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 $Mathdummy$ .

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

2001 IMO (Problems) • Resources
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2001 IMO Problems/Problem 5

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