# The Short, Unhappy Life Of A Doomed Conjecture

So last month amongst the talk about the radius of a circle inscribed in a Pythagorean right triangle I mentioned that I had, briefly, floated a conjecture that might have spun off it. It didn’t, though I promised to describe the chain of thought I had while exploring it, on the grounds that the process of coming up with mathematical ideas doesn’t get described much, and certainly doesn’t get described for the sorts of fiddling little things that make up a trifle like this.

The point from which I started was a question about the radius of a circle inscribed in the right triangle with legs of length 5, 12, and 13. This turns out to have a radius of 2, which is interesting because it’s a whole number. It turns out to be simple to show that for a Pythagorean right triangle, that is, a right triangle whose legs are a Pythagorean triple — like (3, 4, 5), or (5, 12, 13), any where the square of the biggest number is the same as what you get adding together the squares of the two smaller numbers — the inscribed circle has a radius that’s a whole number. For example, the circle you could inscribe in a triangle of sides 3, 4, and 5 would have radius 1. The circle inscribed in a triangle of sides 8, 15, and 17 would have radius 3; so does the circle inscribed in a triangle of sides 7, 24, and 25.

Since I now knew that (and in multiple ways: HowardAt58 had his own geometric solution, and you can also do this algebraically) I started to wonder about the converse. If a Pythagorean right triangle’s inscribed circle has a whole number for a radius, can does knowing a circle has a whole number for a radius tell us anything about the triangle it’s inscribed in? This is an easy way to build new conjectures: given that “if A is true, then B must be true”, can it also be that “if B is true, then A must be true”? Only rarely will that be so — it’s neat when it is — but we might be able to patch something up, like, “if B, C, and D are all simultaneously true, then A must be true”, or perhaps, “if B is true, then at least E must be true”, where E resembles A but maybe doesn’t make such a strong claim. Thus are tiny little advances in mathematics created.

So the starting point of the doomed conjecture is: “if the circle inscribed in a right triangle has a radius that’s a whole number, then the right triangle is Pythagorean”, which falls over and dies right away. Why?

Well, the first problem is one of scaling. Imagine a triangle with a circle inscribed in it. How do you know what the radius of the circle is? You know it because we draw a little line from the center of the circle to the edge and we label it r and somewhere along the line we say that r is 1 or 2 or 1.75 or the square root of pi or whatnot. If we’re trying to make a precise drawing of the figure we might include a legend, and say the space between these two vertical strokes is one unit of distance. Fine, but, how is the drawing different if the space between those two vertical strokes is two units? Or the square of pi units? The drawing looks the same, after all.

And there’s the problem. If I started with the right triangle whose sides have length 1, 1, and the square root of 2 — which isn’t a Pythagorean triplet — then the circle inscribed in it has a length of $1 / \left(2 + \sqrt{2}\right)$ which is about 0.29289. But if I multiplied all the distances in that figure by $2 + \sqrt{2}$ then I’d have a right triangle whose legs are of length $2 + \sqrt{2}, 2 + \sqrt{2}, 2 + 2\sqrt{2}$ and whose inscribed circle has radius exactly 1. And that’s a fine enough right triangle but its legs are certainly not a Pythagorean triplet. In short, if I start with a circle that’s got a radius which isn’t a whole number, I can just scale my whole diagram, without changing the shape of the triangle at all, until the circle has a whole-number radius, and so how can I know anything about the triangle?

This sort of problem, where the attempt to say anything definitive is foiled by the fact you can rescale the problem, has at least one obvious solution: maybe you can say something if you set the scale to your advantage. Suppose that the radius of the circle is exactly 1; what can you say about the triangle it’s inscribed in?

Now, not much of anything, it seems like. If you divide the lengths of the 5-12-13 triangle by 2, you get a right triangle with inscribed circle of radius 1. It’s no longer a Pythagorean triangle, since its sides have length 2.5, 6, and 7.5, and two of those are not whole numbers, but we knew we were giving up the sides of whole number length going in. Similarly, divide the sides of an 8-15-17 or a 7-24-25 triangle by 3 and you get a triangle with inscribed circle radius 1, but what’s that got us?

Well, there is that these sides are all rational numbers. If you can show something is true about rational numbers, and you can scale that, then you’ve also proven it for whole numbers: multiply any three rational numbers by the right number (or a product of it!) and you’ve got three whole numbers again. So surely there’s something to be said about a right triangle, with a circle of you-know-what size, where all three legs have rational length, right?

Yeah; what you can say about that is that it’s a scaled-down version of a Pythagorean right triangle, which is where we came in. We don’t seem to be getting anywhere.

What if we give up the requirement that all three sides of the triangle have a rational length? If we do that, then we get a lot of candidate triangles — for example, there’s the scaled-down version of the triangle with legs length 1, 1, and the square root of 2 mentioned above. But none of these candidates are going to be Pythagorean triples, by definition.

So that’s gotten us nowhere, and is about as far as I can go in the line of “if B, C, and D are true, then A must be true”; I can’t think of conditions to add to “the circle has a radius that’s a whole number” that get me anywhere. Maybe we have to look at something different, along the lines of “if B is true, then E must be true”. For example, if we have a right triangle, with hypotenuse of length c, and sides of length a and b, then the radius r of the inscribed circle is $r = \frac{a\cdot b}{a + b + c}$; can we say anything about the relationship of a, b, and c, if we insist on r being a whole number?

Well, we can say that there are combinations of a, b, and c that are all rational, for which r is rational, which is the next-best thing to being a whole number. Those are the Pythagorean triples we came on the scene with. And we could have a, b, and c all be irrational: an example is the $2 + \sqrt{2}, 2 + \sqrt{2}, 2 + 2\sqrt{2}$ triangle I mentioned earlier. Could it be the case that just one of the sides is irrational? Conceivably, if it’s a or b. How about if two sides are irrational? That’s imaginable too, and at least my imagination finds that easier to envision.

But in the process of trying to find a set of a, b, and c such that exactly one or exactly two of them were irrational, while the circle had a rational radius, I gave in to despair: I couldn’t think of anything interesting that I was getting from this. A right triangle must have zero, one, two, or three sides of irrational length? Nobody needs to go to any effort to prove that.

So this was the point at which I declared my conjecture dead. I can’t find an “E”. It might be interesting to actually identify a case of a triangle with one or with two irrational legs which has a rational-radius incircle but that feels like it’s just a bundle of calculating not likely to get to something at least as interesting as “a Pythagorean triangle has an inscribed circle with a whole-number radius”. Too bad. And yet I admit a nagging feel that I should at least come up with one example of each before declaring the whole project dead.

## Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there.

## 2 thoughts on “The Short, Unhappy Life Of A Doomed Conjecture”

1. ivasallay says:

I’ve come up with a dead conjecture or two before, too. They are still fun to explore.

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1. They can be. I’m just disappointed I got to nowhere much; I was having so much fun with the triangle.

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