Estonian Math Competitions

The ratio of the inradius and the circumradius of a regular $n$-gon is $\cos\frac{180^\circ}{n}$. Hence the ratio of the areas of the incircle and the circumcircle of a regular $n$-gon is $\cos^2\frac{180^\circ}{n}$.

Let a regular $n$-gon and a regular $m$-gon with a common circumcircle be given. W.l.o.g., let $n \le m$ and the area of the common circumcircle be $1$. By the above, the areas of the incircles of these polygons are $\cos^2\frac{180^\circ}{n}$ and $\cos^2\frac{180^\circ}{m}$, respectively. As their sum must equal the area of the common circumcircle, we get the equation

$$\cos^2\frac{180^\circ}{n} + \cos^2\frac{180^\circ}{m} = 1$$

which is equivalent to

$$\cos^2\frac{180^\circ}{n} = \sin^2\frac{180^\circ}{m}$$

As both $n$ and $m$ are larger than $2$, both $\frac{180^\circ}{n}$ and $\frac{180^\circ}{m}$ are less than $90^\circ$, whence the equation reduces to

$$\cos\frac{180^\circ}{n} = \sin\frac{180^\circ}{m}$$

Thus

$$\frac{180^\circ}{n} + \frac{180^\circ}{m} = 90^\circ$$

implying that

$$\frac{1}{n} + \frac{1}{m} = \frac{1}{2}$$

If $n = 3$ then $m = 6$; if $n = 4$ then $m = 4$; if $n > 4$ then $m < 4$, contradicting the assumption $n \le m$.