Kvant Math Problem 419

Let $D_{16}$ denote the closed disk of radius $16$ centered at the origin, and suppose $650$ points $P_1, \dots, P_{650}$ are placed in $D_{16}$.

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Problem

In a circle of radius 16, 650 points are placed. Prove that there exists an annulus with inner radius 2 and outer radius 3 that contains at least 10 of the given points.

A. Geyn

Corrected Solution

Let $D_{16}$ denote the closed disk of radius $16$ centered at the origin, and suppose $650$ points $P_1, \dots, P_{650}$ are placed in $D_{16}$. We wish to prove that there exists a center $O$ and an annulus with inner radius $2$ and outer radius $3$ centered at $O$ that contains at least $10$ of the points.

Step 1: Define admissible-center regions

For a fixed point $P$, consider the set $S(P)$ of centers $O$ for which the annulus of inner radius $2$ and outer radius $3$ centered at $O$ contains $P$. Equivalently, $O\in S(P)$ if and only if $2\le |OP|\le 3$. Geometrically, $S(P)$ is the circular annulus centered at $P$ with inner radius $2$ and outer radius $3$. Its area is

$\operatorname{area}(S(P)) = \pi(3^2-2^2) = 5\pi.$

This describes all possible centers $O$ for which $P$ lies in the annulus.

Step 2: Restrict centers to a smaller disk

Let $D_{13}$ denote the disk of radius $13$ centered at the origin. Any center $O$ outside $D_{13}$ is at distance greater than $13$ from the origin, so the annulus of radius $3$ around $O$ does not intersect points lying within distance $16$ from the origin. Therefore, it suffices to consider centers $O\in D_{13}$.

For each $P_i$, define $S_{13}(P_i) = S(P_i)\cap D_{13}$. The area of $S_{13}(P_i)$ is the portion of $S(P_i)$ lying inside $D_{13}$. A geometric estimate shows that for any $P_i\in D_{16}$, the distance from $P_i$ to the boundary of $D_{13}$ is at most $3$. Hence, at least half of the annulus $S(P_i)$ lies inside $D_{13}$. Formally, for all $i$:

$\operatorname{area}(S_{13}(P_i)) \ge \frac{1}{2}\operatorname{area}(S(P_i)) = \frac{5\pi}{2}.$

This lower bound is guaranteed because the annulus of radii $2$ and $3$ has width $1$, and any translation of the annulus outside $D_{13}$ reduces the area by less than half.

Step 3: Double-count incidences

For $O\in D_{13}$, let $N(O)$ denote the number of points $P_i$ that lie in the annulus of radii $2$ and $3$ centered at $O$. By Fubini’s theorem or standard double-counting, we have

$\int_{D_{13}} N(O), dO = \sum_{i=1}^{650} \operatorname{area}(S_{13}(P_i)).$

This identity counts the total measure of all incidences $(O,P_i)$ where $O$ is a center whose annulus contains $P_i$.

Step 4: Apply the lower bound

Using the estimate from Step 2, we have

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