# 2001 AMC 10 Problems/Problem 11

## Problem

Consider the dark square in an array of unit squares, part of which is shown. The first ring of squares around this center square contains $8$ unit squares. The second ring contains $16$ unit squares. If we continue this process, the number of unit squares in the $100^\text{th}$ ring is $[asy] unitsize(3mm); defaultpen(linewidth(1pt)); fill((2,2)--(2,7)--(7,7)--(7,2)--cycle, mediumgray); fill((3,3)--(6,3)--(6,6)--(3,6)--cycle, gray); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle, black); for(real i=0; i<=9; ++i) { draw((i,0)--(i,9)); draw((0,i)--(9,i)); }[/asy]$ $\textbf{(A)}\ 396 \qquad \textbf{(B)}\ 404 \qquad \textbf{(C)}\ 800 \qquad \textbf{(D)}\ 10,\!000 \qquad \textbf{(E)}\ 10,\!404$

## Solution

### Solution 1

We can partition the $n^\text{th}$ ring into $4$ rectangles: two containing $2n+1$ unit squares and two containing $2n-1$ unit squares.

There are $2(2n+1)+2(2n-1)=4n+2+4n-2=8n$ unit squares in the $n^\text{th}$ ring.

Thus, the $100^\text{th}$ ring has $8 \times 100 = \boxed{\textbf{(C)} 800}$ unit squares.

### Solution 2

We can make the $n^\text{th}$ ring by removing a square of side length $2n-1$ from a square of side length $2n+1$.

This ring contains $(2n+1)^2-(2n-1)^2=(4n^2+4n+1)-(4n^2-4n+1)=8n$ unit squares.

Thus, the $100^\text{th}$ ring has $8 \times 100 = \boxed{\textbf{(C)}\ 800}$ unit squares.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 