2001 IMO Problems/Problem 5

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$ABC$ 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 $B$.


[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.

See also

2001 IMO (Problems) • Resources
Preceded by
Problem 4
1 2 3 4 5 6 Followed by
Problem 6
All IMO Problems and Solutions
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