11. The floor plan of a castle is designed in this way: consider a circumference of 2019 meters with a regular polygon of 2019 vertices inscribed in it. Once the vertices of the polygon are numbered from 1 to 2019 in a clockwise direction, draw circles of length $2019 m$ centered at each point numbered with a perfect square. The floor plan of the castle consists of the union of all the circles drawn. How many meters does the perimeter of the castle measure?
(A) 2692
(B) 4038
(C) 4627
(D) $\frac{29370}{2 \pi}$
(E) $2019 \pi$
11. The answer is (B). Let $\Gamma$ be the initial circle and $C_{1}, C_{2}, \ldots, C_{44}$ the circles drawn with centers at the vertices of the polygon labeled with perfect squares; note that, since the radius $R$ of each circle $C_{i}$ is equal to the radius of $\Gamma$ (all circles have a perimeter of $2019 = 2 \pi R$), $C_{i}$ passes through the center of $\Gamma$, which we will call $O$. We also note that for every $i$ the circles $C_{i}$ and $C_{i+1}$ (in the case $i=44$ the circles $C_{44}$ and $C_{1}$; from now on we will index all elements cyclically, that is, we will set $i+1=1$ for $i=44$) intersect at a point $P_{i}$ outside the circle $\Gamma$. This is because, denoting $O_{i}$ as the centers of the circles, the minimum value for the angle $\widehat{O_{i} \widehat{O O_{i+1}}}$ such that the intersection of $C_{i}$ and $C_{i+1}$ different from $O$ does not lie outside $\Gamma$ is such that $O O_{i}=O P_{i}=O O_{i+1}$ and is therefore $2 \pi / 3$; however, the angle $\widehat{O_{i} O O_{i+1}}$ in our case is at most $\frac{(i+1)^{2}-i^{2}}{2019} \cdot 2 \pi=\frac{2 i+1}{2019} \cdot 2 \pi \leq \frac{89}{2019} \cdot 2 \pi < 2 \pi / 3$.

The perimeter of the castle is thus obtained by summing, for $i$ from 1 to 44, the length of the arc $P_{i} P_{i+1}$ of the circle $C_{i+1}$ that does not contain the point $O$; this length is $R \theta_{i}$, where $\theta_{i}$ is the measure of the angle $P_{i} \widehat{O_{i+1} P}_{i+1}$ in radians. But, since they subtend the same arc of $C_{i+1}$, the angle $P_{i} \widehat{O_{i+1} P_{i+1}}$ is twice the angle $\widehat{P_{i} P_{i+1}}$; and, summing for all $i$ from 1 to 44, we obtain that $\theta_{1}+\ldots+\theta_{44}$ is therefore $4 \pi$ (the angles $\widehat{P_{i} O_{i+1}}$ cover precisely a full angle). The perimeter of the castle is thus $4 \pi R$ and, since we know that $2 \pi R$ is 2019 meters, the correct answer is 4038 meters.

Note that the fact that the centers of the circles were certain vertices of a regular polygon was not crucial to solving this problem: the same result (the value of $4 \pi R$ for the perimeter) could have been shown for any number of circles of radius $R$ that all have a common point, provided that the angle formed by that point and the centers of any two consecutive circles is at most $2 \pi / 3$.