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 .

$$\begin{array}{rl}[ACD]& =\frac{1}{4}\sqrt{(1+b+c)(-1+b+c)(1-b+c)(1+b-c)}\\ & =\frac{1}{4}\sqrt{((b+c{)}^2-1)(1-(b-c{)}^2)}\\ & =\frac{1}{4}\sqrt{(b+c{)}^2+(b-c{)}^2-(b^2-c^2{)}^2-1}\\ & =\frac{1}{4}\sqrt{2(b^2+c^2)-(b^2-c^2{)}^2-1}\\ & =\frac{1}{4}\sqrt{2(x^2+y^2+2)-(x^2-y^2{)}^2-1}\\ & =\frac{1}{4}\sqrt{2((x+y{)}^2-2xy+2)-(x+y{)}^2(x-y{)}^2-1}\\ & =\frac{1}{4}\sqrt{5-4xy-(x-y{)}^2}\\ & =\frac{1}{4}\sqrt{5-(x+y{)}^2}\\ & =\frac{1}{2}\end{array}$$

Total area