Two concentric circles have radii 1 and 4

Question

What is $k+m+n$?

Thank you for helping. Please give a solution for me, if you don't mind.

Answer 1

Let $O$ be the center of the concentric circles, let $A$ be the center of the top light-gray circle, and let $B$ be the center of one of the upper dark-gray circles; let $r$ be the common radius of the smaller circles.

Clearly, symmetry dictates that $\angle AOB = 60^\circ$. Note that $|OA| = 4-r$, and $|OB| = 1+r$, and $|AB| = 2r$. Thus, by the Law of Cosines (and the assumption that $r$ must be postive):

$$\begin{align} |AB|^2 &= |OA|^2 + |OB|^2 - 2 |OA| |OB| \cos 60^\circ \\ \implies \qquad (2r)^2 &= (4-r)^2 + (1+r)^2 - ( 4-r )( 1 + r ) \\ \implies \qquad 0 &= r^2 + 9 r - 13 \\ \implies \qquad r &= \frac{1}{2}\left( -9 + \sqrt{7\cdot 19} \right) \end{align}$$

Thus, $k = -9$, $m = 133$, and $n = 2$, so that $k+m+n = 126$.

Answer 2

as you can see from the picture you simply take some congruent triangles, find a regular triangle and three times use the pythagorean theorem to find 126.