<s>[INST] What is the value of $m+n$ if $\frac{m}{n}$ is the radius of the smallest of the three circles? 

Circles of radii 5, 5, 8 and $\frac{m}{n} $(the smallest circle) are
  mutually externally tangent to all circles, where $m$ and $n$ are
  relatively prime positive  integers. Find $m + n$.

Source: Bangladesh Math Olympiad 2017 Junior category.
I can not figure the radius of the smallest circle. Is there any formula for an internal point in a isosceles triangle which can help me to solve this math?
 [/INST] Let $x$ be a radius of the little circle.
Thus, by Heron we can get an area of the left and of the right triangle:
$$\sqrt{(x+13)x\cdot8\cdot5}=\sqrt{40x(x+13)}.$$
An area of the lower triangle it's
$$\frac{10\sqrt{(x+5)^2-5^2}}{2}=5\sqrt{x^2+10x)}.$$
The area of the full triangle it's
$$\frac{10\sqrt{13^2-5^2}}{2}=60.$$
Thus, $$5\sqrt{x^2+10x}+2\sqrt{40x(x+13)}=60$$ or
$$\sqrt{5(x^2+10x)}+\sqrt{32(x^2+13x)}-12\sqrt5=0.$$
Can you end it now?
I got $$x=\frac{8}{9}.$$
Actually, the expression $\sqrt{5(x^2+10x)}+\sqrt{32(x^2+13x)}$ increases, which says it's enough to substitute $x=\frac{8}{9}.$
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