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

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==Solution==
 
==Solution==
 
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 \angle <math>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>.
 
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 \angle <math>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>.
 +
 +
(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
 
-Fanyuchen20020715

Revision as of 10:15, 8 June 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

Video Solution

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

See Also

2020 AIME II (Problems • Answer Key • Resources)
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