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

This problem needs a solution. If you have a solution for it, please help us out by adding it. 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 reminder 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.

1981 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5
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1981 USAMO Problems/Problem 1

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