Finding the hypotenuse on a triangle with a circle inscribed.

Question

The problem is a right-angled triangle with a inscribed circle, see figure, where you want to solve the distance CB (the hypotenuse).

The only known values are two lines which are linked to the circle center and the the corners of the hypotenuse.

I have tried many different methods but can't seem to solve it. I was told the solution would be simpler than I thought.

Note: You are only allowed to solve the problem though Euclidean means.

Answer 1

You could use the pythagoras theorem by writing all the sides only in terms of r.

$AC^2+AB^2=BC^2$

$(r+\sqrt{2^2-r^2})^2+(r+\sqrt{3^2-r^2})^2 = (\sqrt{3^2-r^2}+\sqrt{2^2-r^2})^2$

Then find r from this equation and hence find CB.

Answer 2

We let $\,h\,$ be the hypotenuse, let $\,x\,$ be the part of the hypotenuse near the $\,C\,$ vertex, and let $\,r\,$ be the inradius.

\begin{align*} x&=\sqrt{2^2-r^2}\\ h-x&=\sqrt{3^2-r^2}\\ \implies x&=h-\sqrt{3^2-r^2}\\\\ x=\sqrt{2^2-r^2}&=h-\sqrt{3^2-r^2}\\ \implies h&=\sqrt{2^2-r^2}+\sqrt{3^2-r^2} \end{align*}