CF 1920C - Partitioning the Array

Rating: 1600
Tags: brute force, math, number theory
Solve time: 48s
Verified: no

Solution

Problem-Type Check

This is a Type B problem, a proof of existence. The requirement is to show that for any placement of 650 points in a disk of radius 16, there exists at least one annulus of inner radius 2 and outer radius 3 containing at least 10 points. The proposed solution attempts to establish this using a universal lower bound on the area of admissible centers, summed over all points, and an averaging argument. The approach is consistent with the requirements of a Type B problem.

Step-by-Step Verification

Step 1: Define admissible-center regions - VALID

The claim that for a fixed point $P$, the set $S(P)$ of centers $O$ with $2\le |OP|\le 3$ is a circular annulus of area $5\pi$ is correct. The computation $\pi(3^2-2^2)=5\pi$ is exact. This step is fully justified.

Step 2: Restrict centers to a smaller disk - UNJUSTIFIED (Justification gap)

The solution defines $D_{13}$ and claims that at least half of each annulus $S(P_i)$ lies inside $D_{13}$ because the annulus has width $1$ and the distance from $P_i$ to the boundary of $D_{13}$ is at most $3$. This is a heuristic geometric argument; it does not rigorously guarantee that $\operatorname{area}(S_{13}(P_i))\ge 5\pi/2$ for all positions of $P_i$. For example, if $P_i$ lies near the edge of $D_{16}$ opposite the origin, a substantial portion of the annulus could lie outside $D_{13}$ and the fraction of area inside $D_{13}$ could be less than $1/2$. This is a justification gap, not a critical error: the conclusion is likely correct, but the argument is insufficiently rigorous. A precise lower bound would require computing the minimal intersection of a circle of radius 3 with a disk of radius 13, which is omitted.

Step 3: Double-count incidences - VALID

The integral identity

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

correctly counts the measure of incidences $(O,P_i)$. This step is standard and fully justified once Step 2 is assumed.

Step 4: Apply the lower bound - UNJUSTIFIED (Justification gap)

The argument relies on the claimed bound $\operatorname{area}(S_{13}(P_i)) \ge 5\pi/2$. Since Step 2 does not rigorously justify this bound, the averaging argument that $\int N(O),dO \ge 650 \cdot 5\pi/2$ is unproven. This is a justification gap, not a critical error: the averaging argument would work if the bound were valid.

Completeness Check

The proposed solution correctly sets up the framework for a universal existence argument using double-counting and averaging. All points are considered, and the final conclusion follows if the area lower bound holds. The only unhandled or unjustified part is the precise geometric estimate that guarantees at least half the area of each annulus lies inside $D_{13}$. There are no other missing cases.

Summary

The solution correctly identifies the structure of the argument and applies double-counting and averaging properly. The primary flaw is the lack of a fully rigorous justification for the lower bound on the area of $S_{13}(P_i)$. The claim that at least half the annulus lies inside the smaller disk is presented as a heuristic geometric estimate rather than a precise computation. This gap leaves the argument logically incomplete, even though the conclusion is almost certainly correct.

VERDICT: FAIL - the solution lacks a rigorous proof that at least half of each point's admissible annulus lies inside the radius-13 disk.