# 2013 IMO Problems/Problem 3

## Problem

Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the excircles opposite $B$ and $C$, respectively. Suppose that the circumcentre of triangle $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$. Prove that triangle $ABC$ is right-angled.

Proposed by Alexander A. Polyansky, Russia

## Solution $[asy] unitsize(2.5cm); void b() { pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10); pen dd=linetype("4 8"); /* Define the excenter */ pair excenter(pair A=(0,0), pair B=(0,0), pair C=(0,0)) { return extension(A,bisectorpoint(C,A,B),B,rotate(90,B)*bisectorpoint(A,B,C)); } /* Draw points */ pair A=D("A",dir(115),N), B=D("B",(-1,0),W), C=D("C",(1,0),E), I=D(incenter(A,B,C)), Ia=D("I_a",excenter(A,B,C),S), Ib=D("I_b",excenter(B,C,A),E), Ic=D("I_c",excenter(C,A,B),W), A1=D("A_1",foot(Ia,B,C),SSW), B1=D("B_1",foot(Ib,C,A),E), C1=D("C_1",foot(Ic,A,B),N), Ma=D("M_a",midpoint(Ib--Ic),N), Mb=D("M_b",midpoint(Ic--Ia),SW), Mc=D("M_c",midpoint(Ia--Ib),SE), V=D("V",circumcenter(Ia,Ib,Ic),SE); /* Draw paths */ D(unitcircle,heavyblue); D(circumcircle(B,C,V),linetype("2 2")+rgb(0.6,0,1)); D(circumcircle(B,C1,V),linetype("2 2")+rgb(0.6,0,1)); D(A--B--C--cycle); D(Ia--Ib--Ic--cycle,gray(0.3)+linewidth(1)); D(A1--B1--C1--cycle); D(I--Ia,dd+red); D(I--Ib,dd+red); D(I--Ic,dd+red); D(V--Ia,dd+heavygreen); D(V--Ib,dd+heavygreen); D(V--Ic,dd+heavygreen); } b(); pathflag=false; b(); [/asy]$ Let the excenters opposite $A,B,C$ be $I_a,I_b,I_c$. Let the midpoint of $\overline{I_bI_c}$ be $M_a$, which lies on $(ABC)$, the nine-point circle of $\triangle I_aI_bI_c$; analogously define $M_b,M_c$. $M_aB=M_aC$ and $BC_1=s-a=B_1C$, so $\triangle M_aBC_1\cong\triangle M_aCB_1$ (SAS), thus $M_a$ is equidistant from $B_1,C_1$, with analogous results for $M_b,M_c$. It follows that the circumcentre of $\triangle A_1B_1C_1$ is one of $M_a,M_b,M_c$; WLOG, suppose it is $M_a$.

By isogonal conjugacy, $I_aA_1,I_bB_1I_cC_1$ concur at the Bevan point $V$ of $\triangle ABC$. $M_aM_b$ is the common perpendicular bisector of $\overline{C_1A_1}$ and $\overline{I_cC}$, so $C_1A_1\parallel I_cC$. $(A_1C_1M_b)$ is the circle on diameter $\overline{VB}$, so by Reim's theorem, $V \in (I_bI_cBC)$.

Hence $\angle I_cI_aI_b=\tfrac{1}{2}\angle I_cVI_b=45^{\circ}\implies\angle CAB=180^{\circ}-2\angle I_cI_aI_b=90^{\circ}$, as required.