Difference between revisions of "2019 AIME II Problems/Problem 15"
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To finish, <cmath>bc= \frac{(b\cos B)(c\cos C)}{\cos B\cos C}=\frac{16\cdot 21}{(2/\sqrt{35})(3/\sqrt{40})}=560\sqrt{14}.</cmath> | To finish, <cmath>bc= \frac{(b\cos B)(c\cos C)}{\cos B\cos C}=\frac{16\cdot 21}{(2/\sqrt{35})(3/\sqrt{40})}=560\sqrt{14}.</cmath> | ||
The requested sum is <math>\boxed{574}</math>. | The requested sum is <math>\boxed{574}</math>. | ||
− | + | - crazyeyemoody907 | |
− | |||
Remark: The proof that <math>a \cos A = PQ</math> can be found here: http://www.irmo.ie/5.Orthic_triangle.pdf | Remark: The proof that <math>a \cos A = PQ</math> can be found here: http://www.irmo.ie/5.Orthic_triangle.pdf |
Revision as of 19:05, 12 December 2023
Contents
[hide]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 . - crazyeyemoody907
Remark: The proof that can be found here: http://www.irmo.ie/5.Orthic_triangle.pdf
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.