# Old Memories About Trisecting Angles

This is just some remembering. When I was in middle school (about 12 or 13 years old), I was convinced I could find a way to trisect an angle. Of course, this is known to be impossible, by an argument that I now (but didn’t, at the time) understand. That didn’t stop me from trying, though. Here’s the construction I spent most of my time on:

I started out with an angle, which is the two outermost lines on the given drawing. Using the compass, I marked off points equally far along the sides. I then constructed their midpoint. Then I drew three overlapping circles centered at those three points, as shown. The intersection points of these circles furthest from the vertex of the angle were used to draw my “trisecting” lines.

Of course, this doesn’t really trisect an angle, since it’s a construction with a compass and straight edge, and no such construction can possibly trisect an angle. But if you didn’t know that, there are a few sanity checks you might perform.

- Does it appears to work? Yes, it does, as it turns out. In fact, I performed this construction fifty or sixty times, and measured the result with a protractor. Each time, it was close enough to trisected that I chalked up the difference to human error. (Having just performed the construction now using the Kig software, to capture the image above, I can now see that the result is visibly off for extremely small angles; but of course these are the ones that I was least able to check without reasonable computer software at the time.)
- Does it work for cases that are easily calculated? Yes! A few easy examples: If you start with a 180 degree angle, you get back 60 degree angles… perfectly trisected. A zero-degree angle is vacuously correct. It’s a slightly more involved argument, but the construction also trisects a 90 degree angle correctly.

At this point, one can’t be blamed for getting a little excited. We have a construction that gives a division of an angle into three parts. The parts always seem to look equal, and are always close enough to be within reasonable measurement error. It works for the cases we’ve checked so far. One can’t be blamed for shifting gears a little here, gaining a little confidence, and looking a little less for a counter-example, and a little more for a proof.

Of course, those familiar with the impossibility of this task will be looking for the case where we start with a 60 degree angle. And, of course, it turns out not to work. In particular, the “trisected” angle works out to about 19.1066 degrees. Close, but not quite.

So what am I really constructing? Well, something pretty ugly. I’m leaving out the math (basically, just pick out a couple triangles, and apply known triangle relations such as the law of sines and thelaw of cosines), but, the answer in Maxima notation (and simplified as best as Maxima can) is -asin((sin((3*%theta-5*%pi)/6)*abs(sin(%theta/2)))/sqrt(-2*sin(%theta/2)*cos((3*%theta-5*%pi)/6)+sin(%theta/2)^2+1)). Yeah, wow. But how could we have fooled ourselves into thinking that mass of complicated stuff is actually theta / 3? Well, take a look at Maxima’s plot of this angle versus theta (that starting angle):

Ifthe construction were correct, that graph would be a straight line passing through (3,1). As it turns out, it may be somewhat difficult to tell the difference in this graph! So while this construction is wrong, it’s also remarkably close.

it’s always nice to be able to _finally_ solve a problem you thought about as a teenager.

Sweet. I spent a month or two believing the ASS Theorem was indeed correct. First I made a little mistake that enabled me to convince myself it was, then spent a day or two working up a proof that contained a rather subtle fallacy.

I greatly annoyed my geometry teacher in this endeavor, he was quite a good teacher actually but he had no interest (ability?) in dissecting my “proof”. A few months later I realized the fallacy.

Unfortunately I don’t have my geometry textbook anymore, but I think I may have relied upon a small error in the formulation of one of the theorems in the book. Either that or I forgot to check the definitions completely… the book would have had to define an “angle” using three non-collinear points, which seems like a silly definition, but maybe it did that.

Draw a line AB with a straightedge. Construct a line perpendicular to AB through the halfway point of AB with a compass. Call the halfway point of AB point C.

Draw a circle with the compass around point C, using CB as the radius of the circle. Call the points of the line perpendicular to AB that intersect with the circle D and E, with D at the top and E at the bottom. Point C will be the halfway point of the line DE. The angle <DCB is 90 degrees.

Now place the compass on point B and trace out a portion of a circle with radius BC, starting at point C and continuing until the arc intersects with the circle drawn using CB as the radius. Call the point of this intersection point F. Connect points CF and BF. The triangle CFB is an equilateral triangle with equal interior angles of 60 degrees each. Furthermore, by extending the arc of radius BC and then tracing an arc of radius DC with D as the center point, one can bisect the 90 degree angle <DCB into two 45 degree angles by drawing a line from point C to the point where the two arcs intersect. Call the point where this line intersects with the circle of radius CB point G.

It stands to reason, that angle <DCB is 90 degrees. It stands to reason that angle <FCB is 60 degrees. It stands to reason that angle <DCF is 30 degrees as <DCB minus <FCB = <DCF. It stands to reason that angle <GCB is 45 degrees (by bisecting DCB). From that one can deduce that <FCB minus <GCB = <FCG which is equal to 15 degrees. Angle <DCF (30 degrees) can also be bisected into two angles of 15 degrees each.

Place the compass on point G (which is at 45 degrees), and draw a circle with radius GF. This circle will intersect the arc subtended by angle <GCB at the 30 degree mark. Call this point H. One can now draw a circle with point H as the center and radius HG to create a new intersection on the arc subtended by angle <HCB. The intersection will be at the 15 degree mark.

In my mind, at least I proved to myself that an angle of 45 degrees can be successfully trisected into 15 degree angles using a straightedge and compass.

BTW, what would be interesting if one could find a construction that could trisect 15 and 60 degrees. The constituent angles would be 5 and 20 degrees. Since they can be easily bisected, one could find 2.5 and 10 degree angles. It seems to me that a large number of angles can be trisected, just not all of them.

Such is the lot of an inquiring mind. Straight lines and curves (secants and the radial section above) are quite something.

It is one thing to construct certain angles out of unit lengths, quite another to take a unit length and divide it by three and then try to assign these 3rd’s to the arc of an angle.

Well, it was fun chasing the goose.

Is trisection of obtuse angle is impossible?

Yes. If you could trisect an obtuse angle, then you’d also be able to trisect any angle, simply by doubling it until it’s obtuse, trisecting that, and then biscting the result the same number of times. So, trisecting an obtuse angle is in general impossible.

I would like to have your email address to send you a method for trisecting an obtuse angle.

There exist a theorem that if an acute angle is trisected you can also trisect its supplementary angle.

Theorem: The trisectrices adjacent to the common side of two supplementary angles determine a precise 60-degree angle

New method about general angle division and angle trisection, on a link

http://www.mathramz.com/math/node/1726

Discovered By

Eng/ Ahmed elsamin

(Arab Republic of Egypt)

http://www.dinbali.com