2020 AIME II Problems/Problem 15

Revision as of 02:21, 9 March 2021 by Justoverrated (talk | contribs) (Solution)

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 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

See Also

2020 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Last Problem
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png