This one comes from solving the problem in a different way. It's based on the triangle inequality.

[The triangle inequality is A + B > C. In other words, the sum of the lengths of any two sides of a triangle must be greater than the length of remaining side.]

You can describe any triangle by writing down the lengths of its sides. For example, a [3,4,5] is a right triangle. Once again, we'll set the longest side to be length 1. Our triangles will be of the form [1,P,Q], with 0 < P <= 1 and 0 < Q <= 1. Then we have the following conditions to meet:

1. 0 < P <= 1
2. 0 < Q <= 1
3. 1 + P > Q (triangle inequality)
4. Q + 1 > P (triangle inequality)
5. P + Q > 1 (triangle inequality)

This might look a little daunting, but it's not hard to solve at all. Let's take it in steps.

The first steps are #1 and #2. They just say that both P and Q are greater than zero and less than or equal to 1. Look at it in terms of the plane, where the horizontal axis represents P and the vertical axis represents Q. Each point (P,Q) represents the triangle [1,P,Q]. The inequalities look like this:

(Remember, a solid line means include the points on the line, while a dotted line means exclude the points.)

The intersection of the red and blue areas satifies both inequalities. From now on we'll only look at the doubly-shaded box.

Inequality #3 is trivial. P+1>Q is already true since P>=Q (Inequality #3).

Inequality #4 is trivial as well. Q+1>P because 1>=P (Inequality #1).

So the only inequality left to satisfy is #5, P+Q>1. That region looks like this:

Taking the intersection of them all and removing redundant triangles gives:

The shaded area satisfies all the inequalities. The next question is: how big is it?

We started with an area that was 1x1. Inequality #5 cut that in half. So the answer is 1/2.

Once again, we can ask where the isosceles and equilateral triangles lie in this region. Each point on the red line corresponds to a isosceles triangle [1,x,x]. Each point on the yellow side is an isosceles triangle [1,1,y]. They meet at the orange point, where we find the equilateral triangle [1,1,1].

## Generalization of the Solution

The inequality approach to this problem can easily be generalized to any shape. The key is generalizing the triangle inequality to more sides. In English, this is:

The sum of the lengths of all the sides but one of any polygon is greater than the length of the remaining side.

This gives n inequalities, where n is the number of sides. Then you must take symmetries into account, but that isn't too hard to do.

## A Potential Conundrum

Astute problem-solvers will have realized by now that this answer is totally different from that on the other solution page. What's going on?

This apparent paradox comes from the fact that the problem isn't well-defined. What we're trying to do is map each triangle shape to a point in the plane. We first choose a mapping, and then look at the resulting set of points. The shape thus defined depends on which mapping you choose. So when we approached the problem in two different ways, and mapped points in two different ways, we got two different answers.

There is a one-to-one correspondence between these two solutions which is readily apparent. Look at the similarities. Each region has three sides, only one of which is part of the solution. The isosceles triangles lie on a line down the center and on one side of each region. The equilateral triangles lie at a vertex.

This apparent paradox doesn't always arise. All circles have the same shape. The same is true with squares. But when the situation becomes more complex, the way we approach the problem defines the solution.

Isn't that a great metaphor for life?

Last updated 3 June 2000