1998 JBMO Problems/Problem 2
Problem 2
Let be a convex pentagon such that , and . Compute the area of the pentagon.
Solutions
Solution 1
Let
Let
Applying cosine rule to we get:
Substituting we get:
From above,
Thus,
So, area of =
Let be the altitude of from .
So
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 =
By
Solution 2
Let . Denote the area of by .
can be found by Heron's formula.
Let .
Total area .
By durianice
Solution 3
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 .
- SomebodyYouUsedToKnow
See Also
1998 JBMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 | ||
All JBMO Problems and Solutions |