Difference between revisions of "1981 USAMO Problems/Problem 1"
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== Solution == | == Solution == | ||
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Let <math>n=3k+1</math>. Multiply throughout by <math>\pi/3n</math>. We get | Let <math>n=3k+1</math>. Multiply throughout by <math>\pi/3n</math>. We get | ||
| − | <math>\frac{\pi}{3} | + | <math>\frac{\pi}{3} = \frac{\pi \times k}{n} + \frac{\pi}{3n}</math> |
Re-arranging, we get | Re-arranging, we get | ||
| − | <math>\frac{\pi}{3} | + | <math>\frac{\pi}{3} - \frac{\pi \times k}{n} = \frac{\pi}{3n}</math> |
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. | 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. | ||
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 | ||
| − | <math>\frac{\pi}{3} | + | <math>\frac{\pi}{3} - \frac{\pi \times k}{n} = \frac{2\pi}{3n}</math> |
Hence in this case, we will have to subtract <math>k</math> times the original angle from <math>\frac{\pi}{3}</math> to get twice the the trisected angle. We can bisect it after that to get the trisected angle. | Hence in this case, we will have to subtract <math>k</math> times the original angle from <math>\frac{\pi}{3}</math> to get twice the the trisected angle. We can bisect it after that to get the trisected angle. | ||
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| + | ==Generalization== | ||
| + | If regular polygons of <math>m</math> sides and <math>n</math> sides can be constructed, where <math>m</math> and <math>n</math> are relatively prime integers greater than or equal to three, then regular polygons of <math>mn</math> sides can be constructed. Indeed, such a polygon can be constructed by first constructing an <math>m</math>-gon, and then creating <math>m</math> distinct <math>n</math>-gons with at least one vertex being a vertex of the <math>m</math>-gon. | ||
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== See Also == | == See Also == | ||
{{USAMO box|year=1981|before=First Question|num-a=2}} | {{USAMO box|year=1981|before=First Question|num-a=2}} | ||
Revision as of 17:16, 17 May 2015
Contents
Problem
Prove that if 
is not a multiple of 
, then the angle 
can be trisected with ruler and compasses.
Solution
Let 
. Multiply throughout by 
. We get
Re-arranging, we get
A way to interpret it is that if we know the value 
, then the reminder angle of subtracting times the given angle from 
gives us 
, the desired trisected angle.
This can be extended to the case when 
where now, the equation becomes
Hence in this case, we will have to subtract times the original angle from to get twice the the trisected angle. We can bisect it after that to get the trisected angle.
Generalization
If regular polygons of 
sides and sides can be constructed, where and are relatively prime integers greater than or equal to three, then regular polygons of 
sides can be constructed. Indeed, such a polygon can be constructed by first constructing an -gon, and then creating distinct -gons with at least one vertex being a vertex of the -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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
