2019 AIME II Problems/Problem 15
Contents
Problem
In acute triangle points and are the feet of the perpendiculars from to and from to , respectively. Line intersects the circumcircle of in two distinct points, and . Suppose , , and . The value of can be written in the form where and are positive integers, and is not divisible by the square of any prime. Find .
Diagram
Solution 1
First we have , and by PoP. Similarly, and dividing these each by gives .
It is known that the sides of the orthic triangle are , and its angles are ,, and . We thus have the three sides of the orthic triangle now. Letting be the foot of the altitude from , we have, in , similarly, we get To finish, The requested sum is .
༺\\ crazyeyemoody9❂7 //༻
Solution 2
Let , , and . Let . Then and .
By Power of a Point theorem, Thus . Then , , and Use the Law of Cosines in to get , which rearranges to Upon simplification, this reduces to a linear equation in , with solution . Then So the final answer is
By SpecialBeing2017
Solution 3
Let , , , and . By Power of a Point, Points and lie on the circle, , with diameter , and pow, so Use Law of Cosines in to get ; since , this simplifies as We get and thus Therefore . So the answer is
By asr41
Solution 4 (Clean)
This solution is directly based of @CantonMathGuy's solution. We start off with a key claim.
Claim. and .
Proof.
Let and denote the reflections of the orthocenter over points and , respectively. Since and we have that is a rectangle. Then, since we obtain (which directly follows from being cyclic); hence , or .
Similarly, we can obtain .
A direct result of this claim is that .
Thus, we can set and , then applying Power of a Point on we get . Also, we can set and and once again applying Power of a Point (but this time to ) we get
.
Hence,
and the answer is . ~rocketsri
Video Solution by MOP 2024
~r00tsOfUnity
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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