1988 USAMO Problems/Problem 4
Problem
is a triangle with incenter . Show that the circumcenters of , , and lie on a circle whose center is the circumcenter of .
Solution
Solution 1
Let the circumcenters of , , and be , , and , respectively. It then suffices to show that , , , , , and are concyclic.
We shall prove that quadrilateral is cyclic first. Let , , and . Then and . Therefore minor arc in the circumcircle of has a degree measure of . This shows that , implying that . Therefore quadrilateral is cyclic.
This shows that point is on the circumcircle of . Analagous proofs show that and are also on the circumcircle of , which completes the proof.
Solution 2
Let denote the midpoint of arc . It is well known that is equidistant from , , and (to check, prove ), so that is the circumcenter of . Similar results hold for and , and hence , , and all lie on the circumcircle of .
Solution 3
Extend to point on . By The Incenter-Excenter Lemma, B, I, A are all concyclic. Thus, L is the circumcenter of triangle . In other words, , so is on . Similarly, we can show that and are on , and thus, are all concyclic. It follows that the circumcenters are equal.
See Also
1988 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
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