# 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 $$R=\frac{xy \sqrt{x^2+y^2}}{2xy}=\frac{\sqrt{x^2+y^2}}{2}$$

• In your circumradius formula, you should have $\frac12\sqrt{x^2+y^2}$. – Blue Dec 3 '18 at 5:20
• how would i solve it? i tried everything i knew – weareallin Dec 3 '18 at 5:25
• You haven't used the fact that the sides are integers. Pythagorean Triples can be expressed as $$x = k ( m^2-n^2 )\qquad y = 2 k m n \qquad z = k ( m^2 + n^2 )$$ for integers $k$, $m$, $n$. Try using that information. – Blue Dec 3 '18 at 5:30
• $$\frac{xy}{2} = 2\left(x+y +\sqrt{x^2+y^2}\right) \implies (xy - 4(x+y))^2 = 16(x^2+y^2)\\ \iff xy(xy-8(x+y)+32) = 0 \implies (x-8)(y-8) = 32$$ You then need to check for what positive $x,y$, $\sqrt{x^2+y^2}$ is also an integer. – achille hui Dec 3 '18 at 5:47

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, $$R=10,13, \text{or}\ 20.5$$

$$\frac{xy}{2} = 2\left(x+y +\sqrt{x^2+y^2}\right) \implies (xy - 4(x+y))^2 = 16(x^2+y^2)\\ \iff xy(xy-8(x+y)+32) = 0 \implies (x-8)(y-8) = 32$$ Since $$x$$ and $$y$$ are integers and $$32 = 2^5$$, the total numbers of factors are $$(5+1)=6$$ which are $$1,2,4,8,16,32$$ . So possible cases avoiding any redundancy for we just want circumradii are as follows: $$\underset{32}{(x-8)}\underset{1}{(y-8)} = 32$$ $$\underset{16}{(x-8)}\underset{2}{(y-8)} = 32$$ $$\underset{8}{(x-8)}\underset{4}{(y-8)} = 32$$ So possible triplets turn out to be $$40,9,41$$; $$24,10,26$$; $$16,12,20$$. So, possible circumradii are $$\frac{41}2,\frac{26}2,\frac{20}2$$.