Right triangle from the inradius and circumradius

Text

A right triangle has circumradius $\dfrac52$ and inradius $1$. Determine the legs and the hypotenuse.

Solution

In a right triangle the hypotenuse is a diameter of the circumscribed circle: $c=2R=5$. The inradius of a right triangle is $r=\dfrac{a+b-c}{2}$, so $1=\frac{a+b-5}{2}\ \Longrightarrow\ a+b=7.$ With Pythagoras $a^2+b^2=25$: $2ab=49-25=24\ \Rightarrow\ ab=12.$ $a,b$ are roots of $t^2-7t+12=0$, i.e. $t=3$ and $t=4$. $\boxed{\text{legs }3\text{ and }4,\quad \text{hypotenuse }5.}$