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

(→‎Solution)
(→‎Solution 2 (Official MAA))
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==Problem==
 
==Problem==
 
Let <math>\triangle ABC</math> be an acute scalene triangle with circumcircle <math>\omega</math>. The tangents to <math>\omega</math> at <math>B</math> and <math>C</math> intersect at <math>T</math>. Let <math>X</math> and <math>Y</math> be the projections of <math>T</math> onto lines <math>AB</math> and <math>AC</math>, respectively. Suppose <math>BT = CT = 16</math>, <math>BC = 22</math>, and <math>TX^2 + TY^2 + XY^2 = 1143</math>. Find <math>XY^2</math>.
 
Let <math>\triangle ABC</math> be an acute scalene triangle with circumcircle <math>\omega</math>. The tangents to <math>\omega</math> at <math>B</math> and <math>C</math> intersect at <math>T</math>. Let <math>X</math> and <math>Y</math> be the projections of <math>T</math> onto lines <math>AB</math> and <math>AC</math>, respectively. Suppose <math>BT = CT = 16</math>, <math>BC = 22</math>, and <math>TX^2 + TY^2 + XY^2 = 1143</math>. Find <math>XY^2</math>.
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==Solution 2 (Official MAA)==
 
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Let <math>M</math> denote the midpoint of <math>\overline{BC}</math>. The critical claim is that <math>M</math> is the orthocenter of <math>\triangle AXY</math>, which has the circle with diameter <math>\overline{AT}</math> as its circumcircle. To see this, note that because <math>\angle BXT = \angle BMT = 90^\circ</math>, the quadrilateral <math>MBXT</math> is cyclic, it follows that
 
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<cmath>\angle MXA = \angle MXB = \angle MTB = 90^\circ - \angle TBM = 90^\circ - \angle A,</cmath> implying that <math>\overline{MX} \perp \overline{AC}</math>. Similarly, <math>\overline{MY} \perp \overline{AB}</math>. In particular, <math>MXTY</math> is a parallelogram.
 
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<asy>
 
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defaultpen(fontsize(8pt));
 
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unitsize(0.8cm);
 
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pair A = (0,0);
 
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pair B = (-1.26,-4.43);
 
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pair C = (-1.26+3.89, -4.43);
 
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pair M = (B+C)/2;
 
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pair O = circumcenter(A,B,C);
 
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pair T = (0.68, -6.49);
 
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pair X = foot(T,A,B);
 
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pair Y = foot(T,A,C);
 
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path omega = circumcircle(A,B,C);
 
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real rad = circumradius(A,B,C);
 
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filldraw(A--B--C--cycle, rgb(0.98,0.81,0.69));
 
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label("$\omega$", O + rad*dir(45), SW);
 
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filldraw(T--Y--M--X--cycle, rgb(173/255,216/255,230/255));
 
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draw(M--T);
 
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draw(X--Y);
 
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draw(B--T--C);
 
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draw(A--X--Y--cycle);
 
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draw(omega);
 
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dot("$X$", X, W);
 
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dot("$Y$", Y, E);
 
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dot("$O$", O, W);
 
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dot("$T$", T, S);
 
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dot("$A$", A, N);
 
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dot("$B$", B, W);
 
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dot("$C$", C, E);
 
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dot("$M$", M, N);
 
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</asy>
 
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Hence, by the Parallelogram Law,
 
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<cmath> TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).</cmath> But <math>TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135</math>. Therefore <cmath>XY^2 = \frac13(2 \cdot 1143-135) = 717.</cmath>
 
  
 
==Video Solution 1==
 
==Video Solution 1==

Revision as of 02:22, 9 March 2021

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 $XY^2$.

Video Solution 1

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

Video Solution 2

https://youtu.be/zXGhABDIANY

See Also

2020 AIME II (Problems • Answer Key • Resources)
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

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