Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Find the answer

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 $b$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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 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: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.