2020 AIME II Problems/Problem 15

Problem

Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$ . The tangents to $\omega$ at $B$ and $C$ intersect at $T$ . Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$ , respectively. Suppose $BT = CT = 16$ , $BC = 22$ , and $TX^2 + TY^2 + XY^2 = 1143$ . Find $XY^2$ .

Solution 1

Let $O$ be the circumcenter of $\triangle ABC$ ; say $OT$ intersects $BC$ at $M$ ; draw segments $XM$ , and $YM$ . We have $MT=3\sqrt{15}$ .

Since $\angle A=\angle CBT=\angle BCT$ , we have $\cos A=\tfrac{11}{16}$ . Notice that $AXTY$ is cyclic, so $\angle XTY=180^{\circ}-A$ , so $\cos XTY=-\cos A$ , and the cosine law in $\triangle TXY$ gives $1143-2XY^2=-\frac{11}{8}\cdot XT\cdot YT.$

Since $\triangle BMT \cong \triangle CMT$ , we have $TM\perp BC$ , and therefore quadrilaterals $BXTM$ and $CYTM$ are cyclic. Let $P$ (resp. $Q$ ) be the midpoint of $BT$ (resp. $CT$ ). So $P$ (resp. $Q$ ) is the center of $(BXTM)$ (resp. $CYTM$ ). Then $\theta=\angle ABC=\angle MTX$ and $\phi=\angle ACB=\angle YTM$ . So $\angle XPM=2\theta$ , so $\frac{\frac{XM}{2}}{XP}=\sin \theta,$ which yields $XM=2XP\sin \theta=BT(=CT)\sin \theta=TY$ . Similarly we have $YM=XT$ .

Ptolemy's theorem in $BXTM$ gives $16TY=11TX+3\sqrt{15}BX,$ while Pythagoras' theorem gives $BX^2+XT^2=16^2$ . Similarly, Ptolemy's theorem in $YTMC$ gives $16TX=11TY+3\sqrt{15}CY$ while Pythagoras' theorem in $\triangle CYT$ gives $CY^2+YT^2=16^2$ . Solve this for $XT$ and $TY$ and substitute into the equation about $\cos XTY$ to obtain the result $XY^2=\boxed{717}$ .

(Notice that $MXTY$ is a parallelogram, which is an important theorem in Olympiad, and there are some other ways of computation under this observation.)

-Fanyuchen20020715

Solution 2 (Official MAA)

Let $M$ denote the midpoint of $\overline{BC}$ . The critical claim is that $M$ is the orthocenter of $\triangle AXY$ , which has the circle with diameter $\overline{AT}$ as its circumcircle. To see this, note that because $\angle BXT = \angle BMT = 90^\circ$ , the quadrilateral $MBXT$ is cyclic, it follows that $\angle MXA = \angle MXB = \angle MTB = 90^\circ - \angle TBM = 90^\circ - \angle A,$ implying that $\overline{MX} \perp \overline{AC}$ . Similarly, $\overline{MY} \perp \overline{AB}$ . In particular, $MXTY$ is a parallelogram. $[asy] defaultpen(fontsize(8pt)); unitsize(0.8cm); pair A = (0,0); pair B = (-1.26,-4.43); pair C = (-1.26+3.89, -4.43); pair M = (B+C)/2; pair O = circumcenter(A,B,C); pair T = (0.68, -6.49); pair X = foot(T,A,B); pair Y = foot(T,A,C); path omega = circumcircle(A,B,C); real rad = circumradius(A,B,C); filldraw(A--B--C--cycle, 0.2*royalblue+white); label("$\omega$", O + rad*dir(45), SW); //filldraw(T--Y--M--X--cycle, rgb(150, 247, 254)); filldraw(T--Y--M--X--cycle, 0.2*heavygreen+white); draw(M--T); draw(X--Y); draw(B--T--C); draw(A--X--Y--cycle); draw(omega); dot("$X$", X, W); dot("$Y$", Y, E); dot("$O$", O, W); dot("$T$", T, S); dot("$A$", A, N); dot("$B$", B, W); dot("$C$", C, E); dot("$M$", M, N); [/asy]$ Hence, by the Parallelogram Law, $TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).$ But $TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135$ . Therefore $XY^2 = \frac13(2 \cdot 1143-135) = 717.$

Solution 3 (Law of Cosines)

Let $H$ be the orthocenter of $\triangle AXY$ .

Lemma 1: $H$ is the midpoint of $BC$ .

Proof: Let $H'$ be the midpoint of $BC$ , and observe that $XBH'T$ and $TH'CY$ are cyclical. Define $H'Y \cap BA=E$ and $H'X \cap AC=F$ , then note that: $\angle H'BT=\angle H'CT=\angle H'XT=\angle H'YT=\angle A.$ That implies that $\angle H'XB=\angle H'YC=90^\circ-\angle A$ , $\angle CH'Y=\angle EH'B=90^\circ-\angle B$ , and $\angle BH'Y=\angle FH'C=90^\circ-\angle C$ . Thus $YH'\perp AX$ and $XH' \perp AY$ ; $H'$ is indeed the same as $H$ , and we have proved lemma 1.

Since $AXTY$ is cyclical, $\angle XTY=\angle XHY$ and this implies that $XHYT$ is a paralelogram. By the Law of Cosines: $XY^2=XT^2+TY^2+2(XT)(TY)\cdot \cos(\angle A)$ $XY^2=XH^2+HY^2+2(XH)(HY) \cdot \cos(\angle A)$ $HT^2=HX^2+XT^2-2(HX)(XT) \cdot \cos(\angle A)$ $HT^2=HY^2+YT^2-2(HY)(YT) \cdot \cos(\angle A).$ We add all these equations to get: $HT^2+XY^2=2(XT^2+TY^2) \qquad (1).$ We have that $BH=HC=11$ and $BT=TC=16$ using our midpoints. Note that $HT \perp BC$ , so by the Pythagorean Theorem, it follows that $HT^2=135$ . We were also given that $XT^2+TY^2=1143-XY^2$ , which we multiply by $2$ to use equation $(1)$ . $2(XT^2+TY^2)=2286-2 \cdot XY^2$ Since $2(XT^2+TY^2)=2(HT^2+TY^2)=HT^2+XY^2$ , we have $135+XY^2=2286-2 \cdot XY^2$ $3 \cdot XY^2=2151.$ Therefore, $XY^2=\boxed{717}$ . ~ MathLuis

Solution 4 (Similarity and median)

Using the Claim (below) we get $\triangle ABC \sim \triangle XTM \sim \triangle YMT.$

Corresponding sides of similar $\triangle XTM \sim \triangle YMT$ is $MT,$ so

$\triangle XTM = \triangle YMT \implies MY = XT, MX = TY \implies XMYT$ – parallelogram.

$4 TD^2 = MT^2 = \sqrt{BT^2 - BM^2} =\sqrt{153}.$ The formula for median $DT$ of triangle $XYT$ is $2 DT^2 = XT^2 + TY^2 – \frac{XY^2}{2},$ $3 \cdot XY^2 = 2XT^2 + 2TY^2 + 2XY^2 – 4 DT^2,$ $3 \cdot XY^2 = 2 \cdot 1143-153 = 2151 \implies XY^2 = \boxed{717}.$

Claim

Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$ . The tangents to $\omega$ at $B$ and $C$ intersect at $T$ . Let $X$ be the projections of $T$ onto line $AB$ . Let M be midpoint BC. Then triangle ABC is similar to triangle XTM.

Proof

$\angle BXT = \angle BMT = 90^o \implies$ the quadrilateral $MBXT$ is cyclic.

$BM \perp MT, TX \perp AB \implies \angle MTX = \angle MBA.$

$\angle CBT = \angle BAC = \frac {\overset{\Large\frown} {BC}}{ 2} \implies \triangle ABC \sim \triangle XTM.$

vladimir.shelomovskii@gmail.com, vvsss

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