circlescontest-mathgeometrytriangles
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?
Let $E$ be on $BC$ so $CE = CD$. Then $\angle DEC = 80^{\circ}$ so $CADE$ is cyclic, so we have by the power of the point with respect to $B$: $$a\cdot (a-d) = b\cdot (b-x)\;\;\;\;\;\;\;\;\;\;\;\;(1)$$
By angle bisector theorem we have $${b-x\over x} = {a\over b}\implies b(b-x)= ax\;\;\;\;\;\;\;\;(2)$$
By (1) and (2) we have $$a(a-d)= ax \implies a-d= x$$
Since we are searching for $d+x = a= 2017$ we are done.
$\triangle ABC \sim \triangle AMM' \sim \triangle PNM'$
$\frac{BC}{AB}=\frac{NM'}{PN}$
$PM'=PM=2PN$
$NM'=\sqrt{(2PN)^2-PN^2}=PN\sqrt3$
$\frac{NM'}{PN}=\sqrt3$
$\frac{BC}{AB}=\sqrt3$
$BC=AB\sqrt3=5\sqrt3$
$a=5,b=3,\color{blue}{a+b=8}$
Best Answer
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 $\frac 89$)