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Difference between revisions of "1981 USAMO Problems/Problem 1"

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<math>\frac{\pi}{3} - \frac{\pi \times k}{n} = \frac{\pi}{3n}</math>
 
<math>\frac{\pi}{3} - \frac{\pi \times k}{n} = \frac{\pi}{3n}</math>
  
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A way to interpret it is that if we know the value <math>k</math>, then the reminder angle of subtracting <math>k</math> times the given angle from <math>\frac{\pi}{3}</math> gives us <math>\frac{\pi}{3n}</math>, the desired trisected angle.
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A way to interpret it is that if we know the value <math>k</math>, then the remainder angle of subtracting <math>k</math> times the given angle from <math>\frac{\pi}{3}</math> gives us <math>\frac{\pi}{3n}</math>, the desired trisected angle.
  
 
This can be extended to the case when <math>n=3k+2</math> where now, the equation becomes
 
This can be extended to the case when <math>n=3k+2</math> where now, the equation becomes
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[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
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[[Category:Geometric Construction Problems]]

Latest revision as of 16:31, 8 July 2019

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.

See Also

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

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