Constructing the Trisection of an Angle
Date: 07/20/2008 at 01:19:35 From: Land Subject: Trisection This method of trisecting an angle works very well for a 60 degree angle: Construct a 60 degree angle with b being the vertex. Construct an arc AC to intersect the sides of the angle at A and C. Extend the arc to form a complete circle. Construct the chord of the arc AC. Bisect the angle ABC. The bisector will intersect the chord AC at E. Using C as the vertex with radius CE with starting point E, construct an arc that intersects the circle at F. Construct the line FE and extend until it intersects the circle at G. Construct the line GB and extend until it intersects the circle at H. Using G as the center and GH as the radius, draw a circle. This circle will intersect FE at L. Construct the line LB and extend until it intersects the original circle at P. The angle PBC is 1/3 of the angle ABC. It is 20 degrees. This method is not exact to the thousandths of a degree, but it is always accurate to a hundredth. I would say that is close enough for folk music but not for a symphony orchestra. Why are so many angles trisectable? Eg, 180, 135, 90, 67.5, 45, 22.5, etc. There are many angles that are trisectable. It is as if there is an ocean of non-trisectability that is punctuated by islands of trisectability bravely poking their heads above the waves. What does Wantzel's proof say about the 90, or the 45? If his proof declares that these angles, like the 60, can't be trisected when, in fact, they can, what does the proof prove? I am skeptical of proofs that prove too much. Why are some angles trisectable and others not? Don't all angles encounter the same constraints about taking the cube root? Or for some angles, do some of the terms drop out leaving only a simplified equation that can be solved? I have never seen this discussed anywhere.
Date: 07/20/2008 at 04:47:49 From: Doctor Jacques Subject: Re: TRISECTION Hi Land, Why are some angles trisectable and others not? Let's answer the second question first. As you apparently know, it is not possible to find a general construction, using only an unmarked ruler and a compass, that will allow you to trisect any angle (there is much "fine print" in the rules that define what is allowed). There is a rigorous proof of this; the proof itself is quite simple, but you have to know something of Galois theory, which is not that simple. I will try to give you an informal argument (NOT a correct proof!) that gives you a feeling of the deep reason behind that. When you trisect an angle x, there are really three solutions: y1 = x/3 y2 = x/3 + 120° y3 = x/3 + 240° You can draw the angle x as AOB in a circle of center O. If you choose any of y1, y2, or y3, and draw the corresponding arc three times starting from A, you always end up in B. From an algebraic point of view, if you know nothing about the angle x, these three solutions are completely equivalent--there is no way to distinguish them. Now, in any construction with ruler and compass, you can create new points in three ways: 1. By intersecting two lines. This gives one new point. 2. By intersecting a line and a circle. This gives two new points. 3. By intersecting two circles. This gives again two points. I'm deliberately ignoring special cases, like parallel lines, line tangent to the circle, or line that does not intersect the circle; once again, this is not a rigorous proof. The point is that every time you use (2) or (3), you multiply the number of possible choices by 2. Therefore, at the end of your construction, the number of possible results will be a power of 2. It is, however, possible that some combinations of choices will result in the same final point. However, as the three solutions of the problem are equivalent, you should expect that the various results of the construction will be split evenly between these three solutions; in other words, the number of possible choices (a power of 2) would be divisible by 3, which is clearly impossible. In terms of Galois theory, the (rigorous) translation of this argument is that it is not possible to solve an irreducible cubic equation only by extracting square roots: trisecting an angle is equivalent to solving a cubic equation, and constructions (2) and (3) are equivalent to extracting square roots. So, why are some angles trisectable? The reason is that for some angles, the cubic equation may not be irreducible: it may have a rational root. In that case, you are left with a quadratic equation that you can solve with square roots. (In this context, a "rational root" means a rational function of the coefficients). From a geometric point of view, this means that, for some angles x, the three solutions (y1,y2,y3) are not equivalent: you can construct at least one of the solutions without using constructions (2) or (3) (technically, you must also be able to draw parallel lines, which may require reformulating the rules). The first case where this happens is when x = 0°. In this case, the solutions are 0°, 120°, and 240°. Note that one solution is the original angle itself: you can certainly construct it without using (2) or (3). This leaves you with 2 other solutions, and this is indeed a power of 2. Another case is when the angle x is a rational multiple (p/q) of 360°, where p and q are integers and q is not a multiple of 3. From now on, we will measure angles in units of full circles, so that an angle of 1 means 360°. If gcd(3,q) = 1, there is an integer r such that 3r = 1 (mod q), i.e., 3r = nq + 1 (You can compute r using the extended Euclidean algorithm). If we take y = rx = (rp)/q, we have: 3y = 3rp/q = p/q + pn = x + pn As pn represents an integer number of full circles, this is the same angle as x: y is one of the solutions, and you can construct the two other solutions by solving quadratic equations (in practice, you would simply draw arcs of 120° and 240° from the first solution). For example, assume that we want to trisect an angle of 1/7 (of a circle). We find easily that: 3*5 = 1 (mod 7) This means that one of the solutions is 5 times the original angle, i.e. y1 = 5/7 of a circle. Indeed, we have: 3*y1 = 3*(5/7) = 15/7 = 1/7 + 2 (full circles) and you can find the other 2 solutions by drawing angles of 120° on either side of y1. Note that in this example, y1 is not the "natural" solution: it lies outside the acute angle x. In this case, the natural solution is obtained by adding 120° = 1/3: y2 = y1 + 1/3 = 5/7 + 1/3 = 22/21 which is the same angle as 1/21, the "natural" solution. Note that this construction requires that you have the original angle at your disposal (you cannot construct 1/21 directly), and that you know q, the denominator of that angle. In summary, the only trisectable angles are (p/q)*360°, where p and q are integers, and q is not a multiple of 3 (the first case corresponds to q=1). Please feel free to write back if you require further assistance. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/
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