# 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

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 $\theta$, 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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TRISECTION OF A GIVEN ANGLE WITH THE HELP OF COMPASS AND RULER - Hi i am a student, according to me, we can trisect an angle. For trisection there are some rules which are following- First of all draw a baseline of a given measurement, after this if we want to trisect a 120 degree angle then as conventionally draw it with the help of compass and ruler but remember one thing that when you are making the initial arc measure it 3 cm. Then use the formula of length of arc ( 2πrθ⁄360 )and divide it by 3 because you want to trisect the given angle. ( take π= 22/7) e.g.- 2πrθ/(360*3) = (2*22*3*120)/(7*360*3) = 44/21 = 2.095 cm approximately

Draw a baseline OX.
Draw an arc AB having radius 3 cm with center O, now draw a 120° angle with the help of compass and ruler (angle AOB = 120°).
Now measure 2.095 cm in your compass and with center B we will cut the previous arc AB on D and with same measurement taking D as center cut an arc DE.
Now join   OE & OD, it is the formation of angle AOE, EOC and DOB.
Angle AOE = 40°, EOC = 40° and DOB = 40°.

With the help of the above method we can trisect any angle (30°, 45°, 60°, 75°, 90°, 120°, 135°, 150°…). We can also divide an angle into four, five… equal parts ( tetra section, pent section….) by this method. Thank you. -Rishi Kr. Sharma

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i dont know. i m just thinking that –  Rishi Sharma May 15 '13 at 3:49
Sorry, that was rude of me. It looks to me that your answer will not qualify, since you take $\pi=22/7$, which is false. Also, you only mention certain angles, not any given angle. Also, it is provable that a $60^\circ$ angle cannot be exactly trisescted - it's not possible to construct a $20^\circ$ angle. –  alex.jordan May 15 '13 at 3:54
i know that pi is not equal to 22/7, it is a irrational no. i also agree that a 60 degree angle cannot be exactly trisected but if we assume that pi=22/7 or 3.1415 then we can divide 60 but there will be some % error. can you please try it? –  Rishi Sharma May 15 '13 at 9:15
@RishiSharma: The problem is not to approximate a trisection of an angle, it is to actually trisect the angle. –  Jim May 24 '13 at 17:16
1. Draw two lines intersecting at a point (point a = (l1p0, l2p0).
2. Draw an arc centered at point a crossing both lines line 1=l1p2, line 2=l2p2.
3. Without adjusting the compass draw short arcs along both lines till you have two more points on each line(l1p2, l1p3 and l2p2, l2p3).
4. Now go back to the point where the first arc crosses one line and make an arc that spans the chord of the first arc draw an arc from one the point where the first arc crosses one line to the other point where the same arc crosses the other line.
5. draw a line from l1p3 to l2p3.
6. Take the compass and draw an arc along this new line length same as chord from l1p1 to l2p1. swing arcs from l1p3 and l2p3.
7. Draw lines from point a to these new points along the chord from l1p3 and l2p3. These will be trisecting lines for any given angle up to 180. At 180 and above you need to use the other side of the lines intersecting point a and perform the same procedure on the side that is less than 180, extending the final lines to the side greater than 180.

I figured this out while waiting for my wife and son at the mall yesterday. I've toyed with the problem for over 50 years.

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Without a picture this is incomprehensible, nevermind that it contradicts a theorem that's been known for almost 200 years. –  Jim May 24 '13 at 17:14
For what it's worth, one way of 'rescuing' this content would be to pose it not as an answer but as a question: assuming that you're willing to trust the last 100+ years of mathematics on this problem, a diagram and a question of 'why doesn't this angle trisection work?' would likely be more favorably received than this answer has. –  Steven Stadnicki May 24 '13 at 17:56

This problem was posed to my college class and likewise we were told it was impossible. Nevertheless, I thought about it overnight and came up with a very simple solution, albeit, it "sort of" cheats. However, if you were stuck on a desert isle with nothing but a compass, straightedge and pencil and your life depended on trisecting an angle, you CAN do it in a very short time.

Here's the trick. Bisect the angle first. Then, bisect the new angle. Then, bisect the new angle....

1/2 - 1/4 + 1/8 - 1/16... = 1/3.

SIMPLE! In only 3 bisections you get a trisection within the accuracy of your instruments.

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By the way, using this same technique I've proven that you can multi-sect any angle as long as the power series converges to that fraction. You can also use combinations of bisects and trisects, quintsects and septsects etc. –  Larry Wakeman Dec 10 '12 at 15:42
(Sorry, I don't have the math font) You represented the summation incorrectly. It should be the sum 1 to infinity of -(-1/2)^n. The fraction at each step looks like 1/3(d +- 1) / d where d is 2, 4, 8, 16, 32... so d +- 1 = 3, 3, 9, 15, 33 which is always divisible by 3. This obviously converges to 1/3. –  Larry Wakeman Dec 10 '12 at 15:51
But the question asks to perform an exact trisection! –  martini Dec 10 '12 at 16:05
(About not having the "math font", the setup here uses a Javascript-based feature called MathJAX, that allows TeX-style formulas. If it's not working with your browser, it might be because Javascript is disabled for the site?) –  hardmath Dec 10 '12 at 16:15
Where does it say "exact"? Anyway, I said it sort of cheats. But you can trisect as accurately as you want, and certainly within the accuracy of your equipment in only 3 to 5 steps. –  Larry Wakeman Dec 10 '12 at 16:37

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

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