1981 USAMO Problems/Problem 1

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


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.


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.

See Also

1981 USAMO (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5
All USAMO Problems and Solutions

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