# 2001 IMO Problems/Problem 5

## Problem $ABC$ is a triangle. $X$ lies on $BC$ and $AX$ bisects angle $A$. $Y$ lies on $CA$ and $BY$ bisects angle $B$. Angle $A$ is $60^{\circ}$. $AB + BX = AY + YB$. Find all possible values for angle $B$.

## Solution $[asy] import cse5; import graph; import olympiad; dotfactor = 3; unitsize(1.5inch); pair A = (0,sqrt(3)), D= (-1, 0), E=(1,0); pair Bb = rotate(40,E)*A; pair B = extension(A,D,E,Bb); pair H = foot(A,D,E); pair X = extension(A,H,B,E); pair Yy = bisectorpoint(A,B,E); pair Y =extension(A,E,B,Yy); pair C = E - (0,0.1); dot("B", B, NW); dot("Y", Y, NE); dot("D", D, W); dot("E", E, E); dot("A",A,N); dot("X",X,S); label("C",E+(0,-0.1),E); draw(A--D--E--cycle); draw(B--Y); draw(B--E); // draw(B--Xx--E,dashed); // draw(Y--Xx, dashed); draw(A--X--D, dashed); [/asy]$

Let $D$ be on extension of $AB$ and $BD=BX$. Let $E$ be on $YC$ and $YE=YB$, then $\[AD=AB+BD=AB+BX=AY+YB=AE\]$ Since $A=60$, $\triangle{ADE}$ is equilateral. Let $\angle{ABY}=x$, then, $\[\angle{YBX}=\angle{BDX}=\angle{BXD}=\angle{YEX}=x\]$ We claim that $X$ must be on $BE$, i.e., $C=E$. If $X$ is not on $BE$, then $\angle{EBX}=\angle{YBX}-\angle{YBE}=\angle{YEX}-\angle{YEB}=\angle{BEX}$, which leads to $BX=EX=DX$, and $\triangle{BDX}$ is equilateral, which is not possible. With that, we have, in $\triangle{ABE}$, $60+2x+x=180$, $x=40$, and $\angle{ABE}=80$.

Solution by $Mathdummy$.