|
|
The Trisection of AnglesIntroductionThe heading above is the title of my book: Since there are no copies available from Vantage Press, I plan to make it available here, if space permits, or somewhere on the web. In the meantime, I will present some ideas here. I am well aware of the "proofs" of impossibility. There are no such geometric proofs. All such "proofs" are based on number theory, and my basic contention is that the mathematical "proofs" of impossibility are based on a mis-application of "number" to Geometry. There are a number of things I have to do before I begin getting the book on these pages. One of those things is to find out how I am going to get the diagrams of the constructions here. I must also do some work on the two preceding sections of my site. After all, the work there was the basis for the work here. In the meantime I will give the "recipe" for a construction that is the basis of the proof of the construction I will present later for the trisection of angles. I could call it the "Rubino Theorem", but for the sake of modesty I shall refer to it as the XYV Relationship. It contains the required 1:3 relationship of angles: i.e., an angle and its triple angle or, conversely, an angle and its trisection. It is relatively simple and intuitive, being composed by the superimposition of what I call a regular X, a regular Y, and a regular V, which I shall now define. In fact, it is so simple and intuitive that it has already been defined sufficiently to visualize it! Try it. Try it on a typewriter. Type a capital X, backspace and type a capital Y, backspace again and type a capital V. Without a typewriter, try to visualize the structure by squeezing this XYV together so that one is on top of the other. You could also hand print them as large as you like on a piece of paper. To be a little more formal, let me define the term "regular" as I have used it above, to simply mean that the line segments composing them are equal. For example, given any pair of intersecting lines, mark off equal line segments from the point of intersection to get a regular X. Actually, you get two X's depending on how you look at it; i.e., if you lay the printed X on it's side you would get another one that is a little shorter and a little fatter - unless of course the X is made from lines intersecting at right angles. Then the two X's would be the same. For the sake of keeping this discussion simple, let's just forget about the right angled X, and the short fat one too. Also, let's focus our attention on the upper angle of the regular X so that we can transform it into a regular Y by simply squeezing the lower legs of the X an equal amount till they coincide. Since the upper angle of both the X and Y thus formed are equal, it is a simple matter to superimpose one upon the other. To superimpose the regular V on both of them is even simpler: just draw the lines connecting the endpoints of the Y. Notice that each of those lines intersect a lower leg of the X. That angle is the critical angle. Then the angles formed by the side of the V with the Y are equal to 1/3 the critical angle. Conversely, the critical angle is 3 times the angles formed by the side of the V with the Y.
THIS IS NOT A TRISECTION I can't emphasize that enough. It is merely a relationship of angles that contains an angle equal to 1/3 another angle. But if I start with the triple angle - the one I called the critical angle above - and I reconstructed an XYV relationship from it then I will have a legitimate trisection. Don't you agree? I would like you to make a commitment about that now because when I do it, the way I do it, some very interesting paradoxes are raised. Don't get me wrong. I'm not going to cheat with neusis constructions. I will only use straight edge and compass. So far I haven't even done anything that requires a compass, since equal lengths could be obtained without them. The XYV "Theorem" In order to raise the level of the XYV relationship to that of a theorem, and in order to make it as simple as possible for you to prove it for yourself, I am going to go over it again in slightly different ways to indicate some alternative proofs. The following may seem a little tedious. Try to look at it as an example of problem solving which starts from a "primitive" level to find "creative" solutions. Given any two intersecting lines, draw two lines that are parallel ( e.g.,
"||") to each other and that intersect the X at points equidistant from the
given point of intersection. (e.g.,"|X|"). O.K., just draw one of the lines. Notice that it forms an isosceles triangle. Then the external angle is equal to the sum of the two equal base angles; i.e., the base angles of the isosceles triangle are each equal to 1/2 the external angle. That's easier than the way you were taught to bisect an angle, isn't it? Just one line. No circles! Just an ability to "measure" equal lengths. But --- you say: the 1/2 angle is not in the middle of the "given" angle! It's easy enough to move the half angle to the middle of a given angle. It can also be done without circles. It can even be done without a "straight edge" by using our shared, common-sense, intuitive abilities. A "line of sight" could substitute for a "straight edge" and a string could be used to measure equalities, or if folded on itself, halves and other fractions. For example, I could walk across the gap at the bottom of a regular X following a line of sight and carrying a string to "measure" the distance, then fold the string and walk half way back. The line of sight from that point to the point of intersection gives the bisector of the angle. There are many ways to bisect an angle and for the present purpose it really doesn't matter how it's done. The important point here is that the two triangles formed by connecting the top and bottom points of the regular Y are isosceles, with their base angles equal to 1/4 of the Y's top angle. Let the following represent a regular Y and 1/2 a regular Y with its leg extended into the upper angle:
" \ / ........... |/"
" .| ............ | " Since the leg of the Y bisected the bottom angle of the X, extending it into the upper angle bisects that angle as well. (vertical angles). Taking the upper angle of the Y as a reference angle, the 1/2 of a regular Y above shows the half angle as an external angle of the isoscles triangle formed by the line connecting the upper and lower lines of the Y (not shown) with each equal to 1/2 the half angle, or 1/4 of the reference angle. I can't think of any way to present a "picture" of this with my present limitations, so you will just have to use your imagination, or draw your own diagrams for the following steps. It should be relatively easy to see that the line connecting the endpoints of the Y will intersect the lower leg of the superimposed X to form an external angle of a triangle containing the quarter angle and the half angle. Since it equals the sum of those angles, it equals 3/4 of the reference angle. Therefor the angle formed by the intersection of the V with the X is three times the quarter angle formed by the intersection of the V with the Y, and, conversely, the latter is 1/3 the former. q.e.d. by the geometric theorems of the sums of angles in a triangle. But ... some people don' like those theorems. So ... I will show another proof. This one should be a little easier, but you will have to draw your own diagrams. Given any two intersecting lines, draw any circle using the point of intersection "I" as center and intersecting the lines at points "A","B","C", and "D". The result above can be viewed as a "regular" X, or as two diameters of a circle. I have oriented my X like the printed X and labeled it clockwise starting from the upper left. Bisect angle CID intersecting the arc CD at "E" with its extension intersecting arc AB at "F". It doesn't matter how the bisection is done; the result will be the same. I have drawn my extension ("IF") of the bisector ("EI") using a dotted line to preserve the appearance of the regular Y (ABIE) thus formed. Note that: Draw line EB intersecting line IC at "G". You could also draw the line EA to form the regular V giving the complete picture of the superimposed regular XYV, but it isn't necessary. Then: angle FEB = 1/3 angle AGB Before I present the proof of the XYV theorem stated above in specific detail, I would like to make some minor simplification. The geometric theorems I will apply to the construction above are those that express the measure of an angle in terms of the intercepted arcs of a circle. Instead of stating that any angle intersecting within a circle is equal to 1/2 the sum of the intercepted arcs, I shall state that such angles are measured by the sum of the intersected arcs and I shall consider an inscribed angle as intercepting two arcs one of which has no measure. The effect is to simplify the halving of the sum, which would double the "value" of the "measure", but would have no effect on the relative comparisons of the angles so "measured". Proof: Angle FEB = arc FB + arc EE (arc EE = 0) Angle FIB = angle DIE = arc FB + arc DE or Angle AGB = angle EGC = arc AF + arc FB + arc EC or Q.E.D.
|