Three pentagons

A regular polygon with $n$ sides of length $s$ has interior angles of $\theta = \pi- \frac{2\pi}{n}$ , an apothem of $a = \frac{s}{2\tan\frac{\pi}{n}}$ , a radius of $R = \frac{s}{2\sin\frac{\pi}{n}}$ , and area of $A = \frac{n}{2}R^2\sin\frac{2\pi}{n}$ . Therefore, if $r$ is the length of the radius of each circle, the regular polygon made from the centers of the circles have sides of $s = 2r$ and therefore an area of $S(n)$ $=$ $\frac{n}{2}R^2\sin\frac{2\pi}{n}$ $=$ $\frac{n}{2}(\frac{2r}{2\sin\frac{\pi}{n}})^2\sin\frac{2\pi}{n}$ or $S(n) = nr^2 \cot \frac{\pi}{n}$

Let $Z$ be the center of all the regular polygons. Then we can find the radius of the small regular polygon by examining $\triangle AQZ$ .

As the radius of one of the circles, $AQ = r$ . As the radius of the regular polygon with sides $2s$ , $AZ = \frac{r}{\sin \frac{\pi}{n}}$ . Finally, as the sum of a right angle (from the tangent line) and half the interior angle of the regular polygon, $\angle AQZ$ $=$ $\frac{\pi}{2} + \frac{1}{2}(\pi - \frac{2\pi}{n})$ $=$ $\pi - \frac{\pi}{n}$ . Using the law of cosines, we have $(\frac{r}{\sin \frac{\pi}{n}})^2 = r^2 + QZ^2 - 2rQZ\cos(\pi - \frac{\pi}{n})$ which after solving using the quadratic formula and simplifying means $QZ = -\cos\frac{\pi}{n} + \cot\frac{\pi}{n}\sqrt{\sin^2\frac{\pi}{n} + 1}$ . Since $QZ$ is the radius of the small regular polygon, its area is $S_m(n)$ $=$ $\frac{n}{2}R^2\sin\frac{2\pi}{n}$ $=$ $\frac{n}{2}(-\cos\frac{\pi}{n} + \cot\frac{\pi}{n}\sqrt{\sin^2\frac{\pi}{n} + 1})^2\sin\frac{2\pi}{n}$ or $S_m(n) = nr^2\cot\frac{\pi}{2}(\cos^2\frac{\pi}{n}(3 - 2\cos^2\frac{\pi}{n}) - 2\cos^2\frac{\pi}{n}\sin\frac{\pi}{n}\sqrt{\sin^2\frac{\pi}{n} + 1})$

We can find the radius of the large regular polygon by examining $\triangle ALZ$ .

As the radius of one of the circles, $AL = r$ . As from before, $AZ = \frac{r}{\sin \frac{\pi}{n}}$ . Finally, $\angle ALZ$ is the interior angle of the large regular polygon minus half the interior angle of the large regular polygon ( $\angle KLZ$ ) minus the difference of the interior angle of the large regular polygon and a right angle from the tangent line ( $\angle ALM = \angle KLM - \angle KLA$ ), which means $\angle ALZ$ $=$ $(\pi - \frac{2\pi}{n}) - \frac{1}{2}(\pi - \frac{2\pi}{n}) - ((\pi - \frac{2\pi}{n}) - \frac{\pi}{2})$ $=$ $\frac{\pi}{n}$ . Using the law of cosines, we have $(\frac{r}{\sin \frac{\pi}{n}})^2 = r^2 + LZ^2 - 2rLZ\cos(\frac{\pi}{n})$ which after solving using the quadratic formula and simplifying means $LZ = \cos\frac{\pi}{n} + \cot\frac{\pi}{n}\sqrt{\sin^2\frac{\pi}{n} + 1}$ . Since $LZ$ is the radius of the large regular polygon, its area is $S_M(n)$ $=$ $\frac{n}{2}R^2\sin\frac{2\pi}{n}$ $=$ $\frac{n}{2}(\cos\frac{\pi}{n} + \cot\frac{\pi}{n}\sqrt{\sin^2\frac{\pi}{n} + 1})^2\sin\frac{2\pi}{n}$ or $S_M(n) = nr^2\cot\frac{\pi}{2}(\cos^2\frac{\pi}{n}(3 - 2\cos^2\frac{\pi}{n}) + 2\cos^2\frac{\pi}{n}\sin\frac{\pi}{n}\sqrt{\sin^2\frac{\pi}{n} + 1})$

Therefore, the ratio $\frac{S_M(n) + S_m(n)}{S(n)} = 2\cos^2\frac{\pi}{n}(3 - 2\cos^2\frac{\pi}{n})$

Since $\lim_{n \to \infty} \cos\frac{\pi}{n} = 1$ , $\lim_{n \to \infty} \frac{S_M(n) + S_m(n)}{S(n)} = 2\cdot1^2(3 - 2\cdot1^2) = \boxed{2}$ .

When $n = 5$ , $\cos\frac{\pi}{5} = \frac{1 + \sqrt{5}}{4}$ , and the ratio is $2(\frac{1 + \sqrt{5}}{4})^2(3 - 2(\frac{1 + \sqrt{5}}{4})^2) = \frac{11 + 3\sqrt{5}}{8}$ , so $U = 11$ , $V = 3$ , $W = 5$ , and $Z = 8$ , and $U + V + W + Z = \boxed{27}$ .

Suppose that the original polygon has side $2$ . Extend the lines $AE$ and $LK$ to meet at $X$ . Then, using the right-angled triangle $ALX$ and using the Sine Rule in the triangle $KEX$ , we deduce that $AX = \sec\theta$ and $EX = \cos\alpha \sec\theta$ , and hence $2 = (1 - \cos\alpha)\sec\theta$ , where $\alpha$ is the exterior angle of the polygon.

Similarly, $LX = \tan\theta$ and $KX = \sin(\theta-\alpha)\sec\theta$ , and hence $LK \; = \; \tan\theta - \sin(\theta-\alpha)\sec\theta \; = \; (1 - \cos\alpha)\tan\theta +\sin\alpha \; = \; \sqrt{(3 - \cos\alpha)(1 + \cos\alpha)} + \sin\alpha$ A similar set of calculations shows that $QR \; = \; \sqrt{(3 - \cos\alpha)(1 + \cos\alpha)} - \sin\alpha$ Thus $\frac{S_M+ S_m}{S} \; = \; \frac{LK^2 + QR^2}{4} \; = \; \tfrac12\big[(3 - \cos\alpha)(1 + \cos\alpha) + \sin^2\alpha\big] \; = \; 2 + \cos\alpha - \cos^2\alpha$ This argument works for any regular polygon, and so $\frac{S_M(n) + S_m(n)}{S(n)} \; = \; 2 + \cos\tfrac{2\pi}{n} - \cos^2\tfrac{2\pi}{n}$ for any integer $n \ge 5$ . The argument can be adjusted to cope with $n=3,4$ - in these cases $\theta < \alpha$ . The formula for the fraction is the same, however. When $n=5$ (this problem) we obtain the ratio $\tfrac{1}{8} (11 + 3 \sqrt{5})$ , giving the answer $\boxed{27}$ . In addition $\lim_{n \to \infty} \frac{S_M(n) + S_m(n)}{S(n)} \; = \; 2$