MathNet: a Global Multimodal Benchmark for Mathematical Reasoning and Retrieval
Paper • 2604.18584 • Published • 8
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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\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... | [
"\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\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\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\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\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\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\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\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\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\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$... | [
"\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\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\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... |
End of preview. Expand in Data Studio
Shaden Alshammari1* Kevin Wen1* Abrar Zainal3* Mark Hamilton1 Navid Safaei4 Sultan Albarakati2 William T. Freeman1† Antonio Torralba1†
1MIT 2KAUST 3HUMAIN 4Bulgarian Academy of Sciences *† equal contribution
Quick Start · Overview · Tasks · Comparison · Dataset Stats · Data Sources · Pipeline · Schema · License · Citation
Note: This is a test for the HF hosting website. The dataset isn’t fully uploaded yet; it will be uploaded on Tuesday, April 21, 2026.
Quick start
from datasets import load_dataset
# Default: all problems
ds = load_dataset("ShadenA/MathNet", split="train")
# Or a specific country / competition-body config
arg = load_dataset("ShadenA/MathNet", "Argentina", split="train")
apmo = load_dataset("ShadenA/MathNet", "Asia_Pacific_Mathematics_Olympiad_APMO", split="train")
row = ds[0]
print(row["competition"], row["year"], row["country"])
print(row["problem_markdown"])
for img in row["images"]:
img.show() # PIL image — renders inline in the HF viewer
Overview
Mathematical problem solving remains a challenging test of reasoning for large language and multimodal models, yet existing benchmarks are limited in size, language coverage, and task diversity. We introduce MathNet, a high-quality, large-scale, multimodal, and multilingual dataset of Olympiad-level math problems together with a benchmark for evaluating mathematical reasoning in generative models and mathematical retrieval in embedding-based systems.
MathNet spans 47 countries, 17 languages, and two decades of competitions, comprising 30,676 expert-authored problems with solutions across diverse domains. Alongside the core dataset, we construct a retrieval benchmark of mathematically equivalent and structurally similar problem pairs curated by human experts.
Three benchmark tasks
| Task | What it measures | |
|---|---|---|
| I | Problem Solving | Generative models on Olympiad problems, graded against expert solutions |
| II | Math-Aware Retrieval | Embedding models' ability to retrieve mathematically equivalent / structurally similar problems |
| III | Retrieval-Augmented Problem Solving | How retrieval quality affects reasoning when similar problems are given as context |
Even state-of-the-art reasoners remain challenged: 78.4% (Gemini-3.1-Pro) and 69.3% (GPT-5) on MathNet-Solve-Test. Embedding models struggle with equivalence retrieval (Recall@1 under 5% for all tested models), and RAG gains are highly sensitive to retrieval quality — expert retrieval lifts DeepSeek-V3.2-Speciale to 97.3% on MathNet-RAG.
How MathNet compares to existing math benchmarks
| Benchmark | Size | Languages | Multimodal | Source | Difficulty |
|---|---|---|---|---|---|
| GSM8K | 8,500 | EN | — | Crowdsourced | Grade school |
| MATH | 12,500 | EN | — | Competitions/textbooks | High school |
| MATH-Vision | 3,040 | EN | ✓ | Math competitions | High school |
| OlympiadBench | 6,142 | EN, ZH | ✓ | Official websites | Olympiad |
| OlympicArena | 3,233 | EN, ZH | ✓ | Official websites | Olympiad |
| Omni-Math | 4,428 | EN | — | AoPS / contest pages | Olympiad |
| OlymMATH | 200 | EN, ZH | — | AoPS / official | Olympiad |
| MathArena | 162 | EN | ✓ | Newly released competitions | Olympiad |
| IMOBench | 460 | EN | — | IMO & national archives | Olympiad |
| MathNet (ours) | 30,676 | 17 (EN, ZH, ES, RU, FR, RO, + 11 more) | ✓ | Official country booklets / international & national contests | Olympiad |
Dataset at a glance
What the figure shows. (a) A mix of national, regional, TST, and international competitions. (b) MathNet solutions are substantially longer than those in prior math benchmarks — long-form proofs, not one-line answers. (c) Problems per year — the corpus has grown steadily since the early 2000s. (d) Coverage across geometry, algebra, combinatorics, number theory, and their sub-topics. (e) 74% English, 26% non-English across 17 languages; Portuguese, Spanish, French, Italian, Serbian, Slovenian, German, Chinese, Romanian, Korean, Dutch, Russian, Mongolian, Macedonian, Polish, and Hungarian all appear.
Topic taxonomy (excerpt)
MathNet ships with a curated olympiad-style taxonomy. Top-level domains include:
- Geometry — plane (triangles, quadrilaterals, circles, concurrency/collinearity, transformations, Miquel/Simson/Brocard, geometric inequalities, combinatorial geometry, analytic methods), solid, differential, non-Euclidean
- Algebra — prealgebra, polynomials, inequalities, functional equations, sequences/series, linear algebra, abstract algebra
- Number Theory — divisibility, primes, modular arithmetic, Diophantine equations, quadratic residues, (p)-adic methods
- Combinatorics — counting, graph theory, extremal / pigeonhole, invariants/monovariants, games, coloring, generating functions
- Calculus / Analysis — limits, inequalities, real analysis, combinatorial analysis
- Probability & Statistics — discrete and continuous
Every problem carries a hierarchical topic path (e.g. Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals) usable for stratified evaluation or curriculum construction.
Data sources
Each year, participating IMO countries contribute original problems for use in their national contests and team selection examinations. MathNet is built from official problem booklets collected from 47 countries spanning 1985–2025 — 1,595 PDF volumes totalling more than 25,000 pages. Unlike prior math benchmarks that rely on community platforms such as AoPS, every problem and solution in MathNet is authored and disseminated by national teams themselves, ensuring expert-level quality, stylistic consistency, and immunity from the noisy or informal annotations that plague crowd-sourced collections.
A meaningful portion of the collection — particularly older national booklets — was physically obtained and scanned by hand by our IMO expert co-authors, who have attended the International Mathematical Olympiad since 2006 and accumulated a personal archive of official competition materials over nearly two decades.
Data pipeline
Extracting aligned problem–solution pairs from a heterogeneous corpus of mathematical documents is non-trivial: some booklets separate problems and solutions into different sections, others interleave them; numbering schemes and naming conventions vary across countries and even within a single document. Regex-based heuristics break down at this scale, so we designed a multi-stage LLM pipeline.
Stage 1 — Document ingestion & segmentation. All booklets are converted to Markdown via dots-ocr, a multilingual document parsing framework designed for both digital typeset PDFs and scanned copies across many languages. Gemini-2.5-Flash then identifies problem and solution segments by outputting only their line numbers, and records authors, hints, remarks, source file, and page numbers for provenance.
Stage 2 — Problem–solution extraction. Given the line segments from Stage 1, GPT-4.1 extracts the corresponding problem and solution in LaTeX-friendly Markdown, together with a surrounding text buffer to handle cases where content spans across context boundaries.
Stage 3 — Extraction verification. Each extracted pair passes three independent checks before being retained:
- Rule-based similarity check — text similarity between the extraction and original OCR output ensures the LLM made only formatting changes and introduced no hallucinated content.
- GPT-4.1-as-judge — GPT-4.1 compares page screenshots against the extracted pair to catch OCR errors, incorrect figure associations, and incomplete solutions.
- Human expert review — low-confidence cases are manually reviewed by annotators. A pair is retained only if all three mechanisms agree.
Provenance (source booklet, page numbers, authors where given) is preserved on every problem.
What this preview contains
A diverse 100-problem slice sampled round-robin across countries, prioritizing problems with figures so the multimodal path is visible end-to-end. Images are embedded in the parquet as HF Image() features — they render inline in the dataset viewer and decode to PIL on load.
Schema
| Column | Type | Notes |
|---|---|---|
unique_id |
string | Stable SHA-256 content hash |
country |
string | Country / regional body of origin |
competition |
string | e.g. IMO 2023, Cono Sur Mathematical Olympiad |
year |
int32 | Year of competition |
section |
string|null | Day / round / level |
problem_number |
string | As printed in the booklet |
problem_markdown |
string | Problem statement (Markdown + LaTeX) |
solutions_markdown |
list<string> | Official / provided solutions |
answers_markdown |
list<string> | Final answers when stated separately |
topics |
list<list<string>> | Hierarchical tags |
topics_flat |
list<string> | Joined A > B > C strings |
language |
string | Source booklet language |
source_booklet |
string | Booklet id (e.g. ARG_2003) |
booklet_source |
string | Upstream collection label |
has_images |
bool | Whether the problem cites figures |
num_images |
int32 | Count of referenced figures |
images |
list<Image> | Inlined bytes, decoded to PIL |
natural_language_description |
string|null | LLM-assisted NL rephrasing |
main_ideas |
list<string> | LLM-assisted key solution ideas |
final_answer |
string|null | LLM-extracted final answer |
problem_type |
string|null | proof, answer, proof and answer, … |
metadata_confidence |
float32 | Self-rated confidence of LLM metadata |
original_problem_markdown |
string|null | Pre-normalization text |
The enriched fields (
natural_language_description,main_ideas,final_answer,problem_type,metadata_confidence) are LLM-assisted and not fully human-audited in the preview. Treat them as convenience annotations, not ground truth.
Configs / splits
One config per country or regional body plus a default all config unioning everything. Each config has a single train split — this is a preview, not the train/test partitioning of MathNet-Solve (which is train: 23,776, test: 6,400, test-hard: 500 in the full release).
Intended uses & limitations
Good for. Olympiad-level reasoning evaluation, multilingual math evaluation, figure-grounded multimodal math, topic-stratified analysis, retrieval benchmarks over mathematical structure, and RL training — the large pool of expert-written solutions provides dense rewards for verifiable-answer problems, while the math-aware similarity pairs open a new axis: rewarding a model for retrieving a structurally equivalent problem is a natural, automatically verifiable signal that does not require a closed-form answer.
Caveats.
- Not contamination-clean. Olympiad problems are indexed widely; assume leakage when evaluating pretrained models.
- Preview schema may change before the full release.
- LLM-assisted metadata is imperfect.
License
With the kind support of IMO President Gregor Dolinar, we reached out to the leaders of all participating countries and obtained their permission to share this dataset publicly. Where a country or contest organization asserts its own copyright, that copyright is retained and takes precedence — see competition, country, and source_booklet on each row. For all remaining problems where no explicit copyright was asserted, the dataset is released under Creative Commons Attribution 4.0 International (CC BY 4.0).
In short: use freely, cite the paper, and respect any explicit rights claimed by the original national team.
If you are a rightsholder with a concern, please open an issue or email shaden@mit.edu.
Citation
@inproceedings{alshammari2026mathnet,
title = {MathNet: A Global Multimodal Benchmark for Mathematical
Reasoning and Retrieval},
author = {Alshammari, Shaden and Wen, Kevin and Zainal, Abrar and
Hamilton, Mark and Safaei, Navid and Albarakati, Sultan and
Freeman, William T. and Torralba, Antonio},
booktitle = {International Conference on Learning Representations},
year = {2026},
url = {https://mathnet.mit.edu}
}
Links
- 🌐 Website & paper: https://mathnet.mit.edu
- 🔭 Browse all 30K problems: https://mathnet.mit.edu/explorer.html
- ✉️ Contact: shaden@mit.edu
© 2026 Massachusetts Institute of Technology · MathNet · ICLR 2026
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