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
See Also
1998 JBMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All JBMO Problems and Solutions |