I was recently asked to prove that a circle inscribed in a right triangle with sides 3, 4 and 5 had a radius of 1. Mathematics never ceases to amaze me! How could a standard 3, 4, 5 triangle be associated with a circle of radius 1? It’s akin to the 9 year old me discovering that the sum of the first n odd numbers is simply n squared.
Anyway, on with the proof. In the triangle above we can assume that AB = 5, AC = 4 and BC = 3. The points P, Q and R are where the sides of the triangle meet the circle, and are tangents. If we assume the radius of the circle is r, then CR = CP = r. Hence, AP = 4 – r, and RB = 3 – r. Since tangents to a circle meeting at a point are equal, AQ = AP = 4 – r and BQ = BR = 3 -r. We can also see that the triangles AOP and AOQ are congruent, as is the case with the triangles BOR and BOQ.
We know AB = 5, which implies AQ + BQ = 5. Substituting, we get (4 – r) + (3 – r) = 5. Re-arranging, 7 – 2r = 5, ie 2r = 2 leading to r = 1.
Seeing this got me thinking about the next right triangle I remember, with sides 5, 12 and 13. What would be the radius of the circle inscribed in that? Amazingly enough, the answer to that is also a surprise. That radius is 2; isn’t mathematics wonderful?
