Difference between revisions of "2020 AIME II Problems/Problem 15"
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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>. | ||
− | < | + | |
− | Proof: Let <math>H'</math> be 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>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 | ||
Revision as of 12:47, 15 April 2021
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
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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