## Wednesday, February 1, 2006

### Ptolemy Saves the Day! Topic: Geometry/Inequalities. Level: Olympiad.

Problem: (1997 MOP) Let $Q$ be a quadrilateral whose side lengths are $a, b, c, d$, in that order. Show that the area of $Q$ does not exceed $\frac{ac + bd}{2}$. Solution: Let $A$ be the area of $Q$. Consider writing $A$ in terms of $x,y,z,w$ as shown in the diagram. A common formula for the area of a triangle is equal to half of the product of two sides and the sine of the angle between them. So we have

$A = \frac{1}{2}(xy\sin{\theta}+wz\sin{\theta}+xw\sin{(180-\theta)}+yz\sin{(180-\theta)})$.

And since $\sin{\theta} = \sin{(180-\theta)}$, we get

$A = \frac{1}{2}\sin{\theta}(xy+wz+xw+yz) = \frac{1}{2}\sin{\theta}(x+z)(y+w) \le \frac{1}{2}(x+z)(y+w)$.

By Ptolemy's Inequality, we know $ac+bd \ge (x+z)(y+w)$, so then

$A \le \frac{1}{2}(x+z)(y+w) \le \frac{ac+bd}{2}$

as desired. QED.

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Comment: There should be a geometric, slice-it-up proof to this problem, but the trigonometric proof works out quite nicely as well. Ptolemy's Inequality is pretty useful; the equality case, when the quadrilateral is cyclic, is probably used the most. Whenever a problem has a cyclic quadrilateral with side lengths, Ptolemy is always worth a try. There is a really nice proof using complex numbers (or vectors) and the triangle inequality; see if you can find it.

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Practice Problem: Find the proof of Ptolemy's Inequality mentioned above.