Trisecting an Angle Using Compass and StraightedgeDate: 04/29/2004 at 14:17:13 From: Joe Subject: Trisect I've been working on trisecting an arbitrary angle, and I can do it with no measuring and only a straightedge and a compass. I can prove it with "Geometer's Sketchpad." I divide the arc of the angle and not the angles as you would in bisecting. My reasoning is that one half the circumference of a circle that is two-thirds the size of another is one third the circumference of the larger. I know the chords do not follow but I have a means of correcting this. This follows for any segment of the circle. It is simple and it checks out with any angle and by double checking it with The Geometer's Sketchpad or using a known angle and checking my answer by figuring the chord and comparing. Where can I submit my effort for confirmation? It has been very frustrating to get anyone to take a look. All I have heard is that it's impossible. Date: 04/29/2004 at 14:28:45 From: Doctor Peterson Subject: Re: Trisect Hi, Joe. Sorry--you can't prove ANYTHING with Geometer's Sketchpad! That just draws things and calculates angles and lengths, but only to some level of accuracy. The kind of proof that is required for a geometric construction, under the rules of the trisection game, is a logical proof that it is EXACTLY correct, which the computer can't do (at least not that program). The following page may help you see what I mean by saying that Sketchpad can't prove anything: a proof is a logical derivation, not just a demonstration using measurements or calculations. What you have done is not a proof, judging by your description. Angle Trisection: Construction vs. Drawing http://mathforum.org/library/drmath/view/55366.html But I sort of collect attempted trisections, and I'd be interested in seeing how accurate yours is compared to others I've seen; if it really shows zero error on Sketchpad for all angles, then it must be pretty interesting! I know it WILL turn out to be wrong in some way, since it has been proven to be impossible: Impossible Constructions http://mathforum.org/dr.math/faq/faq.impossible.construct.html But examining it can still be instructive for me, and it will help you understand what's going on. Please send me instructions for the construction (no picture is necessary--it's better if you make sure the instructions completely describe what you do), and I'll have a look at it and let you know what I think. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 05/24/2004 at 12:58:01 From: Joe Subject: Trisect Thanks for your interest. As per your invite, here is my method. 1. Construct an angle AOB. (In the example it will be larger than 90 degrees for simplicity.) 2. Bisect AOB with ray OC, so that COA = COB. 3. Bisect COB with ray OD, so that COD = DOB; and extend the bisector OD beyond O for later use. 4. On OA use the compass to mark off three equal segments OX, XY, YZ of (arbitrary) unit length. 5. Using O as center, construct arcs through X, Y, and Z that intersect OA and OB; the intersection of the arc through Y will be called U. 6. With the compass set at three units (the length of OZ), find the point V on the "back" side of line OD that is that distance from U; with the compass point at V, construct an arc through U. Let W be the point where this arc crosses ray OC. 7. Construct segment YW. Copy this length to the arc through Z (with radius 3 units), forming a chord ZT of the same length. OT trisects AOB, so that BOT = 1/3 AOB. Date: 05/24/2004 at 23:39:42 From: Doctor Peterson Subject: Re: Trisect Hi, Joe. I took the time tonight to try doing your construction on Sketchpad. It looks like I have it right, because it is pretty close to a trisection. I wonder if you might have Sketchpad set to show angles only to the nearest degree; I have it set for thousandths of a degree, and find that for a 90 degree angle, the error in the construction is about 0.034 degrees (about 2 minutes of angle), and for a 180 degree angle, the construction is off by about 0.45 degrees (almost 30 minutes). That's not bad as approximate trisections go, but it certainly is not a true trisection. I also wanted to further explain why what you have given is not what you need to do to solve the trisection problem. Let's compare it to a bisection of an angle. If you gave me the standard construction for bisecting an angle, and said that you proved it by applying it to many angles and seeing that it was always correct, I would have to tell you that you had not yet demonstrated that you really had a bisection, because, although the construction was correct, you had not PROVED it. To prove it, you have to make a logical demonstration that the construction MUST ALWAYS work. Your proof might look something like this: Construction: Given angle AOB. Construct arc AB with center O, with A and B on the rays of the given angle. Construct a circle with center A passing through B, and one with center B passing through A. Construct the ray OC, where C is one of the points of intersection of the two circles just drawn. It is claimed that OC bisects angle AOB, that is, that angles AOC and BOC are congruent. Proof: Consider triangles AOC and BOC. By the construction of points A and B, corresponding sides AO and BO are congruent. By the construction of the circles with centers at A and B, sides AC and BC are congruent, since they are radii of two circles with the same radius. Trivially, sides OC in both triangles are congruent. Therefore, by the SSS congruence theorem, triangles AOC and BOC are congruent, and so angles AOC and BOC are congruent. That is what you have to present in order to submit a trisection: not a measurement that makes it look right, but a proof that it must be exactly right, no matter what angle you start with. And that is what has been proved to be impossible. So no construction you do will meet the requirements of the "game". Your construction is better than many I've seen, especially for acute angles; but "close" doesn't count here. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 05/25/2004 at 12:38:42 From: Joe Subject: Thank you (Trisect) Thanks so much for the evaluation of my trisection method. I think I used 1/100th accuracy for my Sketchpad drawing. If I go backwards from a known angle and chord and calculate it comes out as accurate as pi. After reading your proof comment I can see that I have a lot to learn. I was not aware that construction was not enough to satisfy the requirement. I hate to accept anything as impossible so I may continue. It keeps my mind alert. Thanks again. Date: 05/25/2004 at 22:21:32 From: Doctor Peterson Subject: Re: Thank you (Trisect) Hi, Joe. I think I referred you earlier to this page: Impossible constructions http://mathforum.org/dr.math/faq/faq.impossible.construct.html As that explains, it has been proved to be impossible to exactly trisect any arbitrary angle using only compass and unmarked straightedge; and as I pointed out, "exact" to a mathematician means something entirely different from "as close as you can measure"; it means "provably exact". "Impossible" is a common thing in math. For example, it is impossible to find two odd numbers whose sum is odd; it is easy to prove that the sum of two odd numbers is always even, so we just accept that. I've never heard of people who, when told this fact, spend their lives trying to find two odd numbers whose sum is odd; but there have been thousands of people who have wasted a lot of time trying to find a way to trisect an angle. Most of them probably have no idea that a valid trisection is not just a drawing but a proof! They just hear that it can't be done, don't understand how that could be proved, and decide to show that it doesn't apply to them. Since it's not as obvious as adding odd numbers, they never realize how silly what they are doing is. To keep your mind alert, I would suggest something more "constructive" than trying to do what has been proved impossible. Take some time to learn the basics of geometric construction and proof, then work through the exercises in a good book on the subject, which will ask you to do constructions that are hard but possible. That way you are giving yourself a major mental challenge, but one that you can really master with effort. It's a lot more rewarding! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 05/26/2004 at 10:40:36 From: Joe Subject: Thank you (Trisect) Thank you, Doctor, for the lesson. I really appreciate it. I came close but no prize. Thanks again. I will put your note about proof under the glass on my desk as a reminder to leave the trisection impossibility alone! Have a nice summer. Sincerely, Joe |
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