2019 AIME II Problems/Problem 15
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
Therefore
By power of point, we have Which are simplified to
Or
(1)
Or
Let Then,
In triangle , by law of cosine
Pluging (1)
Or
Substitute everything by
The quadratic term is cancelled out after simplified
Which gives
Plug back in,
Then
So the final answer is
By SpecialBeing2017
Solution 3
Let and
By power of point, we have and
Therefore, substituting in the values:
Notice that quadrilateral is cyclic.
From this fact, we can deduce that and
Therefore is similar to .
Therefore:
Now using Law of Cosines on we get:
Notice
Substituting and Simplifying:
Now we solve for using regular algebra which actually turns out to be very easy.
We get and from the above relations between the variables we quickly determine , and
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
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
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