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 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
Solution 2
Let . Denote the area of
by
.
can be found by Heron's formula.
Let .
\begin{align*} [ACD]&=\frac{1}{4}\sqrt{(1+b+c)(-1+b+c)(1-b+c)(1+b-c)}\\ &=\frac{1}{4}\sqrt{((b+c)^2-1)(1-(b-c)^2)}\\ &=\frac{1}{4}\sqrt{(b+c)^2+(b-c)^2-(b^2-c^2)^2-1}\\ &=\frac{1}{4}\sqrt{2(b^2+c^2)-(b^2-c^2)^2-1}\\ &=\frac{1}{4}\sqrt{2(x^2+y^2+2)-(x^2-y^2)^2-1}\\ &=\frac{1}{4}\sqrt{2((x+y)^2-2xy+2)-(x+y)^2(x-y)^2-1}\\ &=\frac{1}{4}\sqrt{5-4xy-(x-y)^2}\\ &=\frac{1}{4}\sqrt{5-(x+y)^2}\\ &=\frac{1}{2} \end{align*}
Total area