HMMT February

Geometry Difficulty 5.9 AIME, harder Prove it United States

Problem:
Let $a \neq b$ be positive real numbers and $m, n$ be positive integers. An $m+n$ -gon $P$ has the property that $m$ sides have length $a$ and $n$ sides have length $b$ . Further suppose that $P$ can be inscribed in a circle of radius $a+b$ . Compute the number of ordered pairs $(m, n)$ , with $m, n \leq 100$ , for which such a polygon $P$ exists for some distinct values of $a$ and $b$ .

Solution

Solution:
Letting $x=\frac{a}{a+b}$ , we have to solve
$m \arcsin \frac{x}{2} + n \arcsin \frac{1-x}{2} = \pi$
This is convex in $x$ , so if it is to have a solution, we must find that the LHS exceeds $\pi$ at one of the endpoints. Thus $\max (m, n) \geq 7$ . If $\min (m, n) \leq 5$ we can find a solution by the intermediate value theorem. Also if $\min (m, n) \geq 7$ then
$m \arcsin \frac{x}{2} + n \arcsin \frac{1-x}{2} \geq 14 \arcsin (1 / 4) > \pi$
The inequality $\arcsin (1 / 4) > \frac{\pi}{14}$ can be verified by noting that
$\sin \frac{\pi}{14} < \frac{\pi}{14} < \frac{3.5}{14} = \frac{1}{4}$
The final case is when $\min (m, n) = 6$ . We claim that this doesn't actually work. If we assume that $n=6$ , we may compute the derivative at 0 to be
$\frac{m}{2} - 6 \cdot \frac{1}{\sqrt{3}} = \frac{m-\sqrt{48}}{2} > 0$
so no solution exists.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.