2001 IMO Problems/Problem 5
Contents
[hide]Problem
is a triangle.
lies on
and
bisects angle
.
lies on
and
bisects angle
. Angle
is
.
. Find all possible values for angle
.
Solution1
![[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 .
Solution 2
Refer to the image in Solution 1 without any construction
[img]https://i.imgur.com/IDvALGS.png[/img]
~Lakshya Pamecha
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 |