<s>[INST] A right triangle with integer sides has area equal to twice its perimeter. Find sum of all possible circumradii. 
In right triangle $ABC$, the area is twice the perimeter, and all sides have integer lengths. Compute the sum of all possible circumradii of $ABC$. 

I only have set up an equation 
$$\frac{xy}{2}=2\left(x+y+\sqrt{x^2+y^2}\right)$$ and 
$$R=\frac{xy \sqrt{x^2+y^2}}{2xy}=\frac{\sqrt{x^2+y^2}}{2}$$ 
 [/INST] We can parameterize the sides of a right triangle ABC right-angled at C with integer sides in the following manner:
$$a=k(x^2-y^2),
b=2kxy,
c=k(x^2+y^2)$$
where k is any positive integer, x and y are co-prime integers with $x\not\equiv y\ (\textrm{mod}\ 2)$.
Using the condition, $\frac{ab}{2}=2(a+b+c)$, we obtain $ky(x-y)=4$ whence $k=4, y=1, x=2$ or $k=1,y=4,x=5$ or $k=2, y=2, x=3$. This leads to the triangles $(12,16,20)$, $(10,24,26)$ and $(9,40,41)$. In a right triangle, circumradius is half the hypotenuse. Therefore, $$ R=10,13, \text{or}\ 20.5 $$
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