Difference between revisions of "2020 AIME II Problems/Problem 15"

Revision as of 04:57, 11 August 2020

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

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. $[asy] defaultpen(fontsize(8pt)); unitsize(0.8cm); pair A = (0,0); pair B = (-1.26,-4.43); pair C = (-1.26+3.89, -4.43); pair M = (B+C)/2; pair O = circumcenter(A,B,C); pair T = (0.68, -6.49); pair X = foot(T,A,B); pair Y = foot(T,A,C); path omega = circumcircle(A,B,C); real rad = circumradius(A,B,C); filldraw(A--B--C--cycle, rgb(0.98,0.81,0.69)); label("$\omega$", O + rad*dir(45), SW); filldraw(T--Y--M--X--cycle, rgb(173/255,216/255,230/255)); draw(M--T); draw(X--Y); draw(B--T--C); draw(A--X--Y--cycle); draw(omega); dot("$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 $XY^2 = \frac13(2 \cdot 1143-135) = 717.$

Video Solution 1

https://youtu.be/bz5N-jI2e0U?t=710

Video Solution 2

https://youtu.be/zXGhABDIANY

@@ Line 55: / Line 55: @@
 ==Video Solution 2==
-https://www.youtube.com/watch?v=zXGhABDIANY&t=21s
+https://youtu.be/zXGhABDIANY
 ==See Also==

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