The 3-4-5 triangle may be the most used triangle in geometry, and because of its right angle everyone from builders to teachers knows it. Draw a circle inside it that touches all three sides and the radius comes out neither as 0.97 nor as something under a square root, it comes out as a plain 1. You can see it on the grid paper above, because the circle sits exactly on the first square at the right-angle corner.
The proof in the graphic I shared on Abakcus cuts the triangle into three pieces. The centre of the circle, O, is the only point that sits at the same distance from all three sides, and that distance is r. Join O to the three corners and the big triangle splits into three small ones, BOC, BOA and AOC. Each small triangle has one side of the big triangle as its base and r as its height, because the radius meets the side at a right angle at the point of tangency. Add up the three areas, set them equal to the area of the big triangle, and the calculation closes by itself.
The numerator and the denominator both landing on 12 is the part of this proof I like most, because what looks like a coincidence is really about the triangle itself. The same calculation works for every triangle, and the area of any triangle equals its inradius times its semiperimeter. The 3-4-5 triangle has area 6 and semiperimeter 6, so r cannot be anything else.
The second way needs no area at all. Seen from the right-angle corner B, the circle touches the two legs at F and D. The quadrilateral BDOF has four right angles and two adjacent sides of length r, so it is a square with side r. Since the two tangent segments from a point to a circle are always equal, the two segments from A are 4 − r long and the two from C are 3 − r long. The hypotenuse is just these two pieces laid end to end.
In general, every right triangle has r = (a + b − c) / 2. This formula has a nice side effect. In a right triangle with whole-number sides, a² + b² = c² forces a + b and c to be both odd or both even, so a + b − c is always even and the inradius is always a whole number.
Something is hiding inside the second proof, and the third way comes out of it. Write the tangent segments down one by one and you get 1 from B, 2 from C and 3 from A, and the sides of the triangle are nothing but their pairwise sums, 1 + 2 = 3, 1 + 3 = 4, 2 + 3 = 5. So the 3-4-5 triangle is really built out of 1, 2 and 3, and these three numbers have a small oddity. Their sum and their product are both 6.
This is not an accident. Call the tangent lengths x, y and z and the semiperimeter becomes x + y + z, and Heron's formula gives the square of the area as the semiperimeter times the product of these three lengths. The same area can also be written as r·(x + y + z), so setting the two equal leaves a single equation.
To find how many positive whole-number solutions that last equation has, it is enough to assume x ≤ y ≤ z. The right side is at most 3z, so x·y cannot exceed 3, and the only options left to try are 1·1, 1·2 and 1·3. The first gives the impossible z = 2 + z, the third gives z = 2 and breaks the ordering, and only 1, 2, 3 remains. No other triangle with whole-number sides has an inradius of 1, right-angled or not.
Ask the same question for r = 2 and the answer is no longer unique, there are five triangles, 5-12-13, 6-8-10, 6-25-29, 7-15-20 and 9-10-17. Put 2 in place of r in area = r·s and the area equals the perimeter, and the whole-number triangles whose area equals their perimeter are exactly these five. The 3-4-5 triangle, as the only one whose area equals its semiperimeter, stands alone one step below that list.
Back in right triangles, things get even tidier. Every primitive Pythagorean triplecan be written as m² − n², 2mn and m² + n² with m > n, and putting that into r = (a + b − c) / 2 gives an inradius of n·(m − n). For r = 1 both n and m − n have to be 1, which means m = 2 and n = 1, and there is 3, 4, 5 again.
In the tool below you can change m and n and draw these triangles one by one. Every square on the grid is one unit, so you can count by eye how many squares the circle spans. The tangent segments use the same colours as the figure above, and underneath, x·y·z = r²·(x + y + z) is checked with actual numbers for each triangle.
As m and n grow the triangle stretches, the circle gets bigger, and the square at the right-angle corner turns into a larger square made of unit squares. At m = 2, n = 1 the tool returns to the 3-4-5 triangle with a single unit square in its corner.
