2007 IMO Problems/Problem 2
Problem
Consider five points , and
such that
is a parallelogram and
is a cyclic quadrilateral.
Let
be a line passing through
. Suppose that
intersects the interior of the segment
at
and intersects
line
at
. Suppose also that
. Prove that
is the bisector of
.
Solution
![[asy] import cse5; import graph; import olympiad; dotfactor = 3; unitsize(1.5inch); path circle = Circle(origin, 1); draw(circle); pair D = (-sqrt(3)/2, -0.5), C = (sqrt(3)/2, -0.5); //G = bisectorpoint(C, B, D); pair Ee = rotate(38,C)*D; pair E = IP(C--Ee, circle,1); pair Gg = rotate(76,C)*D; path circle2 = Circle(E, length(C-E)); pair G = IP(C--Gg, circle2, 1); pair F = IP(C--D, circle2, 1); pair Bb = rotate(-104,C)*D; pair B = IP(C--Bb, circle, 1); pair A = extension((-1,B.y),(1,B.y),G,F); draw(circle2, dashed); draw(A--G); draw(C--D--A--B); draw(G--B); draw(E--F); draw(E--C); draw(E--G); dot("$C$", C, dir(30)); dot("$D$", D, SW); dot("$G$", G, SE); dot("$E$", E, SW); dot("$F$", F, SW); dot("$A$", A, SW); dot("$B$", B, SE); draw(D--E,dashed); draw(B--E,dashed); [/asy]](http://latex.artofproblemsolving.com/c/4/d/c4d6cd122415e7b7ccce4bdc0b15896ba36b846d.png)
Since ,
, it suffices to prove
.
Let ,
,
. We have:
so,
Meantime, using Law of Sines on
, we have,
Using Law of Sines on
, and notice that
, we have,
so,
Since
, and
, we have,
. Hence,
or,
There are two possibilities: (1)
, or (2)
. However, (2) would mean
, then
would be a diameter, and
because
is inside the circle, so (2) is not valid. From condition (1), we have
, therefore
.
Solution by Mathdummy
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
2007 IMO (Problems) • Resources | ||
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