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

Assume $O$ to be the center of triangle $ABC$ , $OT$ cross $BC$ at $M$ , link $XM$ , $YM$ . Let $P$ be the middle point of $BT$ and $Q$ be the middle point of $CT$ , so we have $MT=3\sqrt{15}$ . Since $\angle A=\angle CBT=\angle BCT$ , we have $\cos A=\frac{11}{16}$ . Notice that $\angle XTY=180^{\circ}-A$ , so $\cos XYT=-\cos A$ , and this gives us $1143-2XY^2=\frac{-11}{8}XT\cdot YT$ . Since $TM$ is perpendicular to $BC$ , $BXTM$ and $CYTM$ cocycle (respectively), so $\theta_1=\angle ABC=\angle MTX$ and $\theta_2=\angle ACB=\angle YTM$ . So $\angle XPM=2\theta_1$ , so $\frac{\frac{XM}{2}}{XP}=\sin \theta_1$ , which yields $XM=2XP\sin \theta_1=BT(=CT)\sin \theta_1=TY.$ So same we have $YM=XT$ . Apply Ptolemy theorem in $BXTM$ we have $16TY=11TX+3\sqrt{15}BX$ , and use Pythagoras theorem we have $BX^2+XT^2=16^2$ . Same in $YTMC$ and triangle $CYT$ we have $16TX=11TY+3\sqrt{15}CY$ and $CY^2+YT^2=16^2$ . Solve this for $XT$ and $TY$ and submit into the equation about $\cos XYT$ , we can 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

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2020 AIME II Problems/Problem 15

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