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

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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>.
  
−
==Solution==
+
==Solution 1==
−
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>.
+
Let <math>O</math> be the circumcenter of <math>\triangle ABC</math>; say <math>OT</math> intersects <math>BC</math> at <math>M</math>; draw segments <math>XM</math>, and <math>YM</math>. We have <math>MT=3\sqrt{15}</math>.  
 +
 
 +
[[File:Fanyuchen.png|250px|right]]
 +
 
 +
Since <math>\angle A=\angle CBT=\angle BCT</math>, we have <math>\cos A=\tfrac{11}{16}</math>. Notice that <math>AXTY</math> is cyclic, so <math>\angle XTY=180^{\circ}-A</math>, so <math>\cos XTY=-\cos A</math>, and the cosine law in <math>\triangle TXY</math> gives <cmath>1143-2XY^2=-\frac{11}{8}\cdot XT\cdot YT.</cmath>  
 +
 
 +
Since <math>\triangle BMT \cong \triangle CMT</math>, we have <math>TM\perp BC</math>, and therefore quadrilaterals <math>BXTM</math> and <math>CYTM</math> are cyclic. Let <math>P</math> (resp. <math>Q</math>) be the midpoint of <math>BT</math> (resp. <math>CT</math>). So <math>P</math> (resp. <math>Q</math>) is the center of <math>(BXTM)</math> (resp. <math>CYTM</math>). Then <math>\theta=\angle ABC=\angle MTX</math> and <math>\phi=\angle ACB=\angle YTM</math>. So <math>\angle XPM=2\theta</math>, so<cmath>\frac{\frac{XM}{2}}{XP}=\sin \theta,</cmath>which yields <math>XM=2XP\sin \theta=BT(=CT)\sin \theta=TY</math>. Similarly we have <math>YM=XT</math>.  
 +
 
 +
Ptolemy's theorem in <math>BXTM</math> gives <cmath>16TY=11TX+3\sqrt{15}BX,</cmath> while Pythagoras' theorem gives <math>BX^2+XT^2=16^2</math>. Similarly, Ptolemy's theorem in <math>YTMC</math> gives<cmath>16TX=11TY+3\sqrt{15}CY</cmath> while Pythagoras' theorem in <math>\triangle CYT</math> gives <math>CY^2+YT^2=16^2</math>. Solve this for <math>XT</math> and <math>TY</math> and substitute into the equation about <math>\cos XTY</math> to obtain the result <math>XY^2=\boxed{717}</math>.
  
 
(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.)
 
(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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−
filldraw(A--B--C--cycle, rgb(0.98,0.81,0.69));
+
filldraw(A--B--C--cycle, 0.2*royalblue+white);
 
label("$\omega$", O + rad*dir(45), SW);
 
label("$\omega$", O + rad*dir(45), SW);
−
filldraw(T--Y--M--X--cycle, rgb(173/255,216/255,230/255));
+
//filldraw(T--Y--M--X--cycle, rgb(150, 247, 254));
 +
filldraw(T--Y--M--X--cycle, 0.2*heavygreen+white);
 
draw(M--T);  
 
draw(M--T);  
 
draw(X--Y);
 
draw(X--Y);
Line 49: Line 58:
 
</asy>
 
</asy>
 
Hence, by the Parallelogram Law,
 
Hence, by the Parallelogram Law,
−
<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>
+
<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) = \boxed{717}.</cmath>
 +
 
 +
==Solution 3 (Law of Cosines)==
 +
Let <math>H</math> be the orthocenter of <math>\triangle AXY</math>.
 +
 
 +
<b>Lemma 1:</b> <math>H</math> is the midpoint of <math>BC</math>.
 +
 
 +
<b>Proof:</b> Let <math>H'</math> be the midpoint of <math>BC</math>, and observe that <math>XBH'T</math> and <math>TH'CY</math> are cyclical. Define <math>H'Y \cap BA=E</math> and <math>H'X \cap AC=F</math>, then note that:
 +
<cmath>\angle H'BT=\angle H'CT=\angle H'XT=\angle H'YT=\angle A.</cmath>
 +
That implies that <math>\angle H'XB=\angle H'YC=90^\circ-\angle A</math>, <math>\angle CH'Y=\angle EH'B=90^\circ-\angle B</math>, and <math>\angle BH'Y=\angle FH'C=90^\circ-\angle C</math>. Thus <math>YH'\perp AX</math> and <math>XH' \perp AY</math>; <math>H'</math> is indeed the same as <math>H</math>, and we have proved lemma 1.
 +
 
 +
Since <math>AXTY</math> is cyclical, <math>\angle XTY=\angle XHY</math> and this implies that <math>XHYT</math> is a paralelogram.
 +
By the Law of Cosines:
 +
<cmath>XY^2=XT^2+TY^2+2(XT)(TY)\cdot \cos(\angle A)</cmath>
 +
<cmath>XY^2=XH^2+HY^2+2(XH)(HY) \cdot \cos(\angle A)</cmath>
 +
<cmath>HT^2=HX^2+XT^2-2(HX)(XT) \cdot \cos(\angle A)</cmath>
 +
<cmath>HT^2=HY^2+YT^2-2(HY)(YT) \cdot \cos(\angle A).</cmath>
 +
We add all these equations to get:
 +
<cmath>HT^2+XY^2=2(XT^2+TY^2) \qquad (1).</cmath>
 +
We have that <math>BH=HC=11</math> and <math>BT=TC=16</math> using our midpoints. Note that <math>HT \perp BC</math>, so by the Pythagorean Theorem, it follows that <math>HT^2=135</math>. We were also given that <math>XT^2+TY^2=1143-XY^2</math>, which we multiply by <math>2</math> to use equation <math>(1)</math>. <cmath>2(XT^2+TY^2)=2286-2 \cdot XY^2</cmath> Since <math>2(XT^2+TY^2)=2(HT^2+TY^2)=HT^2+XY^2</math>, we have
 +
<cmath>135+XY^2=2286-2 \cdot XY^2</cmath> <cmath>3 \cdot XY^2=2151.</cmath>
 +
Therefore, <math>XY^2=\boxed{717}</math>. ~ MathLuis
 +
 
 +
==Solution 4 (Similarity and median)==
 +
[[File:AIME-II-2020-15a.png|200px|right]]
 +
Using the <i><b>Claim</b></i> (below) we get <math>\triangle ABC \sim \triangle XTM \sim \triangle YMT.</math>
 +
 
 +
Corresponding sides of similar <math>\triangle XTM \sim \triangle YMT</math> is <math>MT,</math> so
 +
 
 +
<math>\triangle XTM = \triangle YMT \implies MY = XT, MX = TY \implies XMYT</math> – parallelogram.
 +
 
 +
<cmath>4 TD^2 = MT^2 = \sqrt{BT^2 - BM^2}  =\sqrt{153}.</cmath>
 +
The formula for median <math>DT</math> of triangle <math>XYT</math> is
 +
<cmath>2 DT^2 = XT^2 + TY^2 – \frac{XY^2}{2},</cmath>
 +
<cmath>3 \cdot XY^2 = 2XT^2 + 2TY^2 + 2XY^2 –  4 DT^2,</cmath>
 +
<cmath>3 \cdot XY^2  = 2 \cdot 1143-153 = 2151 \implies XY^2 = \boxed{717}. </cmath>
 +
 
 +
 
 +
[[File:AIME-II-2020-15b.png|200px|right]]
 +
<i><b>Claim</b></i>
 +
 
 +
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> be the projections of <math>T</math> onto line <math>AB</math>. Let M be midpoint BC. Then triangle ABC is similar to triangle XTM.
 +
 
 +
<i><b>Proof</b></i>
 +
 
 +
<math>\angle BXT = \angle BMT = 90^o \implies</math> the quadrilateral <math>MBXT</math> is cyclic.
 +
 
 +
<math>BM \perp MT, TX \perp AB \implies \angle MTX = \angle MBA.</math>
 +
 
 +
<math>\angle CBT = \angle BAC = \frac {\overset{\Large\frown} {BC}}{ 2} \implies \triangle ABC \sim \triangle XTM.</math>
 +
 
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
  
 
==See Also==
 
==See Also==

Latest revision as of 15:46, 29 January 2023

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 1

Let $O$ be the circumcenter of $\triangle ABC$; say $OT$ intersects $BC$ at $M$; draw segments $XM$, and $YM$. We have $MT=3\sqrt{15}$.

Fanyuchen.png

Since $\angle A=\angle CBT=\angle BCT$, we have $\cos A=\tfrac{11}{16}$. Notice that $AXTY$ is cyclic, so $\angle XTY=180^{\circ}-A$, so $\cos XTY=-\cos A$, and the cosine law in $\triangle TXY$ gives \[1143-2XY^2=-\frac{11}{8}\cdot XT\cdot YT.\]

Since $\triangle BMT \cong \triangle CMT$, we have $TM\perp BC$, and therefore quadrilaterals $BXTM$ and $CYTM$ are cyclic. Let $P$ (resp. $Q$) be the midpoint of $BT$ (resp. $CT$). So $P$ (resp. $Q$) is the center of $(BXTM)$ (resp. $CYTM$). Then $\theta=\angle ABC=\angle MTX$ and $\phi=\angle ACB=\angle YTM$. So $\angle XPM=2\theta$, so\[\frac{\frac{XM}{2}}{XP}=\sin \theta,\]which yields $XM=2XP\sin \theta=BT(=CT)\sin \theta=TY$. Similarly we have $YM=XT$.

Ptolemy's theorem in $BXTM$ gives \[16TY=11TX+3\sqrt{15}BX,\] while Pythagoras' theorem gives $BX^2+XT^2=16^2$. Similarly, Ptolemy's theorem in $YTMC$ gives\[16TX=11TY+3\sqrt{15}CY\] while Pythagoras' theorem in $\triangle CYT$ gives $CY^2+YT^2=16^2$. Solve this for $XT$ and $TY$ and substitute into the equation about $\cos XTY$ to 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, 0.2*royalblue+white); label("$\omega$", O + rad*dir(45), SW); //filldraw(T--Y--M--X--cycle, rgb(150, 247, 254)); filldraw(T--Y--M--X--cycle, 0.2*heavygreen+white); 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 \[XY^2 = \frac13(2 \cdot 1143-135) = \boxed{717}.\]

Solution 3 (Law of Cosines)

Let $H$ be the orthocenter of $\triangle AXY$.

Lemma 1: $H$ is the midpoint of $BC$.

Proof: Let $H'$ be the midpoint of $BC$, and observe that $XBH'T$ and $TH'CY$ are cyclical. Define $H'Y \cap BA=E$ and $H'X \cap AC=F$, then note that: \[\angle H'BT=\angle H'CT=\angle H'XT=\angle H'YT=\angle A.\] That implies that $\angle H'XB=\angle H'YC=90^\circ-\angle A$, $\angle CH'Y=\angle EH'B=90^\circ-\angle B$, and $\angle BH'Y=\angle FH'C=90^\circ-\angle C$. Thus $YH'\perp AX$ and $XH' \perp AY$; $H'$ is indeed the same as $H$, and we have proved lemma 1.

Since $AXTY$ is cyclical, $\angle XTY=\angle XHY$ and this implies that $XHYT$ is a paralelogram. By the Law of Cosines: \[XY^2=XT^2+TY^2+2(XT)(TY)\cdot \cos(\angle A)\] \[XY^2=XH^2+HY^2+2(XH)(HY) \cdot \cos(\angle A)\] \[HT^2=HX^2+XT^2-2(HX)(XT) \cdot \cos(\angle A)\] \[HT^2=HY^2+YT^2-2(HY)(YT) \cdot \cos(\angle A).\] We add all these equations to get: \[HT^2+XY^2=2(XT^2+TY^2) \qquad (1).\] We have that $BH=HC=11$ and $BT=TC=16$ using our midpoints. Note that $HT \perp BC$, so by the Pythagorean Theorem, it follows that $HT^2=135$. We were also given that $XT^2+TY^2=1143-XY^2$, which we multiply by $2$ to use equation $(1)$. \[2(XT^2+TY^2)=2286-2 \cdot XY^2\] Since $2(XT^2+TY^2)=2(HT^2+TY^2)=HT^2+XY^2$, we have \[135+XY^2=2286-2 \cdot XY^2\] \[3 \cdot XY^2=2151.\] Therefore, $XY^2=\boxed{717}$. ~ MathLuis

Solution 4 (Similarity and median)

AIME-II-2020-15a.png

Using the Claim (below) we get $\triangle ABC \sim \triangle XTM \sim \triangle YMT.$

Corresponding sides of similar $\triangle XTM \sim \triangle YMT$ is $MT,$ so

$\triangle XTM = \triangle YMT \implies MY = XT, MX = TY \implies XMYT$ – parallelogram.

\[4 TD^2 = MT^2 = \sqrt{BT^2 - BM^2} =\sqrt{153}.\] The formula for median $DT$ of triangle $XYT$ is \[2 DT^2 = XT^2 + TY^2 – \frac{XY^2}{2},\] \[3 \cdot XY^2 = 2XT^2 + 2TY^2 + 2XY^2 – 4 DT^2,\] \[3 \cdot XY^2 = 2 \cdot 1143-153 = 2151 \implies XY^2 = \boxed{717}.\]


AIME-II-2020-15b.png

Claim

Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$. The tangents to $\omega$ at $B$ and $C$ intersect at $T$. Let $X$ be the projections of $T$ onto line $AB$. Let M be midpoint BC. Then triangle ABC is similar to triangle XTM.

Proof

$\angle BXT = \angle BMT = 90^o \implies$ the quadrilateral $MBXT$ is cyclic.

$BM \perp MT, TX \perp AB \implies \angle MTX = \angle MBA.$

$\angle CBT = \angle BAC = \frac {\overset{\Large\frown} {BC}}{ 2} \implies \triangle ABC \sim \triangle XTM.$

vladimir.shelomovskii@gmail.com, vvsss

See Also

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