problems", "Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments" ]
null
Serbia_MO__md__sr-2020_smo_resenja
Serbia
false
0
Given a convex polyhedron where exactly three edges meet at each vertex, show that you can assign rational numbers to the vertices so that at least one assigned number equals a specified value and the product of the numbers at the vertices of every face equals one.
[ "Constructive labeling using alternating values along a face to make pairwise cancellations in the product", "Case split by parity of the number of vertices on a face (even versus all faces odd)", "If an even face exists, alternate values around it and set all other vertices to one", "If all faces are odd, ch...
null
proof only
0.86
Problem: Дат је конвексан полиедар са бар 5 темена у чијем се сваком темену састају тачно по три ивице. Доказати да је могуће доделити сваком темену тог полиедра неки рационалан број тако да буду задовољени следећи услови: (i) бар један од додељених бројева је једнак 2020; (ii) за сваку страну полиедра, производ броје...
eb501de1ff575da87dc6cf604ee59a6c5f0ccc0e9d6c5f40a9f981eaa29d2f5b
Silk Road Mathematics Competition
XXI SILK ROAD MATHEMATICAL COMPETITION
2,022
null
№1
Convex quadrilateral *ABCD* is inscribed in circle *ω*. Rays *AB* and *DC* intersect at *K*. *L* is chosen on the diagonal *BD* so that $\angle BAC = \angle DAL$. *M* is chosen on the segment *KL* so that $CM \parallel BD$. Prove that the line $BM$ touches $\omega$. (Kungozhin M.) ![](images/SRMC2022_eng_sol_p0_data_5...
[ "Let *N* be a point on the line $AK$ so that $MN \\parallel AL$. Since $\\frac{CM}{DL} = \\frac{NM}{AL} = \\frac{KM}{KL}$ and $\\angle CMN = \\angle DLA$, it follows that $K$ is a center of homothety which sends $\\triangle CMN$ into similar $\\triangle DLA$. On the other hand, $\\triangle DLA$ is similar to $\\tri...
[]
[ [ "Topics", "Geometry", "Plane Geometry", "Quadrilaterals", "Cyclic quadrilaterals" ], [ "Topics", "Geometry", "Plane Geometry", "Circles", "Tangents" ], [ "Topics", "Geometry", "Plane Geometry", "Transformations", "Homothety" ], [ "Topics"...
[ "Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals", "Geometry > Plane Geometry > Circles > Tangents", "Geometry > Plane Geometry > Transformations > Homothety", "Geometry > Plane Geometry > Miscellaneous > Angle chasing" ]
English
SRM_2022
Silk Road Mathematics Competition
true
1
Given a cyclic quadrilateral with two opposite sides extended to meet, and a point on a diagonal chosen so that a certain angle at a vertex matches, pick a point on the line through this intersection and the chosen point so that a line from another vertex is parallel to the diagonal. Show that the line from one vertex ...
[ "Introduce an auxiliary point so that a pair of segments are parallel, enabling a homothety centered at the intersection of two lines to relate two key triangles", "Use similarity of triangles derived from equal inscribed angles in a cyclic configuration", "Establish cyclicity of a quadrilateral via equal direc...
null
proof only
0.9
Convex quadrilateral *ABCD* is inscribed in circle *ω*. Rays *AB* and *DC* intersect at *K*. *L* is chosen on the diagonal *BD* so that $\angle BAC = \angle DAL$. *M* is chosen on the segment *KL* so that $CM \parallel BD$. Prove that the line $BM$ touches $\omega$. (Kungozhin M.) ![](images/SRMC2022_eng_sol_p0_data_5...