*A Geometric Puzzle*

Consider circles for a moment. How many different shapes can a circle have?

It's a simple question, isn't it? All circles have the same shape. They might have different sizes, but the shape is the same.

Now think about squares. How many different shapes can a square have?

If we disregard size and rotation, then there is only one square shape. This leads us to a definition:

*isomorphic*- Two shapes are isomorphic ("same shape") if they have the same shape if we disregard size and rotation
^{1}.

All circles are isomorphic to each other. Likewise, all squares are isomorphic to each other. What about rectangles?

It should be pretty obvious that there are rectangles that aren't isomorphic to each other. Here's an example:

Rectangle X doesn't have the same shape as rectangle Y. As a matter of fact, there are an infinite number of rectangles that aren't isomorphic to each other. To see this, consider rectangles which have a longest side of length 1. If we have two rectangles with other sides *p* and *q*, then the two rectangles are isomorphic only if *p*=*q*. Since there are an infinite number of numbers greater than 0 (the shortest possible side) and less than or equal to 1 (the longest), there are an infinite number of rectangles that aren't isomorphic to each other.

Is there some way we can at least characterize all possible rectangles? As it turns out, there is.

Every rectangle has 4 sides, [*A*, *B*, *A*, *B*]. We don't need to worry about the 3rd and 4th sides, since we know their lengths are the same as 1st and 2nd, respectively. So we're really only concerned with two numbers, *A* and *B*.

Let's look at the relative lengths of the sides. It really doesn't matter which one is greater, only that one is (or else they're equal, and we have a square). Let's assume that *A* is. That means 0 < *B* <= *A*. (*B* has to be greater than 0, since a side can't have 0 length.)

Here's where we do the trick: making the longest side to be length 1. Since we don't care about size, divide both *A* and *B* by the same amount, *A*. Side *A* now has length 1. Our rectangle now has sides [1,*B/A*,1,*B/A*]. Since 0 < *B* <= *A*, 0 < *B/A* <= 1.

We've now characterized all rectangles in terms of a simple inequality. The only thing that needs to vary to create different rectangles is the length of the second and fourth sides. That length can only take one of a fixed set of values (0 < *B/A* <= 1). When we choose a value for *B/A*, we've chosen the shape of a rectangle.

Now we take a leap. What does the set of all possible values *B/A* look like? Since it's only one value, it's one dimensional. Only one thing varies. But what subset of a line do we choose?

That one's easy. Since *B/A* must be greater than 0 (since a side has to have a positive length) and less or equal to 1 (since the longest possible side is 1), it's just a line segment open on one end, closed on the other:

Each point in the set (0,1] gives rise to a unique rectangle shape.

Let's call this set the problem's *shape space*. Each point in a shape space gives rise to a unique thing (in this case, rectangle shape).

Going back to circles and squares, what do their shape spaces look like? The answer is simple: they're nothing but a point. All circles have the same shape, so there's nothing to vary. The same is true for squares.

There's one last thing to note about the rectangle shape space. Take a look at the point 1. We're out on the end here, as far as we can go. The corresponding rectangle is [1,1,1,1]. It's a square. On the other end are the rectangles that are very narrow (but never quite collapse to nothing).

Still with me? Okay, **here's the puzzle**. What does the shape space of triangles look like? Where in the shape space would isosceles and equilateral triangles reside? Remember, rotation and size don't matter.

Think about it for a while. Then come back and take a look at my answer.

1. Technically, two sets in Euclidean N-space are isomorphic if they're invariant under a combination of rotations, translations, and reflections. For this problem, since we're only concerned with shape, we're extending this to uniform scaling in all dimensions.

Last updated 3 June 2000

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All contents ©1997-2002 Mark L. Irons

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