## Task 18/76

Given an annulus and $n$ circles within it such that each of them touches the inner and outer boundary circles and two other circles. How large is $n$ if the ratio of the annulus area $A_{1}$ to the sum of the $n$ circle areas $A_{2}$ is $\sqrt{2}$?
Let $r_{1}$ be the radius of the outer boundary circle and $r_{2}$ the radius of the inner. Then we have

$$
A_{1}=\left(r_{1}^{2}-r_{2}^{2}\right) \pi \quad ; \quad A_{2}=n\left[\frac{1}{4}\left(r_{1}-r_{2}\right)^{2} \pi\right]
$$

From this it follows that

$$
\frac{A_{1}}{A_{2}}=\frac{4\left(r_{1}+r_{2}\right)\left(r_{1}-r_{2}\right)}{n\left(r_{1}-r_{2}\right)\left(r_{1}-r_{2}\right)}=\frac{4\left(r_{1}+r_{2}\right)}{n\left(r_{1}-r_{2}\right)}=\sqrt{2}
$$

The centers of the $n$ circles lie on the vertices of a regular $n$-gon. If we connect the vertices of this $n$-gon to the center of the annulus, we obtain $n$ congruent isosceles triangles with the leg length $\frac{1}{2}\left(r_{1}+r_{2}\right)$ and the base length $\left(r_{1}-r_{2}\right)$; the angle $\alpha$ between the legs is $\frac{360^{\circ}}{n}$. Thus we have

$$
\frac{\frac{1}{2}\left(r_{1}-r_{2}\right)}{\frac{1}{2}\left(r_{1}+r_{2}\right)}=\sin \frac{\alpha}{2}=\sin \frac{180^{\circ}}{n}=\frac{r_{1}-r_{2}}{r_{1}+r_{2}}
$$

Substituting this into the equation above, we get

$$
\frac{4}{n \cdot \sin \frac{180^{\circ}}{n}}=\sqrt{2} \quad ; \quad n \cdot \sin \frac{180^{\circ}}{n}=2 \sqrt{2}
$$

This equation clearly has a solution for $n=4: 4 \cdot \sin 45^{\circ}=2 \sqrt{2}$. It remains to show that this is the only solution.

It is easy to verify that $n=1 ; 2 ; 3$ are not solutions. To prove that there are no solutions for $n \geq 5$, we proceed as follows:

It is well known that the function $\frac{\sin x}{x}$ grows strictly monotonically as $x \rightarrow \infty$ and $\lim _{x \rightarrow \infty} \frac{\sin x}{x}=1$.

Then the function $\frac{\pi \sin x}{x}$ also grows strictly monotonically and

$$
\lim _{x \rightarrow \infty} \frac{\pi \sin x}{x}=\pi
$$

Setting $x=\frac{\pi}{n}$, the function takes the form

$$
n \cdot \sin \frac{\pi}{n}=n \cdot \sin \frac{180^{\circ}}{n}
$$

From the strict monotonicity, it follows immediately that for $n>4$ no further solution can exist.