Triangle Shapes: An Answer

We're interested in finding the phase space of all triangles. In other words, we're looking for a set such that each point of the set corresponds to a unique triangle shape.

One approach to this problem is geometric. Following what we did with rectangles, let's simplify the problem by fixing the length of the longest side. Let's set it to 1, and call this side A. So the length of the other two sides are greater than 0, and less than or equal to 1.

If we fix side A in space, and let the opposite vertex z vary, then each point in the XY plane that is not on the line that contains A defines a triangle. It's easiest to see this by illustration.

[a triangle]

At first it seems that each point (except for those on the line P, which contains side A) defines a unique triangle. However, that's wrong. Look at these triangles:

[four reflected triangles sharing one fixed side]

These triangles are reflections of each other. The pairs defined by z1 and z3 have the same shape, as do z2 and z4. Since we only care about unique shapes, we only need to include one point from each pair. Let's only use points on one side of P.

Now we take the other big step. Since the longest a side can be is 1, that means that we can remove all points with a distance from either endpoint of A that is greater than 1. Let's draw circles of radius 1 around these endpoints:

[unit circles around endpoints of A]

Take the intersection of those two circles, on one side of P. That area looks like this:

[intersected area]

Finally, remove all redundant points. There are few. The only redundant triangles we might find in this region have sides [1,1,x] and [1,x,1]. These are the isosceles triangles whose identical sides are of length 1. Here's what they can look like:

[identical isosceles triangles]

Remove the redundant points on the left arc. We also have to make sure to remove the points on A itself, since the vertices of triangle can't all be on the same line. So, finally, we have:

[the final region]

And that, I submit, is an answer to the question. Using some basic trigonometry, the area of the region is pi/3-sqrt(3)/4.

Special Cases

It's interesting to consider the special cases of triangles: isosceles and equilateral. The latter is easy; the point at the top of the figure has distance 1 from both endpoints of A, which makes a [1,1,1] equilateral triangle.

The isosceles triangles are of two different kinds. There are the isosceles triangles whose base is longer than the sides, and those whose base is shorter than the sides. They meet at the equilateral triangle, whose base is the same length as its sides.

Here's where these triangles lie:

[isosceles triangles]

The red line holds the wide isosceles triangles, while the yellow line holds the narrow isosceles triangles. The orange dot at the top is the equilateral triangle.

A Complication

Now that all that's done, take a look at another answer.

Last updated 3 June 2000
All contents ©1997-2002 Mark L. Irons

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