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Doesn't this imply that you can trisect an arbitrary angle? At lease one that is less than 60 degrees?

Very handsomely done, though.


To clarify, doesn't this imply you can construct a trisection of an arbitrary angle? Every step seems to be possible using just a compass and straightedge.


i see why you might think that, but you can't trisect an arbitrary angle. the reason that this proof doesn't contradict that is that it starts with the equilateral triangle in the middle and builds the triangle from there. if, however, you start with an arbitrary triangle, no compass/straightedge construction can locate that equilateral triangle. the classic counter-example to trisectability is the 60° angle, which can't be trisected. if you try to use this proof to do it, you'd need the 20° angle FIRST, which is exactly the object you were trying to construct.


Oh, true. I guess I didn't think it through.

Anyway, though. Very nice proof for a very excellent theorem.

Your Wife

Hey Polymathematics, I was just wondering if anything huge and lifechanging happened to you lately? Or not so much?


That previous comment really cracks me up.


I've one comment on the Morley proof: You have to add that a & b & c are > 0 !
Otherwise one can choose, for example, c = 30 , a = 30 and b = 0 (all in degrees). Now point B is not defined properly!
The rest of the proof is ok!


how do you trisect a 56 degree angle

Mathlete #1

It is proven impossible to trisect an arbitrary angle.


i'm not sure if mathlete is claiming there's a flaw in the proof or responding to andrew, but i should make it clear that mathlete is correct.

in general, it isn't possible to trisect an arbitrary angle. that means that the equilateral triangle in this proof can't be constructed (given the original triangle) with a compass and straightedge. but that doesn't invalidate the theorem.

Mathlete #1

Can I have the initial of your first name and your last name so that I could cite your work in my research prospectus? It would be great if you could.

Mathlete #1

To clarify, Initial of first name and full last name.

Mathlete #1

Thank you Polymath.

Charlie Huttar

In the third diagram, shouldn't the letter next to point B (giving the size of the angle) be "b"? The text says this angle is b, but the diagram says c.

Charlie Huttar

By the way, this proof was very helpful. Thanks!


I'm glad it was helpful, Charlie. The number next to the B in the third diagram should indeed be a "b", as the text indicates, but I think it is. I do see, though, that the "b" overlaps the diagram a little, and it might look like a "c". Correcting the diagram is tricky, and I'm not going to try it, but thanks for bringing it up. Sorry for any confusion.


wow my discrete math teacher just gave us this one today and i figured out a proof similar to this in that you start with the equilateral triangle in 30 minutes. mine may not have been exactly correct though, somehow i did it only with geometry and no math or variables...

Magdy Essafty

I've a new owen proof, how I show it?

John Trainin

Referring to an earlier comment, and merely dealing with integral numbers of degrees, one can of course trisect any angle for which the trisection of the angle is 3 degrees or a multiple thereof. This follows from the fact that one can construct 15 degrees from an equilateral triangle, and 18 degrees from a regular pentagon.


I do not quite understand your last statement:
If you scale that diagram appropiately, it will result in a triangle ABC that is the same as the arbitrarily given one.

Of course, if the given triangle satisfies Morley's Theorem, then by construction, the triangle constructed is unique (up to scaling), so it will indeed result in a triangle ABC that is the same as the arbitrarily given one (up to scaling).

However, isn't it still possible that some arbitrary given triangle does not satisfy Morley's Theorem, so if you use its angles a, b, c to construct as you did above, you end up with a different triangle with the one you started with?

To summarize, I think there might be something wrong with the logic you are employing.


I retract my previous comment. The proof does indeed make sense.

A disappointed One

I don’t know if this is true, i argued with my friend about this, and he thinks it’s true, maybe it is or maybe it ain’t but i’m not absolutely convinced

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