What is the value of $m+n$ if $\frac{m}{n}$ is the radius of the smallest of the three circles?

Question

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?

Answer 1

Let $r$ denote the radius of the tiny circle (so $r=\frac mn$ in your notation, though it is not clear from the start that $r$ is rational).

Drop the perpendicular from the center of the big circle. We see quickly that it has length $12$.

Drop the perpendicular from the center of the tiny circle. We see that it has length $h$ where $$h^2+5^2=(5+r)^2$$

But in terms of $h,r$ the length of the first perpendicular we looked at is $8+r+h$, so $$8+r+h=12$$

Can you finish from here?

(note: I also get $\frac 89$)

Answer 2

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 $x=\frac{8}{9}.$

Answer 3

A circle tangent to three mutually touching circles is called a Soddy circle. The expression relating the bends (or signed inverse radii) of the four circles is $$ 2(\epsilon_1^2 + \epsilon_2^2 + \epsilon_3^2 + \epsilon_4^2) = (\epsilon_1 + \epsilon_2 + \epsilon_3 + \epsilon_4)^2. $$ Solving this for one of the radii gives $$ r_4=\frac{r_1r_2r_3}{r_1r_2 + r_1r_3+r_2r_3\pm2\sqrt{r_1r_2r_3(r_1+r_2+r_3)}}. $$ Here $$ r_4=\frac{200}{105\pm2\sqrt{200\cdot 18}}=\frac{40}{21\pm 24}=-\frac{40}{3}{\text{ or }}\frac{8}{9}. $$ The positive solution is the radius of the inner circle. (For free, you also get the negative solution, whose magnitude is the radius of the circle tangent to all three other circles on the outside.)

Answer 4

The triangle has a base side of length $10$ and two sides of length $13$ each. The hight this trinagle is $h = \sqrt {13^2-5^2} = 12$. This height consists of the radius of the large circle, the radius $r=\frac mn$ of the smallest circle and the hight of another isoceles triangle with the same base side and two other sides of lengths $5+r$, which is $\sqrt {(5+r)^2-5^2} = \sqrt{10r+r^2}$. Therefore, we have

$$ \begin{align} & \quad 12 = 8 + r + \sqrt{10r+r^2}\\ \Leftrightarrow & \quad 4-r = \sqrt {10r+r^2}\\ \Leftrightarrow & \quad (4-r)^2 = 10r + r^2\\ \Leftrightarrow & \quad 16 - 8r + r^2 = 10r + r^2\\ \Leftrightarrow & \quad 18r = 16\\ \Leftrightarrow & \quad r = \frac{16}{18} = \frac89\\ \end{align} $$

Thus $m=8$ and $n=9$, so $n+m=17$.