Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.0 National olympiad Prove it Ireland

Suppose $ABCD$ is a cyclic quadrilateral, with side lengths $a = AB$ , $b = BC$ , $c = CD$ , $d = DA$ . Prove that its area ( $ABCD$ ) doesn't exceed the following expression
$\frac{ab + ac + ad + bc + bd + cd}{6},$
with equality iff the quadrilateral is a square.

Solution

$(ABCD) = \sqrt{(\sigma - a)(\sigma - b)(\sigma - c)(\sigma - d)},$
whence pairing the factors in the expression $\sqrt{(\sigma - a)(\sigma - b)(\sigma - c)(\sigma - d)}$ in three different ways, and noting that by AM-GM, e.g.,
$\sqrt{(\sigma - a)(\sigma - b)} \le \frac{2\sigma - a - b}{2} = \frac{d + c}{2},$
we see that
$(ABCD) \le \frac{(a + b)(c + d)}{4},$
with equality iff $a = b$ and $c = d$ , that
$(ABCD) \le \frac{(a + d)(b + c)}{4},$
with equality iff $a = d$ and $b = c$ , and that
$(ABCD) \le \frac{(a + c)(b + d)}{4},$
with equality iff $a = c$ and $b = d$ . Consequently, combining these,
$3(ABCD) \le \frac{(a+b)(c+d)}{4} + \frac{(a+c)(b+d)}{4} + \frac{(a+d)(b+c)}{4} = \frac{ab + ac + ad + bc + bd + cd}{2},$
with equality iff $a = b = c = d$ .

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