The catheti of a right triangle with given hypotenuse and radius of inscribed circle

Question

Triangle $\triangle ABC$ is a right triangle with $\measuredangle ACB=90^\circ$. Let $AB=c$ and the radius of the inscribed circle be $r$. Find the catheti and the area of the triangle $\triangle ABC$.

We now that the radius $r=\frac12(a+b-c)$. Furthermore, $c^2=a^2+b^2$. How to find the lengths from here?

Answer 1

You have $a=c_1+r$, $b=c_2+r$ and: $$ \cases{ (c_1+r)^2+(c_2+r)^2=(c_1+c_2)^2\\ c_1+c_2=c\\ } $$ Solving you can find $c_1$ and $c_2$.

Answer 2

Similar to other solution plus a figure:

We have:

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

$BF=BD=c_1$

$AE=AD=c_2$

$AB=c$

$AC=c_2+r$ and $CB=c_1+r$ ⇒ $c_1^2+c_2^2+2r(c_1+c_2)+2r^2=c^2$

$c_1+c_2=c$ ⇒ $c_1^2+c_2^2+2 c_1c_2=c^2$

Subtracting two relations we get:

$$2rc+2r^2-2c_1c_2=0$$

⇒ $c_1c_2=rc+r^2$

$c_1+c_2=c$

Now we construct following quadratic equation to find $c_1$ and $c_2$:

$x^2-(c_1+c_2) x+ c_1c_2=0$

$x^2- c x +r^2+rc=0$

Which gives:

$x=\frac{c+\sqrt{c^2-4r^2-4rc}}{2}=c_1$

$x=\frac{c-\sqrt{c^2-4r^2-4rc}}{2}=c_2$

Example:

$c=105$ and $r=20$ gives:

$c_1=68.5$ and $c_2=36.5$ which results:

$CB=20+68.5=88.5$

$CA=20+36.5=56.5$