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

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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==
 
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Assume <math>O</math> to be the center of triangle <math>ABC</math>, <math>OT</math> cross <math>BC</math> at <math>M</math>, link <math>XM</math>, <math>YM</math>. Let <math>P</math> be the middle point of <math>BT</math> and <math>Q</math> be the middle point of <math>CT</math>, so we have <math>MT=3\sqrt{15}</math>. Since <math>\angle A=\angle CBT=\angle BCT</math>, we have <math>\cos A=\frac{11}{16}</math>. Notice that <math>\angle XTY=180^{\circ}-A</math>, so <math>\cos XYT=-\cos A</math>, and this gives us <math>1143-2XY^2=\frac{-11}{8}XT\cdot YT</math>. Since <math>TM</math> is perpendicular to <math>BC</math>, <math>BXTM</math> and <math>CYTM</math> cocycle (respectively), so <math>\theta_1=\angle ABC=\angle MTX</math> and <math>\theta_2=\angle ACB=\angle YTM</math>. So <math>\angle XPM=2\theta_1</math>, so <cmath>\frac{\frac{XM}{2}}{XP}=\sin \theta_1</cmath>, which yields <math>XM=2XP\sin \theta_1=BT(=CT)\sin \theta_1=TY.</math> So same we have <math>YM=XT</math>. Apply Ptolemy theorem in <math>BXTM</math> we have <math>16TY=11TX+3\sqrt{15}BX</math>, and use Pythagoras theorem we have <math>BX^2+XT^2=16^2</math>. Same in <math>YTMC</math> and triangle <math>CYT</math> we have <math>16TX=11TY+3\sqrt{15}CY</math> and <math>CY^2+YT^2=16^2</math>. Solve this for <math>XT</math> and <math>TY</math> and submit into the equation about <math>\cos XYT</math>, we can obtain the result <math>XY^2=\boxed{717}</math>.
 
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(Notice that <math>MXTY</math> is a parallelogram, which is an important theorem in Olympiad, and there are some other ways of computation under this observation.)
 
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-Fanyuchen20020715
 
  
 
==Solution 2 (Official MAA)==
 
==Solution 2 (Official MAA)==

Revision as of 02:21, 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 $XY^2$.

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 (Problems • Answer Key • Resources)
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