2024 AIME II Problems/Problem 10
Contents
Problem
Let have circumcenter and incenter with , circumradius , and inradius . Find .
Solution
By Euler's formula , we have . Thus, by the Pythagorean theorem, . Let ; notice is isosceles and which is enough to imply that is the midpoint of , and itself is the midpoint of where is the -excenter of . Therefore, and
Note that this problem is extremely similar to 2019 CIME I/14.
Solution 2
Denote . By the given condition, , where is the area of .
Moreover, since , the second intersection of the line and is the reflection of about , denote that as . By Fact 5, .
Thus, we have . Now, we have
~Bluesoul
See also
2024 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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