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

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  • $\begingroup$ In your circumradius formula, you should have $\frac12\sqrt{x^2+y^2}$. $\endgroup$ – Blue Dec 3 '18 at 5:20
  • $\begingroup$ how would i solve it? i tried everything i knew $\endgroup$ – weareallin Dec 3 '18 at 5:25
  • $\begingroup$ 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. $\endgroup$ – Blue Dec 3 '18 at 5:30
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    $\begingroup$ $$\frac{xy}{2} = 2\left(x+y +\sqrt{x^2+y^2}\right) \implies (xy - 4(x+y))^2 = 16(x^2+y^2)\\ \iff xy(xy-8(x+y)+32) = 0 \implies (x-8)(y-8) = 32 $$ You then need to check for what positive $x,y$, $\sqrt{x^2+y^2}$ is also an integer. $\endgroup$ – achille hui Dec 3 '18 at 5:47
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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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$$\frac{xy}{2} = 2\left(x+y +\sqrt{x^2+y^2}\right) \implies (xy - 4(x+y))^2 = 16(x^2+y^2)\\ \iff xy(xy-8(x+y)+32) = 0 \implies (x-8)(y-8) = 32$$ 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 $\frac{41}2,\frac{26}2,\frac{20}2$.

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