Difference between revisions of "2001 IMO Problems/Problem 5"
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==Solution== | ==Solution== | ||
− | {{ | + | <center><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></center> | ||
+ | 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> | ||
+ | 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> | ||
+ | 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. | ||
+ | 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>. | ||
+ | |||
+ | Solution by <math>Mathdummy</math>. | ||
+ | |||
+ | {{alternate solutions}} | ||
==See also== | ==See also== |
Revision as of 00:57, 3 October 2018
Problem
is a triangle.
lies on
and
bisects angle
.
lies on
and
bisects angle
. Angle
is
.
. Find all possible values for angle
.
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]](http://latex.artofproblemsolving.com/a/c/c/acc798c0223d06f7394c721e34c44a6ad158dc27.png)
Let be on extension of
and
. Let
be on
and
, then
Since
,
is equilateral. Let
, then,
We claim that
must be on
, i.e.,
. If
is not on
, then
, which leads to
, and
is equilateral, which is not possible.
With that, we have, in
,
,
, and
.
Solution by .
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 |