Given $372$ points in a circle with a radius of $10$, there is an annulus with radii $2$ and $3$ containing at least $12$ of these points. 
Given $372$ points in a circle with a radius of $10$, show that there is an annulus with inner radius $2$ and outer radius $3$, which contains not less than $12$ of the given points.

My thinking is to try divide the circle in 31 section, each of them contained in some ring with rays 2 and 3. After long thinking I didn't come up with a way to do this. PLEASE HELP ! :)
 A: Let there be $n$ points $P_1,P_2,\ldots,P_n$ in a circle $C$ of radius $r>0$.  Suppose that $a$ and $b$ are real numbers such that $0<a<b$.  Define $D$ to be the circle of radius $r+b$.  For each point $Q$ in $D$, the notation $N(Q)$ represents the number of points $P_j$ that lies in the annulus centered at $Q$ with inner radius $a$ and outer radius $b$.  By selecting $Q$ uniformly randomly in $D$, $N$ is a random variable.  The expected value of $N$ is
$$\mathbb{E}[N]=\frac{1}{\pi (r+b)^2}\,\int_D\,N(Q)\,\text{d}\lambda(Q)\,,$$
where $\lambda$ is the usual two-dimensional Lebesgue measure (or the area measure).
Now, for every $j=1,2,\ldots,n$ and for any point $Q$ in the circle $D$, we define
$$X_j(Q):=\left\{\begin{array}{ll}
1&\text{if }a<|P_jQ|<b\,,\\
0&\text{else}\,.
\end{array}\right.$$
Thus, $N=\sum\limits_{j=1}^n\,X_j$, so
$$\mathbb{E}[N]=\sum_{j=1}^n\,\frac{1}{\pi (r+b)^2}\,\int_D\,X_j(Q)\,\text{d}\lambda(Q)\,.$$
Prove that
$$\int_D\,X_j(Q)\,\text{d}\lambda(Q)=\text{area}(Q)=\pi(b^2-a^2)$$
for each $j=1,2,\ldots,n$.  Therefore,
$$\mathbb{E}[N]=\sum_{j=1}^n\,\frac{\pi(b^2-a^2)}{\pi(r+b)^2}=\left(\frac{b^2-a^2}{(r+b)^2}\right)\,n\,.$$
When $n=372$, $a=2$, $b=3$, and $r=10$, we have
$$\mathbb{E}[N]=\left(\frac{3^2-2^2}{(10+3)^2}\right)\cdot 372>11\,.$$
This proves that there exists a point $Q$ in $D$ for which there are at least $12$ points among the points $P_j$ that lie in the annulus centered at $Q$ with inner radius $2$ and outer radius $3$.
