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

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(Solution 4 (Similarity and median))
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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 <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>.
 +
 
 +
(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.)
  
 
-Fanyuchen20020715
 
-Fanyuchen20020715
  
==Video Solution==
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==Solution 2 (Official MAA)==
https://youtu.be/bz5N-jI2e0U?t=710
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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
 +
<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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 +
 
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filldraw(A--B--C--cycle, rgb(0/255,0/255,255/255));
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label("$\omega$", O + rad*dir(45), SW);
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filldraw(T--Y--M--X--cycle, rgb(0/255,255/255,0/255));
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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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 +
 
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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>
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==Solution 3 (Law of Cosines)==
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Let <math>H</math> be the orthocenter of <math>\triangle AXY</math>.
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<b>Lemma 1:</b> <math>H</math> is the midpoint of <math>BC</math>.
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<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:
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<cmath>\angle H'BT=\angle H'CT=\angle H'XT=\angle H'YT=\angle A.</cmath>
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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.
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By the Law of Cosines:
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<cmath>XY^2=XT^2+TY^2+2(XT)(TY)\cdot \cos(\angle A)</cmath>
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<cmath>XY^2=XH^2+HY^2+2(XH)(HY) \cdot \cos(\angle A)</cmath>
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<cmath>HT^2=HX^2+XT^2-2(HX)(XT) \cdot \cos(\angle A)</cmath>
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<cmath>HT^2=HY^2+YT^2-2(HY)(YT) \cdot \cos(\angle A).</cmath>
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We add all these equations to get:
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<cmath>HT^2+XY^2=2(XT^2+TY^2) \qquad (1).</cmath>
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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:2020 AIME II 15a.png|300px|right]]
 +
Using the <i><b>Lemma</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
 +
 
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<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>
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<cmath>3 \cdot XY^2 = 2XT^2 + 2TY^2 + 2XY^2 –  4 DT^2,</cmath>
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<cmath>3 \cdot XY^2  = 2 \cdot 1143-153 = 2151 \implies XY^2 = \boxed{717}. </cmath>
 +
 
 +
 
 +
[[File:2020 AIME II 15.png|300px|right]]
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<i><b>Lemma</b></i>
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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> 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>
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 +
<math>\angle CBT = \angle BAC = \frac {\overset{\Large\frown} {BC}}{ 2} \implies \triangle ABC \sim \triangle XTM.</math>
 +
 
 +
 
 +
 
 +
'''Shelomovskii, vvsss,  www.deoma-cmd.ru'''
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2020|n=II|num-b=14|after=Last Problem}}
 
{{AIME box|year=2020|n=II|num-b=14|after=Last Problem}}
 +
[[Category:Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:55, 17 June 2022

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

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(0/255,0/255,255/255)); label("$\omega$", O + rad*dir(45), SW); filldraw(T--Y--M--X--cycle, rgb(0/255,255/255,0/255)); 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 \[XY^2 = \frac13(2 \cdot 1143-135) = 717.\]

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)

2020 AIME II 15a.png

Using the Lemma (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 \[2 DT^2 = XT^2 + TY^2 – \frac{XY^2}{2},\] \[3 \cdot XY^2 = 2XT^2 + 2TY^2 + 2XY^2 –  4 DT^2,\] \[3 \cdot XY^2  = 2 \cdot 1143-153 = 2151 \implies XY^2 = \boxed{717}.\]


2020 AIME II 15.png

Lemma

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.

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

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


Shelomovskii, vvsss, www.deoma-cmd.ru

See Also

2020 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Last Problem
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All AIME Problems and Solutions

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