1999 JBMO Problems/Problem 4
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Problem 4
Let be a triangle with
. Also, let
be a point such that
, and let
be the circumcircles of the triangles
and
respectively. Let
and
be diameters in the two circles, and let
be the midpoint of
. Prove that the area of the triangle
is constant (i.e. it does not depend on the choice of the point
).
Solution
Its easy to see that ,
,
are collinear (since
). Applying the sine rule in triangle
, we get
Since
and
are cyclic quadrilaterals,
and
So,
and
Thus,
(the circumcircles
are congruent).
From right triangles and
, we have
So,
Since
is the midpoint of
,
is perpendicular to
and hence
is parallel to
. So area of
and hence is independent of position of
on
.