# Another Reason Why It’s Got To Be 2

To circle back around that inscribed circle problem, about what the radius of the circle that just fits inside a right triangle with sides of length 5, 12, and 13: I’d had an approach for solving it different from HowardAt58’s geometric answer. This isn’t to imply that his answer’s wrong, I should point out: problems can often be solved by several different yet equally valid approaches. (It might almost be the definition of cutting-edge research if it’s a problem there’s only one approach for.)

So here’s another geometry-based approach to finding what the radius of the circle that just fits inside the triangle has to be. We started off with the right triangle, and sides a and b and c; and there’s a circle inscribed in it. This is the biggest circle that’ll fit within the triangle. The circle has some radius, and we’ll just be a little daring and original and use the symbol “r” to stand for that radius. We can draw a line from the center of the circle to the point where the circle touches each of the legs, and that line is going to be of length r, because that’s the way circles work. My drawing, Figure 1, looks a little bit off because I was sketching this out on my iPad and being more exact about all this was just so, so much work.

The next step is to add three more lines to the figure, and this is going to make it easier to see what we want. What we’re adding are liens that go from the center of the circle to each of the corners of the original triangle. This divides the original triangle into three smaller ones, which I’ve lightly colored in as amber (on the upper left), green (on the upper right), and blue (on the bottom). The coloring is just to highlight the new triangles. I know the figure is looking even sketchier; take it up with how there’s no good mathematics-diagram sketching programs for a first-generation iPad, okay?

If we can accept my drawings for what they are already, then, there’s the question of why I did all this subdividing, anyway? The good answer is: looking at this Figure 2, do you see what the areas of the amber, green, and blue triangles have to be? Well, the area of a triangle generally is half its base times its height. A base is the line connecting two of the vertices, and the height is the perpendicular distane between the third vertex and that base. So, for the amber triangle, “a” is obviously a base, and … say, now, isn’t “r” the height?

It is: the radius line is perpendicular to the triangle leg. That’s how inscribed circles work. You can prove this, although you might convince yourself of it more quickly by taking the lid of, say, a mayonnaise jar and a couple of straws. Try laying down the straws so they just touch the jar at one point, and so they cross one another (forming a triangle), and try to form a triangle where the straw isn’t perpendicular to the lid’s radius. That’s not proof, but, it’ll probably leave you confident it could be proven.

So coming back to this: the area of the amber triangle has to be one-half times a times r. And the area of the green triangle has to be one-half times b times r. The area of the blue triangle, yeah, one-half times c times r. This is great except that we have no idea what r is.

But we do know this: the amber triangle, green triangle, and blue triangle together make up the original triangle we started with. So the areas of the amber, green, and blue triangles added together have to equal the area of the original triangle, and we know that. Well, we can calculate that anyway. Call that area “A”. So we have this equation:

$\frac12 ar + \frac12 br + \frac12 cr = A$

Where a, b, and c we know because those are the legs of the triangle, and A we may not have offhand but we can calculate it right away. The radius has to be twice the area of the original triangle divided by the sum of a, b, and c. If it strikes you that this is twice the area of the circle divided by its perimeter, yeah, that it is.

Incidentally, we haven’t actually used the fact that this is a right triangle. All the reasoning done would work if the original triangle were anything — equilateral, isosceles, scalene, whatever you like. If the triangle is a right angle, the area is easy to work out — it’s one-half times a times b — but Heron’s Formula tells us the area of a triangle knowing nothing but the lengths of its three legs. So we have this:

(Right triangle)

$r = \frac{1}{a + b + c} \cdot \left(a\cdot b\right)$

(Arbitrary triangle)

$r = \frac{1}{a + b + c} \cdot 2 \sqrt{p\cdot(p - a)\cdot(p - b)\cdot(p - c)} \mbox{ where } p = \frac12\left(a + b + c\right)$.

Since we started out with a Pythagorean right triangle, with sides 5, 12, and 13, then: a = 5, b = 12, c = 13; a times b is 60; a plus b plus c is 30; and therefore the radius of the inscribed circle is 60 divided by 30, or, 2.

## Author: Joseph Nebus

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

## 6 thoughts on “Another Reason Why It’s Got To Be 2”

1. Neat !
Now show that all pythagorean triangle incircles have whole number radius (easy).

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2. To answer the question above, r = ab/(a+b+c). Since c < a + b for right triangles, it is safe to multiply r by (a+b-c)/(a+b-c) giving r = ab(a+b-c) / ((a+b)^2 -c^2). The denominator simplifies to (a^2+2ab+b^2-c^2) which is 2ab, since c^2 = a^2 + b^2. So r = (a+b-c)/2. Note that c is even if and only if a and b are both even or both odd, so a+b-c is always even, and thus r is always an integer.

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