1989 IMO Problems/Problem 2
is a triangle, the bisector of angle
meets the circumcircle of triangle
in
, points
and
are defined similarly. Let
meet the lines that bisect the two external angles at
and
in
. Define
and
similarly. Prove that the area of triangle
area of hexagon
area of triangle
.
Solution
Notice that since and
substend the same angle in the circle, and so are equal. Thus
is iscoceles, and similarly for triangles
and
. Also, since
is a cyclic quadrilatera,
, and similarly for the other triangles. Thus, the area of triangle
and similarly for the other triangles. Thus, the area of the hexagon is equal to
.
Now we shall find the area of triangle . It is obvious that the points
are collinear since the angle the make at
is
, and similarly for the other points. Thus, it suffices to find the area of each of the triangles
. It is well known that
is the center of the excircle
with radius
. Thus the area of triangle
is
and so the area of triangel
is
.
Now we can prove that area of hexagon
by simplifying. We have
area of hexagon
as desired.
As for the inequality, notice that it is equivalent to
Letting for
positive reals, the inequality becomes
which is true by AM-HM.