You have an angle and you have a pen, paper, compass and a straight edge. You don't know how big the angle is, divide this angle into three equal part using only the material that is listed here? If not possible what other tool is needed (protractors are not allowed)?
The angle trisection problem
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0Welcome to math.SE. Please consider glancing over [FAQ](http://math.stackexchange.com/faq) and please refrain from using imperative style when asking questions. – 2011-08-18
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2Have you looked at [Wikipedia: Angle_trisection](http://en.wikipedia.org/wiki/Angle_trisection) ? – 2011-08-18
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1In general this is not possible, although there are some specific angles that do permit trisection. – 2011-08-18
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0Never heard of it before, but trisection can be done if using of [neusis](http://en.wikipedia.org/wiki/Neusis_construction) is allowed. There is also [Wolfram demonstration](http://demonstrations.wolfram.com/ArchimedessNeusisAngleTrisection/) devoted to trisecting an angle using it. – 2011-08-18
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1@Sasha: "neusis" simply means a *marked* straightedge is allowed. There are also a number of curves that one can use, as well as a carpenter's square, for the trisection of angles. – 2011-08-18
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1Also, there's the tomhawk thingy(Look it up in [link](http://en.wikipedia.org/wiki/Tomahawk_%28geometric_shape%29)), if you have a pen along with your compass and straightedge,you can build this. – 2011-08-18
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0Terry Tao has a recent blog post http://terrytao.wordpress.com/2011/08/10/a-geometric-proof-of-the-impossibility-of-angle-trisection-by-straightedge-and-compass/ proving the impossibility of angle trisection with straight-edge and compass. – 2011-08-18
2 Answers
As was already noted in comments, the problem admits no solution using compass and straight edge. You can render the problem solvable by either allowing neusis or tomahawk.
Here I will explain the Archimedes solution using neusis.
Given an angle $\alpha$, draw a circle centered at its tip point $\mathbf{O}$. Draw a chord $\mathbf{AC}$. Let $\beta = \angle \mathbf{BOC}$, and let $\gamma = \angle \mathbf{BCO}$.
It follows elementary that $\angle \mathbf{OBA} = \beta+\gamma$ and $\angle \mathbf{OAB} = \alpha - \gamma$. Since $\mathbf{OA} = \mathbf{OB}$ as radii, $ \alpha - \gamma = \beta+\gamma$, giving $\gamma = \frac{\alpha - \beta}{2}$. If we further impose $\beta = \gamma$, we get $\beta=\gamma=\frac{\alpha}{3}$.
In this configuration, $\mathbf{CB} = \mathbf{OB}$ as sides opposite to equal angles, which is how the neusis comes in.
One would use the marked ruler, to make $\mathbf{CB}$ equal to the radius of the circle.
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0https://youtu.be/sz6n5YnUGL8 is this video valid? – 2016-06-25