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Difference between revisions of "2001 IMO Problems/Problem 5"

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(→‎Solution)
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==Solution==
 
==Solution==
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{{solution}}
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<center><asy>
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import cse5;
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import graph;
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import olympiad;
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dotfactor = 3;
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unitsize(1.5inch);
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pair A = (0,sqrt(3)), D= (-1, 0), E=(1,0);
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pair Bb = rotate(40,E)*A;
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pair B = extension(A,D,E,Bb);
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pair H = foot(A,D,E);
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pair X = extension(A,H,B,E);
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pair Yy = bisectorpoint(A,B,E);
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pair Y =extension(A,E,B,Yy);
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pair C = E - (0,0.1);
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dot("$B$", B, NW); dot("$Y$", Y, NE);
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dot("$D$", D, W); dot("$E$", E, E);
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dot("$A$",A,N); dot("$X$",X,S);
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label("$C$",E+(0,-0.1),E);
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draw(A--D--E--cycle);
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draw(B--Y);
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draw(B--E);
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// draw(B--Xx--E,dashed);
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// draw(Y--Xx, dashed);
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draw(A--X--D, dashed);
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</asy></center>
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Let <math>D</math> be on extension of <math>AB</math> and <math>BD=BX</math>. Let <math>E</math> be on <math>YC</math> and <math>YE=YB</math>, then <cmath>AD=AB+BD=AB+BX=AY+YB=AE</cmath>
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Since <math>A=60</math>, <math>\triangle{ADE}</math> is equilateral. Let <math>\angle{ABY}=x</math>, then, <cmath>\angle{YBX}=\angle{BDX}=\angle{BXD}=\angle{YEX}=x</cmath>
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We claim that <math>X</math> must be on <math>BE</math>, i.e., <math>C=E</math>. If <math>X</math> is not on <math>BE</math>, then <math>\angle{EBX}=\angle{YBX}-\angle{YBE}=\angle{YEX}-\angle{YEB}=\angle{BEX}</math>, which leads to <math>BX=EX=DX</math>, and <math>\triangle{BDX}</math> is equilateral, which is not possible.
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With that, we have, in <math>\triangle{ABE}</math>, <math>60+2x+x=180</math>, <math>x=40</math>, and <math>\angle{ABE}=80</math>.
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Solution by <math>Mathdummy</math>.
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{{alternate solutions}}
  
 
==See also==
 
==See also==

Revision as of 00:57, 3 October 2018

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.

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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