1998 JBMO Problems/Problem 2
Problem 2
Let be a convex pentagon such that , and . Compute the area of the pentagon.
Solution
Let
Let angle =
Applying cosine rule to triangle we get:
Substituting we get:
From above,
Thus,
So, of triangle =
Let be the altitude of triangle DAC from A.
So
This implies .
Since is a cyclic quadrilateral with , traingle is congruent to . Similarly is a cyclic quadrilateral and traingle is congruent to .
So of triangle + of triangle = of Triangle . Thus of pentagon = of + of + of =
By