# Cutting and rearranging one polygon into another

I think Vladimir Arnold is one of my favorite modern mathematicians. Partly because I find his writings lucid but mostly because every interview or talk I’ve read by him is hilarious. He also took an interest in educational practices which of course resonates with me even though naturally his perspectives are somewhat skewed. One of the books he helped put together and which I came across recently is a set of problems aimed at children called Problems for children: 5 to 15 (in english). When you read it you have to either decide that it’s a joke or more of an idea of how you could set the bar because some of those problems are hard for even for guys like me and I’m 8 years past 15 at this point.

Still some of the problems are really good and especially #13 is a nice example of a problem for which it actually hurts to be proficient in math because you might jump to an answer to quickly. I’ve been slowly going through the problems every other day recently and today I worked with #39 and I thought I would just try and outline my approach to it because it’s a nice problem.

If two polygons have equal areas, then they may be cut into a ﬁnite
number of polygonal parts which may then be rearranged to obtain both the
ﬁrst and second polygons. Prove this!

It is more or less clear that you should really just try and prove that any polygon can be cut and rearranged to a specific polygon such as a square for then you can just reverse the process to move to any polygon from the square. The scheme I followed to prove it was the following one:

1. Find a way to turn rectangles into squares

2. Find a way to turn triangles into rectangles

3. Find a way of turning a polygon into a collection of triangles

4. Find a way of combining a collection of squares into a single square

where in reality the motivation for it all worked in the opposite direction.

Starting out

I like Arnolds book because usually you can find some hint as to how you can approach the problem either from the image associated to the problem or from some other problem which might be more general in nature. If you have a polygon with a certain area $A$ such that the base is $\sqrt{A}$ then it’s height is also $\sqrt{A}$ and just cutting off a corner an rearranging it you get a square with the same area.

Cut and transform rectangle to parallelogram with base $\sqrt{A}$

Now if the you have a general rectangle you can turn it into a paralellogram with base $\sqrt{A}$ simply by drawing a line with that length from one of the corners to the opposite side and lop of a corner, move it to the other side and you get a parallelogram of the desired form. To find the actual distance $\sqrt{A}$ you can use the Euclid construction of the geometric mean of the two sides the special case of which I discussed in the square root post a while back.

Dissect parallelogram to rectangle

To cut and transform a parallelogram into a rectangle is just a special case of the method for transforming the other parallelogram into a square but I list it anyways for chronology.

Cut and transform a triangle to parallelogram

This one didn’t come naturally to me as I originally tried to approach the triangle before the rectangle (for some reason) but once I recalled that cutting the triangle along the midpoints of two sides produced appropriate angles and lengths so that you can rotate the top part to form a parallelogram this one was obvious.

General polygon

Now the case of a general polygon is within my grasp and I like my final solution though I have to admit its not very optimal in terms of the number of cuts you will effectively have to make. The points is that if we have any $(n + 1)$-sided polygon it can always be cut into $n$ triangles each of which can be cut and transformed into a square.

The is likely multiple ways of fitting these pieces together but working specifically with squares I found that combining that cutting and transforming two squares into a cube is intimately related to Pythagoras theorem since several proofs of it consists of cutting two squares placed along the sides of a right triangle and rearranging them to form the square of the hypotenuse. Rather than restating a specific one I’ll just reference cut-the knots excellent collection of proofs where for example Proof #2 will do the trick of designing a procedure for combining two squares.

Final notes: Just to round of the evening I did some searches on this type of problem now that I was done and two results with I find worth to mention was a friendly intruduction from Berkley with animations (including the Pythagoras cut) as well as a site by  Gavin Theobald who has studied a method for finding more minimal cutting methods which though less natural to me is pretty impressive.

Would probably be kind of fun to program a program which dies this sort of thing and animates but I guess that’s a project for another day.