1998 JBMO Problems/Problem 2
Let be a convex pentagon such that , and . Compute the area of the pentagon.
Applying cosine rule to we get:
Substituting we get:
So, area of =
Let be the altitude of from .
This implies .
Since is a cyclic quadrilateral with , is congruent to . Similarly is a cyclic quadrilateral and is congruent to .
So area of + area of = area of . Thus area of pentagon = area of + area of + area of =
Let . Denote the area of by .
can be found by Heron's formula.
Total area .
Construct and to partition the figure into , and .
Rotate with centre such that coincides with and is mapped to . Hence the area of the pentagon is still preserved and it suffices to find the area of the quadrilateral .
Hence = ()=
Since = , = and = , by SSS Congruence, and are congruent, so =
So the area of pentagon .
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