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# Coyoty defies the laws of mathematics.

One of the classic puzzles in mathematics and geometry is the "impossible" trisection of an arbitrary angle using a straight-edge and compass. There are mathematical proofs that it can't be done. Since I learned about this "certainty" in high school, I've been pondering it on and off, determined to do it anyway. In fact, it took the school's valedictorian hours, maybe days, to find the flaws in my first attempt. Almost had it that time.

I finally came up with the solution on April 1, 1994, but I had a problem... While it worked theoretically, doing it physically was a challenge. It seemed to require a non-standardly shaped straight-edge and would not be accepted on that basis, so I backburnered it again until I got around the physical aspects.

Tonight I came up with the solution without realizing it, while explaining my method in a discussion about child prodigies in R.H. Junior's forums.  I was hung up on the straight-edge, but didn't consider the media the diagram is done on.  By using a transparency, not only can I construct the proof physically, but it should qualify as a legal solution.  Now I just got to look for the scholarly prizes to cash in on.

So here it is, my pretty good trick:

It cheats a little, but it does follow the rules, I believe. I have to get a ruling (pun intended) on its "legality", but keep forgetting. It involves moving part of the construction and using it to complete the rest of it. I think it would qualify as the ancient Greeks did geometry, in their heads, but because the movable part has to be constructed separately on a transparent sheet, or as a modification of the straight-edge, it can be questioned. But every part of it uses unruled edges and angles.

The basics are these: Construct an arbitrary angle. Bisect it, with the bisection line extending indefinitely each way. Draw two lines parallel the bisection line, each equally distant on either side. Premise: Somewhere along the area within the parallel lines is an arc that is one-third the original angle. Draw a perpendicular line through the parallel lines, somewhere where the length between the top and bottom lines is smaller than the width of the areas on either side of the parallels' area. Using the angle's radius point and the points where the perpendicular line intersects the outer parallel lines, draw a triangle. Transfer this triangle and the bisection line to the transparent sheet. Lay the sheet so that the bisection line is on the angle's radius point, one point of the perpendicular is on one line of the angle, and the other point of the perpendicular is on the parallel line nearest the angle's line. Anchor the sheet there, or transfer the triangle there. That is the point where the arc of the parallel line area is one third the original angle.. All correlating lines from the triangles formed by the radius point and perpendicular line's points in each section will be equal, and the correlating angles in each of the three sections will be equal.

This method uses the geometry of the diagram itself and not any manufactured tools like the "hatchet" mentioned in the mathematical proofs link above.  Because it uses pure geometry, there's no way to introduce measurement errors.

Did I lose anyone?  At some point I'll try to do a YouTube video illustrating this.
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