problem_id int64 1 978 | question stringlengths 86 2.11k | source stringlengths 19 76 | solution stringlengths 94 14.7k | asymptote_code stringlengths 44 17.8k | solution_image_url stringclasses 16
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1 | Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter
$AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto
lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle
formed by lines $PQ$ and $RS$ is half the size of $\angle XOZ$, where
$O$ is the midpoint of segment $AB$. | 2010 USAJMO Problem 3 | Let $\alpha = \angle BAZ$, $\beta = \angle ABX$.
Since $XY$ is a chord of the circle with diameter $AB$,
$\angle XAY = \angle XBY = \gamma$. From the chord $YZ$,
we conclude $\angle YAZ = \angle YBZ = \delta$.
Triangles $BQY$ and $APY$ are both right-triangles, and share the
angle $\gamma$, therefore they are similar, and so the ratio $PY : YQ = AY : YB$. Now by Thales' theorem the angles $\angle AXB = \angle AYB = \angle AZB$ are all right-angles. Also, $\angle PYQ$,
being the fourth angle in a quadrilateral with 3 right-angles is
again a right-angle. Therefore $\triangle PYQ \sim \triangle AYB$ and
$\angle YQP = \angle YBA = \gamma + \beta$.
Similarly, $RY : YS = AY : YB$, and so $\angle YRS = \angle YAB = \alpha + \delta$.
Now $RY$ is perpendicular to $AZ$ so the direction $RY$ is $\alpha$ counterclockwise from the vertical, and since $\angle YRS = \alpha + \delta$ we see that $SR$ is $\delta$ clockwise from the vertical. (Draw an actual vertical line segment if necessary.)
Similarly, $QY$ is perpendicular to $BX$ so the direction $QY$ is $\beta$ clockwise from the vertical, and since $\angle YQP$ is $\gamma + \beta$ we see that $QY$ is $\gamma$ counterclockwise from the vertical.
Therefore the lines $PQ$ and $RS$ intersect at an angle $\chi = \gamma + \delta$. Now by the central angle theorem $2\gamma = \angle XOY$
and $2\delta = \angle YOZ$, and so $2(\gamma + \delta) = \angle XOZ$,
and we are done.
Note that $RTQY$ is a quadrilateral whose angles sum to 360°; can you find a faster approach using this fact?
Footnote
We can prove a bit more. Namely, the extensions of the segments
$RS$ and $PQ$ meet at a point on the diameter $AB$ that is vertically
below the point $Y$.
Since $YR = AY \sin(\delta)$ and is inclined $\alpha$ counterclockwise
from the vertical, the point $S$ is $AY \sin(\delta) \sin(\alpha)$
horizontally to the right of $Y$.
Now $AS = AY \cos(\delta)$, so $S$ is $AS \sin(\alpha) = AY \cos(\delta)\sin(\alpha)$ vertically above the diameter $AB$. Also,
the segment $SR$ is inclined $\delta$ clockwise from the vertical,
so if we extend it down from $S$ towards the diameter $AB$ it will
meet the diameter at a point which is
$AY \cos(\delta)\sin(\alpha)\tan(\delta) = AY \sin(\delta)\sin(\alpha)$
horizontally to the left of $S$. This places the intersection point
of $RS$ and $AB$ vertically below $Y$.
Similarly, and by symmetry the intersection point of $PQ$ and $AB$
is directly below $Y$ on $AB$, so the lines through $PQ$ and $RS$
meet at a point $T$ on the diameter that is vertically below $Y$.
Footnote to the Footnote
The Footnote's claim is more easily proved as follows.
Note that because $\angle{QPY}$ and $\angle{YAB}$ are both complementary to $\beta + \gamma$, they must be equal. Now, let $PQ$ intersect diameter $AB$ at $T'$. Then $PYT'A$ is cyclic and so $\angle{YT'A} = 180^\circ - \angle{APY} = 90^\circ$. Hence $T'YSB$ is cyclic as well, and so we deduce that $\angle{YST'} = \angle{YBT'} = 90^\circ - \alpha - \delta = \angle{YSR}.$ Hence $S, R, T'$ are collinear and so $T = T'$. This proves the Footnote.
Footnote to the Footnote to the Footnote
The Footnote's claim can be proved even more easily as follows.
Drop an altitude from $Y$ to $AB$ at point $T$. Notice that $P, Q, T$ are collinear because they form the Simson line of $\triangle AXB$ from $Y$. Also notice that $P, Q, T$ are collinear because they form the Simson line of $\triangle AZB$ from $Y$. Since $T$ is at the diameter $AB$, lines $PQ$ and $SR$ must intersect at the diameter.
Footnote to the Fourth Power
There is another, simpler solution using Simson lines. Can you find it? | import olympiad; // Scale unitsize(1inch); real r = 1.75; // Semi-circle: centre O, radius r, diameter A--B. pair O = (0,0); dot(O); label("$O$", O, plain.S); pair A = r * plain.W; dot(A); label("$A$", A, unit(A)); pair B = r * plain.E; dot(B); label("$B$", B, unit(B)); draw(arc(O, r, 0, 180)--cycle); // points X, Y, Z real alpha = 22.5; real beta = 15; real delta = 30; pair X = r * dir(180 - 2*beta); dot(X); label("$X$", X, unit(X)); pair Y = r * dir(2*(alpha + delta)); dot(Y); label("$Y$", Y, unit(Y)); pair Z = r * dir(2*alpha); dot(Z); label("$Z$", Z, unit(Z)); // Feet of perpendiculars from Y pair P = foot(Y, A, X); dot(P); label("$P$", P, unit(P-Y)); dot(P); pair Q = foot(Y, B, X); dot(P); label("$Q$", Q, unit(A-Q)); dot(Q); pair R = foot(Y, B, Z); dot(R); label("$S$", R, unit(R-Y)); dot(R); pair S = foot(Y, A, Z); dot(S); label("$R$", S, unit(B-S)); dot(S); pair T = foot(Y, A, B); dot(T); label("$T$", T, unit(T-Y)); dot(T); // Segments draw(B--X); draw(B--Y); draw(B--R); draw(A--Z); draw(A--Y); draw(A--P); draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S); draw(R--T); draw(P--T); // Right angles draw(rightanglemark(A, X, B, 3)); draw(rightanglemark(A, Y, B, 3)); draw(rightanglemark(A, Z, B, 3)); draw(rightanglemark(A, P, Y, 3)); draw(rightanglemark(Y, R, B, 3)); draw(rightanglemark(Y, S, A, 3)); draw(rightanglemark(B, Q, Y, 3)); // Acute angles import markers; void langle(pair A, pair B, pair C, string l="", real r=40, int n=1, int nm = 0) { string sl = "$\scriptstyle{" + l + "}$"; marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker; markangle(Label(sl), radius=r, n=n, A, B, C, m); } langle(B, A, Z, "\alpha" ); langle(X, B, A, "\beta", n=2); langle(Y, A, X, "\gamma", nm=1); langle(Y, B, X, "\gamma", nm=1); langle(Z, A, Y, "\delta", nm=2); langle(Z, B, Y, "\delta", nm=2); langle(R, S, Y, "\alpha+\delta", r=23); langle(Y, Q, P, "\beta+\gamma", r=23); langle(R, T, P, "\chi", r=15); | [] |
1 | Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter
$AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto
lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle
formed by lines $PQ$ and $RS$ is half the size of $\angle XOZ$, where
$O$ is the midpoint of segment $AB$. | 2010 USAJMO Problem 3 | Let $T$ be the foot of the perpendicular from $Y$ to $\overline{AB}$, let $O$ be the center of the semi-circle.
Since we have a semi-circle, if we were to reflect it over $\overline {AB}$, we would have a full circle, with $\triangle{AXB}$ and $\triangle{AZB}$ inscribed in it. Now, notice that $Y$ is a point on that full circle, so we can say that $T$ lies on the Simson Line $\overline{PQ}$ from $Y$ to $\triangle AXB$ and that it also lies on the Simson line $\overline {RS}$ from $Y$ to $\triangle AZB$. Thus, $T$ lies on two distinct lines in a plane, which means that $T=\overline{PQ}\cap\overline{RS}$. Therefore, it suffices to show that $\angle PTS=\tfrac{1}{2}\angle XOZ$.
Since $m\angle YTA + m\angle YPA = 90^\circ + 90^\circ = 180^\circ$ and $m \angle YTB + m \angle YSB = 90^\circ + 90^\circ = 180^\circ$, we know that $TAPY$ and $TBSY$ are cyclic quadrilaterals.
We use this fact to get \[\angle PTS=\angle PTY+\angle YTS=\angle PAY+\angle YBS=\angle XAY+\angle YBZ. \space \space (1)\] \\
Now note that $\angle XAY$ is the inscribed angle of minor arc $\overset{\huge\frown}{PY}$, and $\angle XOY$ is the central angle of minor arc $\overset{\huge\frown}{AB}$, so $\angle XAY = \frac{\overset{\huge\frown}{PY}}{2} = \frac{\angle XOY}{2}$. Similarly, $\angle YBZ = \frac{\overset{\huge\frown}{YZ}}{2}=\frac{\angle YOZ}{2}$. Thus we can say \[\angle XAY + \angle YBZ = \frac{\angle XOY}{2} + \frac{\angle YOZ}{2}=\frac{\angle XOY + \angle YOZ}{2} = \frac{\angle XOZ}{2}. \space \space (2)\]
Combining statements $(1)$ and $(2)$, we can say that $\angle PTS = \frac{\angle XOZ}{2}$, as desired. $\square$
~thinker123 | currentpicture=new picture; size(12cm); pair O, A, B, X, Y, Z, P, Q, R, SS, T; O=(0, 0); A=(-1, 0); B=(1, 0); X=(Cos(144), Sin(144)); Y=(Cos(105), Sin(105)); Z=(Cos(27), Sin(27)); P=foot(Y, A, X); Q=foot(Y, B, X); R=foot(Y, A, Z); SS=foot(Y, B, Z); T=foot(Y, A, B); dot(O); dot(A); dot(B); dot(X); dot(Y); dot(Z); dot(P); dot(Q); dot(R); dot(SS); dot(T); draw(arc(O, 1, 0, 180)); draw(circumcircle(T, A, Y), dotted); draw(circumcircle(T, B, Y), dotted); draw(A -- B); draw(Z -- O -- X -- A -- Z -- B -- X); draw(A -- Y -- B); draw(P -- T -- SS); draw(P -- Y -- Q); draw(R -- Y -- SS); draw(X -- P); draw(Z -- SS); draw(Y -- T); draw(rightanglemark(Y, T, B, 1.25)); draw(rightanglemark(Y, P, A, 1.25)); draw(rightanglemark(Y, Q, X, 1.25)); draw(rightanglemark(Y, R, Z, 1.25)); draw(rightanglemark(Y, SS, B, 1.25)); draw(rightanglemark(A, X, B, 1.25)); draw(rightanglemark(A, Y, B, 1.25)); draw(rightanglemark(A, Z, B, 1.25)); label("$O$", O, S); label("$A$", A, SW); label("$B$", B, SE); label("$X$", X, (X-B)/length(X-B)); label("$Y$", Y, Y); label("$Z$", Z, (Z-A)/length(Z-A)); label("$P$", P, (P-T)/length(P-T)); label("$Q$", Q, SW); label("$R$", R, SE); label("$S$", SS, (SS-T)/length(SS-T)); label("$T$", T, S); | [] |
2 | Let $ABC$ be a triangle with $\angle A = 90^{\circ}$. Points $D$
and $E$ lie on sides $AC$ and $AB$, respectively, such that $\angle ABD = \angle DBC$ and $\angle ACE = \angle ECB$. Segments $BD$ and
$CE$ meet at $I$. Determine whether or not it is possible for
segments $AB, AC, BI, ID, CI, IE$ to all have integer lengths. | 2010 USAJMO Problem 6 | We know that angle $BIC = 135^{\circ}$, as the other two angles in triangle $BIC$ add to $45^{\circ}$. Assume that only $AB, AC, BI$, and $CI$ are integers. Using the Law of Cosines on triangle BIC,
$BC^2 = BI^2 + CI^2 - 2BI\cdot CI \cdot \cos 135^{\circ}$. Observing that $BC^2 = AB^2 + AC^2$ is an integer and that $\cos 135^{\circ} = -\frac{\sqrt{2}}{2},$ we have
\[BC^2 - BI^2 - CI^2 = BI\cdot CI\cdot \sqrt{2}\]
and therefore,
\[\sqrt{2} = \frac{BC^2 - BI^2 - CI^2}{BI\cdot CI}\]
The LHS ($\sqrt{2}$) is irrational, while the RHS is the quotient of the division of two integers and thus is rational. Clearly, there is a contradiction. Therefore, it is impossible for $AB, AC, BI$, and $CI$ to all be integers, which invalidates the original claim that all six lengths are integers, and we are done. | import olympiad; // Scale unitsize(1inch); // Shape real h = 1.75; real w = 2.5; // Points void ldot(pair p, string l, pair dir=p) { dot(p); label(l, p, unit(dir)); } pair A = origin; ldot(A, "$A$", plain.SW); pair B = w * plain.E; ldot(B, "$B$", plain.SE); pair C = h * plain.N; ldot(C, "$C$", plain.NW); pair D = extension(B, bisectorpoint(C, B, A), A, C); ldot(D, "$D$", D-B); pair E = extension(C, bisectorpoint(A, C, B), A, B); ldot(E, "$E$", E-C); pair I = extension(B, D, C, E); ldot(I, "$I$", A-I); // Segments draw(A--B); draw(B--C); draw(C--A); draw(C--E); draw(B--D); // Angles import markers; draw(rightanglemark(B, A, C, 4)); markangle(Label("$\scriptstyle{\frac{\theta}{2}}$"), radius=40, I, B, E); markangle(Label("$\scriptstyle{\frac{\theta}{2}}$"), radius=40, C, B, I); markangle(Label("$\scriptstyle{\frac{\pi}{4} - \frac{\theta}{2}}$"), radius=40, I, C, B); markangle(Label("$\scriptstyle{\frac{\pi}{4} - \frac{\theta}{2}}$"), radius=40, D, C, I); markangle(Label("$\scriptstyle{\frac{3\pi}{4}}$"), radius=10, B, I, C); | [] |
3 | For a point $P = (a, a^2)$ in the coordinate plane, let $\ell(P)$ denote the line passing through $P$ with slope $2a$. Consider the set of triangles with vertices of the form $P_1 = (a_1, a_1^2)$, $P_2 = (a_2, a_2^2)$, $P_3 = (a_3, a_3^2)$, such that the intersections of the lines $\ell(P_1)$, $\ell(P_2)$, $\ell(P_3)$ form an equilateral triangle $\triangle$. Find the locus of the center of $\triangle$ as $P_1P_2P_3$ ranges over all such triangles. | 2011 USAJMO Problem 3 | Solution 1
Note that the lines $l(P_1), l(P_2), l(P_3)$ are \[y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,\] respectively. It is easy to deduce that the three points of intersection are \[\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).\] The slopes of each side of this equilateral triangle are \[2a_1,2a_2,2a_3,\] and we want to find the locus of \[\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).\] Define the three complex numbers $w_n = 1+2a_ni$ for $n=1,2,3$. Then note that the slope - that is, the imaginary part divided by the real part - of all $w_n^3$ is constant, say it is $k$. Then for $n=1,2,3$,
\begin{align*} \frac{\Im(w_n^3)}{\Re(w_n^3)} &= \frac{\Im((1+2a_ni)^3)}{\Re((1+2a_ni)^3)}\\ &= \frac{\Im(1+6a_ni-12a_n^2-8a_n^3i)}{\Re(1+6a_ni-12a_n^2-8a_n^3i)}\\ &= \frac{6a_n-8a_n^3}{1-12a_n^2}\\ &= k.\\ \end{align*}
Rearranging, we get that\[8a_n^3 -12ka_n^2-6a_n+k=0,\]or\[a_n^3-\frac{3ka_n^2}2-\frac{3a_n}4+\frac k8=0.\]Note that this is a cubic, and the roots are $a_1,a_2$ and $a_3$ which are all distinct, and so there are no other roots. Using Vieta's, we get that
\[a_1+a_2+a_3=\frac{3k}2,\]and\[a_1a_2+a_2a_3+a_3a_1=-\frac34.\]
Obviously all values of $k$ are possible, and so our answer is the line \[\boxed{y=-\frac{1}{4}.}\] $\blacksquare$
~ cocohearts
Solution 2
Note that all the points $P=(a,a^2)$ belong to the parabola $y=x^2$ which we will denote $p$. This parabola has a focus $F=\left(0,\frac{1}{4}\right)$ and directrix $y=-\frac{1}{4}$ which we will denote $d$. We will prove that the desired locus is $d$.
First note that for any point $P$ on $p$, the line $\ell(P)$ is the tangent line to $p$ at $P$. This is because $\ell(P)$ contains $P$ and because $[\frac{d}{dx}] x^2=2x$. If you don't like calculus, you can also verify that $\ell(P)$ has equation $y=2a(x-a)+a^2$ and does not intersect $y=x^2$ at any point besides $P$. Now for any point $P$ on $p$ let $P'$ be the foot of the perpendicular from $P$ onto $d$. Then by the definition of parabolas, $PP'=PF$. Let $q$ be the perpendicular bisector of $\overline{P'F}$. Since $PP'=PF$, $q$ passes through $P$. Suppose $K$ is any other point on $q$ and let $K'$ be the foot of the perpendicular from $K$ to $d$. Then in right $\Delta KK'P'$, $KK'$ is a leg and so $KK'<KP'=KF$. Therefore $K$ cannot be on $p$. This implies that $q$ is exactly the tangent line to $p$ at $P$, that is $q=\ell(P)$. So we have proved Lemma 1: If $P$ is a point on $p$ then $\ell(P)$ is the perpendicular bisector of $\overline{P'F}$.
We need another lemma before we proceed. Lemma 2: If $F$ is on the circumcircle of $\Delta XYZ$ with orthocenter $H$, then the reflections of $F$ across $\overleftrightarrow{XY}$, $\overleftrightarrow{XZ}$, and $\overleftrightarrow{YZ}$ are collinear with $H$.
Proof of Lemma 2: Say the reflections of $F$ and $H$ across $\overleftrightarrow{YZ}$ are $C'$ and $J$, and the reflections of $F$ and $H$ across $\overleftrightarrow{XY}$ are $A'$ and $I$. Then we angle chase $\angle JYZ=\angle HYZ=\angle HXZ=\angle JXZ=m(JZ)/2$ where $m(JZ)$ is the measure of minor arc $JZ$ on the circumcircle of $\Delta XYZ$. This implies that $J$ is on the circumcircle of $\Delta XYZ$, and similarly $I$ is on the circumcircle of $\Delta XYZ$. Therefore $\angle C'HJ=\angle FJH=m(XF)/2$, and $\angle A'HX=\angle FIX=m(FX)/2$. So $\angle C'HJ = \angle A'HX$. Since $J$, $H$, and $X$ are collinear it follows that $C'$, $H$ and $A'$ are collinear. Similarly, the reflection of $F$ over $\overleftrightarrow{XZ}$ also lies on this line, and so the claim is proved.
Now suppose $A$, $B$, and $C$ are three points of $p$ and let $\ell(A)\cap\ell(B)=X$, $\ell(A)\cap\ell(C)=Y$, and $\ell(B)\cap\ell(C)=Z$. Also let $A''$, $B''$, and $C''$ be the midpoints of $\overline{A'F}$, $\overline{B'F}$, and $\overline{C'F}$ respectively. Then since $\overleftrightarrow{A''B''}\parallel \overline{A'B'}=d$ and $\overleftrightarrow{B''C''}\parallel \overline{B'C'}=d$, it follows that $A''$, $B''$, and $C''$ are collinear. By Lemma 1, we know that $A''$, $B''$, and $C''$ are the feet of the altitudes from $F$ to $\overline{XY}$, $\overline{XZ}$, and $\overline{YZ}$. Therefore by the Simson Line Theorem, $F$ is on the circumcircle of $\Delta XYZ$. If $H$ is the orthocenter of $\Delta XYZ$, then by Lemma 2, it follows that $H$ is on $\overleftrightarrow{A'C'}=d$. It follows that the locus described in the problem is a subset of $d$.
Since we claim that the locus described in the problem is $d$, we still need to show that for any choice of $H$ on $d$ there exists an equilateral triangle with center $H$ such that the lines containing the sides of the triangle are tangent to $p$. So suppose $H$ is any point on $d$ and let the circle centered at $H$ through $F$ be $O$. Then suppose $A$ is one of the intersections of $d$ with $O$. Let $\angle HFA=3\theta$, and construct the ray through $F$ on the same halfplane of $\overleftrightarrow{HF}$ as $A$ that makes an angle of $2\theta$ with $\overleftrightarrow{HF}$. Say this ray intersects $O$ in a point $B$ besides $F$, and let $q$ be the perpendicular bisector of $\overline{HB}$. Since $\angle HFB=2\theta$ and $\angle HFA=3\theta$, we have $\angle BFA=\theta$. By the inscribed angles theorem, it follows that $\angle AHB=2\theta$. Also since $HF$ and $HB$ are both radii, $\Delta HFB$ is isosceles and $\angle HBF=\angle HFB=2\theta$. Let $P_1'$ be the reflection of $F$ across $q$. Then $2\theta=\angle FBH=\angle C'HB$, and so $\angle C'HB=\angle AHB$. It follows that $P_1'$ is on $\overleftrightarrow{AH}=d$, which means $q$ is the perpendicular bisector of $\overline{FP_1'}$.
Let $q$ intersect $O$ in points $Y$ and $Z$ and let $X$ be the point diametrically opposite to $B$ on $O$. Also let $\overline{HB}$ intersect $q$ at $M$. Then $HM=HB/2=HZ/2$. Therefore $\Delta HMZ$ is a $30-60-90$ right triangle and so $\angle ZHB=60^{\circ}$. So $\angle ZHY=120^{\circ}$ and by the inscribed angles theorem, $\angle ZXY=60^{\circ}$. Since $ZX=ZY$ it follows that $\Delta ZXY$ is and equilateral triangle with center $H$.
By Lemma 2, it follows that the reflections of $F$ across $\overleftrightarrow{XY}$ and $\overleftrightarrow{XZ}$, call them $P_2'$ and $P_3'$, lie on $d$. Let the intersection of $\overleftrightarrow{YZ}$ and the perpendicular to $d$ through $P_1'$ be $P_1$, the intersection of $\overleftrightarrow{XY}$ and the perpendicular to $d$ through $P_2'$ be $P_2$, and the intersection of $\overleftrightarrow{XZ}$ and the perpendicular to $d$ through $P_3'$ be $P_3$. Then by the definitions of $P_1'$, $P_2'$, and $P_3'$ it follows that $FP_i=P_iP_i'$ for $i=1,2,3$ and so $P_1$, $P_2$, and $P_3$ are on $p$. By lemma 1, $\ell(P_1)=\overleftrightarrow{YZ}$, $\ell(P_2)=\overleftrightarrow{XY}$, and $\ell(P_3)=\overleftrightarrow{XZ}$. Therefore the intersections of $\ell(P_1)$, $\ell(P_2)$, and $\ell(P_3)$ form an equilateral triangle with center $H$, which finishes the proof.
--Killbilledtoucan
Solution 3
Note that the lines $l(P_1), l(P_2), l(P_3)$ are \[y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,\] respectively. It is easy to deduce that the three points of intersection are \[\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).\] The slopes of each side of this equilateral triangle are \[2a_1,2a_2,2a_3,\] and we want to find the locus of \[\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).\] We know that \[2a_1=\tan(\theta), 2a_2=\tan (\theta + 120), 2a_3=\tan (\theta-120)\] for some $\theta.$ Therefore, we can use the tangent addition formula to deduce \[\frac{a_1+a_2+a_3}{3}=\frac{\tan(\theta)+\tan (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}\] and \begin{align*} \frac{a_1a_2+a_2a_3+a_3a_1}{3}&=\frac{\tan\theta (\tan(\theta-120)+\tan(\theta+120))+\tan(\theta-120)\tan(\theta+120)}{12}\\ &=\frac{9\tan^2\theta-3}{12(1-3\tan^2\theta)}\\ &=-\frac{1}{4}.\end{align*} Now we show that $\frac{a_1+a_2+a_3}{3}$ can be any real number. Let's say \[\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}=k\] for some real number $k.$ Multiplying both sides by $2-\tan^2\theta$ and rearranging yields a cubic in $\tan\theta.$ Clearly this cubic has at least one real solution. As $\tan \theta$ can take on any real number, all values of $k$ are possible, and our answer is the line \[\boxed{y=-\frac{1}{4}.}\] Of course, as the denominator could equal 0, we must check $\tan \theta=\pm \frac{1}{\sqrt{3}}.$ \[3\tan \theta-\tan^3\theta=k(2-6\tan^2\theta).\] The left side is nonzero, while the right side is zero, so these values of $\theta$ do not contribute to any values of $k.$ So, our answer remains the same. $\blacksquare$ ~ Benq
Work in progress: Solution 4 (Clean algebra)
It can be easily shown that the center of $\triangle$ has coordinates $\left(\frac{a_{1}+a_{2}+a_{3}}{3},\frac{a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}}{3}\right)$.
Without loss of generality, let $a_{1}>a_{2}>a_{3}$. Notice that $\ell(P_2)$ is a $60^{\circ}$ clockwise rotation of $\ell(P_1)$, $\ell(P_3)$ is a $60^{\circ}$ clockwise rotation of $\ell(P_2)$, and $\ell(P_1)$ is a $60^{\circ}$ clockwise rotation of $\ell(P_3)$. By definition, $\arctan(2a_{i})$ is the (directed) angle from the x-axis to $\ell(P_{i})$. Remember that the range of $\arctan(x)$ is $(-90^{\circ},90^{\circ})$. We have \begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}
Taking the tangent of both sides of each equation and rearranging, we get \begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*} We add these equations to get \[\sqrt{3}(3+4(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}))=0.\] We solve for $a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}$ to get \[a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}=-\frac{3}{4}.\] So, the y-coordinate of $\triangle$ is $-\frac{1}{4}$.
We will prove that the x-coordinate of $\triangle$ can be any real number. If $2a_{1}$ tends to infinity, then $2a_{2}$ tends to $\frac{\sqrt{3}}{2}$ and $2a_{3}$ tends to $-\frac{\sqrt{3}}{2}$. So, $\frac{a_{1}+a_{2}+a_{3}}{3}$ can be arbitrarily large. Similarly, if we let $2a_{3}$ tend to negative infinity, then $2a_{1}$ tends to $\frac{\sqrt{3}}{2}$ and $2a_{2}$ tends to $-\frac{\sqrt{3}}{2}$. So, $\frac{a_{1}+a_{2}+a_{3}}{3}$ can be arbitrarily small. Since $\frac{a_{1}+a_{2}+a_{3}}{3}$ is continuous, it can take any real value. So, the locus is the line $y=-\frac{1}{4}$. | Label f; f.p=fontsize(6); xaxis(-2,2); yaxis(-2,2); real f(real x) { return x^2; } draw(graph(f,-sqrt(2),sqrt(2))); real f(real x) { return (2*sqrt(3)/3)*x-1/3; } draw(graph(f,-5*sqrt(3)/6,2)); real f(real x) { return (-sqrt(3)/9)*x-1/108; } draw(graph(f,-2,2)); real f(real x) { return (-5*sqrt(3)/3)*x-25/12; } draw(graph(f,-49*sqrt(3)/60,-sqrt(3)/60)); | [] |
4 | Points $A$, $B$, $C$, $D$, $E$ lie on a circle $\omega$ and point $P$ lies outside the circle. The given points are such that (i) lines $PB$ and $PD$ are tangent to $\omega$, (ii) $P$, $A$, $C$ are collinear, and (iii) $\overline{DE} \parallel \overline{AC}$. Prove that $\overline{BE}$ bisects $\overline{AC}$. | 2011 USAJMO Problem 5 | This is the solution from EGMO Problem 1.43 page 242
Let $O$ be the center of the circle, and let $M$ be the midpoint of $AC$. Let $\theta$ denote the circle with diameter $OP$. Since $\angle OBP = \angle OMP = \angle ODP = 90^\circ$, $B$, $D$, and $M$ all lie on $\theta$.
Since quadrilateral $BOMP$ is cyclic, $\angle BMP = \angle BOP$. Triangles $BOP$ and $DOP$ are congruent, so $\angle BOP = \angle BOD/2 = \angle BED$, so $\angle BMP = \angle BED$. Because $AC$ and $DE$ are parallel, $M$ lies on $BE$ (using Euclid's Parallel Postulate).
-Evan Chen (vEnhance) | // Block 1
import graph;
unitsize(2 cm);
pair A, B, C, D, E, M, O, P;
path circ;
O = (0,0);
circ = Circle(O,1);
B = dir(100);
D = dir(240);
P = extension(B, B + rotate(90)*(B), D, D + rotate(90)*(D));
C = dir(-40);
A = intersectionpoint((P--(P + 0.9*(C - P))),circ);
E = intersectionpoint((D + 0.1*(C - A))--(D + C - A),circ);
M = (A + C)/2;
draw(circ);
draw(P--B);
draw(P--D);
draw(P--C);
draw(B--E);
draw(D--E);
draw(O--B);
draw(O--D);
draw(O--M);
draw(O--P);
draw(Circle((O + P)/2, abs(O - P)/2),dashed);
draw(D--M);
dot("$A$", A, NE);
dot("$B$", B, NE);
dot("$C$", C, SE);
dot("$D$", D, S);
dot("$E$", E, S);
dot("$M$", M, NE);
dot("$O$", O, dir(0));
dot("$P$", P, W);
label("$\theta$", (O + P)/2 + abs(O - P)/2*dir(120), NW);
// Block 2
import graph; unitsize(2 cm); pair A, B, C, D, E, M, O, P; path circ; O = (0,0); circ = Circle(O,1); B = dir(100); D = dir(240); P = extension(B, B + rotate(90)*(B), D, D + rotate(90)*(D)); C = dir(-40); A = intersectionpoint((P--(P + 0.9*(C - P))),circ); E = intersectionpoint((D + 0.1*(C - A))--(D + C - A),circ); M = (A + C)/2; draw(circ); draw(P--B); draw(P--D); draw(P--C); draw(B--E); draw(D--E); draw(O--B); draw(O--D); draw(O--M); draw(O--P); draw(Circle((O + P)/2, abs(O - P)/2),dashed); draw(D--M); dot("$A$", A, NE); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, S); dot("$E$", E, S); dot("$M$", M, NE); dot("$O$", O, dir(0)); dot("$P$", P, W); label("$\theta$", (O + P)/2 + abs(O - P)/2*dir(120), NW); | [] |
5 | Given a triangle $ABC$, let $P$ and $Q$ be points on segments $\overline{AB}$ and $\overline{AC}$, respectively, such that $AP = AQ$. Let $S$ and $R$ be distinct points on segment $\overline{BC}$ such that $S$ lies between $B$ and $R$, $\angle{BPS} = \angle{PRS}$, and $\angle{CQR} = \angle{QSR}$. Prove that $P$, $Q$, $R$, $S$ are concyclic (in other words, these four points lie on a circle). | 2012 USAJMO Problem 1 | Since $\angle BPS = \angle PRS$, the circumcircle of triangle $PRS$ is tangent to $AB$ at $P$. Similarly, since $\angle CQR = \angle QSR$, the circumcircle of triangle $QRS$ is tangent to $AC$ at $Q$.
For the sake of contradiction, suppose that the circumcircles of triangles $PRS$ and $QRS$ are not the same circle. Since $AP = AQ$, $AP^2 = AQ^2$, so the power of A with respect to both circumcircles is the same. Thus, $A$ lies on the radical axis of both circles. However, both circles pass through $R$ and $S$, so the radical axis of both circles is $RS$. Hence, $A$ lies on $RS$, which is a contradiction.
Therefore, the two circumcircles are the same circle. In other words, $P$, $Q$, $R$, and $S$ all lie on the same circle. | // Block 1
import markers;
unitsize(0.5 cm);
pair A, B, C, O, P, Q, R, S;
A = (2,12);
B = (0,0);
C = (14,0);
P = intersectionpoint(A--B,Circle(A,8));
Q = intersectionpoint(A--C,Circle(A,8));
O = extension(P, P + rotate(90)*(A - P), Q, Q + rotate(90)*(A - Q));
S = intersectionpoint(B--C,arc(O, abs(O - P), 180, 270));
R = intersectionpoint(B--C,arc(O, abs(O - P), 270, 360));
draw(A--B--C--cycle);
draw(Circle(O, abs(O - P)));
draw(S--P--R);
draw(S--Q--R);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$P$", P, W);
label("$Q$", Q, NE);
label("$R$", R, SE);
label("$S$", S, SW);
markangle(1, B, P, S, radius=4mm, marker(markinterval(stickframe(n=1,2mm),true)));
markangle(1, P, R, S, radius=4mm, marker(markinterval(stickframe(n=1,2mm),true)));
markangle(1, R, Q, C, radius=4mm, marker(markinterval(stickframe(n=2,2mm),true)));
markangle(1, R, S, Q, radius=4mm, marker(markinterval(stickframe(n=2,2mm),true)));
// Block 2
import markers; unitsize(0.5 cm); pair A, B, C, O, P, Q, R, S; A = (2,12); B = (0,0); C = (14,0); P = intersectionpoint(A--B,Circle(A,8)); Q = intersectionpoint(A--C,Circle(A,8)); O = extension(P, P + rotate(90)*(A - P), Q, Q + rotate(90)*(A - Q)); S = intersectionpoint(B--C,arc(O, abs(O - P), 180, 270)); R = intersectionpoint(B--C,arc(O, abs(O - P), 270, 360)); draw(A--B--C--cycle); draw(Circle(O, abs(O - P))); draw(S--P--R); draw(S--Q--R); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$P$", P, W); label("$Q$", Q, NE); label("$R$", R, SE); label("$S$", S, SW); markangle(1, B, P, S, radius=4mm, marker(markinterval(stickframe(n=1,2mm),true))); markangle(1, P, R, S, radius=4mm, marker(markinterval(stickframe(n=1,2mm),true))); markangle(1, R, Q, C, radius=4mm, marker(markinterval(stickframe(n=2,2mm),true))); markangle(1, R, S, Q, radius=4mm, marker(markinterval(stickframe(n=2,2mm),true))); | [] |
6 | Let $P$ be a point in the plane of triangle $ABC$, and $\gamma$ a line passing through $P$. Let $A'$, $B'$, $C'$ be the points where the reflections of lines $PA$, $PB$, $PC$ with respect to $\gamma$ intersect lines $BC$, $AC$, $AB$, respectively. Prove that $A'$, $B'$, $C'$ are collinear. | 2012 USAJMO Problem 6 | By the sine law on triangle $AB'P$,
\[\frac{AB'}{\sin \angle APB'} = \frac{AP}{\sin \angle AB'P},\]
so
\[AB' = AP \cdot \frac{\sin \angle APB'}{\sin \angle AB'P}.\]
Similarly,
\begin{align*} B'C &= CP \cdot \frac{\sin \angle CPB'}{\sin \angle CB'P}, \\ CA' &= CP \cdot \frac{\sin \angle CPA'}{\sin \angle CA'P}, \\ A'B &= BP \cdot \frac{\sin \angle BPA'}{\sin \angle BA'P}, \\ BC' &= BP \cdot \frac{\sin \angle BPC'}{\sin \angle BC'P}, \\ C'A &= AP \cdot \frac{\sin \angle APC'}{\sin \angle AC'P}. \end{align*}
Hence,
\begin{align*} &\frac{AB'}{B'C} \cdot \frac{CA'}{A'B} \cdot \frac{BC'}{C'A} \\ &= \frac{\sin \angle APB'}{\sin \angle AB'P} \cdot \frac{\sin \angle CB'P}{\sin \angle CPB'} \cdot \frac{\sin \angle CPA'}{\sin \angle CA'P} \cdot \frac{\sin \angle BA'P}{\sin \angle BPA'} \cdot \frac{\sin \angle BPC'}{\sin \angle BC'P} \cdot \frac{\sin \angle AC'P}{\sin \angle APC'}. \end{align*}
Since angles $\angle AB'P$ and $\angle CB'P$ are supplementary or equal, depending on the position of $B'$ on $AC$,
\[\sin \angle AB'P = \sin \angle CB'P.\]
Similarly,
\begin{align*} \sin \angle CA'P &= \sin \angle BA'P, \\ \sin \angle BC'P &= \sin \angle AC'P. \end{align*}
By the reflective property, $\angle APB'$ and $\angle BPA'$ are supplementary or equal, so
\[\sin \angle APB' = \sin \angle BPA'.\]
Similarly,
\begin{align*} \sin \angle CPA' &= \sin \angle APC', \\ \sin \angle BPC' &= \sin \angle CPB'. \end{align*}
Therefore,
\[\frac{AB'}{B'C} \cdot \frac{CA'}{A'B} \cdot \frac{BC'}{C'A} = 1,\]
so by Menelaus's theorem, $A'$, $B'$, and $C'$ are collinear. | import graph; import geometry; unitsize(0.5 cm); pair[] A, B, C; pair P, R; A[0] = (2,12); B[0] = (0,0); C[0] = (14,0); P = (4,5); R = 5*dir(70); A[1] = extension(B[0],C[0],P,reflect(P + R,P - R)*(A[0])); B[1] = extension(C[0],A[0],P,reflect(P + R,P - R)*(B[0])); C[1] = extension(A[0],B[0],P,reflect(P + R,P - R)*(C[0])); draw((P - R)--(P + R),red); draw(A[1]--B[1]--C[1]--cycle,blue); draw(A[0]--B[0]--C[0]--cycle); draw(A[0]--P); draw(B[0]--P); draw(C[0]--P); draw(P--A[1]); draw(P--B[1]); draw(P--C[1]); draw(A[1]--B[0]); draw(A[1]--B[0]); label("$A$", A[0], N); label("$B$", B[0], S); label("$C$", C[0], SE); dot("$A'$", A[1], SW); dot("$B'$", B[1], NE); dot("$C'$", C[1], W); dot("$P$", P, SE); label("$\gamma$", P + R, N); | [] |
7 | In triangle $ABC$, points $P,Q,R$ lie on sides $BC,CA,AB$ respectively. Let $\omega_A$, $\omega_B$, $\omega_C$ denote the circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively. Given the fact that segment $AP$ intersects $\omega_A$, $\omega_B$, $\omega_C$ again at $X,Y,Z$ respectively, prove that $YX/XZ=BP/PC$. | 2013 USAJMO Problem 3 | In this solution, all lengths and angles are directed.
Firstly, it is easy to see by that $\omega_A, \omega_B, \omega_C$ concur at a point $M$ (the Miquel point). Let $XM$ meet $\omega_B, \omega_C$ again at $D$ and $E$, respectively. Then by Power of a Point, we have \[XM \cdot XE = XZ \cdot XP \quad\text{and}\quad XM \cdot XD = XY \cdot XP\] Thusly \[\frac{XY}{XZ} = \frac{XD}{XE}\] But we claim that $\triangle XDP \sim \triangle PBM$. Indeed, \[\measuredangle XDP = \measuredangle MDP = \measuredangle MBP = - \measuredangle PBM\] and \[\measuredangle DXP = \measuredangle MXY = \measuredangle MXA = \measuredangle MRA = 180^\circ - \measuredangle MRB = \measuredangle MPB = -\measuredangle BPM\]
Therefore, $\frac{XD}{XP} = \frac{PB}{PM}$. Analogously we find that $\frac{XE}{XP} = \frac{PC}{PM}$ and we are done.
courtesy v_enhance, minor clarification by integralarefun | /* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(11cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (-1.0, 3.0); pair B = (-3.0, -3.0); pair C = (4.0, -3.0); pair P = (-0.6698198198198195, -3.0); pair Q = (1.1406465288818244, 0.43122416534181074); pair R = (-1.6269590345062048, 1.119122896481385); path w_A = circumcircle(A,Q,R); path w_B = circumcircle(B,P,R); path w_C = circumcircle(P,Q,C); pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5)); pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5)); pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5)); pair X = (2)*(foot(O_A,A,P))-A; pair Y = (2)*(foot(O_B,A,P))-P; pair Z = (2)*(foot(O_C,A,P))-P; pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P; pair D = (2)*(foot(O_B,X,M))-M; pair E = (2)*(foot(O_C,X,M))-M; /* Draw objects */ draw(A--B, rgb(0.6,0.6,0.0)); draw(B--C, rgb(0.6,0.6,0.0)); draw(C--A, rgb(0.6,0.6,0.0)); draw(w_A, rgb(0.4,0.4,0.0)); draw(w_B, rgb(0.4,0.4,0.0)); draw(w_C, rgb(0.4,0.4,0.0)); draw(A--P, rgb(0.0,0.2,0.4)); draw(D--E, rgb(0.0,0.2,0.4)); draw(P--D, rgb(0.0,0.2,0.4)); draw(P--E, rgb(0.0,0.2,0.4)); draw(P--M, rgb(0.4,0.2,0.0)); draw(R--M, rgb(0.4,0.2,0.0)); draw(Q--M, rgb(0.4,0.2,0.0)); draw(B--M, rgb(0.0,0.2,0.4)); draw(C--M, rgb(0.0,0.2,0.4)); draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8)); draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8)); draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8)); draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(P); dot(Q); dot(R); dot(X); dot(Y); dot(Z); dot(M); dot(D); dot(E); /* Label points */ label("$A$", A, lsf * dir(110)); label("$B$", B, lsf * unit(B-M)); label("$C$", C, lsf * unit(C-M)); label("$P$", P, lsf * unit(P-M) * 1.8); label("$Q$", Q, lsf * dir(90) * 1.6); label("$R$", R, lsf * unit(R-M) * 2); label("$X$", X, lsf * dir(-60) * 2); label("$Y$", Y, lsf * dir(45)); label("$Z$", Z, lsf * dir(5)); label("$M$", M, lsf * dir(M-P)*2); label("$D$", D, lsf * dir(150)); label("$E$", E, lsf * dir(5)); | [] |
7 | In triangle $ABC$, points $P,Q,R$ lie on sides $BC,CA,AB$ respectively. Let $\omega_A$, $\omega_B$, $\omega_C$ denote the circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively. Given the fact that segment $AP$ intersects $\omega_A$, $\omega_B$, $\omega_C$ again at $X,Y,Z$ respectively, prove that $YX/XZ=BP/PC$. | 2013 USAJMO Problem 3 | We will use some construction arguments to solve the problem. Let $\angle BAC=\alpha,$ $\angle ABC=\beta,$ $\angle ACB=\gamma,$ and let $\angle APB=\theta.$ We construct lines through the points $Q,$ and $R$ that intersect with $\triangle ABC$ at the points $Q$ and $R,$ respectively, and that intersect each other at $T.$ We will construct these lines such that $\angle CQV=\angle ARV=\theta.$
Now we let the intersections of $AP$ with $RV$ and $QU$ be $Y'$ and $Z',$ respectively. This construction is as follows.
We know that $\angle BRY'=180^\circ-\angle ARY'=180^\circ-\theta.$ Hence, we have,
\begin{align*} \angle BRY'+\angle BPY' &=180^\circ-\theta+\theta\\ &=180^\circ. \end{align*}
Since the opposite angles of quadrilateral $RY'PB$ add up to $180^\circ,$ it must be cyclic. Similarly, we can also show that quadrilaterals $CQZ'P,$ and $AQTR$ are also cyclic.
Since points $Y'$ and $Z'$ lie on $AP,$ we know that,
\[Y'=\omega_B\cap AP\]
and that
\[Z'=\omega_C\cap AP.\]
Hence, the points $Y'$ and $Z'$ coincide with the given points $Y$ and $Z,$ respectively.
Since quadrilateral $AQTR$ is also cyclic, we have,
\begin{align*} \angle Y'TZ' &=180^\circ-\angle RTQ\\ &=180^\circ-(180^\circ-\angle RAQ)\\ &=\angle RAQ\\ &=\alpha. \end{align*}
Similarly, since quadrilaterals $CQZ'P,$ and $AQTR$ are also cyclic, we have,
\begin{align*} \angle TY'Z' &=180^\circ-\angle RY'P\\ &=180^\circ-(180^\circ-\angle RBP)\\ &=\angle RBP\\ &=\beta, \end{align*}
and,
\begin{align*} \angle Y'Z'T &=180^\circ-\angle PZ'Q\\ &=180^\circ-(180^\circ-\angle PCQ)\\ &=\angle PCQ\\ &=\gamma. \end{align*}
Since these three angles are of $\triangle TY'Z',$ and they are equal to corresponding angles of $\triangle ABC,$ by AA similarity, we know that $\triangle TY'Z'\sim \triangle ABC.$
We now consider the point $X=\omega_c\cap AC.$ We know that the points $A,$ $Q,$ $T,$ and $R$ are concyclic. Hence, the points $A,$ $T,$ $X,$ and $R$ must also be concyclic.
Hence, quadrilateral $AQTX$ is cyclic.
Since the angles $\angle ART$ and $\angle AXT$ are inscribed in the same arc $\overarc{AT},$ we have,
\begin{align*} \angle AXT &=\angle ART\\ &=\theta. \end{align*}
Consider by this result, we can deduce that the homothety that maps $ABC$ to $TY'Z'$ will map $P$ to $X.$ Hence, we have that,
\[Y'X/XZ'=BP/PC.\]
Since $Y'=Y$ and $Z'=Z$ hence,
\[YX/XZ=BP/PC,\]
as required. | // Block 1
import graph; size(12cm); real labelscalefactor = 1.9; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle); draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); draw((-7.61,-5)--(7.09,-5)); draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5)); dot((-3.6988888888888977,6.426666666666669)); label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); dot((-7.61,-5)); label("$B$", (-7.61,-5), SW * labelscalefactor); dot((7.09,-5)); label("$C$", (7.09,-5), SE * labelscalefactor); dot((-2.958888888888898,-5)); label("$P$", (-2.958888888888898,-5), S * labelscalefactor); dot((0.5968131669050584,1.8770271258031248)); label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); dot((-5.053354907372894,2.4694710603912564)); label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); dot((-3.143912404905382,-2.142970212141873)); label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); dot((-3.413789986031826,2.0243286531799747)); label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); dot((-3.3284001481939356,0.7057864725120093)); label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); dot((1.7922953932137468,0.6108747864253139)); label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); dot((-5.8024625203461,-5)); label("$U$", (-5.8024625203461,-5), S * labelscalefactor); dot((-0.10264330299819162,1.125351256231488)); label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor);
// Block 2
import graph; size(12cm); real labelscalefactor = 1.9; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle); draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); draw((-7.61,-5)--(7.09,-5)); draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5)); draw((-3.3284001481939356,0.7057864725120093)--(-0.10264330299819162,1.125351256231488)); draw((-3.3284001481939356,0.7057864725120093)--(-5.053354907372894,2.4694710603912564)); draw((-3.6988888888888977,6.426666666666669)--(-0.10264330299819162,1.125351256231488)); dot((-3.6988888888888977,6.426666666666669)); label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); dot((-7.61,-5)); label("$B$", (-7.61,-5), SW * labelscalefactor); dot((7.09,-5)); label("$C$", (7.09,-5), SE * labelscalefactor); dot((-2.958888888888898,-5)); label("$P$", (-2.958888888888898,-5), S * labelscalefactor); dot((0.5968131669050584,1.8770271258031248)); label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); dot((-5.053354907372894,2.4694710603912564)); label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); dot((-3.143912404905382,-2.142970212141873)); label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); dot((-3.413789986031826,2.0243286531799747)); label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); dot((-3.3284001481939356,0.7057864725120093)); label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); dot((1.7922953932137468,0.6108747864253139)); label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); dot((-5.8024625203461,-5)); label("$U$", (-5.8024625203461,-5), S * labelscalefactor); dot((-0.10264330299819162,1.125351256231488)); label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); | [] |
7 | In triangle $ABC$, points $P,Q,R$ lie on sides $BC,CA,AB$ respectively. Let $\omega_A$, $\omega_B$, $\omega_C$ denote the circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively. Given the fact that segment $AP$ intersects $\omega_A$, $\omega_B$, $\omega_C$ again at $X,Y,Z$ respectively, prove that $YX/XZ=BP/PC$. | 2013 USAJMO Problem 3 | We begin again by noting that the three circumcircles intersect at point $M$ by Miquel's theorem. In addition, we state that the angle $\angle MQC = \alpha$, hence $\angle MPC = \angle MZC = 180 - \alpha$, as well as $\angle AQM$, from which follows that $\angle ARM = \alpha$, so $\angle BRM = 180 - \alpha$, and $\angle BPM = \alpha$.
We shall prove that the triangles $\triangle MZC$ and $\triangle XMP$ and $\triangle YMB$ are similar, which will imply a rotational homotethy with angle $\angle APB$ about the point $M$, that takes $Y,X,Z$ to $B,P,C$, thus proving the problem. (In essence, just imagine we rotate $YMZX$ around M and lengthen things out and get $MCPB$ - the ratios will remain identical.)
We do this by angle chasing. Denote angle $\angle ZMP = \beta$. From the angles labeled before, we now know $\angle ZPB = \alpha-\beta = \angle ZMC$. In addition, $\angle ZPM = \angle ZCM$. So the angles in triangle $\triangle ZMC$ are $\alpha-\beta$, $\beta$ and thus $180-\alpha$, with $\alpha-\beta$ at the point $M$. In addition, $\angle YRM = \angle YPM = \beta$, so $\angle ARY = \alpha-\beta$, so $\angle YMB = \alpha - \beta$. Since $\angle YBM = \angle YPM = \beta$, the triangle $\triangle YMB$ has angles $\alpha-\beta, \beta, 180-\alpha$ also, with $\alpha-\beta$ at $M$. Finally, In triangle $\triangle XPM$, we already know the angle $\angle XPM$ to be $\beta$; we also can find that $\angle AQM = 180-\alpha$, so $\angle AXM = \alpha$, so $\angle MXP = 180-\alpha$, so $\angle XMP = \alpha-\beta$. Thus, the three triangles are similar have a common point $M$, which proves that there is a rotational homotethy around $M$ that maps $Y, X, Z$ to $B, P, C$ as desired.
Solution by SimilarTriangle. | /* DRAGON 0.0.9.6 */ import olympiad; import cse5; size(11cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (-1.0, 3.0); pair B = (-3.0, -3.0); pair C = (4.0, -3.0); pair P = (-0.6698198198198195, -3.0); pair Q = (1.1406465288818244, 0.43122416534181074); pair R = (-1.6269590345062048, 1.119122896481385); path w_A = circumcircle(A,Q,R); path w_B = circumcircle(B,P,R); path w_C = circumcircle(P,Q,C); pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5)); pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5)); pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5)); pair X = (2)*(foot(O_A,A,P))-A; pair Y = (2)*(foot(O_B,A,P))-P; pair Z = (2)*(foot(O_C,A,P))-P; pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P; /* Draw objects */ draw(A--B, rgb(0.6,0.6,0.0)); draw(B--C, rgb(0.6,0.6,0.0)); draw(C--A, rgb(0.6,0.6,0.0)); draw(w_A, rgb(0.4,0.4,0.0)); draw(w_B, rgb(0.4,0.4,0.0)); draw(w_C, rgb(0.4,0.4,0.0)); draw(A--P, rgb(0.0,0.2,0.4)); draw(P--M, rgb(0.0,0.2,0.4)); draw(R--M, rgb(0.4,0.2,0.0)); draw(Q--M, rgb(0.4,0.2,0.0)); draw(B--M, rgb(0.0,0.2,0.4)); draw(C--M, rgb(0.0,0.2,0.4)); draw(Z--M, rgb(0.0,0.2,0.4)); draw(X--M, rgb(0.0,0.2,0.4)); draw(Y--M, rgb(0.0,0.2,0.4)); draw(C--Z, rgb(0.0,0.2,0.4)); draw(B--Y, rgb(0.0,0.2,0.4)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(P); dot(Q); dot(R); dot(X); dot(Y); dot(Z); dot(M); /* Label points */ label("$A$", A, lsf * dir(110)); label("$B$", B, lsf * unit(B-M)); label("$C$", C, lsf * unit(C-M)); label("$P$", P, lsf * unit(P-M) * 1.8); label("$Q$", Q, lsf * dir(90) * 1.6); label("$R$", R, lsf * unit(R-M) * 2); label("$X$", X, lsf * dir(-60) * 2); label("$Y$", Y, lsf * dir(45)); label("$Z$", Z, lsf * dir(5)); label("$M$", M, lsf * dir(M-P)*2); | [] |
8 | Let $\triangle{ABC}$ be a non-equilateral, acute triangle with $\angle A=60^{\circ}$, and let $O$ and $H$ denote the circumcenter and orthocenter of $\triangle{ABC}$, respectively.
(a) Prove that line $OH$ intersects both segments $AB$ and $AC$.
(b) Line $OH$ intersects segments $AB$ and $AC$ at $P$ and $Q$, respectively. Denote by $s$ and $t$ the respective areas of triangle $APQ$ and quadrilateral $BPQC$. Determine the range of possible values for $s/t$. | 2014 USAJMO Problem 2 | Lemma: $H$ is the reflection of $O$ over the angle bisector of $\angle BAC$ (henceforth 'the' reflection)
Proof: Let $H'$ be the reflection of $O$, and let $B'$ be the reflection of $B$.
Then reflection takes $\angle ABH'$ to $\angle AB'O$.
$\Delta ABB'$ is equilateral, and $O$ lies on the perpendicular bisector of $\overline{AB}$
It's well known that $O$ lies strictly inside $\Delta ABC$ (since it's acute), meaning that $\angle ABH' = \angle AB'O = 30^{\circ},$ from which it follows that $\overline{BH'} \perp \overline{AC}$ . Similarly, $\overline{CH'} \perp \overline{AB}$. Since $H'$ lies on two altitudes, $H'$ is the orthocenter, as desired.
End Lemma
So $\overline{OH}$ is perpendicular to the angle bisector of $\angle OAH$, which is the same line as the angle bisector of $\angle BAC$, meaning that $\Delta APQ$ is equilateral.
Let its side length be $s$, and let $PH=t$, where $0 < t < s, t \neq s/2$ because $O$ lies strictly within $\angle BAC$, as must $H$, the reflection of $O$. Also, it's easy to show that if $O=H$ in a general triangle, it's equilateral, and we know $\Delta ABC$ is not equilateral. Hence H is not on the bisector of $\angle BAC \implies t \neq s/2$. Let $\overrightarrow{BH}$ intersect $\overline{AC}$ at $P_B$.
Since $\Delta HP_BQ$ and $BP_BA$ are 30-60-90 triangles, $AB=2AP_B=2(s-QP_B)=2(s-HQ/2)=2s-HQ=2s-(s-t)=s+t$
Similarly, $AC=2s-t$
The ratio $\frac{[APQ]}{[ABC]-[APQ]}$ is $\frac{AP \cdot AQ}{AB \cdot AC - AP \cdot AQ} = \frac{s^2}{(s+t)(2s-t)-s^2}$
The denominator equals $(1.5s)^2-(.5s-t)^2-s^2$ where $.5s-t$ can equal any value in $(-.5s, .5s)$ except $0$. Therefore, the denominator can equal any value in $(s^2, 5s^2/4)$, and the ratio is any value in $\boxed{\left(\frac{4}{5},1\right)}.$
Note: It's easy to show that for any point $H$ on $\overline{PQ}$ except the midpoint, Points B and C can be validly defined to make an acute, non-equilateral triangle. | // Block 1
import olympiad;
unitsize(1inch);
pair A,B,C,O,H,P,Q,i1,i2,i3,i4;
//define dots
A=3*dir(50);
B=(0,0);
C=right*2.76481496;
O=circumcenter(A,B,C);
H=orthocenter(A,B,C);
i1=2*O-H;
i2=2*i1-O;
i3=2*H-O;
i4=2*i3-H;
//These points are for extending PQ. DO NOT DELETE!
P=intersectionpoint(i2--i4,A--B);
Q=intersectionpoint(i2--i4,A--C);
//draw
dot(P);
dot(Q);
draw(P--Q);
dot(A);
dot(B);
dot(C);
dot(O);
dot(H);
draw(A--B--C--cycle);
//label
label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("$P$",P,NW);
label("$Q$",Q,NE);
label("$O$",O,N);
label("$H$",H,N);
//change O and H label positions if interfering with other lines to be added
//further editing: ABCPQOH are the dots to be further used. i1,i2,i3,i4 are for drawing assistance and are not to be used
// Block 2
import olympiad; unitsize(1inch); pair A,B,C,O,H,P,Q,i1,i2,i3,i4; //define dots A=3*dir(50); B=(0,0); C=right*2.76481496; O=circumcenter(A,B,C); H=orthocenter(A,B,C); i1=2*O-H; i2=2*i1-O; i3=2*H-O; i4=2*i3-H; //These points are for extending PQ. DO NOT DELETE! P=intersectionpoint(i2--i4,A--B); Q=intersectionpoint(i2--i4,A--C); //draw dot(P); dot(Q); draw(P--Q); dot(A); dot(B); dot(C); dot(O); dot(H); draw(A--B--C--cycle); //label label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$P$",P,NW); label("$Q$",Q,NE); label("$O$",O,N); label("$H$",H,N); //change O and H label positions if interfering with other lines to be added //further editing: ABCPQOH are the dots to be further used. i1,i2,i3,i4 are for drawing assistance and are not to be used | [] |
9 | Let $ABC$ be a triangle with incenter $I$, incircle $\gamma$ and circumcircle $\Gamma$. Let $M,N,P$ be the midpoints of sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ and let $E,F$ be the tangency points of $\gamma$ with $\overline{CA}$ and $\overline{AB}$, respectively. Let $U,V$ be the intersections of line $EF$ with line $MN$ and line $MP$, respectively, and let $X$ be the midpoint of arc $BAC$ of $\Gamma$.
(a) Prove that $I$ lies on ray $CV$.
(b) Prove that line $XI$ bisects $\overline{UV}$. | 2014 USAJMO Problem 6 | (a)
Solution 1: We will prove this via contradiction: assume that line $IC$ intersects line $MP$ at $Q$ and line $EF$ and $R$, with $R$ and $Q$ not equal to $V$. Let $x = \angle A/2 = \angle IAE$ and $y = \angle C/2 = \angle ICA$. We know that $\overline{MP} \parallel \overline{AC}$ because $MP$ is a midsegment of triangle $ABC$; thus, by alternate interior angles (A.I.A) $\angle MVE = \angle FEA = (180^\circ - 2x) / 2 = 90^\circ - x$, because triangle $AFE$ is isosceles. Also by A.I.A, $\angle MQC = \angle QCA = y$. Furthermore, because $AI$ is an angle bisector of triangle $AFE$, it is also an altitude of the triangle; combining this with $\angle QIA = x + y$ from the Exterior Angle Theorem gives $\angle FRC = 90 - x - y$. Also, $\angle VRQ = \angle FRC = 90 - x - y$ because they are vertical angles. This completes part (a).
Solution 2: First we show that the intersection $V'$of $MP$ with the internal angle bisector of $C$ is the same as the intersection $V''$ of $EF$ with the internal angle bisector of $C.$ Let $D$ denote the intersection of $AB$ with the internal angle bisector of $C,$ and let $a,b,c$ denote the side lengths of $BC, AC, AB.$ By Menelaus on $V'', F, E,$ with respect to $\triangle ADC,$ \[\frac{AF}{FD}\cdot \frac{DV''}{V''C}\cdot \frac{CE}{EA}=-1\]
\[\frac{DV''}{V''C}=-\frac{\frac{bc}{a+b}+\frac{a-b-c}{2}}{\frac{a+b-c}{2}}=\frac{b-a}{b+a}.\] Similarly, \[\frac{BP}{PD}\cdot \frac{DV'}{V'C}\cdot \frac{CM}{MB}=-1\] \[\frac{DV'}{V'C}=-\frac{\frac{ac}{a+b}-\frac{c}{2}}{\frac{c}{2}}=\frac{b-a}{b+a}.\] Since $V'$ and $V''$ divide $CD$ in the same ratio, they must be the same point. Now, since $\frac{b-a}{b+a}>-1,$ $I$ lies on ray $CV.$ $\blacksquare$
Solution 3: By the Iran Lemma, we know $CI, EF, MP$ concur, so Part $\text{A}$ follows easily.
(b)
Solution 1: Using a similar argument to part (a), point U lies on line $BI$. Because $\angle MVC = \angle VCA = \angle MCV$, triangle $VMC$ is isosceles. Similarly, triangle $BMU$ is isosceles, from which we derive that $VM = MC = MB = MU$. Hence, triangle $VUM$ is isosceles.
Note that $X$ lies on both the circumcircle and the perpendicular bisector of segment $BC$. Let $D$ be the midpoint of $UV$; our goal is to prove that points $X$, $D$, and $I$ are collinear, which equates to proving $X$ lies on ray $ID$.
Because $MD$ is also an altitude of triangle $MVU$, and $MD$ and $IA$ are both perpendicular to $EF$, $\overline{MD} \parallel \overline{IA}$. Furthermore, we have $\angle VMD = \angle UMD = x$ because $APMN$ is a parallelogram. (incomplete)
Solution 2: Let $I_A$, $I_B$, and $I_C$ be the excenters of $ABC$. Note that the circumcircle of $ABC$ is the nine-point circle of $I_AI_BI_C$. Since $AX$ is the external angle bisector of $\angle BAC$, $X$ is the midpoint of $I_BI_C$. Now $UV$ and $I_BI_C$ are parallel since both are perpendicular to the internal angle bisector of $\angle BAC$. Since $IX$ bisects $I_BI_C$, it bisects $UV$ as well.
Solution 3: Let $X'$ be the antipode of $X$ with respect to the circumcircle of triangle $ABC$. Then, by the Incenter-Excenter lemma, $X'$ is the center of a circle containing $B$, $I$, and $C$. Because $XX'$ is a diameter, $XB$ and $XC$ are tangent to the aforementioned circle; thus by a well-known symmedian lemma (the Tangent Construction Lemma), $XI$ is the $I$-symmedian of triangle $IBC$. From part (a); we know that $BVUC$ is cyclic with diameter $BC$ (we can derive a similar argument for point $U$); thus, from similar triangles, we know that $XI$ is also the median of triangle $VIU$, and we are done.
Solution 4: Let $M_b, M_c$ be the midpoints of arcs $CA, AB$ respectively, and let $T_a$ be the tangency point between the $A$-mixtilinear incircle of $ABC$ and $\Gamma$. It's well-known that $T_a \in XI$, $T_aX$ bisects $M_bM_c$, $AI \perp EF$, and $AI \perp M_bM_c$.
Now, it's easy to see $EF \parallel M_bM_c$, so $IUV$ and $IM_bM_c$ are homothetic at $I$. But $T_aX \equiv IX$ bisects $M_bM_c$, so Part $\text{B}$ follows directly from the homothety.
~ ike.chen | // Block 1
unitsize(5cm);
import olympiad;
pair A, B, C, I, M, N, P, E, F, U, V, X, R;
A = dir(190);
B = dir(120);
C = dir(350);
I = incenter(A, B, C);
label("$A$", A, W);
label("$B$", B, dir(90));
label("$C$", C, dir(0));
dot(I); label("$I$", I, SSE);
draw(A--B--C--cycle);
real r, R;
r = inradius(A, B, C);
R = circumradius(A, B, C);
path G, g;
G = circumcircle(A, B, C);
g = incircle(A, B, C);
draw(G); draw(g);
label("$\Gamma$", dir(35), dir(35));
label("$\gamma$", 2/3 * dir(125));
M = (B+C)/2;
N = (A+C)/2;
P = (A+B)/2;
label("$M$", M, NE);
label("$N$", N, SE);
label("$P$", P, W);
E = tangent(A, I, r, 1);
F = tangent(A, I, r, 2);
label("$E$", E, SW);
label("$F$", F, WNW);
U = extension(E, F, M, N);
V = intersectionpoint(P--M, F--E);
label("$U$", U, S);
label("$V$", V, NE);
draw(P--M--U--F);
X = dir(235);
label("$X$", X, dir(235));
draw(X--I, dashed);
draw(C--V, dashed);
draw(A--I);
label("$x^\circ$", A + (0.2,0), dir(90));
label("$y^\circ$", C + (-0.4,0), dir(90));
// Block 2
unitsize(5cm); import olympiad; pair A, B, C, I, M, N, P, E, F, U, V, X, R; A = dir(190); B = dir(120); C = dir(350); I = incenter(A, B, C); label("$A$", A, W); label("$B$", B, dir(90)); label("$C$", C, dir(0)); dot(I); label("$I$", I, SSE); draw(A--B--C--cycle); real r, R; r = inradius(A, B, C); R = circumradius(A, B, C); path G, g; G = circumcircle(A, B, C); g = incircle(A, B, C); draw(G); draw(g); label("$\Gamma$", dir(35), dir(35)); label("$\gamma$", 2/3 * dir(125)); M = (B+C)/2; N = (A+C)/2; P = (A+B)/2; label("$M$", M, NE); label("$N$", N, SE); label("$P$", P, W); E = tangent(A, I, r, 1); F = tangent(A, I, r, 2); label("$E$", E, SW); label("$F$", F, WNW); U = extension(E, F, M, N); V = intersectionpoint(P--M, F--E); label("$U$", U, S); label("$V$", V, NE); draw(P--M--U--F); X = dir(235); label("$X$", X, dir(235)); draw(X--I, dashed); draw(C--V, dashed); draw(A--I); label("$x^\circ$", A + (0.2,0), dir(90)); label("$y^\circ$", C + (-0.4,0), dir(90)); | [] |
10 | The isosceles triangle $\triangle ABC$, with $AB=AC$, is inscribed in the circle $\omega$. Let $P$ be a variable point on the arc $\stackrel{\frown}{BC}$ that does not contain $A$, and let $I_B$ and $I_C$ denote the incenters of triangles $\triangle ABP$ and $\triangle ACP$, respectively.
Prove that as $P$ varies, the circumcircle of triangle $\triangle PI_BI_C$ passes through a fixed point. | 2016 USAJMO Problem 1 | We claim that $M$ (midpoint of arc $BC$) is the fixed point.
We would like to show that $M$, $P$, $I_B$, $I_C$ are cyclic.
We extend $PI_B$ to intersect $\omega$ again at R.
We extend $PI_C$ to intersect $\omega$ again at S.
We invert around a circle centered at $P$ with radius $1$ (for convenience).
(I will denote X' as the reflection of X for all the points)
The problem then becomes: Prove $I_B'$, $I_C'$, and $M'$ are collinear.
Now we look at triangle $\triangle PR'S'$. We apply Menelaus (the version where all three points lie outside the triangle).
It suffices to show that
\[\dfrac{PI_B'}{I_B'R'} \cdot \dfrac{R'M'}{M'S'} \cdot \dfrac{S'I_C'}{I_C'P} = 1\]
By inversion, we know $PX' = \dfrac{1}{PX}$ for any point $X$ and $X'Y' = \dfrac{XY}{PX \cdot PY}$ for any points $X$ and $Y$.
Plugging this into our Menelaus equation we obtain that it suffices to show
\[\dfrac{\dfrac{1}{PI_B}}{\dfrac{RI_B}{PI_B \cdot PR}} \cdot \dfrac{\dfrac{RM}{PR \cdot PM}}{\dfrac{SM}{PS \cdot PM}} \cdot \dfrac{\dfrac{SI_C}{PI_C \cdot PS}}{\dfrac{1}{PI_C}} = 1\]
We cancel out the like terms and rewrite. It suffices to show
\[\dfrac{RM \cdot SI_C}{SM \cdot RI_B} = 1\]
We know that $AM$ is the diameter of $\omega$ because $\triangle ABC$ is isosceles and $AM$ is the angle bisector. We also know $\angle RMA = \dfrac{\angle ACB}{2} = \dfrac{\angle ABC}{2} = \angle SMA$ so $R$ and $S$ are symmetric with respect to $AM$ so $RM = SM$.
Thus, it suffices to show $\dfrac{SI_C}{RI_B} = 1$.
This is obvious because $RI_B = RA = SA = SI_C$.
Therefore we are done. $\blacksquare$ | // Block 1
size(8cm);
pair A = dir(90);
pair B = dir(-10);
pair C = dir(190);
pair P = dir(-70);
pair U = incenter(A,B,P);
pair V = incenter(A,C,P);
pair M = dir(-90);
draw(circle((0,0),1));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$I_B$", U, NE);
dot("$I_C$", V, NW);
dot("$M$", M, dir(M));
draw(A--B--C--A);
draw(circumcircle(P,U,V));
// Block 2
size(8cm); pair A = dir(90); pair B = dir(-10); pair C = dir(190); pair P = dir(-70); pair U = incenter(A,B,P); pair V = incenter(A,C,P); pair M = dir(-90); draw(circle((0,0),1)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P$", P, dir(P)); dot("$I_B$", U, NE); dot("$I_C$", V, NW); dot("$M$", M, dir(M)); draw(A--B--C--A); draw(circumcircle(P,U,V)); | [] |
10 | The isosceles triangle $\triangle ABC$, with $AB=AC$, is inscribed in the circle $\omega$. Let $P$ be a variable point on the arc $\stackrel{\frown}{BC}$ that does not contain $A$, and let $I_B$ and $I_C$ denote the incenters of triangles $\triangle ABP$ and $\triangle ACP$, respectively.
Prove that as $P$ varies, the circumcircle of triangle $\triangle PI_BI_C$ passes through a fixed point. | 2016 USAJMO Problem 1 | Let $M$ be the midpoint of arc $BC$. Let $D$ be the midpoint of arc $AB$. Let $E$ be the midpoint of arc $AC$.
Then, $P, I_B$, and $D$ are collinear and $P, I_C$, and $E$ are collinear.
We'll prove $MPI_B I_C$ is cyclic. (Intuition: we'll show that $M$ is the Miquel's point of quadrilateral $DE I_C I_B$.
$D$ is the center of the circle $A I_B B$ (the $P-$ excenter of $PAB$ is also on the same circle). Therefore $D I_B = DB$. Similarly $E I_C = EC$. Since $AB=AC$, $DB=EC$. Therefore $D I_B = E I_C$. Obviously $ME = MD$ and $\angle MEI_C = \angle MEP = \angle MDP = \angle MDI_B$. Thus by SAS, $\triangle MEI_C \cong \triangle MDI_B$.
Hence $\angle I_B M I_C = \angle DME = \angle DPE = \angle I_B P I_C$, so $MPI_B I_C$ is cyclic and we are done.
These problems are copyrighted © by the Mathematical Association of America.
See also
2016 USAJMO (Problems • Resources)
First Problem
Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions | // Block 1
size(8cm);
pair A = dir(90);
pair B = dir(-10);
pair C = dir(190);
pair P = dir(-70);
pair U = incenter(A,B,P);
pair V = incenter(A,C,P);
pair M = dir(-90);
pair D = dir(40);
pair E = dir(140);
draw(circle((0,0),1));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$P$", P, dir(P));
dot("$I_B$", U, NE);
dot("$I_C$", V, NW);
dot("$M$", M, dir(M));
draw(A--B--C--A--P--B ^^ P--C ^^ E--D--P--E--M--V ^^ D--M--U--V);
// Block 2
size(8cm); pair A = dir(90); pair B = dir(-10); pair C = dir(190); pair P = dir(-70); pair U = incenter(A,B,P); pair V = incenter(A,C,P); pair M = dir(-90); pair D = dir(40); pair E = dir(140); draw(circle((0,0),1)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$P$", P, dir(P)); dot("$I_B$", U, NE); dot("$I_C$", V, NW); dot("$M$", M, dir(M)); draw(A--B--C--A--P--B ^^ P--C ^^ E--D--P--E--M--V ^^ D--M--U--V); | [] |
11 | Let $\triangle ABC$ be an acute triangle, with $O$ as its circumcenter. Point $H$ is the foot of the perpendicular from $A$ to line $\overleftrightarrow{BC}$, and points $P$ and $Q$ are the feet of the perpendiculars from $H$ to the lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{AC}$, respectively.
Given that \[AH^2=2\cdot AO^2,\]prove that the points $O,P,$ and $Q$ are collinear. | 2016 USAJMO Problem 5 | It is well-known that $AH\cdot 2AO=AB\cdot AC$ (just use similar triangles or standard area formulas). Then by Power of a Point,
\[AP\cdot AB=AH^2=AQ\cdot AC\] Consider the transformation $X\mapsto \Psi(X)$ which dilates $X$ from $A$ by a factor of $\dfrac{AB}{AQ}=\dfrac{AC}{AP}$ and reflects about the $A$-angle bisector. Then $\Psi(O)$ clearly lies on $AH$, and its distance from $A$ is \[AO\cdot\frac{AB}{AQ}=AO\cdot\frac{AB}{\frac{AH^2}{AC}}=AO\cdot\frac{AB\cdot AC}{AH^2}=\frac{AO\cdot AH\cdot 2AO}{AH^2}=\frac{2AO^2}{AH}=AH\] so $\Psi(O)=H$, hence we conclude that $O,P,Q$ are collinear, as desired. | // Block 1
size(8cm);
pair O=(0,0);
pair A=dir(110);
pair B=dir(-29);
pair C=dir(209);
pair H=foot(A,B,C);
pair P=foot(H,A,B);
pair Q=foot(H,A,C);
draw(A--B--C--A--H--P);
draw(circle(O,1));
draw(Q--H);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$H$", H, S);
dot("$P$", P, NE);
dot("$Q$", Q, NW);
dot("$O$", O, S);
// Block 2
size(8cm); pair O=(0,0); pair A=dir(110); pair B=dir(-29); pair C=dir(209); pair H=foot(A,B,C); pair P=foot(H,A,B); pair Q=foot(H,A,C); draw(A--B--C--A--H--P); draw(circle(O,1)); draw(Q--H); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$H$", H, S); dot("$P$", P, NE); dot("$Q$", Q, NW); dot("$O$", O, S); | [] |
12 | ($*$) Let $ABC$ be an equilateral triangle and let $P$ be a point on its circumcircle. Let lines $PA$ and $BC$ intersect at $D$; let lines $PB$ and $CA$ intersect at $E$; and let lines $PC$ and $AB$ intersect at $F$. Prove that the area of triangle $DEF$ is twice that of triangle $ABC$. | 2017 USAJMO Problem 3 | We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\sqrt{3})$. Using the facts $\triangle{CBP} \sim \triangle{CFB}$ and $\triangle{BCP} \sim \triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \frac{1}{2x}$, and $E = (\frac{1}{x}+1,-\frac{\sqrt{3}}{x})$.
The slope of $FE$ is
\[k = \frac{-\frac{\sqrt{3}}{x} + x\sqrt{3}}{2+\frac{1}{x}+x}\]
It is well-known that $\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\frac{1}{k}$. Since $O=(0,\frac{1}{\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \frac{k}{\sqrt{3}}$, i.e., $D = (\frac{k}{\sqrt{3}},0)$. Using shoelace,
\[2[\triangle{FDE}] = \frac{k}{\sqrt{3}}(-x\sqrt{3})+(-x-1)(-\frac{\sqrt{3}}{x})- (-x\sqrt{3})(\frac{1}{x}+1) - (-\frac{\sqrt{3}}{x})\frac{k}{\sqrt{3}}\]
\[= 2\sqrt{3} + \sqrt{3}(\frac{1}{x}+x) + k(\frac{1}{x} - x)\]
\[= 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+(\frac{1}{x}+x)^2-(\frac{1}{x}-x)^2} {2+\frac{1}{x}+x})\]
\[= 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+4}{2+\frac{1}{x}+x})\]
\[= 4\sqrt{3}\]
So $[\triangle{FDE}] = 2\sqrt{3} = 2[\triangle{ABC}]$. Q.E.D
By Mathdummy. | // Block 1
import cse5;
import graph;
import olympiad;
size(3inch);
pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), O = (0, sqrt(3));
pair P = (-1, -sqrt(11)+sqrt(3));
path circle = Circle(O, 2sqrt(3));
pair D = extension(A,P,B,C);
pair E1 = extension(A,C,B,P);
pair F=extension(A,B,C,P);
draw(circle, black);
draw(A--B--C--cycle);
draw(B--E1--C);
draw(C--F--B);
draw(A--P);
draw(D--E1--F--cycle, dashed);
pair G = extension(O,D,F,E1);
draw(O--G,dashed);
label("A", A, N);
label("B", B, W);
label("C", C, E);
label("P", P, S);
label("D", D, NW);
label("E", E1, SE);
label("F", F, SW);
dot("O", O, SE);
// Block 2
import cse5; import graph; import olympiad; size(3inch); pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), O = (0, sqrt(3)); pair P = (-1, -sqrt(11)+sqrt(3)); path circle = Circle(O, 2sqrt(3)); pair D = extension(A,P,B,C); pair E1 = extension(A,C,B,P); pair F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label("A", A, N); label("B", B, W); label("C", C, E); label("P", P, S); label("D", D, NW); label("E", E1, SE); label("F", F, SW); dot("O", O, SE); | [] |
13 | Let $O$ and $H$ be the circumcenter and the orthocenter of an acute triangle $ABC$. Points $M$ and $D$ lie on side $BC$ such that $BM = CM$ and $\angle BAD = \angle CAD$. Ray $MO$ intersects the circumcircle of triangle $BHC$ in point $N$. Prove that $\angle ADO = \angle HAN$. | 2017 USAJMO Problem 5 | (original diagram by integralarefun)
It's well known that the reflection of $H$ across $\overline{BC}$, $H'$, lies on $(ABC)$. Then $(BHC)$ is just the reflection of $(BH'C)$ across $\overline{BC}$, which is equivalent to the reflection of $(ABC)$ across $\overline{BC}$. Reflect points $A$ and $N$ across $\overline{BC}$ to points $A'$ and $N'$, respectively. Then $N'$ is the midpoint of minor arc $\overarc{BC}$, so $A, D, N'$ are collinear in that order. It suffices to show that $\angle AA'N'=\angle ADO$.
Claim: $\triangle AA'N' \sim \triangle ADO$. The proof easily follows.
Proof: Note that $\angle BAA'=\angle CAO=90^{\circ}-\angle ABC$. Then we have $\angle A'AN'=\angle BAD-\angle BAA'=\angle CAD-\angle CAO=\angle DAO$. So, it suffices to show that \[\frac{AA'}{AN'}=\frac{AD}{AO}\rightarrow AA'\cdot AO=AN'\cdot AD.\] Notice that $\triangle ABA' \sim \triangle AOC$, so that \[\frac{AB}{AA'}=\frac{AO}{AC}\rightarrow AA'\cdot AO=AB\cdot AC.\] Therefore, it suffices to show that \[AB\cdot AC=AN'\cdot AD\rightarrow \frac{AB}{AN'}=\frac{AD}{AC}.\] But it is easy to show that $\triangle BAN'\sim \triangle DAC$, implying the result. $\blacksquare$ | // Block 1
import olympiad;
unitsize(100);
pair pA = dir(120);
pair pB = dir(225);
pair pC = dir(315);
pair pO = origin;
pair pH = orthocenter(pA, pB, pC);
pair pM = midpoint(pB--pC);
pair dD = bisectorpoint(pB, pA, pC);
pair pD = extension(pA, dD, pB, pC);
pair pN = intersectionpoints(pM--(3*pO-2*pM), circumcircle(pB, pH, pC))[0];
pair dHprime = foot(pH, pB, pC);
pair pHprime = 2*dHprime-pH;
pair dAprime = foot(pA, pB, pC);
pair pAprime = 2*dAprime-pA;
pair dNprime = foot(pN, pB, pC);
pair pNprime = 2*dNprime-pN;
draw(pA--pB--pC--cycle);
draw(unitcircle, blue);
draw(pH--pHprime, magenta);
draw(pA--pD, red);
draw(circumcircle(pB, pH, pC), blue);
draw(pA--pH, blue);
draw(pA--pN, magenta);
draw(pA--pO, blue);
draw(pD--pNprime, magenta);
dot("\(A\)", pA, NW);
dot("\(B\)", pB, W);
dot("\(C\)", pC, E);
dot("\(O\)", pO, SW, blue);
dot("\(H\)", pH, SW, blue);
dot("\(N\)", pN, NE, magenta);
dot("\(M\)", pM, S, red);
dot("\(D\)", pD, S, red);
dot("\(H'\)", pHprime, SW, magenta);
dot("\(A'\)", pAprime, SW);
dot("\(N'\)", pNprime, S, magenta);
// Block 2
import olympiad; unitsize(100); pair pA = dir(120); pair pB = dir(225); pair pC = dir(315); pair pO = origin; pair pH = orthocenter(pA, pB, pC); pair pM = midpoint(pB--pC); pair dD = bisectorpoint(pB, pA, pC); pair pD = extension(pA, dD, pB, pC); pair pN = intersectionpoints(pM--(3*pO-2*pM), circumcircle(pB, pH, pC))[0]; pair dHprime = foot(pH, pB, pC); pair pHprime = 2*dHprime-pH; pair dAprime = foot(pA, pB, pC); pair pAprime = 2*dAprime-pA; pair dNprime = foot(pN, pB, pC); pair pNprime = 2*dNprime-pN; draw(pA--pB--pC--cycle); draw(unitcircle, blue); draw(pH--pHprime, magenta); draw(pA--pD, red); draw(circumcircle(pB, pH, pC), blue); draw(pA--pH, blue); draw(pA--pN, magenta); draw(pA--pO, blue); draw(pD--pNprime, magenta); dot("\(A\)", pA, NW); dot("\(B\)", pB, W); dot("\(C\)", pC, E); dot("\(O\)", pO, SW, blue); dot("\(H\)", pH, SW, blue); dot("\(N\)", pN, NE, magenta); dot("\(M\)", pM, S, red); dot("\(D\)", pD, S, red); dot("\(H'\)", pHprime, SW, magenta); dot("\(A'\)", pAprime, SW); dot("\(N'\)", pNprime, S, magenta); | [] |
13 | Let $O$ and $H$ be the circumcenter and the orthocenter of an acute triangle $ABC$. Points $M$ and $D$ lie on side $BC$ such that $BM = CM$ and $\angle BAD = \angle CAD$. Ray $MO$ intersects the circumcircle of triangle $BHC$ in point $N$. Prove that $\angle ADO = \angle HAN$. | 2017 USAJMO Problem 5 | Suppose ray $OM$ intersects the circumcircle of $BHC$ at $N'$, and let the foot of the A-altitude of $ABC$ be $E$. Note that $\angle BHE=90-\angle HBE=90-90+\angle C=\angle C$. Likewise, $\angle CHE=\angle B$. So, $\angle BHC=\angle BHE+\angle CHE=\angle B+\angle C$.
$BHCN'$ is cyclic, so $\angle BN'C=180-\angle BHC=180-\angle B-\angle C=\angle A$. Also, $\angle BAC=\angle A$. These two angles are on different circles and have the same measure, but they point to the same line $BC$! Hence, the two circles must be congruent. (This is also a well-known result)
We know, since $M$ is the midpoint of $BC$, that $OM$ is perpendicular to $BC$. $AH$ is also perpendicular to $BC$, so the two lines are parallel. $AN$ is a transversal, so $\angle HAN=\angle ANO$. We wish to prove that $\angle ANO=\angle ADO$, which is equivalent to $AOND$ being cyclic.
Now, assume that ray $OM$ intersects the circumcircle of $ABC$ at a point $P$. Point $P$ must be the midpoint of $\stackrel{\frown}{BC}$. Also, since $AD$ is an angle bisector, it must also hit the circle at the point $P$. The two circles are congruent, which implies $MN=MP\implies ND=DP\implies$ NDP is isosceles. Angle ADN is an exterior angle, so $\angle ADN=\angle DNP+\angle DPO=2\angle DPO$.
Assume WLOG that $\angle B>\angle C$. So, $\angle DPO=\angle APO=\frac{\angle B+\angle C}{2}-\angle C=\frac{\angle B-\angle C}{2}$.
In addition, $\angle AON=\angle AOP=\angle AOB+\angle BOP=2\angle C+\angle A$. Combining these two equations, $\angle AON+\angle ADN=\angle B-\angle C+2\angle C+\angle A=\angle A+\angle B+\angle C=180$.
Opposite angles sum to $180$, so quadrilateral $AOND$ is cyclic, and the condition is proved.
-william122 | // Block 1
size(9cm);
pair A = dir(130);
pair B = dir(220);
pair C = dir(320);
draw(unitcircle, lightblue);
pair P = dir(-90);
pair Q = dir(90);
pair D = extension(A, P, B, C);
pair O = origin;
pair M = extension(B, C, O, P);
pair N = 2*M-P;
draw(A--B--C--cycle, lightblue);
draw(A--P--Q, lightblue);
draw(A--N--D--O--A, lightblue);
draw(A--D--N--O--cycle, red);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$D$", D, dir(225));
dot("$O$", O, dir(315));
dot("$M$", M, dir(315));
dot("$N$", N, dir(315));
// Block 2
size(9cm); pair A = dir(130); pair B = dir(220); pair C = dir(320); draw(unitcircle, lightblue); pair P = dir(-90); pair Q = dir(90); pair D = extension(A, P, B, C); pair O = origin; pair M = extension(B, C, O, P); pair N = 2*M-P; draw(A--B--C--cycle, lightblue); draw(A--P--Q, lightblue); draw(A--N--D--O--A, lightblue); draw(A--D--N--O--cycle, red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$D$", D, dir(225)); dot("$O$", O, dir(315)); dot("$M$", M, dir(315)); dot("$N$", N, dir(315)); | [] |
14 | $(*)$ Let $ABC$ be a triangle with $\angle ABC$ obtuse. The $A$-excircle is a circle in the exterior of $\triangle ABC$ that is tangent to side $\overline{BC}$ of the triangle and tangent to the extensions of the other two sides. Let $E$, $F$ be the feet of the altitudes from $B$ and $C$ to lines $AC$ and $AB$, respectively. Can line $EF$ be tangent to the $A$-excircle? | 2019 USAJMO Problem 4 | We claim that the answer is no. We proceed with contradiction. Suppose that $EF$ is indeed tangent to the a-excenter. Define the point of tangency to be $X$. Let $G$ to be the intersection of the a-excircle with the extension of $AB$ and $H$ to be the intersection of the a-excircle with the extension of $AC$. Define $I_a$ to be the a-excenter. It is a well-known fact that $I_a$ lies on the angle bisector of $\angle BAC$.
For convenience, let $AB = c, AC = b, BC = a$, $\angle CAB = A, \angle CBA = B, \angle ACB = C$.
Notice that by Power of a Point:
\[FX = FG.\]
\[XE = EH.\]
Therefore, adding, we see that:
\[FE = FX + XE = FG + EH.\]
It would be rather nice if we could re-write $FG$. Indeed, we can:
\[FG = AG - FG = \frac{r_a}{\tan \left( \frac{A}{2} \right)} - b \cos (A)\]
and similarly for $FH$:
\[FH = AH - AF = \frac{r_a}{\tan \left( \frac{A}{2} \right)} - c \cos (A)\]
We now seek to re-write $FE$. Using the law of cosines:
\[FE^2 = AF^2 + AE^ 2- 2AF \cdot AE \cos (A) = \left(c\cos (A) \right)^2 + \left(b\cos (A) \right)^2 - 2 (bc \cos ^2 (A)) \cos (A) = \cos ^2 (A) (c^2 + b^2 - 2bc \cos (A)) = \cos ^2 (A) \cdot a^2\]
Therefore,
\[FE = a \cos (A)\]
Putting this all together, we see that:
\[a \cos (A) = \frac{r_a}{\tan \left(\frac{a}{2} \right)} - b \cos (A) + \frac{r_a}{\tan \left(\frac{a}{2} \right)} - c \cos (A) \Longleftrightarrow\]
\[(a+b+c) \cos (A) = \frac{2r_a}{\tan \left(\frac{a}{2} \right)}\]
Now, we seek to write $\tan \left(\frac{a}{2} \right)$ in terms of the side lengths of the triangle. Notice that:
\[2\cos ^2 \left(\frac{A}{2} \right)-1 = \cos (A) \Longleftrightarrow\]
\[2\cos ^2 \left(\frac{A}{2} \right) = 1 + \frac{b^2 + c^2-a^2}{2bc} = \frac{b^2 + 2bc + c^2 - a^2}{2bc} = \frac{(b+c)^2-a^2}{2bc} = \frac{(a+b+c)(b+c-a)}{2bc} = \frac{(2s)(2s-2a)}{2bc}\]
where $s$ is the semi-perimeter. We get that:
\[\cos \left(\frac{A}{2} \right) = \sqrt{\frac{s(s-a)}{bc}}\]
Using the fact that $\cos^2 \left(\frac{A}{2} \right) + \sin^2 \left(\frac{A}{2} \right) = 1$, we have that:
\[\sin^2 \left(\frac{A}{2} \right) = 1-\frac{b^2 + 2bc + c^2 - a^2}{4bc} = \frac{(b-c)^2 - a^2}{4bc} = \frac{(2s-2c)(2s-2b)}{4bc}\]
Therefore,
\[\tan \left( \frac{A}{2} \right) = \sqrt{ \frac{(s-b)(s-c)}{s(s-a)} }\]
Returning to our original problem:
\[\frac{2r_a}{\sqrt{ \frac{(s-b)(s-c)}{s(s-a)} }} = (a+b+c) \cos (A)\]
It is a well-known fact that $r_a = \sqrt{\frac{s(s-b)(s-c)}{s-a}}$, so:
\[\frac{2\sqrt{\frac{s(s-b)(s-c)}{s-a}}}{\sqrt{ \frac{(s-b)(s-c)}{s(s-a)} }} = 2s \cos (A)\] which implies that:
\[\cos (A) =1 \Longleftrightarrow A = 0\]
which is a contradiction. Hence, our original assumption that $EF$ is tangent to the a-excircle is incorrect. $\blacksquare$
~AopsUser101 | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = 0.92, xmax = 28.12, ymin = -14.66, ymax = 5.16; /* image dimensions */
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902);
/* draw figures */
draw((3.52,1.38)--(5.8,-4.7), linewidth(1) + rvwvcq);
draw((5.8,-4.7)--(10.3,-4.9), linewidth(1) + wvvxds);
draw((10.3,-4.9)--(3.52,1.38), linewidth(1) + rvwvcq);
draw(circle((9.325581455949688,-7.26800412402583), 2.399418339060914), linewidth(1) + sexdts);
draw((5.8,-4.7)--(7.0789360558927,-8.1104961490472), linewidth(1) + rvwvcq);
draw((10.3,-4.9)--(10.956076160142976,-5.507692962492314), linewidth(1) + rvwvcq);
draw((8.503617697888226,-3.2360942688404215)--(6.711506849315068,-7.130684931506849), linewidth(1) + wvvxds);
/* dots and labels */
dot((3.52,1.38),dotstyle);
label("$A$", (3.6,1.58), NE * labelscalefactor);
dot((5.8,-4.7),dotstyle);
label("$B$", (5.88,-4.5), NE * labelscalefactor);
dot((10.3,-4.9),dotstyle);
label("$C$", (10.38,-4.7), NE * labelscalefactor);
dot((9.325581455949688,-7.26800412402583),dotstyle);
label("$I_a$", (9.4,-7.06), NE * labelscalefactor);
dot((7.0789360558927,-8.1104961490472),linewidth(4pt) + dotstyle);
label("$G$", (7.16,-7.96), NE * labelscalefactor);
dot((10.956076160142976,-5.507692962492314),linewidth(4pt) + dotstyle);
label("$H$", (11.04,-5.34), NE * labelscalefactor);
dot((8.503617697888226,-3.2360942688404215),dotstyle);
label("E", (8.58,-3.04), NE * labelscalefactor);
dot((6.711506849315068,-7.130684931506849),dotstyle);
label("$F$", (6.8,-6.94), NE * labelscalefactor);
dot((7.139200885553699,-6.201226130819783),dotstyle);
label("$X$", (7.22,-6), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
// Block 2
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 0.92, xmax = 28.12, ymin = -14.66, ymax = 5.16; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); /* draw figures */ draw((3.52,1.38)--(5.8,-4.7), linewidth(1) + rvwvcq); draw((5.8,-4.7)--(10.3,-4.9), linewidth(1) + wvvxds); draw((10.3,-4.9)--(3.52,1.38), linewidth(1) + rvwvcq); draw(circle((9.325581455949688,-7.26800412402583), 2.399418339060914), linewidth(1) + sexdts); draw((5.8,-4.7)--(7.0789360558927,-8.1104961490472), linewidth(1) + rvwvcq); draw((10.3,-4.9)--(10.956076160142976,-5.507692962492314), linewidth(1) + rvwvcq); draw((8.503617697888226,-3.2360942688404215)--(6.711506849315068,-7.130684931506849), linewidth(1) + wvvxds); /* dots and labels */ dot((3.52,1.38),dotstyle); label("$A$", (3.6,1.58), NE * labelscalefactor); dot((5.8,-4.7),dotstyle); label("$B$", (5.88,-4.5), NE * labelscalefactor); dot((10.3,-4.9),dotstyle); label("$C$", (10.38,-4.7), NE * labelscalefactor); dot((9.325581455949688,-7.26800412402583),dotstyle); label("$I_a$", (9.4,-7.06), NE * labelscalefactor); dot((7.0789360558927,-8.1104961490472),linewidth(4pt) + dotstyle); label("$G$", (7.16,-7.96), NE * labelscalefactor); dot((10.956076160142976,-5.507692962492314),linewidth(4pt) + dotstyle); label("$H$", (11.04,-5.34), NE * labelscalefactor); dot((8.503617697888226,-3.2360942688404215),dotstyle); label("E", (8.58,-3.04), NE * labelscalefactor); dot((6.711506849315068,-7.130684931506849),dotstyle); label("$F$", (6.8,-6.94), NE * labelscalefactor); dot((7.139200885553699,-6.201226130819783),dotstyle); label("$X$", (7.22,-6), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ | [] |
15 | Consider an $n$-by-$n$ board of unit squares for some odd positive integer $n$. We say that a collection $C$ of identical dominoes is a maximal grid-aligned configuration on the board if $C$ consists of $(n^2-1)/2$ dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: $C$ then covers all but one square on the board. We are allowed to slide (but not rotate) a domino on the board to cover the uncovered square, resulting in a new maximal grid-aligned configuration with another square uncovered. Let $k(C)$ be the number of distinct maximal grid-aligned configurations obtainable from $C$ by repeatedly sliding dominoes. Find the maximum value of $k(C)$ as a function of $n$. | 2023 USAJMO Problem 3 | To start off, we put the initial non-covered square in a corner (marked by the shaded square). Let's consider what happens when our first domino slides over the empty square. We will call such a move where we slide a domino over the uncovered square a "step":
When the vertically-oriented domino above the shaded square moved down to cover the shaded square, it had uncovered a new square two squares above the original one. Next, we consider the second another direction the uncovered square can move to:
In this case, the horizontally-oriented domino besides the shaded square moved to cover the previous shaded square. The new uncovered square, as a result, would be either two squares to the left, or to the right of the previous shaded square.
So, to summarize, during each "move", the uncovered square can move either two steps up, down, left, or right. We can make a path for which the shaded square takes:
We put the original shaded square was in the corner, so, the possible number of squares that cound be uncovered is $\left(\dfrac{n+1}{2}\right)^2 = \frac{n^2+2n+1}{4}$, all of which can be connected by a single path. Let those $\frac{n^2+2n+1}{4}$ belong to a set $S$ of squares.
If the initial uncovered square was not an element of $S$, then it could either be $1$ or $\sqrt{2}$ away from a square in $S$. If the initial square was a distance of $1$ from a element of $S$, then there would be $\left(\dfrac{n+1}{2}\right)\left(\dfrac{n-1}{2}\right)$ possible uncovered squares:
If the initial square was a distance of $\sqrt{2}$ away from a square in $S$, then there would be $\left(\dfrac{n-1}{2}\right)^2$ possible squares that could be uncovered:
So, that means the maximal number of squares that can be uncovered is $\frac{n^2+2n+1}{4}$, thus making a maxmial of $k(C)=\boxed{\frac{n^2+2n+1}{4}}$ achievable configurations from $C$ ($C$'s uncovered square must belong in $S$).
~ sml1809
Remark: I believe that this solution does not address the problem when two configurations lead to the same placement of the uncovered square. To address this problem, we can prove that the paths don't make a cycle,
~bebebe | // Block 1
size(5cm);
draw((0,0)--(0,3.2));
draw((1,0)--(1,3.2));
draw((0,0)--(1.2,0));
draw((0,1)--(1.2,1));
draw((0,2)--(1.2,2));
draw((0,3)--(1.2,3));
fill(origin--(1,0)--(1,1)--(0,1)--cycle, grey);
draw((0.1,1.1)--(0.1,2.9)--(0.9,2.9)--(0.9,1.1)--cycle,dashed);
draw((0.1,0.1)--(0.1,1.9)--(0.9,1.9)--(0.9,0.1)--cycle);
// Block 2
size(5cm);
draw((0,0)--(0,3.2));
draw((1,0)--(1,3.2));
draw((0,0)--(1.2,0));
draw((0,1)--(1.2,1));
draw((0,2)--(1.2,2));
draw((0,3)--(1.2,3));
fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, grey);
draw((0.1,2.9)--(0.1,2.1)--(1.9,2.1)--(1.9,2.9)--cycle);
draw((1.1,2.9)--(1.1,2.1)--(2.9,2.1)--(2.9,2.9)--cycle,dashed);
draw((0.1,0.1)--(0.1,1.9)--(0.9,1.9)--(0.9,0.1)--cycle);
// Block 3
size(5cm);
draw((0,0)--(0,5));
draw((1,0)--(1,5));
draw((2,0)--(2,5));
draw((3,0)--(3,5));
draw((4,0)--(4,5));
draw((5,0)--(5,5));
draw((0,0)--(5,0));
draw((0,1)--(5,1));
draw((0,2)--(5,2));
draw((0,3)--(5,3));
draw((0,4)--(5,4));
draw((0,5)--(5,5));
draw((0.5,0.5)--(0.5,2.5)--(2.5,2.5)--(2.5,0.5)--(4.5,0.5)--(4.5,2.5)--(4.5,4.5)--(2.5,4.5)--(0.5,4.5));
dot((0.5,0.5));
dot((0.5,2.5));
dot((2.5,2.5));
dot((2.5,0.5));
dot((4.5,0.5));
dot((4.5,2.5));
dot((4.5,4.5));
dot((2.5,4.5));
dot((0.5,4.5));
// Block 4
draw((0,0)--(0,5));
draw((1,0)--(1,5));
draw((2,0)--(2,5));
draw((3,0)--(3,5));
draw((4,0)--(4,5));
draw((5,0)--(5,5));
draw((0,0)--(5,0));
draw((0,1)--(5,1));
draw((0,2)--(5,2));
draw((0,3)--(5,3));
draw((0,4)--(5,4));
draw((0,5)--(5,5));
dot((0.5,1.5));
dot((0.5,3.5));
dot((2.5,1.5));
dot((2.5,3.5));
dot((4.5,1.5));
dot((4.5,3.5));
// Block 5
draw((0,0)--(0,5));
draw((1,0)--(1,5));
draw((2,0)--(2,5));
draw((3,0)--(3,5));
draw((4,0)--(4,5));
draw((5,0)--(5,5));
draw((0,0)--(5,0));
draw((0,1)--(5,1));
draw((0,2)--(5,2));
draw((0,3)--(5,3));
draw((0,4)--(5,4));
draw((0,5)--(5,5));
dot((1.5,1.5));
dot((1.5,3.5));
dot((3.5,1.5));
dot((3.5,3.5));
// Block 6
size(5cm); draw((0,0)--(0,3.2)); draw((1,0)--(1,3.2)); draw((0,0)--(1.2,0)); draw((0,1)--(1.2,1)); draw((0,2)--(1.2,2)); draw((0,3)--(1.2,3)); fill(origin--(1,0)--(1,1)--(0,1)--cycle, grey); draw((0.1,1.1)--(0.1,2.9)--(0.9,2.9)--(0.9,1.1)--cycle,dashed); draw((0.1,0.1)--(0.1,1.9)--(0.9,1.9)--(0.9,0.1)--cycle);
// Block 7
size(5cm); draw((0,0)--(0,3.2)); draw((1,0)--(1,3.2)); draw((0,0)--(1.2,0)); draw((0,1)--(1.2,1)); draw((0,2)--(1.2,2)); draw((0,3)--(1.2,3)); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, grey); draw((0.1,2.9)--(0.1,2.1)--(1.9,2.1)--(1.9,2.9)--cycle); draw((1.1,2.9)--(1.1,2.1)--(2.9,2.1)--(2.9,2.9)--cycle,dashed); draw((0.1,0.1)--(0.1,1.9)--(0.9,1.9)--(0.9,0.1)--cycle);
// Block 8
size(5cm); draw((0,0)--(0,5)); draw((1,0)--(1,5)); draw((2,0)--(2,5)); draw((3,0)--(3,5)); draw((4,0)--(4,5)); draw((5,0)--(5,5)); draw((0,0)--(5,0)); draw((0,1)--(5,1)); draw((0,2)--(5,2)); draw((0,3)--(5,3)); draw((0,4)--(5,4)); draw((0,5)--(5,5)); draw((0.5,0.5)--(0.5,2.5)--(2.5,2.5)--(2.5,0.5)--(4.5,0.5)--(4.5,2.5)--(4.5,4.5)--(2.5,4.5)--(0.5,4.5)); dot((0.5,0.5)); dot((0.5,2.5)); dot((2.5,2.5)); dot((2.5,0.5)); dot((4.5,0.5)); dot((4.5,2.5)); dot((4.5,4.5)); dot((2.5,4.5)); dot((0.5,4.5));
// Block 9
draw((0,0)--(0,5)); draw((1,0)--(1,5)); draw((2,0)--(2,5)); draw((3,0)--(3,5)); draw((4,0)--(4,5)); draw((5,0)--(5,5)); draw((0,0)--(5,0)); draw((0,1)--(5,1)); draw((0,2)--(5,2)); draw((0,3)--(5,3)); draw((0,4)--(5,4)); draw((0,5)--(5,5)); dot((0.5,1.5)); dot((0.5,3.5)); dot((2.5,1.5)); dot((2.5,3.5)); dot((4.5,1.5)); dot((4.5,3.5));
// Block 10
draw((0,0)--(0,5)); draw((1,0)--(1,5)); draw((2,0)--(2,5)); draw((3,0)--(3,5)); draw((4,0)--(4,5)); draw((5,0)--(5,5)); draw((0,0)--(5,0)); draw((0,1)--(5,1)); draw((0,2)--(5,2)); draw((0,3)--(5,3)); draw((0,4)--(5,4)); draw((0,5)--(5,5)); dot((1.5,1.5)); dot((1.5,3.5)); dot((3.5,1.5)); dot((3.5,3.5)); | [] |
16 | Let $ABCD$ be a cyclic quadrilateral with $AB = 7$ and $CD = 8$. Points $P$ and $Q$ are selected on segment $AB$ such that $AP = BQ = 3$. Points $R$ and $S$ are selected on segment $CD$ such that $CR = DS = 2$. Prove that $PQRS$ is a cyclic quadrilateral. | 2024 USAJMO Problem 1 | First, let $E$ and $F$ be the midpoints of $AB$ and $CD$, respectively. It is clear that $AE=BE=3.5$, $PE=QE=0.5$, $DF=CF=4$, and $SF=RF=2$. Also, let $O$ be the circumcenter of $ABCD$.
By properties of cyclic quadrilaterals, we know that the circumcenter of a cyclic quadrilateral is the intersection of its sides' perpendicular bisectors. This implies that $OE\perp AB$ and $OF\perp CD$. Since $E$ and $F$ are also bisectors of $PQ$ and $RS$, respectively, if $PQRS$ is indeed a cyclic quadrilateral, then its circumcenter is also at $O$. Thus, it suffices to show that $OP=OQ=OR=OS$.
Notice that $PE=QE$, $EO=EO$, and $\angle QEO=\angle PEO=90^\circ$. By SAS congruency, $\Delta QOE\cong\Delta POE\implies QO=PO$. Similarly, we find that $\Delta SOF\cong\Delta ROF$ and $OS=OR$. We now need only to show that these two pairs are equal to each other.
Draw the segments connecting $O$ to $B$, $Q$, $C$, and $R$.
Also, let $r$ be the circumradius of $ABCD$. This means that $AO=BO=CO=DO=r$. Recall that $\angle BEO=90^\circ$ and $\angle CFO=90^\circ$. Notice the several right triangles in our figure.
Let us apply Pythagorean Theorem on $\Delta BEO$. We can see that $EO^2+EB^2=BO^2\implies EO^2+3.5^2=r^2\implies EO=\sqrt{r^2-12.25}.$
Let us again apply Pythagorean Theorem on $\Delta QEO$. We can see that $QE^2+EO^2=QO^2\implies0.5^2+r^2-12.25=QO^2\implies QO=\sqrt{r^2-12}.$
Let us apply Pythagorean Theorem on $\Delta CFO$. We get $CF^2+OF^2=OC^2\implies4^2+OF^2=r^2\implies OF=\sqrt{r^2-16}$.
We finally apply Pythagorean Theorem on $\Delta RFO$. This becomes $OF^2+FR^2=OR^2\implies r^2-16+2^2=OR^2\implies OR=\sqrt{r^2-12}$.
This is the same expression as we got for $QO$. Thus, $OQ=OR$, and recalling that $OQ=OP$ and $OR=OS$, we have shown that $OP=OQ=OR=OS$, and we are done.
~Technodoggo | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/wiki/User:Azjps/geogebra */
import graph; size(12cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -12.19, xmax = 24.94, ymin = -15.45, ymax = 6.11; /* image dimensions */
pen wrwrwr = rgb(0.38,0.38,0.38);
/* draw figures */
draw(circle((2.92,-3.28), 5.90), linewidth(2) + wrwrwr);
draw((-2.52,-1.01)--(3.46,2.59), linewidth(2) + wrwrwr);
draw((7.59,-6.88)--(-0.29,-8.22), linewidth(2) + wrwrwr);
draw((3.46,2.59)--(7.59,-6.88), linewidth(2) + wrwrwr);
draw((-0.29,-8.22)--(-2.52,-1.01), linewidth(2) + wrwrwr);
draw((0.03,0.52)--(1.67,-7.89), linewidth(2) + wrwrwr);
draw((5.61,-7.22)--(0.89,1.04), linewidth(2) + wrwrwr);
/* dots and labels */
dot((2.92,-3.28),dotstyle);
label("$O$", (2.43,-3.56), NE * labelscalefactor);
dot((-2.52,-1.01),dotstyle);
label("$A$", (-2.91,-0.91), NE * labelscalefactor);
dot((3.46,2.59),linewidth(4pt) + dotstyle);
label("$B$", (3.49,2.78), NE * labelscalefactor);
dot((7.59,-6.88),dotstyle);
label("$C$", (7.82,-7.24), NE * labelscalefactor);
dot((-0.29,-8.22),linewidth(4pt) + dotstyle);
label("$D$", (-0.53,-8.62), NE * labelscalefactor);
dot((0.03,0.52),linewidth(4pt) + dotstyle);
label("$P$", (-0.13,0.67), NE * labelscalefactor);
dot((0.89,1.04),linewidth(4pt) + dotstyle);
label("$Q$", (0.62,1.16), NE * labelscalefactor);
dot((5.61,-7.22),linewidth(4pt) + dotstyle);
label("$R$", (5.70,-7.05), NE * labelscalefactor);
dot((1.67,-7.89),linewidth(4pt) + dotstyle);
label("$S$", (1.75,-7.73), NE * labelscalefactor);
dot((0.46,0.78),linewidth(4pt) + dotstyle);
label("$E$", (0.26,0.93), NE * labelscalefactor);
dot((3.64,-7.55),linewidth(4pt) + dotstyle);
label("$F$", (3.73,-7.39), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
// Block 2
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/wiki/User:Azjps/geogebra */
import graph; size(12cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -12.19, xmax = 24.94, ymin = -15.45, ymax = 6.11; /* image dimensions */
pen wrwrwr = rgb(0.38,0.38,0.38);
/* draw figures */
draw(circle((2.92,-3.28), 5.90), linewidth(2) + wrwrwr);
draw((-2.52,-1.01)--(3.46,2.59), linewidth(2) + wrwrwr);
draw((7.59,-6.88)--(-0.29,-8.22), linewidth(2) + wrwrwr);
draw((3.46,2.59)--(7.59,-6.88), linewidth(2) + wrwrwr);
draw((-0.29,-8.22)--(-2.52,-1.01), linewidth(2) + wrwrwr);
draw((0.03,0.52)--(1.67,-7.89), linewidth(2) + wrwrwr);
draw((5.61,-7.22)--(0.89,1.04), linewidth(2) + wrwrwr);
draw((0.46,0.78)--(2.92,-3.28), linewidth(2) + wrwrwr);
draw((2.92,-3.28)--(3.64,-7.55), linewidth(2) + wrwrwr);
draw((2.92,-3.28)--(7.59,-6.88), linewidth(2) + wrwrwr);
draw((5.61,-7.22)--(2.92,-3.28), linewidth(2) + wrwrwr);
draw((2.92,-3.28)--(3.46,2.59), linewidth(2) + wrwrwr);
draw((2.92,-3.28)--(0.89,1.04), linewidth(2) + wrwrwr);
/* dots and labels */
dot((2.92,-3.28),dotstyle);
label("$O$", (2.43,-3.56), NE * labelscalefactor);
dot((-2.52,-1.01),dotstyle);
label("$A$", (-2.91,-0.91), NE * labelscalefactor);
dot((3.46,2.59),linewidth(1pt) + dotstyle);
label("$B$", (3.49,2.78), NE * labelscalefactor);
dot((7.59,-6.88),dotstyle);
label("$C$", (7.82,-7.24), NE * labelscalefactor);
dot((-0.29,-8.22),linewidth(1pt) + dotstyle);
label("$D$", (-0.53,-8.62), NE * labelscalefactor);
dot((0.03,0.52),linewidth(1pt) + dotstyle);
label("$P$", (-0.13,0.67), NE * labelscalefactor);
dot((0.89,1.04),linewidth(1pt) + dotstyle);
label("$Q$", (0.62,1.16), NE * labelscalefactor);
dot((5.61,-7.22),linewidth(1pt) + dotstyle);
label("$R$", (5.70,-7.05), NE * labelscalefactor);
dot((1.67,-7.89),linewidth(1pt) + dotstyle);
label("$S$", (1.75,-7.73), NE * labelscalefactor);
dot((0.46,0.78),linewidth(1pt) + dotstyle);
label("$E$", (0.26,0.93), NE * labelscalefactor);
dot((3.64,-7.55),linewidth(1pt) + dotstyle);
label("$F$", (3.73,-7.39), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
// Block 3
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/wiki/User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.19, xmax = 24.94, ymin = -15.45, ymax = 6.11; /* image dimensions */ pen wrwrwr = rgb(0.38,0.38,0.38); /* draw figures */ draw(circle((2.92,-3.28), 5.90), linewidth(2) + wrwrwr); draw((-2.52,-1.01)--(3.46,2.59), linewidth(2) + wrwrwr); draw((7.59,-6.88)--(-0.29,-8.22), linewidth(2) + wrwrwr); draw((3.46,2.59)--(7.59,-6.88), linewidth(2) + wrwrwr); draw((-0.29,-8.22)--(-2.52,-1.01), linewidth(2) + wrwrwr); draw((0.03,0.52)--(1.67,-7.89), linewidth(2) + wrwrwr); draw((5.61,-7.22)--(0.89,1.04), linewidth(2) + wrwrwr); /* dots and labels */ dot((2.92,-3.28),dotstyle); label("$O$", (2.43,-3.56), NE * labelscalefactor); dot((-2.52,-1.01),dotstyle); label("$A$", (-2.91,-0.91), NE * labelscalefactor); dot((3.46,2.59),linewidth(4pt) + dotstyle); label("$B$", (3.49,2.78), NE * labelscalefactor); dot((7.59,-6.88),dotstyle); label("$C$", (7.82,-7.24), NE * labelscalefactor); dot((-0.29,-8.22),linewidth(4pt) + dotstyle); label("$D$", (-0.53,-8.62), NE * labelscalefactor); dot((0.03,0.52),linewidth(4pt) + dotstyle); label("$P$", (-0.13,0.67), NE * labelscalefactor); dot((0.89,1.04),linewidth(4pt) + dotstyle); label("$Q$", (0.62,1.16), NE * labelscalefactor); dot((5.61,-7.22),linewidth(4pt) + dotstyle); label("$R$", (5.70,-7.05), NE * labelscalefactor); dot((1.67,-7.89),linewidth(4pt) + dotstyle); label("$S$", (1.75,-7.73), NE * labelscalefactor); dot((0.46,0.78),linewidth(4pt) + dotstyle); label("$E$", (0.26,0.93), NE * labelscalefactor); dot((3.64,-7.55),linewidth(4pt) + dotstyle); label("$F$", (3.73,-7.39), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
// Block 4
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/wiki/User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.19, xmax = 24.94, ymin = -15.45, ymax = 6.11; /* image dimensions */ pen wrwrwr = rgb(0.38,0.38,0.38); /* draw figures */ draw(circle((2.92,-3.28), 5.90), linewidth(2) + wrwrwr); draw((-2.52,-1.01)--(3.46,2.59), linewidth(2) + wrwrwr); draw((7.59,-6.88)--(-0.29,-8.22), linewidth(2) + wrwrwr); draw((3.46,2.59)--(7.59,-6.88), linewidth(2) + wrwrwr); draw((-0.29,-8.22)--(-2.52,-1.01), linewidth(2) + wrwrwr); draw((0.03,0.52)--(1.67,-7.89), linewidth(2) + wrwrwr); draw((5.61,-7.22)--(0.89,1.04), linewidth(2) + wrwrwr); draw((0.46,0.78)--(2.92,-3.28), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(3.64,-7.55), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(7.59,-6.88), linewidth(2) + wrwrwr); draw((5.61,-7.22)--(2.92,-3.28), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(3.46,2.59), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(0.89,1.04), linewidth(2) + wrwrwr); /* dots and labels */ dot((2.92,-3.28),dotstyle); label("$O$", (2.43,-3.56), NE * labelscalefactor); dot((-2.52,-1.01),dotstyle); label("$A$", (-2.91,-0.91), NE * labelscalefactor); dot((3.46,2.59),linewidth(1pt) + dotstyle); label("$B$", (3.49,2.78), NE * labelscalefactor); dot((7.59,-6.88),dotstyle); label("$C$", (7.82,-7.24), NE * labelscalefactor); dot((-0.29,-8.22),linewidth(1pt) + dotstyle); label("$D$", (-0.53,-8.62), NE * labelscalefactor); dot((0.03,0.52),linewidth(1pt) + dotstyle); label("$P$", (-0.13,0.67), NE * labelscalefactor); dot((0.89,1.04),linewidth(1pt) + dotstyle); label("$Q$", (0.62,1.16), NE * labelscalefactor); dot((5.61,-7.22),linewidth(1pt) + dotstyle); label("$R$", (5.70,-7.05), NE * labelscalefactor); dot((1.67,-7.89),linewidth(1pt) + dotstyle); label("$S$", (1.75,-7.73), NE * labelscalefactor); dot((0.46,0.78),linewidth(1pt) + dotstyle); label("$E$", (0.26,0.93), NE * labelscalefactor); dot((3.64,-7.55),linewidth(1pt) + dotstyle); label("$F$", (3.73,-7.39), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | [] |
17 | A given convex pentagon $ABCDE$ has the property that the area of each of the five triangles $ABC$, $BCD$, $CDE$, $DEA$, and $EAB$ is unity. Show that all pentagons with the above property have the same area, and calculate that area. Show, furthermore, that there are infinitely many non-congruent pentagons having the above area property. | 1972 USAMO Problem 5 | Lemma: Convex pentagon $A_0A_1A_2A_3A_4$ has the property that $[A_0A_1A_2] = [A_1A_2A_3] = [A_2A_3A_4] = [A_3A_4A_0] = [A_4A_0A_1]$ if and only if $\overline{A_{n - 1}A_{n + 1}}\parallel\overline{A_{n - 2}A_{n + 2}}$ for $n = 0, 1, 2, 3, 4$ (indices taken mod 5).
Proof: For the "only if" direction, since $[A_0A_1A_2] = [A_1A_2A_3]$, $A_0$ and $A_3$ are equidistant from $\overline{A_1A_2}$, and since the pentagon is convex, $\overline{A_0A_3}\parallel\overline{A_1A_2}$. The other four pairs of parallel lines are established by a symmetrical argument, and the proof for the other direction is just this, but reversed.
Let $A'B'C'D'E'$ be the inner pentagon, labeled so that $A$ and $A'$ are opposite each other, and let $a, b, c, d, e$ be the side lengths of the inner pentagon labeled opposite their corresponding vertices. From the lemma, pentagon $A'B'C'D'E'$ is similar to pentagon $ABCDE$ with $AB = m(A'B')$ and parallelogram $ABCB'$ (and cyclic) has area 2. Supposing
\begin{align*} P &= [ABCDE] \\ Q &= [A'B'C'D'E'] \\ R &= \sum_{\text{cyc}}[AC'D'] \\ S &= \sum_{\text{cyc}}[ABD'], \end{align*}
we have
\begin{align*} \sum_{\text{cyc}}[ABCB'] &= 5Q + 3R + 2S \\ 10 &= 3Q + R + 2P. \end{align*}
Since $BCDC'$ and $CDED'$ are parallelograms, $BD' = EC' = (m - 1)a$. Triangle $EB'C'$ is similar to triangle $ECB$, so $me = BC = \left(\frac{2m - 1}{m - 1}\right)e$, and with the requirement that $m > 1$,
\[m = \frac{2m - 1}{m - 1}\Rightarrow m = \frac{3 + \sqrt{5}}{2}.\]
Now, we compute that $[AC'D'] = \frac{1}{2m - 1} = \sqrt{5} - 2$, and similar computation for the other four triangles gives $R = 5\sqrt{5} - 10$. From the aforementioned pentagon similarity, $Q = P/m^2 = \left(\frac{7 - 3\sqrt{5}}{2}\right)P$. Solving for $P$, we have
\begin{align*} 10 &= 3\left(\frac{7 - 3\sqrt{5}}{2}\right)P + (5\sqrt{5} - 10) + 2P \\ 20 - 5\sqrt{5} &= \left(\frac{25 - 9\sqrt{5}}{2}\right)P \\ P &= \frac{40 - 10\sqrt{5}}{25 - 9\sqrt{5}} \\ &= \frac{5 + \sqrt{5}}{2}. \end{align*}
To show that there are infinitely many pentagons with the given property, we start with triangle $A'B'C'$ and construct the original pentagon using the required length ratios. This triangle has three continuous degrees of freedom, and the condition that the equal areas be specifically unity means that there are two continuous degrees of freedom in which the configuration can vary, which implies the desired result.
Alternatively, one can shear a regular pentagon, a transformation which preserves areas and the property of a pair of lines being parallel. | // Block 1
size(120);
defaultpen(fontsize(10));
pathpen = black;
pair A=MP("A",dir(90),dir(90)), B=MP("B",dir(90-72),dir(90-72)), C=MP("C",dir(90-2*72),dir(90-2*72)), D=MP("D",dir(90-3*72),dir(90-3*72)), E=MP("E",dir(90-4*72),dir(90-4*72));
D(A--B--C--D--E--cycle);
D(A--C--E--B--D--cycle);
pair Ap = MP("A'",IP(B--D,C--E),dir(270)), Bp = MP("B'",IP(A--D,C--E),dir(270-72)), Cp = MP("C'",IP(A--D,B--E),dir(270-72*2)), Dp = MP("D'",IP(A--C,B--E),dir(270-72*3)), Ep = MP("E'",IP(A--C,B--D),dir(270-72*4));
// Block 2
size(120); defaultpen(fontsize(10)); pathpen = black; pair A=MP("A",dir(90),dir(90)), B=MP("B",dir(90-72),dir(90-72)), C=MP("C",dir(90-2*72),dir(90-2*72)), D=MP("D",dir(90-3*72),dir(90-3*72)), E=MP("E",dir(90-4*72),dir(90-4*72)); D(A--B--C--D--E--cycle); D(A--C--E--B--D--cycle); pair Ap = MP("A'",IP(B--D,C--E),dir(270)), Bp = MP("B'",IP(A--D,C--E),dir(270-72)), Cp = MP("C'",IP(A--D,B--E),dir(270-72*2)), Dp = MP("D'",IP(A--C,B--E),dir(270-72*3)), Ep = MP("E'",IP(A--C,B--D),dir(270-72*4)); | [] |
18 | Consider the two triangles $\triangle ABC$ and $\triangle PQR$ shown in Figure 1. In $\triangle ABC$, $\angle ADB = \angle BDC = \angle CDA = 120^\circ$. Prove that $x=u+v+w$. | 1974 USAMO Problem 5 | Solution 1
We rotate figure $PRQM$ by a clockwise angle of $\pi/3$ about $Q$ to obtain figure $RR'QM'$:
Evidently, $MM'Q$ is an equilateral triangle, so triangles $MRM'$ and $ABC$ are congruent. Also, triangles $PMQ$ and $RM'Q$ are congruent, since they are images of each other under rotations. Then
\[[ABC] + \frac{b^2 \sqrt{3}}{4} = [MRM'] + [MM'Q] = [QMR] + [RM'Q] = [QMR] + [PMQ] .\]
Then by symmetry,
\[3 [ABC] + \frac{(a^2+b^2+c^2)\sqrt{3}}{4} = 2\bigl( [PMQ] + [QMR] + [RMP] \bigr) = 2 [PRQ] .\]
But $ABC$ is composed of three smaller triangles. The one with sides $w,v,a$ has area $\tfrac{1}{2} wv \sin 120^\circ = \frac{wv \sqrt{3}}{4}$. Therefore, the area of $ABC$ is
\[\frac{(wv+vu+uw)\sqrt{3}}{4} .\]
Also, by the Law of Cosines on that small triangle of $ABC$, $a^2 = w^2 + wv+ v^2$, so by symmetry,
\[\frac{(a^2 + b^2 + c^2)\sqrt{3}}{4} = \frac{\bigl[2(u^2+w^2+v^2) + wv + vu + uw \bigr] \sqrt{3}}{4}.\]
Therefore
\begin{align*} \frac{(u+v+w)^2 \sqrt{3}}{2} &= 3 \frac{(wv+vu+uw)\sqrt{3}}{4} + \frac{\bigl[2(u^2+w^2+v^2) + wv + vu + uw \bigr] \sqrt{3}}{4} \\ &= 3 [ABC] + \frac{(a^2+b^2+c^2)\sqrt{3}}{4} = 2[PQR] . \end{align*}
But the area of triangle $PQR$ is $x^2 \sqrt{3}/4$. It follows that $u+v+w=x$, as desired. $\blacksquare$
Solution 2
Rotate $\triangle ABC$ $60$ degrees clockwise about $A$ to get $\triangle AB'C'$. Observe that $\triangle ADD'$ is equilateral, which means $D'D=AD=u$. Also, $B',D',D,C$ are collinear because $\angle B'D'A + \angle AD'D = 120+60=180$ and $\angle CDA + \angle DD'A = 120+60=180$. The resulting $\triangle B'AC$ has side lengths $b,c,u+v+w$ and the angle opposite side $u+v+w$ has magnitude $A+60$.
If we perform the rotation about points $B$ and $C$, we get two triangles. One has side lengths $a,c,u+v+w$ and the angle opposite side $u+v+w$ has magnitude $B+60$, and the other has side lengths $b,c,u+v+w$ and the angle opposite side $u+v+w$ has magnitude $C+60$.
These three triangles fit together because $(A+60)+(B+60)+(C+60) = 360$. The result is an equilateral triangle of side length $u+v+w$.
Solution 3
As in the first solution, we rotate and establish that $\triangle MRM' \cong \triangle ABC$.
Let $X$ and $Y$ be points on $\overline{RQ}$ and $\overline{PQ}$, respectively, such that $M$ lies on $\overline{XY}$ and $\overline{XY}\parallel \overline{PR}$. We note that $m\angle RXM = 120^\circ$. The rotation then takes $Y$ to $X$, so $m\angle RXM'=m\angle PYM = 120^\circ$.
It follows that $RX=BD=v$, $MX=AD=u$, $M'X=CD=w$.
Since $m\angle MXM' = 120^\circ$ and $m\angle MQM' = 60^\circ$, $MXM'Q$ is cyclic.
By Ptolemy's theorem,
\begin{align*} (MX)(M'Q) + (XM')(QM) &= (MM')(XQ) \\ XQ &= MX + M'X \\ &= u+w. \end{align*}
Finally, $RQ = RX+XQ = u+v+w$, as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page. | // Block 1
size(300); defaultpen(1); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); pair RR=R+Q-P, MM= rotate(-60,Q)*M; draw(P--R--RR--Q--P--M--MM--RR); draw(Q--R--M--Q--MM--R); label("$P$",P,W); label("$Q$",Q,E); label("$R$",R,W); label("$M$",M,NW); label("$R'$",RR,NE); label("$M'$",MM,ESE);
// Block 2
size(200); defaultpen(fontsize(8)); pair A=6*expi(5/9*pi), B=3*expi(5/9*pi+2*pi/3), C=4*expi(5/9*pi+4*pi/3), D=(0,0); path triabc = D--A--B--C--D--B..A--C; draw(triabc); label("$A$",A,( 0, 1)); label("$B$",B,(-1,-1));label("$b$",(C+A)/2,( 1, 1)); label("$C$",C,( 1,-1));label("$w$",(C+D)/2,( 0, 1)); label("$D$",D,( 1, 1)); transform rot60a = rotate(-60,A); pair A1 = rot60a*A, B1 = rot60a*B, C1 = rot60a*C, D1 = rot60a*D; path triabc1 = D1--A1--B1--C1--D1--B1..A1--C1; draw(triabc1, linetype("8 8")); label("$B'$",B1,(-1, 0));label("$v$",(B1+D1)/2,( 0, 1)); label("$C'$",C1,( 0,-1));label("$c$",(A1+B1)/2,( 0, 1)); label("$D'$",D1,(-1,-1));label("$u$",(D1+D)/2,( 0, -1)); draw(A--B1--C--A, red+1.5); dot(A^^B1^^D1^^D^^C);
// Block 3
size(400); defaultpen(fontsize(8)); picture transC; pair A=6*expi(5/9*pi), B=3*expi(5/9*pi+2*pi/3), C=4*expi(5/9*pi+4*pi/3), D=(0,0); path triabc = D--A--B--C--D--B..A--C; draw(triabc); label("$A$",A,( 0, 1));label("$u$",(A+D)/2,( 1, 0)); label("$B$",B,(-1,-1)); label("$C$",C,( 1,-1));label("$c$",(A+B)/2,(-1, 1)); label("$D$",D,( 1, 1)); transform rot60b = rotate(-60,B); pair A1 = rot60b*A, B1 = rot60b*B, C1 = rot60b*C, D1 = rot60b*D; path triabc1 = D1--A1--B1--C1--D1--B1..A1--C1; draw(triabc1, linetype("8 8")); label("$A'$",A1,( 1, 0));label("$a$",(B1+C1)/2,(-1, 0)); label("$C'$",C1,( 0,-1));label("$w$",(C1+D1)/2,( 1, 0)); label("$D'$",D1,(-1,-1));label("$v$",(D1+D)/2,( 1, 1)); draw(A--B--C1--A, red+1.5); dot(A^^B^^D1^^D^^C1); draw(transC, triabc); label(transC, "$A$",A,( 0, 1));label(transC, "$a$",(B+C)/2,( 0,-1)); label(transC, "$B$",B,(-1,-1));label(transC, "$v$",(B+D)/2,( 0, 1)); label(transC, "$C$",C,( 1,-1)); label(transC, "$D$",D,( 0,-1)); transform rot60c = rotate(-60,C); pair A2 = rot60c*A, B2 = rot60c*B, C2 = rot60c*C, D2 = rot60c*D; path triabc2 = D2--A2--B2--C2--D2--B2..A2--C2; draw(transC, triabc2, linetype("8 8")); label(transC, "$A'$",A2,( 1, 0));label(transC, "$u$",(A2+D2)/2,( 1,-1)); label(transC, "$b$",(C2+A2)/2,( 1,-1)); label(transC, "$w$",(D+D2)/2,( 1,-1)); label(transC, "$D'$",D2,( 0, 1)); draw(transC, A2--B--C--A2, red+1.5); dot(transC, C^^B^^D2^^D^^A2); add(shift(15*right)*transC);
// Block 4
size(300); defaultpen(1); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); pair RR=R+Q-P, MM= rotate(-60,Q)*M; draw(P--R--RR--Q--P--M--MM--RR); draw(Q--R--M--Q--MM--R); label("$P$",P,W); label("$Q$",Q,E); label("$R$",R,W); label("$M$",M,NW); label("$R'$",RR,NE); label("$M'$",MM,ESE); | [] |
19 | Let $A,B,C,D$ denote four points in space and $AB$ the distance between $A$ and $B$, and so on. Show that
\[AC^2+BD^2+AD^2+BC^2\ge AB^2+CD^2.\] | 1975 USAMO Problem 2 | Solution 1
If we project points $A,B,C,D$ onto the plane parallel to $\overline{AB}$ and $\overline{CD}$, $AB$ and $CD$ stay the same but $BC, AC, AD, BD$ all decrease, making the inequality sharper. Thus, it suffices to prove the inequality when $A,B,C,D$ are coplanar:
Let $AD=a, AC=b, BC=c, BD=d, AB=m, CD=n$. We wish to prove that $a^2+b^2+c^2+d^2\ge m^2+n^2$. Let us fix $\triangle BCD$ and the length $AB$ and let $A$ vary on the circle centered at $B$ with radius $m$. If we find the minimum value of $a^2+b^2$, which is the only variable quantity, and prove that it is larger than $m^2+n^2-c^2-d^2$, we will be done.
First, we express $a^2+b^2$ in terms of $c,d,m,\theta,\phi$, using the Law of Cosines:
\begin{align*} a^2+b^2 &= c^2+d^2+2m^2-2cm\cos(\theta)-2dm\cos(\phi-\theta) \\ (a^2+b^2-c^2-d^2-2m^2)^2 &= 4m^2(c^2\cos^2(\theta)+d^2\cos^2(\phi-\theta)+2cd\cos(\theta)\cos(\phi-\theta)) \end{align*}
$a^2+b^2$ is a function of $\theta$, so we take the derivative with respect to $\theta$ and obtain that $a^2+b^2$ takes a minimum when
\begin{align*} c\sin(\theta)-d\sin(\phi-\theta) &= 0 \\ c^2\sin^2(\theta)+d^2\sin^2(\phi-\theta)-2cd\sin(\theta)\sin(\phi-\theta) &= 0 \\ (a^2+b^2-c^2-d^2-2m^2)^2 &= 4m^2(c^2+d^2+2cd(\cos(\theta)\cos(\phi-\theta)-\sin(\theta)\sin(\phi-\theta))) \\ &= 4m^2(c^2+d^2+2cd\cos{\phi})\\ &= 4m^2(2c^2+2d^2-n^2) \end{align*}
Define $p=a^2+b^2$ and $q=c^2+d^2$:
\begin{align*} (p-q-2m^2)^2 &= 4m^2(2q-n^2) \\ p^2+q^2+4m^4-4m^2p+4m^2q-2pq &= 8m^2q-4m^2n^2 \\ p^2+q^2+4m^4-4m^2p-4m^2q-2pq &= -4m^2n^2 \\ p^2-2pq+q^2-4m^2(p+q) &= -4m^2(m^2+n^2) \\ \frac{(p-q)^2}{m^2} &= p+q-m^2-n^2\geq 0 \\ a^2+b^2+c^2+d^2 &\geq m^2+n^2 \\ \end{align*}
Solution 2
Let
\begin{align*} A &= (0,0,0) \\ B &= (1,0,0) \\ C &= (a,b,c) \\ D &= (x,y,z). \end{align*}
It is clear that every other case can be reduced to this.
Then, with the distance formula and expanding,
\begin{align*} AC^2 + BD^2 + AD^2 + BC^2 - AB^2 - CD^2 &= x^2-2x+1+y^2+z^2+a^2-2a+b^2+c^2+2ax+2by+2cz \\ &= (x+a-1)^2 + (y+b)^2 + (z+c)^2. \\ &\geq 0, \end{align*}
which rearranges to the desired inequality.
Solution 3
Because the distances are all squared, we must only prove the result in one dimension, and then we can just add up the three individual inequalities for the $x$, $y$, and $z$ dimension to get the desired result. Let $x_a$, $x_b$, $x_c$, and $x_d$ be the positions of $A$, $B$, $C$, and $D$ respectively. Then we must show that,
\begin{align*} (x_a - x_c)^2 + (x_b-x_d)^2 + (x_a - x_d)^2 + (x_b - x_c)^2 &\geq (x_a - x_b)^2 + (x_c-x_d)^2 \\ x_a^2 + x_b^2 + x_c^2 + x_d^2 &\geq 2x_a x_c + 2x_b x_d + 2x_a x_d + 2x_b x_c - 2x_a x_b - 2x_c x_d \\ (x_a + x_b)^2 + (x_c + x_d)^2 &\geq 2(x_a +x_b)(x_c + x_d)\\ (x_a + x_b - x_c - x_d)^2 &\geq 0. \end{align*}
So we are done.
Solution 4 (Vector bash)
Let $a$, $b$, $c$, $d$ correspond to the position vectors of points A, B, C, and D, respectively, with respect to an arbitrary origin O. Let us also for simplicity define $a^2 = a \cdot a = ||a||^2$, where $||a||$ is the magnitude of vector $a$. Because squares are non-negative, $a^2$ is non-negative for all vectors $a$. Thus,
\[(a + b - c - d)^2 \ge 0\]
Because dot product is linear, we expand to obtain
\[a^2 + b^2 + c^2 + d^2 + 2a \cdot b + 2 c \cdot d - 2 a \cdot c - 2 a \cdot d - 2 b \cdot c - 2 c \cdot d \ge 0,\]
from which we add $a^2 + b^2 + c^2 + d^2$ to both sides, rearrange, and complete the square to get
\[(a-c)^2 + (a-d)^2 + (b-c)^2 + (b-d)^2 \ge (a-b)^2 + (c-d)^2.\]
As $(a-b)^2 = ||a-b||^2 = ||AB||^2 = AB^2$ and likewise for the others,
\[AC^2 + AD^2 + BC^2 + BD^2 \ge AB^2 + CD^2,\]
which is what we wanted to prove.
NOTES:
1. Equality holds when the vector equality $a + b = c + d$ holds, which occurs when A, B, C, and D are the vertices of a (planar) parallelogram, in that order.
2. The algebra employed here is almost identical to the algebra used in Solution 3, which means that Solution 3 is just a simplification of this solution.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page. | // Block 1
defaultpen(fontsize(8)); pair A=(2,4), B=(0,0), C=(4,0), D=(4,2); label("A",A,(0,1));label("D",D,(1,0));label("B",B,(-1,-1));label("C",C,(1,-1)); axialshade(A--C--D--cycle, lightgray, A, gray, D); draw(A--B--C--A--D--C);draw(B--D, linetype("8 8")); label("$m$",(A+B)/2,(-1,1));label("$n$",(C+D)/2,(1,0)); label("$c$",(B+C)/2,(0,-1));label("$b$",(A+C)/2,(-1,-1)); label("$a$",(A+D)/2,(1,1));label("$d$",(B+D)/2,(-1,1));
// Block 2
size(200); defaultpen(fontsize(8)); pair A=(8,3), B=(4,-5), C=(10,0), D=(0,0); draw(A--C--B--D--A--B);draw(C--D);draw(anglemark(A,B,D,40));draw(anglemark(C,B,A,55,60)); label("A",A,(0,1));label("D",D,(-1,0));label("B",B,(0,-1));label("C",C,(1,0)); label("$m$",(A+B)/2,(1,0));label("$n$",(C+D)/2,(0,1)); label("$c$",(B+C)/2,(1,-1));label("$b$",(A+C)/2,(-1,-1)); label("$a$",(A+D)/2,(0,1));label("$d$",(B+D)/2,(-1,-1)); label("$\phi-\theta$",anglemark(A,B,D,40),(1,5));label("$\theta$",anglemark(C,B,A,55),(8,9)); | [] |
20 | (a) Suppose that each square of a $4\times 7$ chessboard, as shown above, is colored either black or white. Prove that with any such coloring, the board must contain a rectangle (formed by the horizontal and vertical lines of the board such as the one outlined in the figure) whose four distinct unit corner squares are all of the same color.
(b) Exhibit a black-white coloring of a $4\times 6$ board in which the four corner squares of every rectangle, as described above, are not all of the same color. | 1976 USAMO Problem 1 | There are many ways to prove the first part, we will show one.
Consider the first row. It contains at least four cells of the same color, say white. Pick any four such cells. Let's now consider these four columns only. If any of the remaining three rows contains two white cells in some of these columns, we are done. Otherwise, we have the following situation: three rows times four columns, and in each row there is at most one white cell. We can now pick any two rows and then find two columns such that all four cells where they intersect are black, and we are done. (Note that this proof in fact shows that already in a $3\times 7$ board there must be such a rectangle.)
For the second part, consider the board shown below. Its six columns contain the six different permutations of two white and two black squares. From this observation it is immediately obvious that no two columns agree on two cells of the same color. | // Block 1
void fillsq(int x, int y){
fill((x,y)--(x+1,y)--(x+1,y+1)--(x,y+1)--cycle, mediumgray);
}
int i;
fillsq(0,0);fillsq(0,1);
fillsq(1,0);fillsq(1,2);
fillsq(2,0);fillsq(2,3);
fillsq(3,1);fillsq(3,2);
fillsq(4,1);fillsq(4,3);
fillsq(5,2);fillsq(5,3);
for(i=0; i<=6; ++i){draw((i,0)--(i,4),black+0.5);}
for(i=0; i<=4; ++i){draw((0,i)--(6,i),black+0.5);}
// Block 2
void fillsq(int x, int y){ fill((x,y)--(x+1,y)--(x+1,y+1)--(x,y+1)--cycle, mediumgray); } int i; fillsq(0,0);fillsq(0,1); fillsq(1,0);fillsq(1,2); fillsq(2,0);fillsq(2,3); fillsq(3,1);fillsq(3,2); fillsq(4,1);fillsq(4,3); fillsq(5,2);fillsq(5,3); for(i=0; i<=6; ++i){draw((i,0)--(i,4),black+0.5);} for(i=0; i<=4; ++i){draw((0,i)--(6,i),black+0.5);} | [] |
20 | (a) Suppose that each square of a $4\times 7$ chessboard, as shown above, is colored either black or white. Prove that with any such coloring, the board must contain a rectangle (formed by the horizontal and vertical lines of the board such as the one outlined in the figure) whose four distinct unit corner squares are all of the same color.
(b) Exhibit a black-white coloring of a $4\times 6$ board in which the four corner squares of every rectangle, as described above, are not all of the same color. | 1976 USAMO Problem 1 | We will prove the first part by the pigeonhole principle. To form a rectangle, two squares have in one column have to be colored the same color as the corresponding squares in another column. In the example given, the two second and third squares of the fourth and seventh columns are all white. When coloring this rectangle, every column can be entirely one color, have 3 squares be of one color, or have 2 squares be of one color. Coloring each column with two white squares and two black squares will minimize the number of squares, since a smaller fraction of the column is one color (if more than 2 squares in the same column are the same color, then there is more than one coloring in another column that can create a rectangle). There are ${4}\choose{2}$, or $6$ different ways a column can be colored with two white and two black squares. We need the same coloring in two different columns to create a rectangle, and this is guaranteed with 7 columns by the pigeonhole principle.
For the second part, we can draw a rectangle with each column having a different coloring with 2 black and 2 white squares. One such rectangle is shown below:
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page. | // Block 1
void fillsq(int x, int y){
fill((x,y)--(x+1,y)--(x+1,y+1)--(x,y+1)--cycle, mediumgray);
}
int i;
fillsq(0,3);fillsq(0,2);
fillsq(1,3);fillsq(1,1);
fillsq(2,3);fillsq(2,0);
fillsq(3,2);fillsq(3,1);
fillsq(4,2);fillsq(4,0);
fillsq(5,1);fillsq(5,0);
for(i=0; i<=6; ++i){draw((i,0)--(i,4),black+0.5);}
for(i=0; i<=4; ++i){draw((0,i)--(6,i),black+0.5);}
// Block 2
void fillsq(int x, int y){ fill((x,y)--(x+1,y)--(x+1,y+1)--(x,y+1)--cycle, mediumgray); } int i; fillsq(0,3);fillsq(0,2); fillsq(1,3);fillsq(1,1); fillsq(2,3);fillsq(2,0); fillsq(3,2);fillsq(3,1); fillsq(4,2);fillsq(4,0); fillsq(5,1);fillsq(5,0); for(i=0; i<=6; ++i){draw((i,0)--(i,4),black+0.5);} for(i=0; i<=4; ++i){draw((0,i)--(6,i),black+0.5);} | [] |
21 | $P$ lies between the rays $OA$ and $OB$. Find $Q$ on $OA$ and $R$ on $OB$ collinear with $P$ so that $\frac{1}{PQ} + \frac{1}{PR}$ is as large as possible. | 1979 USAMO Problem 4 | Let $r = OP, x = \angle OPR, a = \angle POR,$ and $b = \angle POQ.$ Then $\angle ORP = \pi - x - a$ and $\angle OQP = x - b.$ Using the Law of Sines on $\triangle OPR$ gives
\[PR = \sin a * \frac{r}{\sin(\pi - x - a)} = \sin a * \frac{r}{\sin(x + a)},\]
and using the Law of Sines on $\triangle OPQ$ gives
\[PQ = \sin b * \frac{r}{\sin(x - b)}.\]
Note that $r, a,$ and $b$ are given constants.
Hence,
\[\frac{1}{PR} + \frac{1}{PQ} = \frac{1}{r} (\frac{\sin(x + a)}{\sin a} + \frac{\sin(x - b)}{\sin b}) = \frac{\sin(x+a)\sin b + \sin(x-b)\sin a}{r \sin a \sin b} = \frac{\sin x\cos a\sin b + \sin a\cos x\sin b + \sin x\cos b\sin a - \cos x\sin b\sin a}{r \sin a \sin b} = \frac{\sin x(\sin b \cos a + \sin a \cos b)}{r \sin a \sin b}\]
Clearly, this quantity is maximized when $\sin x = 1.$ Because $x$ must be less than $\pi$, $\frac{1}{PQ} + \frac{1}{PR}$ is as large as possible when $x = \frac{\pi}{2},$ or when line $QR$ is perpendicular to line $PO$. | // Block 1
pair O = (0,0), A = (14,28), Q = (20,40), B = (16,0), R = (25,0), P = (23,16);
dot(O); dot(A); dot(Q); dot(B); dot(R); dot(P);
label("O", O, S);
label("A", A, W);
label("Q", Q, W);
label("B", B, S);
label("R", R, S);
label("P", P, E);
draw(O--R--Q--O); draw(O--P);
label("r", O--P, N);
// Block 2
pair O = (0,0), A = (14,28), Q = (20,40), B = (16,0), R = (25,0), P = (23,16); dot(O); dot(A); dot(Q); dot(B); dot(R); dot(P); label("O", O, S); label("A", A, W); label("Q", Q, W); label("B", B, S); label("R", R, S); label("P", P, E); draw(O--R--Q--O); draw(O--P); label("r", O--P, N); | [] |
22 | In triangle $ABC$, angle $A$ is twice angle $B$, angle $C$ is obtuse, and the three side lengths $a, b, c$ are integers. Determine, with proof, the minimum possible perimeter. | 1991 USAMO Problem 1 | Solution 1
(diagram by integralarefun)
After drawing the triangle, also draw the angle bisector of $\angle A$, and let it intersect $\overline{BC}$ at $D$. Notice that $\triangle ADC\sim \triangle BAC$, and let $AD=x$. Now from similarity,
\[x=\frac{bc}{a}\]
However, from the angle bisector theorem, we have
\[BD=\frac{ac}{b+c}\]
but $\triangle ABD$ is isosceles, so
\[x=BD\Longrightarrow \frac{bc}{a}=\frac{ac}{b+c}\Longrightarrow a^2=b(b+c)\]
so all sets of side lengths which satisfy the conditions also meet the boxed condition.
Notice that $\text{gcd}(a, b, c)=1$ or else we can form a triangle by dividing $a, b, c$ by their greatest common divisor to get smaller integer side lengths, contradicting the perimeter minimality. Since $a$ is squared, $b$ must also be a square because if it isn't, then $b$ must share a common factor with $b+c$, meaning it also shares a common factor with $c$, which means $a, b, c$ share a common factor—a contradiction. Thus we let $b = x^2, b+c = y^2$, so $a = xy$, and we want the minimal pair $(x,y)$.
By the Law of Cosines,
\[b^2 = a^2 + c^2 - 2ac\cos B\]
Substituting $a^2 = b^2 + bc$ yields $\cos B = \frac{b+c}{2a} = \frac{y}{2x}$. Since $\angle C > 90^{\circ}$, $0^{\circ} < \angle B < 30^{\circ} \Longrightarrow \sqrt{3} < \frac{y}{x} < 2$. For $x \le 3$ there are no integer solutions. For $x = 4$, we have $y = 7$ that works, so the side lengths are $(a, b, c)=(28, 16, 33)$ and the minimal perimeter is $\boxed{77}$.
Alternate Solution
In $\triangle ABC$ let $\angle B = \beta, \angle A = 2\beta, \angle C = 180^{\circ} - 3\beta$. From the law of sines, we have
\[\frac{a}{\sin 2\beta} = \frac{b}{\sin \beta} = \frac{c} {\sin (180^{\circ} - 3\beta)} = \frac{c}{\sin 3\beta}\]
Thus the ratio
\[b : a : c = \sin\beta : \sin 2\beta : \sin 3\beta\]
We can simplify
\[\frac{\sin 2\beta}{\sin\beta} = \frac{2\sin\beta\cos\beta}{\sin\beta} = 2\cos\beta\]
Likewise,
\[\frac{\sin 3\beta}{\sin\beta} = \frac{\sin 2\beta\cos\beta + \sin\beta\cos 2\beta}{\sin\beta} = \frac{2\sin\beta\cos^2\beta + \sin\beta(\cos^2\beta - \sin^2\beta)}{\sin\beta}\]
\[= {2 \cos^2 \beta + \cos^2 \beta - \sin^2 \beta} = 4\cos^2 \beta - 1\]
Letting $\gamma = \cos\beta$, rewrite
\[b : a : c = 1 : 2\gamma : 4\gamma^2 - 1\]
We find that to satisfy the conditions for an obtuse triangle, $\beta \in (0^\circ, 30^\circ)$ and therefore $\gamma \in \left(\frac{\sqrt{3}}{2}, 1\right)$.
The rational number with minimum denominator (in order to minimize scaling to obtain integer solutions) above $\frac{\sqrt{3}}{2}$ is $\frac{7}{8}$, which also has a denominator divisible by 2 (to take advantage of the coefficients of 2 and 4 in the ratio and further minimize scaling).
Inserting $\gamma = \frac{7}{8}$ into the ratio, we find $b : a : c = 1 : \frac{7}{4} : \frac{33}{16}$. When scaled minimally to obtain integer side lengths, we find
\[b, a, c = 16, 28, 33\]
and that the perimeter is $\boxed{77}$.
(note by integralarefun: The part of the solution about finding $\gamma$ is not rigorous and would likely require further proof in an actual test.)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page. | import olympiad; pair A, B, C, D, extensionAC; real angleABC; path braceBC; A = (0, 0); B = (2, 0); D = (1, .5); angleABC = atan(.5); //y = 4x/3 and x+2y = 2 (sides AC and BC, respectively) intersect here: C = (6/11, 8/11); braceBC = brace(C, B, .1); label("$\mathsf{A}$", A, W); label("$\mathsf{B}$", B, E); label("$\mathsf{C}$", C, N); label("$\mathsf{D}$", D, S); label("$\mathsf{a}$", braceBC, NE); label("$\mathsf{b}$", A--C, NW); label("$\mathsf{c}$", A--B, S); label("$\mathsf{x}$", A--D, N); draw(A--B--C--cycle); draw(A--D); draw(anglemark(C, B, A)); draw(anglemark(B, A, D)); draw(anglemark(D, A, C)); draw(braceBC); | [] |
23 | Let $\, D \,$ be an arbitrary point on side $\, AB \,$ of a given triangle $\, ABC, \,$ and let $\, E \,$ be the interior point where $\, CD \,$ intersects the external common tangent to the incircles of triangles $\, ACD \,$ and $\, BCD$. As $\, D \,$ assumes all positions between $\, A \,$ and $\, B \,$, prove that the point $\, E \,$ traces the arc of a circle. | 1991 USAMO Problem 5 | Let the incircle of $ACD$ and the incircle of $BCD$ touch line $AB$ at points $D_a,D_b$, respectively; let these circles touch $CD$ at $C_a$, $C_b$, respectively; and let them touch their common external tangent containing $E$ at $T_a,T_b$, respectively, as shown in the diagram below.
We note that
\[CE = CC_a - EC_a = CC_b - EC_b = \frac{CC_a + CC_b - (EC_a + EC_b)}{2} .\]
On the other hand, since $EC_a$ and $ET_a$ are tangents from the same point to a common circle, $EC_a = T_aE$, and similarly $EC_b = ET_b$, so
\[EC_a + EC_b = T_aE + ET_b = T_a T_b .\]
On the other hand, the segments $T_a T_b$ and $D_a D_b$ evidently have the same length, and $D_a D_b = D_aD + DD_b$, so $EC_a + EC_b = D_aD + DD_b$. Thus
\[CE = \frac{CC_a + CC_b - (EC_a + EC_b)}{2} = \frac{CC_a + CC_b - D_aD - DD_b}{2} .\]
If we let $s_a$ be the semiperimeter of triangle $ACD$, then $CC_a = s_a - AD$, and $D_aD = s_a - AC$, so
\[CC_a - D_aD = (s_a - AD) - (s_a - AC) = AC - AD .\]
Similarly,
\[CC_b - DD_b = BC - DB,\]
so that
\begin{align*} CE &= \frac{CC_a + CC_b - D_aD - DD_b}{2} = \frac{AC + BC - (AD+DB)}{2} \\ &= \frac{AC + BC - AB}{2} . \end{align*}
Thus $E$ lies on the arc of the circle with center $C$ and radius $(AB+BC-AB)/2$ intercepted by segments $CA$ and $CB$. If we choose an arbitrary point $X$ on this arc and let $D$ be the intersection of lines $CX$ and $AB$, then $X$ becomes point $E$ in the diagram, so every point on this arc is in the locus of $E$. $\blacksquare$ | // Block 1
size(220);
defaultpen(1);
pair A=(0,0), B=(220,0), C=(18.7723,118.523);
pair D=(72.6,0);
pair Ia=incenter(A,D,C), Ib=incenter(B,D,C);
pair Ta=(24.9758,52.5775),Tb=(86.6196,67.4129);
pair E=IntersectionPoint((Ta--Tb),(C--D));
path Oa=circle(Ia,inradius(A,D,C));
path Ob=circle(Ib,inradius(B,D,C));
pair Da=IP(Oa,A--B), Db=IP(Ob,A--B);
pair Ca=IP(Oa,C--D), Cb=IP(Ob,C--D);
draw(D--C--A--B--C);
draw(Ta--Tb);
draw(Oa);
draw(Ob);
dot(A,linewidth(4));
dot(B,linewidth(4));
dot(C,linewidth(4));
dot(D,linewidth(4));
dot(E,linewidth(4));
dot(Ta,linewidth(4));
dot(Tb,linewidth(4));
dot(Ca,linewidth(4));
dot(Cb,linewidth(4));
dot(Da,linewidth(4));
dot(Db,linewidth(4));
label("\(A\)",A,SW);
label("\(B\)",B,SE);
label("\(C\)",C,W);
label("\(D\)",D,S);
label("\(E\)",E,NNE);
label("\(T_a\)",Ta,N);
label("\(T_b\)",Tb,WNW);
label("\(D_a\)",Da,S);
label("\(D_b\)",Db,S);
label("\(C_a\)",Ca,WSW);
label("\(C_b\)",Cb,ENE);
// Block 2
size(220); defaultpen(1); pair A=(0,0), B=(220,0), C=(18.7723,118.523); pair D=(72.6,0); pair Ia=incenter(A,D,C), Ib=incenter(B,D,C); pair Ta=(24.9758,52.5775),Tb=(86.6196,67.4129); pair E=IntersectionPoint((Ta--Tb),(C--D)); path Oa=circle(Ia,inradius(A,D,C)); path Ob=circle(Ib,inradius(B,D,C)); pair Da=IP(Oa,A--B), Db=IP(Ob,A--B); pair Ca=IP(Oa,C--D), Cb=IP(Ob,C--D); draw(D--C--A--B--C); draw(Ta--Tb); draw(Oa); draw(Ob); dot(A,linewidth(4)); dot(B,linewidth(4)); dot(C,linewidth(4)); dot(D,linewidth(4)); dot(E,linewidth(4)); dot(Ta,linewidth(4)); dot(Tb,linewidth(4)); dot(Ca,linewidth(4)); dot(Cb,linewidth(4)); dot(Da,linewidth(4)); dot(Db,linewidth(4)); label("\(A\)",A,SW); label("\(B\)",B,SE); label("\(C\)",C,W); label("\(D\)",D,S); label("\(E\)",E,NNE); label("\(T_a\)",Ta,N); label("\(T_b\)",Tb,WNW); label("\(D_a\)",Da,S); label("\(D_b\)",Db,S); label("\(C_a\)",Ca,WSW); label("\(C_b\)",Cb,ENE); | [] |
24 | Prove
\[\frac{1}{\cos 0^\circ \cos 1^\circ} + \frac{1}{\cos 1^\circ \cos 2^\circ} + \cdots + \frac{1}{\cos 88^\circ \cos 89^\circ} = \frac{\cos 1^\circ}{\sin^2 1^\circ}.\] | 1992 USAMO Problem 2 | Solution 1
Consider the points $M_k = (1, \tan k^\circ)$ in the coordinate plane with origin $O=(0,0)$, for integers $0 \le k \le 89$.
Evidently, the angle between segments $OM_a$ and $OM_b$ is $(b-a)^\circ$, and the length of segment $OM_a$ is $1/\cos a^\circ$. It then follows that the area of triangle $M_aOM_b$ is $\tfrac{1}{2} \sin(b-a)^\circ \cdot OM_a \cdot OM_b = \tfrac{1}{2} \sin(b-a)^\circ / (\cos a^\circ \cdot \cos b^\circ)$. Therefore
\begin{align*} \sum_{k=0}^{88} \frac{\tfrac{1}{2} \sin 1^\circ}{ \cos k^\circ \cos k+1^\circ} &= \sum_{k=0}^{88} [ M_k O M_{k+1} ] \\ &= [ M_0 O M_{89} ] \\ &= \frac{ \tfrac{1}{2} \sin 89^\circ}{\cos 89^\circ} = \frac{\tfrac{1}{2} \cos 1^\circ}{\sin 1^\circ} , \end{align*}
so
\[\sum_{k=0}^{88} \frac{1}{\cos k^\circ \cos (k+1)^\circ} = \frac{\cos 1^\circ}{ \sin^2 1^\circ},\]
as desired. $\blacksquare$
Solution 2
First multiply both sides of the equation by $\sin 1$, so the right hand side is $\frac{\cos 1}{\sin 1}$. Now by rewriting $\sin 1=\sin((k+1)-k)=\sin(k+1)\cos(k)+\sin(k)\cos(k+1)$, we can derive the identity $\tan(n+1)-\tan(n)=\frac{\sin 1}{\cos(n)\cos(n+1)}$. Then the left hand side of the equation simplifies to $\tan 89-\tan 0=\tan 89=\frac{\sin 89}{\cos 89}=\frac{\cos 1}{\sin 1}$ as desired.
Solution 3
Multiply by $\sin{1}$. We get:
$\frac {\sin{1}}{\cos{0}\cos{1}} + \frac {\sin{1}}{\cos{1}\cos{2}} + ... + \frac {\sin{1}}{\cos{88}\cos{89}} = \frac {\cos{1}}{\sin{1}}$
we can write this as:
$\frac {\sin{1 - 0}}{\cos{0}\cos{1}} + \frac {\sin{2 - 1}}{\cos{1}\cos{2}} + ... + \frac {\sin{89 - 88}}{\cos{88}\cos{89}} = \frac {\cos{1}}{\sin{1}}$
This is an identity $\tan{a} - \tan{b} = \frac {\sin{(a - b)}}{\cos{a}\cos{b}}$
Therefore;
$\sum_{i = 1}^{89}[\tan{k} - \tan{(k - 1)}] = \tan{89} - \tan{0} = \cot{1}$, because of telescoping.
but since we multiplied $\sin{1}$ in the beginning, we need to divide by $\sin{1}$. So we get that:
$\frac {1}{\cos{0}\cos{1}} + \frac {1}{\cos{1}\cos{2}} + \frac {1}{\cos{2}\cos{3}} + .... + \frac {1}{\cos{88}\cos{89}} = \frac {\cos{1}}{\sin^2{1}}$
as desired. QED
Solution 4
Let $S = \frac{1}{\cos 0^\circ\cos 1^\circ} + \frac{1}{\cos 1^\circ\cos 2^\circ} + ... + \frac{1}{\cos 88^\circ\cos 89^\circ}$.
Multiplying by $\sin 1^\circ$ gives
\[S \sin 1^\circ = \frac{\sin(1^\circ-0^\circ)}{\cos 0^\circ\cos 1^\circ} + ... + \frac{\sin(89^\circ-88^\circ)}{\cos 88^\circ\cos 89^\circ}\]
Notice that $\frac{\sin((x+1^\circ)-x)}{\cos 0^\circ\cos 1^\circ} = \tan (x+1^\circ) - \tan x$ after expanding the sine, and so
\[S \sin 1^\circ = \left(\tan 1^\circ - \tan 0^\circ\right) + \cdots + \left(\tan 89^\circ - \tan 88^\circ\right) = \tan 89^\circ - \tan 0^\circ = \cot 1^\circ = \frac{\cos 1^\circ}{\sin 1^\circ},\] so \[S = \frac{\cos 1^\circ}{\sin^21^\circ}.\] | size(200); defaultpen(1); pair O=(0,0), a=expi(0), b=expi(1/6), c=expi(2/6), d=expi(3/6), y=expi(32/30), z= expi(34/30); pair A=a, B=b/b.x, C= c/c.x, D=d/d.x, Y=y/y.x, Z=z/z.x, E=(D+Y)/2; pair X=(O+E)/2; draw(O--A--Z); draw(B--O--C--D--O--Y--Z--O); label("\(O\)",O,SW); label("\(M_0\)",A,ESE); label("\(M_1\)",B,ESE); label("\(M_2\)",C,ESE); label("\(M_3\)",D,ESE); label("\(\vdots\)",E,ESE); label("\(M_{88}\)",Y,ESE); label("\(M_{89}\)",Z,ESE); label("\(\ddots\)",X); | [] |
25 | Let $ABCD$ be a convex quadrilateral such that diagonals $AC$ and $BD$ intersect at right angles, and let $E$ be their intersection. Prove that the reflections of $E$ across $AB$, $BC$, $CD$, $DA$ are concyclic. | 1993 USAMO Problem 2 | Diagram
Work
Let $X$, $Y$, $Z$, $W$ be the foot of the altitude from point $E$ of $\triangle AEB$, $\triangle BEC$, $\triangle CED$, $\triangle DEA$.
Note that reflection of $E$ over all the points of $XYZW$ is similar to $XYZW$ with a scale of $2$ with center $E$. Thus, if $XYZW$ is cyclic, then the reflections are cyclic.
$\angle EWA$ is right angle and so is $\angle EXA$. Thus, $EXAW$ is cyclic with $EA$ being the diameter of the circumcircle.
Follow that, $\angle EWX\cong\angle EAX\cong \angle EAB$ because they inscribe the same angle.
Similarly $\angle EWZ\cong \angle EDC$, $\angle EYX\cong \angle EBA$, $\angle EYZ\cong \angle ECD$.
Futhermore, $m\angle XYZ+m\angle XWZ= m\angle EWX+m\angle EYX+m\angle EYZ+m\angle EWZ=$$360^\circ-m\angle CED-m\angle AEB=180^\circ$.
Thus, $\angle XYZ$ and $\angle XWZ$ are supplementary and follows that, $XYZW$ is cyclic.
$\mathbb{Q.E.D}$ | import olympiad; defaultpen(0.8pt+fontsize(12pt)); pair E; E=(0,0); label('$E$',E,N); pair A,B,C,D; A=(9,0); B=(0,13); C=(-13,0); D=(0,-11); draw(A--B--C--D--cycle,blue); label('$A$',A,E); label('$B$',B,N); label('$C$',C,W); label('$D$',D,S); pair T,R,S,Q; T=reflect(A, B)*E; R=reflect(C, B)*E; S=reflect(C, D)*E; Q=reflect(A, D)*E; pair W,X,Y,Z; W=extension(A,D,E,Q); X=extension(A,B,E,T); Y=extension(C,B,E,R); Z=extension(C,D,E,S); draw(W--X--Y--Z--cycle,red); label('$X$',X,NE); label('$Y$',Y,NW); label('$Z$',Z, SW); label('$W$',W,SE); dot(A); dot(B); dot(C); dot(D); dot(E); dot(W); dot(X); dot(Y); dot(Z); | [] |
26 | A convex hexagon $ABCDEF$ is inscribed in a circle such that $AB=CD=EF$ and diagonals $AD,BE$, and $CF$ are concurrent. Let $P$ be the intersection of $AD$ and $CE$. Prove that $\frac{CP}{PE}=(\frac{AC}{CE})^2$. | 1994 USAMO Problem 3 | Let the diagonals $AD$, $BE$, $CF$ meet at $Q$.
First, let's show that the triangles $\triangle AEC$ and $\triangle QED$ are similar.
$\angle ACE=\angle ADE$ because $A$,$C$,$D$ and $E$ all lie on the circle, and $\angle ADE=\angle QDE$. $\angle AEB=\angle CED$ because $AB=CD$, and $A$,$B$,$C$,$D$ and $E$ all lie on the circle. Then,
$\angle AEB=\angle CED \rightarrow \angle AEB+\angle BEC=\angle CED+\angle BEC \rightarrow \angle AEC=\angle QED$
Therefore, $\triangle AEC$ and $\triangle QED$ are similar, so $AC/CE=QD/DE$.
Next, let's show that $\triangle AEC$ and $\triangle CDQ$ are similar.
$\angle AEC=\angle ADC$ because $A$,$C$,$D$ and $E$ all lie on the circle, and $\angle ADC=\angle CDQ$. $\angle EAD=\angle ECD$ because $A$,$C$,$D$ and $E$ all lie on the circle. $\angle DAC=\angle ECF$ because $CD=EF$, and $A$,$C$,$D$,$E$ and $F$ all lie on the circle. Then,
$\angle EAC=\angle EAD+\angle DAC=\angle ECD+\angle ECF=\angle DCQ$
Therefore, $\triangle AEC$ and $\triangle CDQ$ are similar, so $AC/CE=CQ/QD$.
Lastly, let's show that $\triangle CPQ$ and $\triangle EPD$ are similar.
Because $CD=EF$ and $C$,$D$,$E$ and $F$ all lie on the circle, $CF$ is parallel to $DE$. So, $\triangle CPQ$ and $\triangle EPD$ are similar, and $CQ/DE=CP/PE$.
Putting it all together, $\frac{CP}{PE}=\frac{CQ}{DE}=\frac{AC}{CE}\cdot \frac{QD}{DE}=(\frac{AC}{CE})^2$.
Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol943.html | // Block 1
pair A,B,C,D,E,F,P,Q;
A=(-0.96,0.28);
B=(-0.352,0.936);
C=(0,1);
D=(4/5,3/5);
E=(4/5,-3/5);
F=(0,-1);
P=IntersectionPoint(A--D,C--E);
Q=IntersectionPoint(A--D,C--F);
draw(A--B);
draw(B--C);
draw(C--D);
draw(D--E,green);
draw(E--F);
draw(F--A);
draw(A--C,red);
draw(A--Q);
draw(A--E,red);
draw(B--Q);
draw(C--E,red);
draw(C--F);
draw(Q--E,green);
draw(Q--D,green);
draw(circle((0,0),1));
label("\(A\)",A,W);
label("\(B\)",B,N);
label("\(C\)",C,N);
label("\(D\)",D,NE);
label("\(E\)",E,SE);
label("\(F\)",F,S);
label("\(P\)",(0.3,0.8),S);
label("\(Q\)",(-0.15,0.4),S);
// Block 2
pair A,B,C,D,E,F,P,Q;
A=(-0.96,0.28);
B=(-0.352,0.936);
C=(0,1);
D=(4/5,3/5);
E=(4/5,-3/5);
F=(0,-1);
P=IntersectionPoint(A--D,C--E);
Q=IntersectionPoint(A--D,C--F);
draw(A--B);
draw(B--C);
draw(C--D,green);
draw(D--E);
draw(E--F);
draw(F--A);
draw(A--C,red);
draw(A--Q);
draw(A--E,red);
draw(B--E);
draw(C--E,red);
draw(Q--F);
draw(Q--C,green);
draw(Q--D,green);
draw(circle((0,0),1));
label("\(A\)",A,W);
label("\(B\)",B,N);
label("\(C\)",C,N);
label("\(D\)",D,NE);
label("\(E\)",E,SE);
label("\(F\)",F,S);
label("\(P\)",(0.3,0.8),S);
label("\(Q\)",(-0.15,0.4),S);
// Block 3
pair A,B,C,D,E,F,P,Q;
A=(-0.96,0.28);
B=(-0.352,0.936);
C=(0,1);
D=(4/5,3/5);
E=(4/5,-3/5);
F=(0,-1);
P=IntersectionPoint(A--D,C--E);
Q=IntersectionPoint(A--D,C--F);
draw(A--B);
draw(B--C);
draw(C--D);
draw(D--E,green);
draw(E--F);
draw(F--A);
draw(A--C);
draw(A--Q);
draw(A--E);
draw(B--E);
draw(P--E,green);
draw(Q--F);
draw(C--P,red);
draw(Q--P,red);
draw(C--Q,red);
draw(D--P,green);
draw(circle((0,0),1));
label("\(A\)",A,W);
label("\(B\)",B,N);
label("\(C\)",C,N);
label("\(D\)",D,NE);
label("\(E\)",E,SE);
label("\(F\)",F,S);
label("\(P\)",(0.3,0.8),S);
label("\(Q\)",(-0.15,0.4),S);
// Block 4
pair A,B,C,D,E,F,P,Q; A=(-0.96,0.28); B=(-0.352,0.936); C=(0,1); D=(4/5,3/5); E=(4/5,-3/5); F=(0,-1); P=IntersectionPoint(A--D,C--E); Q=IntersectionPoint(A--D,C--F); draw(A--B); draw(B--C); draw(C--D); draw(D--E,green); draw(E--F); draw(F--A); draw(A--C,red); draw(A--Q); draw(A--E,red); draw(B--Q); draw(C--E,red); draw(C--F); draw(Q--E,green); draw(Q--D,green); draw(circle((0,0),1)); label("\(A\)",A,W); label("\(B\)",B,N); label("\(C\)",C,N); label("\(D\)",D,NE); label("\(E\)",E,SE); label("\(F\)",F,S); label("\(P\)",(0.3,0.8),S); label("\(Q\)",(-0.15,0.4),S);
// Block 5
pair A,B,C,D,E,F,P,Q; A=(-0.96,0.28); B=(-0.352,0.936); C=(0,1); D=(4/5,3/5); E=(4/5,-3/5); F=(0,-1); P=IntersectionPoint(A--D,C--E); Q=IntersectionPoint(A--D,C--F); draw(A--B); draw(B--C); draw(C--D,green); draw(D--E); draw(E--F); draw(F--A); draw(A--C,red); draw(A--Q); draw(A--E,red); draw(B--E); draw(C--E,red); draw(Q--F); draw(Q--C,green); draw(Q--D,green); draw(circle((0,0),1)); label("\(A\)",A,W); label("\(B\)",B,N); label("\(C\)",C,N); label("\(D\)",D,NE); label("\(E\)",E,SE); label("\(F\)",F,S); label("\(P\)",(0.3,0.8),S); label("\(Q\)",(-0.15,0.4),S);
// Block 6
pair A,B,C,D,E,F,P,Q; A=(-0.96,0.28); B=(-0.352,0.936); C=(0,1); D=(4/5,3/5); E=(4/5,-3/5); F=(0,-1); P=IntersectionPoint(A--D,C--E); Q=IntersectionPoint(A--D,C--F); draw(A--B); draw(B--C); draw(C--D); draw(D--E,green); draw(E--F); draw(F--A); draw(A--C); draw(A--Q); draw(A--E); draw(B--E); draw(P--E,green); draw(Q--F); draw(C--P,red); draw(Q--P,red); draw(C--Q,red); draw(D--P,green); draw(circle((0,0),1)); label("\(A\)",A,W); label("\(B\)",B,N); label("\(C\)",C,N); label("\(D\)",D,NE); label("\(E\)",E,SE); label("\(F\)",F,S); label("\(P\)",(0.3,0.8),S); label("\(Q\)",(-0.15,0.4),S); | [] |
27 | Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$, $\angle MBA=20^\circ$ , $\angle MAC= 40^\circ$ and $\angle MCA=30^\circ$. Prove that the triangle is isosceles. | 1996 USAMO Problem 5 | Solution 1
Clearly, $\angle AMB = 150^\circ$ and $\angle AMC = 110^\circ$. Now by the Law of Sines on triangles $ABM$ and $ACM$, we have \[\frac{AB}{\sin 150^\circ} = \frac{AM}{\sin 20^\circ}\] and \[\frac{AC}{\sin 110^\circ} = \frac{AM}{\sin 30^\circ}.\] Combining these equations gives us \[\frac{AB}{AC} = \frac{\sin 150^\circ \sin 30^\circ}{\sin 20^\circ \sin 110^\circ}.\] Without loss of generality, let $AB = \sin 150^\circ \sin 30^\circ = \frac{1}{4}$ and $AC = \sin 20^\circ \sin 110^\circ$. Then by the Law of Cosines, we have
\begin{align*} BC^2 &= AB^2 + AC^2 - 2(AB)(BC)\cos\angle BAC\\ &= \frac{1}{16} + \sin^2 20^\circ\sin^2 110^\circ - 2\left(\frac{1}{4}\right)\sin 20^\circ\sin 110^\circ\cos 50^\circ \\ &= \frac{1}{16} + \sin^2 20^\circ \sin^2 110^\circ - \frac{1}{2}\sin 20^\circ\sin 110^\circ\sin 40^\circ \\ &= \frac{1}{16} + \sin^2 20^\circ \sin^2 110^\circ - \sin 20^\circ\sin 110^\circ\sin 20^\circ\cos 20^\circ \\ &= \frac{1}{16} \end{align*}
Thus, $AB = BC$, our desired conclusion.
Solution 2
By the law of sines, $\frac{BM}{sin(10^\circ)}=\frac{AM}{sin(20^\circ)}$ and $\frac{CM}{sin(40^\circ)}=\frac{AM}{sin(30^\circ)}$, so $\frac{BM}{CM}=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}$.
Let $\angle MBC=x$. Then, $\angle MCB=80^\circ-x$. By the law of sines, $\frac{BM}{CM}=\frac{sin(80^\circ-x)}{sin(x)}$.
Combining, we have $\frac{sin(80^\circ-x)}{sin(x)}=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}$. From here, we can use the given trigonometric identities at each step:
\begin{align*} \frac{sin(80^\circ-x)}{sin(x)}&=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}\\ sin(80^\circ-x)sin(20^\circ)sin(40^\circ)&=sin(10^\circ)sin(30^\circ)sin(x)\\ sin(80^\circ-x)sin(20^\circ)sin(40^\circ)&=\frac{1}{2}sin(10^\circ)sin(x)&[sin(30^\circ)=1/2]\\ sin(80^\circ-x)sin(30^\circ-10^\circ)sin(30^\circ+10^\circ)&=\frac{1}{2}sin(10^\circ)sin(x)\\ sin(80^\circ-x)(cos^2(10^\circ)-cos^2(30^\circ))&=\frac{1}{2}sin(10^\circ)sin(x)&[sin(A-B)sin(A+B)=cos^2 B-cos^2 A]\\ sin(80^\circ-x)(cos^2(10^\circ)-\frac{3}{4})&=\frac{1}{2}sin(10^\circ)sin(x)&[cos(30^\circ)=\frac{\sqrt{3}}{2}]\\ sin(80^\circ-x) \frac{4cos^3(10^\circ)-3cos(10^\circ)}{4cos(10^\circ)}&=\frac{1}{2}sin(10^\circ)sin(x)\\ sin(80^\circ-x) \frac{cos(30^\circ)}{4cos(10^\circ)}&=\frac{1}{2}sin(10^\circ)sin(x)&[cos(3A)=4cos^3 A-3cos A]\\ sin(80^\circ-x)cos(30^\circ)&=2sin(10^\circ)cos(10^\circ)sin(x)\\ sin(80^\circ-x)cos(30^\circ)&=sin(20^\circ)sin(x)&[sin(2A)=2sin A cos A ]\\ sin(80^\circ-x)sin(60^\circ)&=sin(20^\circ)sin(x)&[cos(30^\circ)=sin(60^\circ)]\\ \frac{1}{2}(cos(20^\circ-x)-cos(140^\circ-x))&=\frac{1}{2}(cos(20^\circ-x)-cos(20^\circ+x))&[sin A sin B=\frac{1}{2}(cos(A-B)-cos(A+B))]\\ cos(140^\circ-x)&=cos(20^\circ+x) \end{align*}
The only acute angle satisfying this equality is $x=60^\circ$. Therefore, $\angle ACB=80^\circ-x+30^\circ=50^\circ$ and $\angle BAC=10^\circ+40^\circ=50^\circ$. Thus, $\triangle ABC$ is isosceles.
Solution 3
If $\angle{MBC} = x$ then by Angle Sum in a Triangle we have $\angle{MCB} = 80^\circ - x$. By Trig Ceva we have
\[\sin 10^\circ \sin x \sin 30^\circ = \sin (80^\circ - x) \sin 40^\circ \sin 20^\circ.\]
Because $\dfrac{\sin x}{\sin (80^\circ - x)}$ is monotonic increasing over $(0, \dfrac{\pi}{2})$, there is only one solution $0 \le x \le \dfrac{\pi}{2}$ to the equation. We claim it is $x = 60^\circ$, which will make $ABC$ isosceles with $\angle{A} = \angle{C}$.
Notice that
\[\sin 20^\circ \sin 20^\circ \sin 40^\circ = 2 \sin 10^\circ \cos 10^\circ \sin 20^\circ \sin 40^\circ\]
\[= \sin 10^\circ (\sin 10^\circ + \frac{1}{2}) \sin 40^\circ\]
\[= \sin 10^\circ (\frac{1}{2} \sin 40^\circ + \frac{1}{2} (\cos 30^\circ - \cos 50^\circ))\]
\[= \sin 10^\circ \frac{1}{2} \cos 30^\circ\]
\[= \sin 10^\circ \sin 30^\circ \sin 60^\circ,\]
as desired. | pair A,B,C,M; A=(0,0); B=(1,2); C=(2,0); M=(0.8,1.1); draw(A--B); draw(B--C); draw(C--A); draw(A--M); draw(B--M); draw(C--M); label("\(A\)",A,SW); label("\(B\)",B,N); label("\(C\)",C,SE); label("\(M\)",M,NE); | [] |
28 | To clip a convex $n$-gon means to choose a pair of consecutive sides $AB, BC$ and to replace them by three segments $AM, MN,$ and $NC,$ where $M$ is the midpoint of $AB$ and $N$ is the midpoint of $BC$. In other words, one cuts off the triangle $MBN$ to obtain a convex $(n+1)$-gon. A regular hexagon $P_6$ of area $1$ is clipped to obtain a heptagon $P_7$. Then $P_7$ is clipped (in one of the seven possible ways) to obtain an octagon $P_8$, and so on. Prove that no matter how the clippings are done, the area of $P_n$ is greater than $\frac{1}{3}$, for all $n\ge6$. | 1997 USAMO Problem 4 | $\textbf{Claim:}$ It is impossible to choose two non-adjacent sides and clip a whole part of it off.
$\textbf{Proof:}$ If you clip adjacent sides, you can cut off at most up to the blue lines; Clipping more is impossible due to the degrees getting larger and larger and more and more circular.
Thus, after infinitely many clips, the hexagon bounded by the blue lines is left, so after finitely many clips, the area left is more than that hexagon.
The side length of that hexagon is $\frac{\sqrt{3}}{3}$ of the large one, because of 30-30-120 triangles. Thus, the area is $\frac13$ of the larger one, so we are done. | // Block 1
size(200);
draw((1, 0)--(0.5, 0.866)--(-0.5, 0.866)--(-1, 0)--(-0.5, -0.866)--(0.5, -0.866)--(1, 0));
draw((1, 0)--(-0.5, 0.866)--(-0.5, -0.866)--(1, 0), blue);
draw((-1, 0)--(0.5, -0.866)--(0.5, 0.866)--(-1, 0), blue);
// Block 2
size(200); draw((1, 0)--(0.5, 0.866)--(-0.5, 0.866)--(-1, 0)--(-0.5, -0.866)--(0.5, -0.866)--(1, 0)); draw((1, 0)--(-0.5, 0.866)--(-0.5, -0.866)--(1, 0), blue); draw((-1, 0)--(0.5, -0.866)--(0.5, 0.866)--(-1, 0), blue); | [] |
29 | Let ${\cal C}_1$ and ${\cal C}_2$ be concentric circles, with ${\cal C}_2$ in the interior of ${\cal C}_1$. From a point $A$ on ${\cal C}_1$ one draws the tangent $AB$ to ${\cal C}_2$ ($B\in {\cal C}_2$). Let $C$ be the second point of intersection of $AB$ and ${\cal C}_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects ${\cal C}_2$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$. | 1998 USAMO Problem 2 | First, $AD=\frac{AB}{2}=\frac{AC}{4}$. Because $E$,$F$ and $B$ all lie on a circle, $AE \cdot AF=AB \cdot AB=\frac{AB}{2} \cdot 2AB=AD \cdot AC$. Therefore, $\triangle ACF \sim \triangle AED$, so $\angle ACF = \angle AED$. Thus, quadrilateral $CFED$ is cyclic, and $M$ must be the center of the circumcircle of $CFED$, which implies that $MC=\frac{CD}{2}$. Putting it all together,
$\frac{AM}{MC}=\frac{AC-MC}{MC}=\frac{AC-\frac{CD}{2}}{\frac{CD}{2}}=\frac{AC-\frac{AC-AD}{2}}{\frac{AC-AD}{2}}=\frac{AC-\frac{3AC}{8}}{\frac{3AC}{8}}=\frac{\frac{5AC}{8}}{\frac{3AC}{8}}=\frac{5}{3}$
Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol982.html | // Block 1
pair O,A,B,C,D,E,F,DEb,CFb,Fo,M;
O=(0,0);
A=(1.732,1);
B=(0,1);
C=(-1.732,1);
D=(0.866,1);
Fo=(-1,-0.5);
path AC,AF,DE,CF,DEbM,CFbM,C1,C2;
C1=circle(O,2);
C2=circle(O,1);
E=intersectionpoints(A--Fo,C2)[0];
F=intersectionpoints(A--Fo,C2)[1];
DEb=((D.x+E.x)/2.0,(D.y+E.y)/2.0);
CFb=((C.x+F.x)/2.0,(C.y+F.y)/2.0);
M=(-0.433,1);
path AC=A--C;
path AF=A--F;
path DEbM=DEb--M;
path CFbM=CFb--M;
path DE=D--E;
path CF=C--F;
draw(AC);
draw(AF);
draw(DE);
draw(CF);
draw(DEbM);
draw(CFbM);
draw(C1);
draw(C2);
label("\(A\)",A,NE);
label("\(B\)",B,N);
label("\(C\)",C,NW);
label("\(D\)",D,N);
label("\(E\)",E,SE);
label("\(F\)",F,SW);
label("\(M\)",M,N);
// Block 2
pair O,A,B,C,D,E,F,DEb,CFb,Fo,M; O=(0,0); A=(1.732,1); B=(0,1); C=(-1.732,1); D=(0.866,1); Fo=(-1,-0.5); path AC,AF,DE,CF,DEbM,CFbM,C1,C2; C1=circle(O,2); C2=circle(O,1); E=intersectionpoints(A--Fo,C2)[0]; F=intersectionpoints(A--Fo,C2)[1]; DEb=((D.x+E.x)/2.0,(D.y+E.y)/2.0); CFb=((C.x+F.x)/2.0,(C.y+F.y)/2.0); M=(-0.433,1); path AC=A--C; path AF=A--F; path DEbM=DEb--M; path CFbM=CFb--M; path DE=D--E; path CF=C--F; draw(AC); draw(AF); draw(DE); draw(CF); draw(DEbM); draw(CFbM); draw(C1); draw(C2); label("\(A\)",A,NE); label("\(B\)",B,N); label("\(C\)",C,NW); label("\(D\)",D,N); label("\(E\)",E,SE); label("\(F\)",F,SW); label("\(M\)",M,N); | [] |
30 | Let $n \geq 5$ be an integer. Find the largest integer $k$ (as a function of $n$) such that there exists a convex $n$-gon $A_{1}A_{2}\dots A_{n}$ for which exactly $k$ of the quadrilaterals $A_{i}A_{i+1}A_{i+2}A_{i+3}$ have an inscribed circle. (Here $A_{n+j} = A_{j}$.) | 1998 USAMO Problem 6 | Lemma: If quadrilaterals $A_iA_{i+1}A_{i+2}A_{i+3}$ and $A_{i+2}A_{i+3}A_{i+4}A_{i+5}$ in an equiangular $n$-gon are tangential, and $A_iA_{i+3}$ is the longest side quadrilateral $A_iA_{i+1}A_{i+2}A_{i+3}$ for all $i$, then quadrilateral $A_{i+1}A_{i+2}A_{i+3}A_{i+4}$ is not tangential.
Proof:
If quadrilaterals $A_iA_{i+1}A_{i+2}A_{i+3}$ and $A_{i+2}A_{i+3}A_{i+4}A_{i+5}$ are tangential, then $A_iA_{i+3}$ must have side length of $j+y-u$, and $A_{i+2}A_{i+5}$ must have side length of $h + y - n$ (One can see this from what is known as walk-around). Suppose quadrilateral $A_{i+1}A_{i+2}A_{i+3}A_{i+4}$ is tangential. Then, again, we see that $A_{i+1}A_{i+4}$ must have side length $u + n - y$. We assumed by lemma that $A_iA_{i+3} > A_{i}A_{i+1}$ for all $i$, so we have $A_iA_{i+3} > j$, $A_{i+1}A_{i+4} > y$, and $A_{i+2}A_{i+5} > h$. If we add up the side lengths $A_iA_{i+3} + A_{i+1}A_{i+4} + A_{i+2}A_{i+5}$, we get:
\[A_iA_{i+3} + A_{i+1}A_{i+4} + A_{i+2}A_{i+5} = j + y - u + h + y - n + u + n - y\]
\[A_iA_{i+3} + A_{i+1}A_{i+4} + A_{i+2}A_{i+5} = j + h + y\]
However, by the lemma, we assumed that $A_iA_{i+3} > j$, $A_{i+1}A_{i+4} > y$, and $A_{i+2}A_{i+5} > h$. Adding these up, we get:
\[A_iA_{i+3} + A_{i+1}A_{i+4} + A_{i+2}A_{i+5} > j + h + y,\]
which is a contradiction. Thus, quadrilateral $A_{i+1}A_{i+2}A_{i+3}A_{i+4}$ is not tangential, proving the lemma.
By lemma, the maximum number of quadrilaterals in a $n$-gon occurs when the tangential quadrilaterals alternate, giving us $k = \lfloor \frac{n}{2} \rfloor$.
Note that one can find the ratio of side of an equiangular $n$-gon in order for alternating quadrilaterals to be tangential.
Since exterior angles of a equiangular $n$-gon have degree measure $\frac{2pi}{n}$, one can write the equation:
\[2x = y + y + 2x \cos \frac{2pi}{n}\]
\[y = x \left( 1- \cos \frac{2pi}{n} \right)\]
\[\frac{y}{x} = \frac{1}{1- \cos \frac{2pi}{n}}\]
Thus, we can find the ratio of sides of an equiangular $n$-gon which fits the maximum to be $1 : 1- \cos \frac{2\pi}{n}$. Note that if $n$ is even, we can easily alternate them, but if $n$ is odd, we must have two adjacent sides be the same length, and that length must be the larger side in the ratio of adjacent sides. The proof is left as an exercise for the reader. | // Block 1
import geometry;
size(10cm);
pair A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U;
A = (-1,0);
B = (1,0);
draw(Circle(A,1)^^Circle(B,1));
C = (sqrt(2)/2-1,sqrt(2)/2);
D = (-sqrt(3)/2 - 1, .5);
E = (-sqrt(3)/2 - 1, -.5);
F = (-1,-1);
G = (1,-1);
H = (sqrt(3)/2 + 1, -.5);
I = (sqrt(3)/2 + 1, .5);
J = (1-sqrt(2)/2, sqrt(2)/2);
K = (-1-2/sqrt(3), 0);
L = extension(K,E,F,G);
M = (1+2/sqrt(3), 0);
N = extension(M,H,F,G);
O = extension(K,D,C,N);
P = extension(M,I,L,J);
Q = midpoint(F--G);
R = midpoint(K--O);
S = midpoint(P--M);
T = midpoint(O--C);
U = midpoint(J--P);
draw(O--K--L--N--M--P--L^^K--M^^O--N);
label("$A_i$", O, NW);
label("$A_{i+1}$", K, W);
label("$A_{i+2}$", L, SW);
label("$A_{i+3}$", N, SE);
label("$A_{i+4}$", M, dir(0));
label("$A_{i+5}$", P, NE);
label("$j$", R, W);
label("$u$", E, SW);
label("$y$", Q, S);
label("$n$", H, SE);
label("$h$", S, NE);
label("$j + y - u$", T, NE);
label("$h + y - n$", U, SW);
// Block 2
import geometry;
size(10cm);
pair A, B, C, D, E, F;
A = (0,0);
B = (1,0);
C = (1+(1-cos(2pi/9))*cos(2pi/9), (1-cos(2pi/9))*sin(2pi/9));
D = (-(1-cos(2pi/9))*cos(2pi/9), (1-cos(2pi/9))*sin(2pi/9));
E = midpoint(D--A);
F = midpoint(A--B);
draw(A--B--C--D--A);
label("$A_i$", D, NW);
label("$A_{i+1}$", A, SW);
label("$A_{i+2}$", B, SE);
label("$A_{i+3}$", C, NE);
label("$x$", E, W);
label("$y$", F, S);
// Block 3
import geometry; size(10cm); pair A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U; A = (-1,0); B = (1,0); draw(Circle(A,1)^^Circle(B,1)); C = (sqrt(2)/2-1,sqrt(2)/2); D = (-sqrt(3)/2 - 1, .5); E = (-sqrt(3)/2 - 1, -.5); F = (-1,-1); G = (1,-1); H = (sqrt(3)/2 + 1, -.5); I = (sqrt(3)/2 + 1, .5); J = (1-sqrt(2)/2, sqrt(2)/2); K = (-1-2/sqrt(3), 0); L = extension(K,E,F,G); M = (1+2/sqrt(3), 0); N = extension(M,H,F,G); O = extension(K,D,C,N); P = extension(M,I,L,J); Q = midpoint(F--G); R = midpoint(K--O); S = midpoint(P--M); T = midpoint(O--C); U = midpoint(J--P); draw(O--K--L--N--M--P--L^^K--M^^O--N); label("$A_i$", O, NW); label("$A_{i+1}$", K, W); label("$A_{i+2}$", L, SW); label("$A_{i+3}$", N, SE); label("$A_{i+4}$", M, dir(0)); label("$A_{i+5}$", P, NE); label("$j$", R, W); label("$u$", E, SW); label("$y$", Q, S); label("$n$", H, SE); label("$h$", S, NE); label("$j + y - u$", T, NE); label("$h + y - n$", U, SW);
// Block 4
import geometry; size(10cm); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (1+(1-cos(2pi/9))*cos(2pi/9), (1-cos(2pi/9))*sin(2pi/9)); D = (-(1-cos(2pi/9))*cos(2pi/9), (1-cos(2pi/9))*sin(2pi/9)); E = midpoint(D--A); F = midpoint(A--B); draw(A--B--C--D--A); label("$A_i$", D, NW); label("$A_{i+1}$", A, SW); label("$A_{i+2}$", B, SE); label("$A_{i+3}$", C, NE); label("$x$", E, W); label("$y$", F, S); | [] |
31 | Find the smallest positive integer $n$ such that if $n$ squares of a $1000 \times 1000$ chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board. | 2000 USAMO Problem 4 | We claim that $n = 1999$ is the smallest such number. For $n \le 1998$, we can simply color any of the $1998$ squares forming the top row and the left column, but excluding the top left corner square.
We now show that no configuration with no colored right triangles exists for $n = 1999$. We call a row or column filled if all $1000$ of its squares are colored. Then any of the remaining $999$ colored squares must share a column or row, respectively, with one of the colored squares in a filled row or column. These two squares, and any other square in the filled row or column, form a colored right triangle, giving us a contradiction. Hence, no filled row or column may exist.
Let $m$ be the number of columns with $1$ colored square. Then there are $1999-m$ colored squares in the remaining columns, and in each of these $< 1999-m$ columns that have a colored square must have at least two colored squares in them. These two colored squares will form a triangle with any other colored square in either of the rows containing the colored squares. Hence, each of the $1999-m$ colored squares must be placed in different rows, but as there are only $1000$ rows, the inequality $1999 - m \le 1000 \Longrightarrow m \ge 999$ holds. If $m = 1000$, then each column only has $1$ colored square, leaving no place for the remaining $999$, contradiction. If $m = 999$, then each of the $1000$ rows has $1$ black square, leaving no place for the other $999$, contradiction. Hence $n = \boxed{1999}$ is the minimal value. | // Block 1
for(int i = 0; i < 10; ++i){
for(int j = 0; j < 10; ++j){
if((i == 0 || j == 9) && !(j-i == 9)) fill(shift(i,j)*unitsquare,rgb(0.3,0.3,0.3));
else draw(shift(i,j)*unitsquare);
}
}
// Block 2
for(int i = 0; i < 10; ++i){ for(int j = 0; j < 10; ++j){ if((i == 0 || j == 9) && !(j-i == 9)) fill(shift(i,j)*unitsquare,rgb(0.3,0.3,0.3)); else draw(shift(i,j)*unitsquare); } } | [] |
32 | Let $A_1A_2A_3$ be a triangle and let $\omega_1$ be a circle in its plane passing through $A_1$ and $A_2.$ Suppose there exist circles $\omega_2, \omega_3, \dots, \omega_7$ such that for $k = 2, 3, \dots, 7,$ $\omega_k$ is externally tangent to $\omega_{k - 1}$ and passes through $A_k$ and $A_{k + 1},$ where $A_{n + 3} = A_{n}$ for all $n \ge 1$. Prove that $\omega_7 = \omega_1.$ | 2000 USAMO Problem 5 | Solution 1
Let the circumcenter of $\triangle ABC$ be $O$, and let the center of $\omega_k$ be $O_k$. $\omega_k$ and $\omega_{k-1}$ are externally tangent at the point $A_k$, so $O_k, A_k, O_{k-1}$ are collinear.
$O$ is the intersection of the perpendicular bisectors of $\overline{A_1A_2}, \overline{A_2A_3}, \overline{A_3A_1}$, and each of the centers $O_k$ lie on the perpendicular bisector of the side of the triangle that determines $\omega_k$. It follows from $OA_k = OA_{k+1}, O_kA_k = O_kA_{k+1}, OO_k = OO_k \Longrightarrow \triangle OA_kO_k \cong \triangle OA_{k+1}O_k$ that $\angle OA_kO_k = \angle OA_{k+1}O_k$.
Since $O, A_k$, and the perpendicular bisector of $\overline{A_kA_{k+1}}$ are fixed, the angle $OA_kO_k$ determines the position of $O_k$ (since $O_k$ lies on the perpendicular bisector). Let $\theta_k = m\angle OA_kO_k$; then, $\theta_i = \theta_j$ and $i \equiv j \pmod{3}$ together imply that $O_i \equiv O_j$.
Now $\theta_1 = \angle OA_1O_1 = \angle OA_2O_1 = 180 - \angle OA_2O_2 = 180 - \theta_2$ (due to collinearility). Hence, we have the recursion $\theta_k = 180 - \theta_{k-1}$, and so $\theta_k = \theta_{k-2}$. Thus, $\theta_{1} = \theta_{7}$.
$\theta_{1} = \theta_{7}$ implies that $O_1 \equiv O_7$, and circles $\omega_1$ and $\omega_7$ are the same circle since they have the same center and go through the same two points.
Solution 2
Using the collinearity of certain points and the fact that $A_k A_{k+1} O_k$ is isosceles, we quickly deduce that
\[\angle{A_1 A_2 O_1} = 180^\circ - \angle{A_2} - (180^\circ - \angle{A_3}) + (180^\circ - \angle{A_1}) - (180^\circ - \angle{A_2}) + (180^\circ - \angle{A_3}) - (180^\circ - \angle{A_1}) + \angle{A_7 A_8 O_7}.\]
From ASA Congruence we deduce that $A_1 A_2 O_1$ and $A_7 A_8 O_7$ are congruent triangles, and so $O_1 A_1 = O_7 A_7$, that is $\omega_1 = \omega_7$. | size(300); pathpen = linewidth(0.7); pen t = linetype("2 2"); pair A = (0,0), B=3*expi(1), C=(3.5)*expi(0); /* arbitrary points */ pair O=circumcenter(A,B,C), O1 = O + 5*( ((B+C)/2) - O ), O2 = IP(O -- O + 100*( ((A+C)/2) - O ), O1 -- O1 + 10*( C - O1 )); D(MP("A_3",A,SW)--MP("A_1",B,N)--MP("A_2",C,SE)--cycle); D(MP("O",O,NW)); D(MP("O_1",O1,E)); D(MP("O_2",O2)); D(O--B--O1--C--O--A--O2--C, linetype("4 4") + linewidth(0.7)); D(O1--O--O2,linetype("4 4") + linewidth(0.6)); D(CP(O1,C),t);D(CP(O2,C),t);D(circumcircle(A,B,C),t); | [] |
33 | Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2 = BD_1$ and $CE_2 = AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ = D_2P$. | 2001 USAMO Problem 2 | Solution 1
It is well known that the excircle opposite $A$ is tangent to $\overline{BC}$ at the point $D_2$. (Proof: let the points of tangency of the excircle with the lines $BC, AB, AC$ be $D_3, F,G$ respectively. Then $AB+BD_3=AB + BF=AF = AG = AC + CG=AC + CD_3$. It follows that $2CD_3 = AB + BC - AC$, and $CD_3 = s-b = BD_1 = CD_2$, so $D_3 \equiv D_2$.)
Now consider the homothety that carries the incircle of $\triangle ABC$ to its excircle. The homothety also carries $Q$ to $D_2$ (since $A,Q,D_2$ are collinear), and carries the tangency points $E_1$ to $G$. It follows that $\frac{AQ}{QD_2} = \frac{AE_1}{E_1G} = \frac{s-a}{E_1C + CD_2} = \frac{s-a}{CD_1 + BD_1} = \frac{s-a}{a}$.
By Menelaus' Theorem on $\triangle ACD_2$ with segment $\overline{BE_2}$, it follows that $\frac{CE_2}{E_2A} \cdot \frac{AP}{PD_2} \cdot \frac{BD_2}{BC} = 1 \Longrightarrow \frac{AP}{PD_2} = \frac{(b - (s-a)) \cdot a}{(a-(s-b)) \cdot AE_1} = \frac{a}{s-a}$. It easily follows that $AQ = D_2P$. $\blacksquare$
Solution 2
The key observation is the following lemma.
Lemma: Segment $D_1Q$ is a diameter of circle $\omega$.
Proof: Let $I$ be the center of circle $\omega$, i.e., $I$ is the incenter of triangle $ABC$. Extend segment $D_1I$ through $I$ to intersect circle $\omega$ again at $Q'$, and extend segment $AQ'$ through $Q'$ to intersect segment $BC$ at $D'$. We show that $D_2 = D'$, which in turn implies that $Q = Q'$, that is, $D_1Q$ is a diameter of $\omega$.
Let $l$ be the line tangent to circle $\omega$ at $Q'$, and let $l$ intersect the segments $AB$ and $AC$ at $B_1$ and $C_1$, respectively. Then $\omega$ is an excircle of triangle $AB_1C_1$. Let $\mathbf{H}_1$ denote the dilation with its center at $A$ and ratio $AD'/AQ'$. Since $l\perp D_1Q'$ and $BC\perp D_1Q'$, $l\parallel BC$. Hence $AB/AB_1 = AC/AC_1 = AD'/AQ'$. Thus $\mathbf{H}_1(Q') = D'$, $\mathbf{H}_1(B_1) = B$, and $\mathbf{H}_1(C_1) = C$. It also follows that an excircle $\Omega$ of triangle $ABC$ is tangent to the side $BC$ at $D'$.
It is well known that \[CD_1 = \frac{1}{2}(BC + CA - AB).\] We compute $BD'$. Let $X$ and $Y$ denote the points of tangency of circle $\Omega$ with rays $AB$ and $AC$, respectively. Then by equal tangents, $AX = AY$, $BD' = BX$, and $D'C = YC$. Hence \[AX = AY = \frac{1}{2}(AX + AY) = \frac{1}{2}(AB + BX + YC + CA) = \frac{1}{2}(AB + BC + CA).\] It follows that \[BD' = BX = AX - AB = \frac{1}{2}(BC + CA - AB).\] Combining these two equations yields $BD' = CD_1$. Thus \[BD_2 = BD_1 - D_2D_1 = D_2C - D_2D_1 = CD_1 = BD',\] that is, $D' = D_2$, as desired. $\blacksquare$
Now we prove our main result. Let $M_1$ and $M_2$ be the respective midpoints of segments $BC$ and $CA$. Then $M_1$ is also the midpoint of segment $D_1D_2$, from which it follows that $IM_1$ is the midline of triangle $D_1QD_2$. Hence \[QD_2 = 2IM_1\] and $AD_2\parallel M_1I$. Similarly, we can prove that $BE_2\parallel M_2I$.
2001usamo2-2.png
Let $G$ be the centroid of triangle $ABC$. Thus segments $AM_1$ and $BM_2$ intersect at $G$. Define transformation $\mathbf{H}_2$ as the dilation with its center at $G$ and ratio $-1/2$. Then $\mathbf{H}_2(A) = M_1$ and $\mathbf{H}_2(B) = M_2$. Under the dilation, parallel lines go to parallel lines and the intersection of two lines goes to the intersection of their images. Since $AD_2\parallel M_1I$ and $BE_2\parallel M_2I$, $\mathbf{H}_2$ maps lines $AD_2$ and $BE_2$ to lines $M_1I$ and $M_2I$, respectively. It also follows that $\mathbf{H}_2(I) = P$ and \[\frac{IM_1}{AP} = \frac{GM_1}{AG} = \frac{1}{2}\] or \[AP = 2IM_1.\] This yields \[AQ = AP - QP = 2IM_1 - QP = QD_2 - QP = PD_2,\] as desired.
Note: We used directed lengths in our calculations to avoid possible complications caused by the different shapes of triangle $ABC$.
Solution 3
Here is a rather nice solution using barycentric coordinates:
Let $A$ be $(1,0,0)$, $B$ be $(0,1,0)$, and $C$ be $(0,0,1)$. Let the side lengths of the triangle be $a,b,c$ and the semi-perimeter $s$.
Now, \[CD_1=s-c, BD_1=s-b, AE_1=s-a, CE_1=s-c.\] Thus, \[CD_2=s-b, BD_2=s-c, AE_2=s-c, CE_2=s-a.\]
Therefore, $D_2=(0:s-b:s-c)$ and $E_2=(s-a:0:s-c).$ Clearly then, the non-normalized coordinates of $P=(s-a:s-b:s-c).$
Normalizing, we have that \[D_2=\left(0,\frac{s-b}{a},\frac{s-c}{a}\right), E_2=\left(\frac{s-a}{b},0,\frac{s-c}{b}\right), P=\left(\frac{s-a}{s},\frac{s-b}{s},\frac{s-c}{s}\right).\]
Now, we find the point $Q'$ inside the triangle on the line $AD_2$ such that $AQ'=D_2P$. It is then sufficient to show that this point lies on the incircle.
$P$ is the fraction $\frac{s-a}{s}$ of the way "up" the line segment from $D_2$ to $A$. Thus, we are looking for the point that is $\frac{s-a}{s}$ of the way "down" the line segment from $A$ to $D_2$, or, the fraction $1-\frac{s-a}{s}$ of the way "up".
Thus, $Q'$ has normalized $x$-coordinate $1-\frac{s-a}{s}=\frac{a}{s}$.
As the line $AD_2$ has equation $(s-c)y=(s-b)z$, it can easily be found that \[Q'=\left(\frac{a^2}{as}, \frac{(s-a)(s-b)}{as}, \frac{(s-a)(s-c)}{as}\right)=(a^2:(s-a)(s-b):(s-a)(s-c)).\]
Recalling that the equation of the incircle is \[a^2yz+b^2xz+c^2xy+(x+y+z)[(s-a)^2x+(s-b)^2y+(s-c)^2z]=0.\] We must show that this equation is true for $Q'$'s values of $x,y,z$.
Plugging in our values, this means showing that
\[a^2(s-a)^2(s-b)(s-c)+a^2(s-a)[b^2(s-c)+c^2(s-b)]+as[a^2(s-a)^2+(s-a)(s-b)^3+(s-a)(s-c)^3]=0.\]
Dividing by $a(s-a)$, this is just
\[a(s-a)(s-b)(s-c)+a[b^2(s-c)+c^2(s-b)]+a^2s(s-a)+s(s-b)^3+s(s-c)^3=0.\]
Plugging in the value of $s:$
\[\frac{a(-a+b+c)(a-b+c)(a+b-c)}{8}+\frac{ab^2(a+b-c)}{2}+\frac{ac^2(a-b+c)}{2}+\frac{a^2(a+b+c)(-a+b+c)}{4}+\frac{(a+b+c)(a-b+c)^3}{16}+\frac{(a+b+c)(a+b-c)^3}{16}=0.\]
\[2a(-a+b+c)(a-b+c)(a+b-c)+8ab^2(a+b-c)+8ac^2(a-b+c)+4a^2(a+b+c)(-a+b+c)+(a+b+c)(a-b+c)^3+(a+b+c)(a+b-c)^3=0\]
\[2a[(-a+b+c)(a-b+c)(a+b-c)+4b^2(a+b-c)+4c^2(a-b+c)]+(a+b+c)[4a^2(-a+b+c)+(a-b+c)^3+(a+b-c)^3]=0\]
The first bracket is just
\[-a^3-b^3-c^3+a^2b+a^2c+b^2c+b^2a+c^2a+c^2b-2abc+4ab^2+4b^3-4b^2c+4ac^2-4bc^2+4c^3\]
\[=-a^3+3b^3+3c^3+a^2b+a^2c-3b^2c+5b^2a+5c^2a-3c^2b-2abc\]
and the second bracket is \[-2a^3+4a^2b+4a^2c+6ab^2+6ac^2-12abc.\]
Dividing everything by $2a$ gives
\[-a^3+3b^3+3c^3+a^2b+a^2c-3b^2c+5b^2a+5c^2a-3c^2b-2abc+(a+b+c)(-a^2+2ab+2ac+3b^2+3c^2-6bc),\] which is $0$, as desired.
As $Q'$ lies on the incircle and $AD_2$, $Q'=Q$, and our proof is complete.
Solution 4
We again use Barycentric coordinates. As before, let $A$ be $(1,0,0)$, $B$ be $(0,1,0)$, and $C$ be $(0,0,1)$. Also
\[D_{1}=\left( 0,\frac{s-c}{a},\frac{s-b}{a} \right), D_2=\left(0,\frac{s-b}{a},\frac{s-c}{a}\right), E_2=\left(\frac{s-a}{b},0,\frac{s-c}{b}\right), P=\left(\frac{s-a}{s},\frac{s-b}{s},\frac{s-c}{s}\right).\]
Now, consider a point $Q'$ for which $AQ'=PD_{2}$. Then working component-vise, we get $Q'+P=A+D_{2}$ from which we can easily get the coordinates of $Q'$ as;
\[Q'=\left(\frac{a}{s},\frac{(s-a)(s-b)}{sa},\frac{(s-a)(s-c)}{sa}\right)\]
It suffices to show that $Q'=Q$.
Let $I=\left(\frac{a}{2s},\frac{b}{2s},\frac{c}{2s}\right)$ be the incenter of triangle $ABC$. We claim that $I$ is the midpoint of $Q'D_{1}$. Indeed,
\[\frac{1}{2}\left(0+\frac{a}{s}\right)=\frac{a}{2s}\]
\[\frac{1}{2}\left(\frac{(s-a)(s-b)}{sa}+\frac{s-c}{a}\right)=\frac{(s-a)(s-b)+s(s-c)}{2sa}=\frac{ab}{2sa}=\frac{b}{2s}\]
\[\frac{1}{2}\left(\frac{(s-a)(s-c)}{sa}+\frac{s-b}{a}\right)=\frac{c}{2s}\]
Hence the claim has been proved.
Since $I$ is the center of $\omega$ and the midpoint of $Q'D_{1}$, thus $Q'$ is the point diametrically opposite to $D_{1}$, and hence it lies on $\omega$ and closer to $A$ off the 2 points. Thus $Q=Q'$ as desired. | pathpen = linewidth(0.7); size(300); pen d = linetype("4 4") + linewidth(0.6); pair B=(0,0), C=(10,0), A=7*expi(1),O=D(incenter(A,B,C)),D1 = D(MP("D_1",foot(O,B,C))),E1 = D(MP("E_1",foot(O,A,C),NE)),E2 = D(MP("E_2",C+A-E1,NE)); /* arbitrary points */ /* ugly construction for OA */ pair Ca = 2C-A, Cb = bisectorpoint(Ca,C,B), OA = IP(A--A+10*(O-A),C--C+50*(Cb-C)), D2 = D(MP("D_2",foot(OA,B,C))), Fa=2B-A, Ga=2C-A, F=MP("F",D(foot(OA,B,Fa)),NW), G=MP("G",D(foot(OA,C,Ga)),NE); D(OA); D(MP("A",A,N)--MP("B",B,NW)--MP("C",C,NE)--cycle); D(incircle(A,B,C)); D(CP(OA,D2),d); D(B--Fa,linewidth(0.6)); D(C--Ga,linewidth(0.6)); D(MP("P",IP(D(A--D2),D(B--E2)),NNE)); D(MP("Q",IP(incircle(A,B,C),A--D2),SW)); clip((-20,-10)--(-20,20)--(20,20)--(20,-10)--cycle); | [] |
34 | Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$, respectively. Lines $AB$ and $DE$ intersect at $F$, while lines $BD$ and $CF$ intersect at $M$. Prove that $MF = MC$ if and only if $MB\cdot MD = MC^2$. | 2003 USAMO Problem 4 | Solution 1
Extend segment $DM$ through $M$ to $G$ such that $FG\parallel CD$.
Then $MF = MC$ if and only if quadrilateral $CDFG$ is a parallelogram, or, $FD\parallel CG$. Hence $MC = MF$ if and only if $\angle GCD = \angle FDA$, that is, $\angle FDA + \angle CGF = 180^\circ$.
Because quadrilateral $ABED$ is cyclic, $\angle FDA = \angle ABE$. It follows that $MC = MF$ if and only if
\[180^\circ = \angle FDA + \angle CGF = \angle ABE + \angle CGF,\]
that is, quadrilateral $CBFG$ is cyclic, which is equivalent to
\[\angle CBM = \angle CBG = \angle CFG = \angle DCF = \angle DCM.\]
Because $\angle DMC = \angle CMB$, $\angle CBM = \angle DCM$ if and only if triangles $BCM$ and $CDM$ are similar, that is
\[\frac{CM}{BM} = \frac{DM}{CM},\]
or $MB\cdot MD = MC^2$.
Solution 2
We first assume that $MB\cdot MD = MC^2$. Because $\frac{MC}{MD} = \frac{MB}{MC}$ and $\angle CMD = \angle BMC$, triangles $CMD$ and $BMC$ are similar. Consequently, $\angle MCD = \angle MBC$. This exact condition can also be visualized through power of M with respect to the circumcircle of triangle $BDC$ and getting the angle condition from the alternate segment theorem.
Because quadrilateral $ABED$ is cyclic, $\angle DAE = \angle DBE$. Hence
\[\angle FCA = \angle MCD = \angle MBC = \angle DBE = \angle DAE = \angle CAE,\]
implying that $AE\parallel CF$, so $\angle AEF = \angle CFE$. Because quadrilateral $ABED$ is cyclic, $\angle ABD = \angle AED$. Hence
\[\angle FBM = \angle ABD = \angle AED = \angle AEF = \angle CFE = \angle MFD.\]
Because $\angle FBM = \angle DFM$ and $\angle FMB = \angle DMF$, triangles $BFM$ and $FDM$ are similar. Consequently, $\frac{FM}{DM} = \frac{BM}{FM}$, or $FM^2 = BM\cdot DM = CM^2$. Therefore $MC^2 = MB\cdot MD$ implies $MC = MF$.
Now we assume that $MC = MF$. Applying Ceva's Theorem to triangle $BCF$ and cevians $BM, CA, FE$ gives
\[\frac{BA}{AF}\cdot\frac{FM}{MC}\cdot\frac{CE}{EB} = 1,\]
implying that $\frac{BA}{AF} = \frac{BE}{EC}$, so $AE\parallel CF$.
Consequently, $\angle DCM = \angle DAE$. Because quadrilateral $ABED$ is cyclic, $\angle DAE = \angle DBE$. Hence
\[\angle DCM = \angle DAE = \angle DBE = \angle CBM.\]
Because $\angle CBM = \angle DCM$ and $\angle CMB = \angle DMC$, triangles $BCM$ and $CDM$ are similar. Consequently, $\frac{CM}{DM} = \frac{BM}{CM}$, or $CM^2 = BM\cdot DM$.
Combining the above, we conclude that $MF = MC$ if and only if $MB\cdot MD = MC^2$.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page. | // Block 1
defaultpen(fontsize(10)+0.6); size(250); var theta=22, r=0.58; pair B=origin, A=dir(theta), C=A+(rotate(78)*0.8*A), O=IP(CR(B,r),CR(A,r)); path c=CR(O,r); pair D=IP(c,A--C), E=IP(c,B--C), F=extension(A,B,D,E), M=extension(B,D,C,F), G=extension(D,M,F,F+C-D); draw(A--B--C--A^^E--F--C^^A--F^^B--M^^E--M); draw(c); draw(F--G--C^^M--G,gray+0.4); dot("$A$",A,dir(F-E)); dot("$B$",B,2*dir(B-A)); dot("$C$",C,1.5*dir(C-A)); dot("$D$",D,2.5*dir(250)); dot("$E$",E,2.5*dir(C-A)); dot("$F$",F,dir(F-E)); dot("$M$",M,2.5*dir(255)); dot("$G$",G,dir(G-M));
// Block 2
defaultpen(fontsize(10)+0.6); size(250); var theta=22, r=0.58; pair B=origin, A=dir(theta), C=A+(rotate(78)*0.8*A), O=IP(CR(B,r),CR(A,r)); path c=CR(O,r); pair D=IP(c,A--C), E=IP(c,B--C), F=extension(A,B,D,E), M=extension(B,D,C,F), G=extension(D,M,F,F+C-D); draw(A--B--C--A^^E--F--C^^A--F^^B--M^^E--M); draw(c); draw(A--E,gray+0.4); dot("$A$",A,dir(F-E)); dot("$B$",B,2*dir(B-A)); dot("$C$",C,1.5*dir(C-A)); dot("$D$",D,2.5*dir(250)); dot("$E$",E,2.5*dir(C-A)); dot("$F$",F,dir(F-E)); dot("$M$",M,2.5*dir(255)); | [] |
35 | At the vertices of a regular hexagon are written six nonnegative integers whose sum is 2003. Bert is allowed to make moves of the following form: he may pick a vertex and replace the number written there by the absolute value of the difference between the numbers written at the two neighboring vertices. Prove that Bert can make a sequence of moves, after which the number 0 appears at all six vertices. | 2003 USAMO Problem 6 | Assume the original numbers are $a,b,c,d,e,f$. Since $a+b+c+d+e+f$ is odd, either $a+c+e$ or $b+d+f$ must be odd. WLOG let $a+c+e$ be odd and $a\ge c\ge e \ge 0$.
Case 1
$a,c,e>0$. Define Operation A as the sequence of moves from Step 1 to Step 3, shown below:
Notice that Operation A changes the numbers $a,c,e$ to $c-e,c,a-c$ and they are all nonnegative, since $a\ge c\ge e$. Their sum changes from $a+c+e$ to $a+c-e$; it decreases as long as $e\ne 0$. If we repeat Operation A enough times, its sum will decrease and eventually we will arrive at a point where at least one of the numbers in the positions originally occupied by $a,c,e$ has become a 0.
Case 2
$a,c>0$ and $e=0$. Define Operation B as the sequence of moves from Step 1 to Step 3, shown below:
where in Step 3, we take the nonnegative choice of $2c-a$ or $a-2c$. $a,c,0$ is changed to either $c,2c-a,0$ or $c,a-2c,0$. If we have $c,2c-a,0$, their sum is $3c-a$ and this is less than $a+c+0$ (the original sum) unless $a=c$, but $a\ne c$ since the original sum $a+c+0$ is odd by assumption. If we have $c,a-2c,0$, their sum is $a-c$, which is less than $a+c$. Operation B applied repeatedly will cause either $a$ or $c$ to become $0$.
Case 3
$a>0$ and $c=e=0$. Define Operation C as the sequence of moves from Step 1 to Step 4, shown below: | // Block 1
size(300); defaultpen(fontsize(9)); label("$d$",expi(0),(0,0)); label("$c$",expi(pi/3),(0,0),red); label("$b$",expi(2*pi/3),(0,0)); label("$a$",expi(pi),(0,0),red); label("$f$",expi(4*pi/3),(0,0)); label("$e$",expi(5*pi/3),(0,0),red); label("Step 1",(0,-2),(0,0)); label("$c-e$",(5,0)+expi(0),(0,0)); label("$c$",(5,0)+expi(pi/3),(0,0),red); label("$a-c$",(5,0)+expi(2*pi/3),(0,0)); label("$a$",(5,0)+expi(pi),(0,0),red); label("$a-e$",(5,0)+expi(4*pi/3),(0,0)); label("$e$",(5,0)+expi(5*pi/3),(0,0),red); label("Step 2",(5,-2),(0,0)); label("$c-e$",(10,0)+expi(0),(0,0)); label("$c$",(10,0)+expi(pi/3),(0,0),red); label("$a-c$",(10,0)+expi(2*pi/3),(0,0)); label("$c-e$",(10,0)+expi(pi),(0,0),red); label("$a-e$",(10,0)+expi(4*pi/3),(0,0)); label("$a-c$",(10,0)+expi(5*pi/3),(0,0),red); label("Step 3",(10,-2),(0,0));
// Block 2
size(300); defaultpen(fontsize(9)); label("$d$",expi(0),(0,0)); label("$c$",expi(pi/3),(0,0),red); label("$b$",expi(2*pi/3),(0,0)); label("$a$",expi(pi),(0,0),red); label("$f$",expi(4*pi/3),(0,0)); label("$0$",expi(5*pi/3),(0,0),red); label("Step 1",(0,-2),(0,0)); label("$c$",(5,0)+expi(0),(0,0)); label("$c$",(5,0)+expi(pi/3),(0,0),red); label("$a-c$",(5,0)+expi(2*pi/3),(0,0)); label("$a$",(5,0)+expi(pi),(0,0),red); label("$a$",(5,0)+expi(4*pi/3),(0,0)); label("$0$",(5,0)+expi(5*pi/3),(0,0),red); label("Step 2",(5,-2),(0,0)); label("$c$",(10,0)+expi(0),(0,0)); label("$\begin{array}{l}2c-a\\ a-2c\end{array}$",(10,0)+expi(pi/3),(0,0),red); label("$a-c$",(10,0)+expi(2*pi/3),(0,0)); label("$c$",(10,0)+expi(pi),(0,0),red); label("$a$",(10,0)+expi(4*pi/3),(0,0)); label("$0$",(10,0)+expi(5*pi/3),(0,0),red); label("Step 3",(10,-2),(0,0));
// Block 3
size(400); defaultpen(fontsize(9)); label("$d$",expi(0),(0,0)); label("$0$",expi(pi/3),(0,0),red); label("$b$",expi(2*pi/3),(0,0)); label("$a$",expi(pi),(0,0),red); label("$f$",expi(4*pi/3),(0,0)); label("$0$",expi(5*pi/3),(0,0),red); label("Step 1",(0,-2),(0,0)); label("$0$",(5,0)+expi(0),(0,0)); label("$0$",(5,0)+expi(pi/3),(0,0),red); label("$a$",(5,0)+expi(2*pi/3),(0,0)); label("$a$",(5,0)+expi(pi),(0,0),red); label("$a$",(5,0)+expi(4*pi/3),(0,0)); label("$0$",(5,0)+expi(5*pi/3),(0,0),red); label("Step 2",(5,-2),(0,0)); label("$0$",(10,0)+expi(0),(0,0)); label("$0$",(10,0)+expi(pi/3),(0,0),red); label("$a$",(10,0)+expi(2*pi/3),(0,0)); label("$0$",(10,0)+expi(pi),(0,0),red); label("$a$",(10,0)+expi(4*pi/3),(0,0)); label("$0$",(10,0)+expi(5*pi/3),(0,0),red); label("Step 3",(10,-2),(0,0)); label("$0$",(15,0)+expi(0),(0,0)); label("$0$",(15,0)+expi(pi/3),(0,0),red); label("$0$",(15,0)+expi(2*pi/3),(0,0)); label("$0$",(15,0)+expi(pi),(0,0),red); label("$0$",(15,0)+expi(4*pi/3),(0,0)); label("$0$",(15,0)+expi(5*pi/3),(0,0),red); label("Step 4",(15,-2),(0,0)); | [] |
36 | (Zuming Feng) Let $ABC$ be an acute-angled triangle, and let $P$ and $Q$ be two points on side $BC$. Construct point $C_1$ in such a way that convex quadrilateral $APBC_1$ is cyclic, $QC_1 \parallel CA$, and $C_1$ and $Q$ lie on opposite sides of line $AB$. Construct point $B_1$ in such a way that convex quadrilateral $APCB_1$ is cyclic, $QB_1 \parallel BA$, and $B_1$ and $Q$ lie on opposite sides of line $AC$. Prove that points $B_1, C_1,P$, and $Q$ lie on a circle. | 2005 USAMO Problem 3 | Solution 1
Let $B_1'$ be the second intersection of the line $C_1A$ with the circumcircle of $APC$, and let $Q'$ be the second intersection of the circumcircle of $B_1' C_1P$ and line $BC$. It is enough to show that $B_1'=B_1$ and $Q' =Q$. All our angles will be directed, and measured mod $\pi$.
Since points $C_1, P, Q', B_1'$ are concyclic and points $C_1, A,B_1'$ are collinear, it follows that
\[\angle C_1 Q' P \equiv \angle C_1 B_1' P \equiv \angle A B_1' P .\]
But since points $A, B_1', P, C$ are concyclic,
\[\angle AB_1'P \equiv \angle ACP .\]
It follows that lines $AC$ and $C_1 Q'$ are parallel. If we exchange $C$ with $B$ and $C_1$ with $B_1'$ in this argument, we see that lines $AB$ and $B_1' Q'$ are likewise parallel.
It follows that $Q'$ is the intersection of $BC$ and the line parallel to $AC$ and passing through $C_1$. Hence $Q' = Q$. Then $B_1'$ is the second intersection of the circumcircle of $APC$ and the line parallel to $AB$ passing through $Q$. Hence $B_1' = B_1$, as desired. $\blacksquare$
Motivation: One can notice that if you take such a $B_1'$ then $PQB_1'C_1$ is cyclic, and that similarly $PQB_1C_1'$ is also cyclic. One gets the intuition that only one such circle should exist where the other chord passes through A, and so sets up a ghost point for $Q$, which works. ~cocohearts
Solution 2
Lemma. $B_1, A, C_1$ are collinear.
Suppose they are not collinear. Let line $B_1 A$ intersect circle $ABP$ (i.e. the circumcircle of $ABP$) again at $C_2$ distinct from $C_1$. Because $\angle C_2 B_1 Q = \angle C_2 AB = \angle C_2 PB = 180^\circ - \angle C_2 PQ$, we have that $B_1 C_2 PQ$ is cyclic. Hence $\angle C_2 QP = \angle C_2 B_1 P = \angle P B_1 A = \angle C$, so $C_2 Q // AC$. Then $C_2$ must be the other intersection of the parallel to $AC$ through $Q$ with circle $ABP$. Then $C_2$ is on segment $C_1 Q$, so $C_2$ is contained in triangle $ABQ$. However, line $AB_1$ intersects this triangle only at point $A$ because $B_1$ lies on arc $AC$ not containing $P$ of circle $APC$, a contradiction. Hence, $B_1, A, C_1$ are collinear, as desired.
As a result, we have $\angle C_1 B_1 Q = \angle C_1 AB = \angle C_1 PB = 180^\circ - \angle C_1 PQ$, so $B_1 C_1 PQ$ is cyclic, as desired.
Solution 3
Due to the parallel lines, $m\angle C_1QB_1=m\angle A.$ Therefore, it suffices to prove that \[m\angle C_1PB_1=m\angle C_1BA+m\angle ACB_1=m\angle A.\] Note that $m\angle BC_1A+m\angle AB_1C=180^{\circ}$ by the cyclic quadrilaterals. Now, the condition simplifies to proving $C_1,A,B_1$ collinear.
Use barycentric coordinates. Let \[A=(1,0,0), B=(0,1,0), C=(0,0,1), P=(0,p,1-p), Q=(0,q,1-q).\] By the parallel lines, \[C_1=(1-q-z',q,z'), B_1=(q-y',y',1-q)\] for some $y',z'.$ As $A$ is a vertex of the reference triangle, we must prove that $q(1-q)=y'z'.$
Now we find the circumcircle of $\triangle APC.$ Let its equation be \[-a^2yz-b^2xz-c^2xy+(x+y+z)(ux+vy+wz)=0.\] Substituting in $A,C$ gives $u=w=0.$ Substituting in $P$ yields \[-a^2p(1-p)+vp=0\implies v = a^2(1-p).\] Substituting in $B_1$ yields
\[0=-a^2y'(1-q)-b^2(q-y')(1-q)-c^2y'(q-y')+a^2(1-p)y'=0.\]
\[c^2y'^2+(a^2q-a^2p+b^2(1-q)-c^2q)y'-b^2q(1-q)=0\]
Similarly, the circumcircle of $\triangle APB$ is \[-a^2yz-b^2xz-c^2xy+(x+y+z)a^2pz.\] Now, we substitute $C_1$ into the equation
\[0=-a^2qz'-b^2z'(1-q-z')-c^2q(1-q-z')+a^2pz'=0\]
\[b^2 z'^2+(a^2p-a^2q-b^2(1-q)+c^2q)z'-c^2q(1-q)=0\]
Let $k=a^2p-a^2q-b^2(1-q)+c^2q,$ and note that both $y',z'$ are negative. By the quadratic formula, we have
\[y'=\frac{k-\sqrt{k^2+4b^2c^2q(1-q)}}{2c^2}, z'=\frac{-k-\sqrt{k^2+4b^2c^2q(1-q)}}{2b^2}\]
Multiplying these two, we have \[y'z'=\frac{4b^2c^2q(1-q)}{4b^2c^2}=q(1-q),\] as desired. $\blacksquare$
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page. | size(300); defaultpen(1); pair A=(2,5), B=(-1,0), C=(5,0); pair C1=(.5,5.7); path O1=circumcircle(A,B,C1); pair P=IntersectionPoint(O1,B--C,1); path O2=circumcircle(A,P,C); pair B1=IntersectionPoint(O2,C1--5A-4C1,0); path O=circumcircle(B1,C1,P); pair Q=IntersectionPoint(O,B--C,1); draw(C1--P--A--B--C--A); draw(P--B1--C1--Q--B1); draw(O1,dashed+linewidth(.7)); draw(O2,dashed+linewidth(.7)); draw(O,dotted); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$P$",P,S); label("$Q'$",Q,S); label("$C_1$",C1,N); label("$B_1'$",B1,E); | [] |
37 | (Kiran Kedlaya, Sungyoon Kim) Let $ABC$ be an acute triangle with $\omega$, $\Omega$, and $R$ being its incircle, circumcircle, and circumradius, respectively. The circle $\omega_A$ is tangent internally to $\Omega$ at $A$ and externally tangent to $\omega$. Circle $\Omega_A$ is internally tangent to $\Omega$ at $A$ and tangent internally to $\omega$. Let $P_A$ and $Q_A$ denote the centers of $\omega_A$ and $\Omega_A$, respectively. Define points $P_B$, $Q_B$, $P_C$, $Q_C$ analogously. Prove that
\[8P_AQ_A \cdot P_BQ_B \cdot P_CQ_C \le R^3,\]
with equality if and only if triangle $ABC$ is equilateral. | 2007 USAMO Problem 6 | Solution 1
Lemma.
\[P_{A}Q_{A}=\frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}}\]
Proof. Note $P_{A}$ and $Q_{A}$ lie on $AO$ since for a pair of tangent circles, the point of tangency and the two centers are collinear.
Let $\omega$ touch $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. Note $AE=AF=s-a$. Consider an inversion, $\mathcal{I}$, centered at $A$, passing through $E$, $F$. Since $IE\perp AE$, $\omega$ is orthogonal to the inversion circle, so $\mathcal{I}(\omega)=\omega$. Consider $\mathcal{I}(\omega_{A})=\omega_{A}'$. Note that $\omega_{A}$ passes through $A$ and is tangent to $\omega_{A}$, hence $\omega_{A}'$ is a line that is tangent to $\omega$. Furthermore, $\omega_{A}'\perp AO$ because $\omega_{A}$ is symmetric about $OA$, so the inversion preserves that reflective symmetry. Since it is a line that is symmetric about $AO$, it must be perpendicular to $AO$. Likewise, $\mathcal{I}(\Omega_{A})=\Omega_{A}'$ is the other line tangent to $\omega$ and perpendicular to $AO$.
Let $\omega_{A} \cap AO=X$ and $\omega_{A}' \cap AO=X'$ (second intersection).
Let $\Omega_{A} \cap AO=Y$ and $\Omega_{A}' \cap AO=Y'$ (second intersection).
Evidently, $AX=2AP_{A}$ and $AY=2AQ_{A}$. We want:
\[\star=AQ_{A}-AP_{A}=\frac{1}{2}(AY-AX)=\frac{(s-a)^{2}}{2}\left(\frac{1}{AY'}-\frac{1}{AX'}\right)\]
by inversion. Note that $\omega_{A}' || \Omega_{A}'$, and they are tangent to $\omega$, so the distance between those lines is $2r=AX'-AY'$. Drop a perpendicular from $I$ to $AO$, touching at $H$. Then $AH=AI\cos\angle OAI=AI\cos\frac{1}{2}|\angle B-\angle C|$. Then $AX', AY' = AI\cos\frac{1}{2}|\angle B-\angle C|\pm r$. So $AX'\cdot AY' = AI^{2}\cos^{2}\frac{1}{2}|\angle B-\angle C|-r^{2}$.
\begin{align*} \star &= \frac{(s-a)^{2}}{2}\cdot\frac{AX' - AY'}{AY'\cdot AX'} \\ &= \frac{r(s-a)^{2}}{AI^{2}\cos^{2}\frac{1}{2}|\angle B-\angle C|-r^{2}} \\ &= \frac{\frac{(s-a)^{2}}{r}}{\left(\frac{AI}{r}\right)^{2}\cos^{2}\frac{1}{2}|\angle B-\angle C|-1}. \end{align*}
Note that $\frac{AI}{r} = \frac{1}{\sin\frac{A}{2}}$. Applying the double angle formulas and $1-\cos\angle A=\frac{2(s-b)(s-c)}{bc}$, we get
\begin{align*} \star &= \frac{\frac{(s-a)^{2}}{r}}{\frac{1+\cos (\angle B-\angle C)}{1-\cos A}+1} \\ &= \frac{\frac{(s-a)^{2}}{r}\cdot (1-\cos \angle A)}{\cos(\angle B-\angle C)+\cos(\pi-\angle B-\angle C)} \\ &= \frac{(s-a)^{2}(1-\cos \angle A)}{2r\sin \angle B\sin \angle C} \\ &= \frac{(s-a)^{2}(s-b)(s-c)}{rbc\sin\angle B\sin\angle C} \\ P_{A}Q_{A} &= \frac{4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}} \end{align*}
End Lemma
The problem becomes:
\begin{align*} 8\prod \frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}} &\leq R^{3} \\ \frac{2^{9}R^{6}\left[\prod(s-a)\right]^{4}}{r^{3}a^{4}b^{4}c^{4}} &\leq R^{3} \\ \left(\frac{4AR}{abc}\right)^{4}\cdot\left(\frac{A}{rs}\right)^{4}2rR^{2} &\leq R^{3} \\ 2r &\leq R, \end{align*}
which is true because $OI^{2}=R(R-2r)$, equality is when the circumcenter and incenter coincide. As before, $\angle OAI=\frac{1}{2}|\angle B-\angle C|=0$, so, by symmetry, $\angle A=\angle B=\angle C$. Hence the inequality is true iff $\triangle ABC$ is equilateral.
Comment: It is much easier to determine $AP_{A}$ by considering $\triangle IAP_{A}$. We have $AI$, $\angle IAO$, $IP_{A}=r+r(P_{A})$, and $AP_{A}=r(P_{A})$. However, the inversion is always nice to use. This also gives an easy construction for $w_{A}$ because the tangency point is collinear with the intersection of $w_{A}'$ and $w$.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page. | size(400); defaultpen(fontsize(8)); pair A=(2,8), B=(0,0), C=(13,0), I=incenter(A,B,C), O=circumcenter(A,B,C), p_a, q_a, X, Y, X1, Y1, D, E, F; real r=abs(I-foot(I,A,B)), R=abs(A-O), a=abs(B-C), b=abs(A-C), c=abs(A-B), x=(((b+c-a)/2)^2)/(r^2+4*r*R+((b+c-a)/2)^2), y=((b+c-a)/2)^2/(r^2+((b+c-a)/2)^2); p_a=x*(O-A)+A; q_a=y*(O-A)+A; X=intersectionpoint(Circle(p_a,x*abs(O-A)), A+(O-A)*0.01--O); X1=intersectionpoint(intersectionpoint(incircle(A,B,C),I--I+O-A)+abs(A-O)*expi(angle(A-O)-pi/2)--intersectionpoint(incircle(A,B,C),I--I+O-A)-abs(A-O)*expi(angle(A-O)-pi/2), q_a--q_a+O-A); Y=intersectionpoint(Circle(q_a,y*abs(O-A)), O--2*O-A); Y1=intersectionpoint(intersectionpoint(incircle(A,B,C),I--I+A-O)+abs(A-O)*expi(angle(A-O)-pi/2)--intersectionpoint(incircle(A,B,C),I--I+A-O)-abs(A-O)*expi(angle(A-O)-pi/2), O--A); draw(intersectionpoint(incircle(A,B,C),I--I+A-O)+abs(A-O)*expi(angle(A-O)-pi/2)--intersectionpoint(incircle(A,B,C),I--I+A-O)-abs(A-O)*expi(angle(A-O)-pi/2),blue+0.7); draw(intersectionpoint(incircle(A,B,C),I--I+O-A)+abs(A-O)*expi(angle(A-O)-pi/2)--intersectionpoint(incircle(A,B,C),I--I+O-A)-0.5*abs(A-O)*expi(angle(A-O)-pi/2),red+0.7); draw(A--B--C--A); draw(Circle(A,abs(foot(I,A,B)-A))); draw(incircle(A,B,C)); draw(I--A--O--Y); draw(Circle(p_a,x*abs(O-A)),red+0.7); draw(Circle(q_a,y*abs(O-A)),blue+0.7); label("$A$",A,(-1,1)); label("$I$",I,(-1,0));label("$O$",O,(-1,-1)); label("$P_A$",p_a,(0.5,1));label("$Q_A$",q_a,(1,0)); label("$X$",X,(1,0));label("$Y$",Y,(1,-1)); label("$X'$",X1,(1,0));label("$Y'$",Y1,(0,1)); label("$\omega$",I+r*expi(pi/12), (1,0)); label("$\omega_A$",p_a+x*abs(O-A)*expi(pi/6), (1,1)); label("$\Omega_A$",q_a+y*abs(O-A)*expi(pi/6), (1,0)); label("$\Omega_A'$",intersectionpoint(incircle(A,B,C),I--I+A-O)-abs(A-O)*expi(angle(A-O)-pi/2), (0,2)); label("$\omega_A'$",intersectionpoint(incircle(A,B,C),I--I+O-A)-0.5*abs(A-O)*expi(angle(A-O)-pi/2), (0,2)); dot(p_a^^q_a^^A^^I^^O^^X^^Y^^X1^^Y1); | [] |
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