# 1981 USAMO Problems/Problem 1

## Problem

Prove that if $n$ is not a multiple of $3$, then the angle $\frac{\pi}{n}$ can be trisected with ruler and compasses.

## Solution

Let $n=3k+1$. Multiply throughout by $\pi/3n$. We get $\frac{\pi}{3} = \frac{\pi \times k}{n} + \frac{\pi}{3n}$

Re-arranging, we get $\frac{\pi}{3} - \frac{\pi \times k}{n} = \frac{\pi}{3n}$

A way to interpret it is that if we know the value $k$, then the remainder angle of subtracting $k$ times the given angle from $\frac{\pi}{3}$ gives us $\frac{\pi}{3n}$, the desired trisected angle.

This can be extended to the case when $n=3k+2$ where now, the equation becomes $\frac{\pi}{3} - \frac{\pi \times k}{n} = \frac{2\pi}{3n}$

Hence in this case, we will have to subtract $k$ times the original angle from $\frac{\pi}{3}$ to get twice the the trisected angle. We can bisect it after that to get the trisected angle.

## Generalization

If regular polygons of $m$ sides and $n$ sides can be constructed, where $m$ and $n$ are relatively prime integers greater than or equal to three, then regular polygons of $mn$ sides can be constructed. Indeed, such a polygon can be constructed by first constructing an $m$-gon, and then creating $m$ distinct $n$-gons with at least one vertex being a vertex of the $m$-gon.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 