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== |
− | + | 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, | + | 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( | + | //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)== | ==Solution 3 (Law of Cosines)== | ||
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<cmath>HT^2=HY^2+YT^2-2(HY)(YT) \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: | We add all these equations to get: | ||
− | <cmath>HT^2+XY^2=2( | + | <cmath>HT^2+XY^2=2(XT^2+TY^2) \qquad (1).</cmath> |
− | We | + | 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>135+XY^2=2286-2 \cdot XY^2</cmath> <cmath>3 \cdot XY^2=2151.</cmath> |
Therefore, <math>XY^2=\boxed{717}</math>. ~ MathLuis | 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 16:46, 29 January 2023
Contents
Problem
Let be an acute scalene triangle with circumcircle
. The tangents to
at
and
intersect at
. Let
and
be the projections of
onto lines
and
, respectively. Suppose
,
, and
. Find
.
Solution 1
Let be the circumcenter of
; say
intersects
at
; draw segments
, and
. We have
.
Since , we have
. Notice that
is cyclic, so
, so
, and the cosine law in
gives
Since , we have
, and therefore quadrilaterals
and
are cyclic. Let
(resp.
) be the midpoint of
(resp.
). So
(resp.
) is the center of
(resp.
). Then
and
. So
, so
which yields
. Similarly we have
.
Ptolemy's theorem in gives
while Pythagoras' theorem gives
. Similarly, Ptolemy's theorem in
gives
while Pythagoras' theorem in
gives
. Solve this for
and
and substitute into the equation about
to obtain the result
.
(Notice that 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 denote the midpoint of
. The critical claim is that
is the orthocenter of
, which has the circle with diameter
as its circumcircle. To see this, note that because
, the quadrilateral
is cyclic, it follows that
implying that
. Similarly,
. In particular,
is a parallelogram.
Hence, by the Parallelogram Law,
But
. Therefore
Solution 3 (Law of Cosines)
Let be the orthocenter of
.
Lemma 1: is the midpoint of
.
Proof: Let be the midpoint of
, and observe that
and
are cyclical. Define
and
, then note that:
That implies that
,
, and
. Thus
and
;
is indeed the same as
, and we have proved lemma 1.
Since is cyclical,
and this implies that
is a paralelogram.
By the Law of Cosines:
We add all these equations to get:
We have that
and
using our midpoints. Note that
, so by the Pythagorean Theorem, it follows that
. We were also given that
, which we multiply by
to use equation
.
Since
, we have
Therefore,
. ~ MathLuis
Solution 4 (Similarity and median)
Using the Claim (below) we get
Corresponding sides of similar is
so
– parallelogram.
The formula for median
of triangle
is
Claim
Let be an acute scalene triangle with circumcircle
. The tangents to
at
and
intersect at
. Let
be the projections of
onto line
. Let M be midpoint BC. Then triangle ABC is similar to triangle XTM.
Proof
the quadrilateral
is cyclic.
vladimir.shelomovskii@gmail.com, vvsss
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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