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Geometry Difficulty 4.8 AIME Prove it Saudi Arabia

Let $ABCDE$ be a pentagon with $\widehat{A} = \widehat{B} = \widehat{C} = \widehat{D} = 120^\circ$ . Prove that
$4AC \cdot BD \ge 3AE \cdot ED.$

Solution

It is clear that $AB \parallel ED$ and $AE \parallel CD$ . Let $F$ be the intersection point of lines $AB$ and $CD$ . The inequality is equivalent to
$\frac{AC}{AF} \cdot \frac{BD}{FD} \ge \frac{3}{4}. \quad (1)$
Using the Law of Sines in triangles $ACF$ and $BDF$ , we get that (1) is equivalent to
$\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{\sin \widehat{ACF}} \cdot \frac{\sqrt{3}}{\sin \widehat{DBF}} \ge \frac{3}{4},$
so $\sin \widehat{ACF} \cdot \sin \widehat{DBF} \le 1$ , which is clearly true.
We have equality if and only if $\widehat{ACF} = \widehat{DBF} = 90^\circ$ , hence $BC = AB = CD = BF = CF$ . This means $AEDF$ is a rhombus, and $B$ and $C$ are the midpoints of $AF$ and $DF$ respectively.

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