1988 USAMO Problems/Problem 4


$\Delta ABC$ 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 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.

See Also

1988 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5
All USAMO Problems and Solutions

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