1988 USAMO Problems/Problem 4

Problem

$\Delta ABC$ is a triangle with incenter $I$ . Show that the circumcenters of $\Delta IAB$ , $\Delta IBC$ , and $\Delta ICA$ lie on a circle whose center is the circumcenter of $\Delta ABC$ .

Solution

Solution 1

Let the circumcenters of $\Delta IAB$ , $\Delta IBC$ , and $\Delta ICA$ be $O_c$ , $O_a$ , and $O_b$ , respectively. It then suffices to show that $A$ , $B$ , $C$ , $O_a$ , $O_b$ , and $O_c$ are concyclic.

We shall prove that quadrilateral $ABO_aC$ is cyclic first. Let $\angle BAC=\alpha$ , $\angle CBA=\beta$ , and $\angle ACB=\gamma$ . Then $\angle ICB=\gamma/2$ and $\angle IBC=\beta/2$ . Therefore minor arc $\overarc{BIC}$ in the circumcircle of $IBC$ has a degree measure of $\beta+\gamma$ . This shows that $\angle CO_aB=\beta+\gamma$ , implying that $\angle BAC+\angle BO_aC=\alpha+\beta+\gamma=180^{\circ}$ . Therefore quadrilateral $ABO_aC$ is cyclic.

This shows that point $O_a$ is on the circumcircle of $\Delta ABC$ . Analagous proofs show that $O_b$ and $O_c$ are also on the circumcircle of $ABC$ , which completes the proof. $\blacksquare$

Solution 2

Let $M$ denote the midpoint of arc $AC$ . It is well known that $M$ is equidistant from $A$ , $C$ , and $I$ (to check, prove $\angle IAM = \angle AIM = \frac{\angle BAC + \angle ABC}{2}$ ), so that $M$ is the circumcenter of $AIC$ . Similar results hold for $BIC$ and $CIA$ , and hence $O_c$ , $O_a$ , and $O_b$ all lie on the circumcircle of $ABC$ .

Solution 3

Extend $CI$ to point $L$ on $(ABC)$ . By The Incenter-Excenter Lemma, B, I, A are all concyclic. Thus, L is the circumcenter of triangle $IAB$ . In other words, $L=O_c$ , so $O_c$ is on $(ABC)$ . Similarly, we can show that $O_a$ and $O_b$ are on $(ABC)$ , and thus, $A,B,C,O_a,O_b,O_c$ are all concyclic. It follows that the circumcenters are equal.

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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