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2001 Iran NMO (Round 3) Problems/Problem 5

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Problem

In triangle $ABC$ , let $I$ be the incenter and $I_a$ the excenter opposite $A$ . Suppose that $\overline{II_a}$ meets $\overline{BC}$ and the circumcircle of triangle $ABC$ at $A'$ and $M$ , respectively. Let $N$ be the midpoint of arc $MBA$ of the circumcircle of triangle $ABC$ . Let lines $NI$ and $NI_a$ intersect the circumcircle of triangle $ABC$ again at $S$ and $T$ , respectively. Prove that $S$ , $T$ , and $A'$ are collinear.

Solution

We will use directed angles mod $\pi$ , and directed arcs mod $2\pi$ .

Since $\widehat{MN}\equiv \widehat{NA}$ , it follows that $\angle I_a TS \equiv \angle NTS \equiv \tfrac{1}{2} (\widehat{NA} + \widehat{AS}) \equiv \tfrac{1}{2} (\widehat{MN} + \wideha...$ It follows that quadrilateral $I_aITS$ is cyclic.

On the other hand, $\angle IBI_a \equiv \angle ICI_a \equiv \tfrac{\pi}{2}$ , so quadrilateral $IBCI_a$ is cyclic.

Now, since $II_a$ is the radical axis of the circumcircles of $I_aITS$ and $IBCI_a$ , $BC$ is the radical axis of the circumcircles of $IBCI_a$ and $ABC$ , and $TS$ is the circumcircle of $ABC$ and $I_aITS$ , these three lines concur at $A'$ , as desired. $\blacksquare$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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