Difference between revisions of "2020 AIME I Problems/Problem 15"
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== Problem == | == Problem == | ||
+ | Let <math>\triangle ABC</math> be an acute triangle with circumcircle <math>\omega,</math> and let <math>H</math> be the intersection of the altitudes of <math>\triangle ABC.</math> Suppose the tangent to the circumcircle of <math>\triangle HBC</math> at <math>H</math> intersects <math>\omega</math> at points <math>X</math> and <math>Y</math> with <math>HA=3,HX=2,</math> and <math>HY=6.</math> The area of <math>\triangle ABC</math> can be written as <math>m\sqrt{n},</math> where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n.</math> | ||
== Solution == | == Solution == |
Revision as of 16:42, 12 March 2020
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Contents
[hide]Problem
Let be an acute triangle with circumcircle and let be the intersection of the altitudes of Suppose the tangent to the circumcircle of at intersects at points and with and The area of can be written as where and are positive integers, and is not divisible by the square of any prime. Find
Solution
The following is a power of a point solution to this menace of a problem:
Let points be what they appear as in the diagram below. Note that is not insignificant; from here, we set by PoP and trivial construction. Now, is the reflection of over . Note , and therefore by Pythagorean theorem we have . Consider . We have that , and therefore we are ready to PoP with respect to . Setting , we obtain by PoP on , and furthermore, we have . Now, we get , and from we take However, squaring and manipulating with yields that and from here, since we get the area to be . ~awang11's sol
Solution 2
See Also
2020 AIME I (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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