Trigonometry/Converting One Triangle into Another

The Problem

 * Given two triangles of the same area, can we cut one up into a finite number of pieces which can then be rearranged to make the other?

Thought Process
If we can cut a triangle up and rearrange the pieces to make a rectangle, then we're doing well. Rectangles are easy to work with. If we can get a number of small rectangles, somehow make them into rectangles with one side of length one, then stack them all up on that side into one rectangular bar with one side of one, then we've won. Why? Because the process of cutting and rearranging is reversible. To spell that out:


 * We can cut and rearrange the first triangle to make a rectangle with the same area and one side of length one. We can do the same for the other.  If we superimpose the cut-lines of the two ways of dividing up this unit rectangle we have pieces which can be rearranged to make either the first triangle or the second.

Well, we can cut any triangle into two right triangles, and each of those right triangles can be cut and reassembled to make a rectangle. So we can certainly get two rectangles out of each triangle. We can cut these up into lots of small rectangles.

Now, how do we convert those rectangles to rectangles with one unit side?

A little experimentation shows that the way to go is via parallelograms. We can make cuts at an angle to get the length of side that we want. And we can cut parallelograms up and reassemble them to make rectangles, whilst preserving the length of one side. Once we've hit on the idea of using parallelograms, and looked at ways of slicing and reassembling them, we see that we don't really need to do anything special with rectangles. Here's how we do it.

Move I
First take one triangle of area A, and cut it into two pieces and reassemble as a parallelogram with the same base, using move I, like so:

We've taken the top triangle of half the dimensions and rotated it 180 degrees.

Move II
Now given any parallelogram we can use move II, a cut and swapping of the order of the two pieces that preserves the base and shifts the top edge left or right by up to the length of the base:

Changing the length of One Side
By repeating move II enough times we can make the length of the side that isn't the base anything we like, as long that is as it is at least $$\displaystyle A/(\text{length of base})$$. That's not a very stringent restriction. We're aiming to get a parallelogram that has one pair of sides the length of one triangle's base and the other pair of sides the length of the other triangle's base.


 * Why do we want to do that? Because then we have a parallelogram which could either have been formed from the first triangle or from the second.

What could go wrong? The only thing that could go wrong is if the side we are trying to get is $$\displaystyle < A/(\text{length of base})$$. To avoid that we position our original triangles so that each has its longest side as its base. Then by using move II one or more times:
 * (a) We do not change the base length, which is the same as that of the original triangle.
 * (b) We get the other pair of sides to be the length of the base of the second triangle.

Just to confirm that we're not trying for too short a side length, we note that in each triangle the longest side must be at least the length of a side of an equilateral triangle of area A. That's because the equilateral triangle does the best job of keeping its longest side short for a given area. In the equilateral triangle each side is $$\displaystyle \sqrt{\frac{4A}{3}}$$. Those longest sides of our two triangles must both be at least $$\displaystyle \sqrt{\frac{4A}{3}}$$. The product of the longest side lengths is  $$\displaystyle \geq {\frac{4A}{3}} > A$$ and we're not trying for too short a side.

Half way There
We now have a parallelogram with its base the length of the base of one triangle and the other side the length of the base of the other.

We're exactly half way through converting one triangle into the other. We can flip the parallelogram so that the other side is the base. Now doing the steps in reverse, finishing with move I in reverse, we have reassembled our pieces into the second triangle.


 * Back to Hilbert's Third Problem