Friday puzzle -- triangle division

Re: Friday puzzle

If into 2 pieces, I think not ...

Any [finite] number of pieces

You are right for two pieces, but can it be done with more pieces?

Re: Friday puzzle -- triangle division

Based on a bit of playing with it, I am going to say no, it is not possible, but I don't know how to prove it.

Re: Friday puzzle -- triangle division

I also think the answer is no. Here might be the start of a proof.

Label the vertices of the triangle A, B, C, starting from the top left, so that C is the obtuse angle. Now divide triangle ABC into two triangles by drawing a line from C to line AB so that the there is a new vertex D on line AB. D can be located so that triangle CDB is acute. But since angle CDB is acute, this means that angle CDA must be obtuse since the sum of angles CDA and CDB must = 180. So triangle CDA is obtuse. Dividing the new obtuse triangle ADC will produce the same result.

Subtle error in proof

If you can find the error, can you fix it or does it make you rethink your conclusion?

Half an answer

Larry is right in the sense that he has proven that the triangle can't be divided into acute triangles starting with a cut from the vertex of the obtuse angle to the opposite side.

But it can be done!

This is not the other half

If you start at either acute angle and draw a line to the opposite side you still have obtuse triangles. So the answer must be to start on a side. You can construct right angle triangles by drawing perpendiculars to the adjacent side. But altering the right angle to make it acute automatically creates an obtuse angle adjacent to it. The best I can do with Alex's example is to create four right angle triangles and one acute triangle.

Last hints

1) There must be at cut that goes through the vertex at the obtuse angle, to create acute angles at that vertex.

2) Larry's observation about a cut from C to AB is valid, in that it creates another obtuse triangle leaving you with the same problem you started with (or it could leave you with two right triangles, but that leads to an analogous never-ending cycle).

Will post answer tomorrow p.m.

Re: Last hints

Extend a line from the obtuse angle toward the opposite side, but stop at an interior point (X). The point X has to be along a line from the obtuse angle such that when the obtuse angle is divided in two, the resultant angles are each acute. Starting from X, create a pentagon (not sure if regular pentagon or not) by extending a line to side CA, to side CB, and two lines to side AB (using my initial notation. This pentagon shape that is interior to the original triangle now consists of five triangles. When this is done, the original triangle will now be divided into the five triangles that make up the pentagon shape, plus two more triangles that include the other two acute angles of the original triangle. I think all seven triangles are acute. If the pentagon is a regular pentagon, its interior angles will all be acute (72*). The two angles that resulted from dividing the original obtuse angle will be acute. It seems like all angles are acute, but I can not prove it yet. If the sides of the pentagon are extended, a five sided star will be created, part of which includes the original triangle.

Kudos to Larry!!!

That is the solution to a really tough one.

Here's a picture of the solution