Difference between revisions of "2020 AIME II Problems/Problem 15"
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− | filldraw(A--B--C--cycle, rgb(0 | + | filldraw(A--B--C--cycle, rgb(0/255,0/255,255/255)); |
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(0/255,255/255,0/255)); |
draw(M--T); | draw(M--T); | ||
draw(X--Y); | draw(X--Y); | ||
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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> | <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> | ||
− | ==Solution 3 ( | + | ==Solution 3 (Law of Cosines)== |
Let <math>H</math> be the orthocenter of <math>\triangle AXY</math>. | Let <math>H</math> be the orthocenter of <math>\triangle AXY</math>. | ||
− | <b>Lemma 1:</b> <math>H</math> is the midpoint of <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> | + | <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> | <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. | 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. | ||
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Since <math>AXTY</math> is cyclical, <math>\angle XTY=\angle XHY</math> and this implies that <math>XHYT</math> is a paralelogram. | 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: | By the Law of Cosines: | ||
− | <cmath>XY^2=XT^2+TY^2+2(XT)(TY)\cdot cos(\angle A)</cmath> | + | <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>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=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> | + | <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: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 | ||
+ | |||
+ | <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:2020 AIME II 15.png|300px|right]] | ||
+ | <i><b>Lemma</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> | ||
+ | |||
+ | |||
+ | |||
+ | '''Shelomovskii, vvsss, www.deoma-cmd.ru''' | ||
==See Also== | ==See Also== |
Revision as of 15:55, 17 June 2022
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
Assume to be the center of triangle , cross at , link , . Let be the middle point of and be the middle point of , so we have . Since , we have . Notice that , so , and this gives us . Since is perpendicular to , and cocycle (respectively), so and . So , so , which yields So same we have . Apply Ptolemy theorem in we have , and use Pythagoras theorem we have . Same in and triangle we have and . Solve this for and and submit into the equation about , we can 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 Lemma (below) we get
Corresponding sides of similar is so
– parallelogram.
The formula for median of triangle is
Lemma
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.
Shelomovskii, vvsss, www.deoma-cmd.ru
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
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