# Trisect unknown angle using pencil, straight edge & compass; Prove validity of technique

This question was posed by my high school geometry teacher, for extra credit:

Is it possible, using only a pencil, a straightedge (not a ruler) and a compass to trisect an angle of unknown value? If so, describe the technique, and mathematically prove its validity. If not, mathematically prove the impossibility.

I recall wondering at the time, why he was posing this question to a freshman high school geometry class, and if he genuinely expected any of us to come up with an answer. Nothing more was said about it and to my knowledge, nobody submitted an answer. But the question stuck in my mind, and I would often spend hours puzzling over it, trying over and over to come up with a valid solution.

I'll be 60 years old this July, and I'm no closer to an answer now than I was that day, so many years ago. If anyone knows the answer, I'd appreciate a simple explanation, as I am no mathematician.

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Is the Wikipedia article simple enough? – Chris Eagle Jan 7 '11 at 4:43
Terry Tao posted on his webpage a few days back on this. You can look it up here. terrytao.wordpress.com/2011/08/10/… – user17762 Aug 15 '11 at 12:24
Can you fold the paper? – p1l0t Mar 9 '15 at 13:59

The question, given the "unknown value", is equivalent to asking whether it is possible to trisect a general angle using only straightedge and compass.

This is known as the problem of trisecting the angle, one of the three classic problems of greek geometry, together with the problem of duplicating the cube (given a line segment, construct using only compass and straightedge a second segment such that the cube with edge length the second segment has double the volume of the cube with edge length the original line segment) and squaring the circle (given a line segment, construct a second line segment using only straightedge and compass whose square has the same area as a circle with radius the original line segment). The use of "pencil" is implicit in the statements.

The three problems are also known as the three classical impossibilities, because it is impossible to come up with such a technique. This does not mean nobody has been able to find one, but that it has been proven that no such techniques exist.

Below, everytime I say "construct", I mean "using only pencil, a straightedge, and a compass". The straightedge lets you extend any line segment, and construct a line segment through two given points; the compass lets you draw a circle with center a given point and going through another given point. You can determine points by intersecting things you can draw.

Think in terms of cartesian coordinates, place a point in the plane which will be the origin, and assume that you can mark off a distance of $1$ (that is, construct a line segment of length $1$). If you consider all line segments with one end at the origin that you can construct, the collection of all numbers $(a,b)$ that are the other endpoint of these "constructible line segments" form what are called the "constructible numbers." It is easy to see that if you have two constructible numbers, you can also construct their sum, difference, product, and quotient, just using a straightedge. A little harder is to show that if you have a constructible number $a\gt 0$, then you can also find its square root $\sqrt{a}$ if you also use the compass (the idea is that the compass will let you construct a curve that is given by a quadratic equation, the straightedge will let you construct a curve that is given by a linear equation, and playing one against the other you can obtain square roots). You can then repeat this process and obtain yet more constructible numbers, and more, and more, using these operations (addition, subtraction, multiplication, division, and taking square roots of nonnegative numbers already obtained).

Turns out that these are the only numbers you can construct using straightedge and compass. So, for example, one cannot construct the number $\sqrt[3]{2}$, because it cannot be obtained from $0$ and $1$ through a sequence of additions, subtractions, multiplications, divisions, and taking square roots (this is not perhaps obvious; it is nonetheless true; it has to do with the fact that the polynomial with rational coefficients, smallest degree, and leading coefficient $1$ which has $\sqrt[3]{2}$ as a root is $x^3-2$, which is of degree $3$; if it were constructible, the smallest such polynomial would have to have degree a power of $2$). Since this is the length you would need to duplicate the cube if you are given a segment of length $1$, which is constructible, duplicating the cube is in general impossible.

Squaring the circle is likewise impossible, because if you are given a line segment of length $1$, then you would need to construct a line segment of length $\sqrt{\pi}$. If you could, then you would also be able to construct $\pi$; but $\pi$ is a transcendental number (proven by von Lindermann in 1882), which means it cannot be constructed by the processes described above.

Dealing with the trisection of the angle is a bit more involved, but not much. If you can construct an angle of $\alpha$ radians, then you can construct a right triangle with angle $\alpha$, and with that triangle you can construct a segment of length $\cos(\alpha)$. Conversely, if you can construct a segment of length $\cos(\alpha)$, and then you draw a perpendicular to it of length $\sqrt{1-\cos^2(\alpha)}$ (which can be done), then you get a right triangle which has an angle of $\alpha$ radians.

Now, since we can construct equilateral triangles using straightedge and compass, and the inner angle of an equilateral triangle is $\frac{\pi}{3}$ radians, then the angle of $\frac{\pi}{3}$ is constructible, so someone could possibly "give you" that angle to trisect. In order to be able to trisect this angle, you need to be able to construct an angle of $\frac{\pi}{9}$ radians (20 degrees). So if trisecting the angle were possible, you would be able to construct a line segment of length $\cos(20^{\circ})=\beta$. However, using the triple angle formula $$\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta)$$ we have that $4\beta^3 - 3\beta - \frac{1}{2} = 0$ (set $\theta=\frac{\pi}{9}$ in the above), so that the polynomial with integer coefficients, of smallest degree, and leading coefficient $1$, that has $\cos(20^{\circ})$ as a root is $x^3 - \frac{3}{4}x - \frac{1}{8}$. As with duplication of the cube, because the degree of this polynomial is not a power of $2$, the roots cannot be constructed at all.

This means that you cannot even construct an angle of $20^{\circ}$, which means that you would never be able to trisect an angle of $60^{\circ}$; in particular, there can be no general method for trisecting a given angle (since you can be given an angle of $60^{\circ}$, but you cannot even construct the angle of $20^{\circ}$, even knowing that this is the angle you are supposed to construct). So it is also impossible to trisect the angle.

If you allow yourself other tools, such as a ruler instead of a straightedge, then duplicating the cube and trisecting the angle become possible; it is also possible to do it with "origami" (by folding the paper you are given in different ways). But if you restrict yourself to the classic tools of straightedge and compass, it cannot be done.

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If you are allowed to mark the straightedge with the pencil, Archimedes showed how to do it. Just search for "Archimedes angle trisection".

Here is one URL: https://en.wikipedia.org/wiki/Angle_trisection

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## protected by J. M.Jun 2 '13 at 16:56

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