# 2001 Iran NMO (Round 3) Problems/Problem 5

## 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

$[asy] size(300); defaultpen(1); pair A=(1,2.7), B=(0,0), C=(3,0); path O=circumcircle(A,B,C); pair I=incenter(A,B,C); pair BB=A+5(B-A); pair II=bisectorpoint(BB,B,C); pair Ia=IntersectionPoint(B--(B+6(II-B)),A--(A+5(I-A))); pair Aa=IntersectionPoint(A--Ia,B--C); pair M=IntersectionPoint(A--Ia,O,1); pair NN=bisectorpoint(A,C,M); pair N=IntersectionPoint(C--(5(NN-C)+C),O,1); pair S=IntersectionPoint(N--(N+3(I-N)),O,1); pair T=IntersectionPoint(N--(2Ia-N),O,1); draw(O); draw(A--B--C--A--Ia); draw(Ia--N--S); draw(S--T,dashed); draw(B--I--C--Ia--B,dashed); draw(circumcircle(I,B,C),dotted); draw(circumcircle(I,T,S),dotted); label("A",A,N); label("B",B,SW); label("C",C,SE); label("I",I,NE); label("I_a",Ia,SE); label("A'",Aa,SE); label("M",M,SW); label("N",N,W); label("S",S,E); label("T",T,SW); [/asy]$

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

Since $\widehat{MN}\equiv \widehat{NA}$, it follows that $$\angle I_a TS \equiv \angle NTS \equiv \tfrac{1}{2} (\widehat{NA} + \widehat{AS}) \equiv \tfrac{1}{2} (\widehat{MN} + \widehat{AS}) \equiv \angle MIS \equiv \angle I_a IS .$$ 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. $\blacksquare$