# 1990 AJHSME Problems/Problem 25

## Problem

How many different patterns can be made by shading exactly two of the nine squares? Patterns that can be matched by flips and/or turns are not considered different. For example, the patterns shown below are not considered different.

$[asy] fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,gray); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle,gray); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,linewidth(1)); draw((2,0)--(2,3),linewidth(1)); draw((0,1)--(3,1),linewidth(1)); draw((1,0)--(1,3),linewidth(1)); draw((0,2)--(3,2),linewidth(1)); fill((6,0)--(8,0)--(8,1)--(6,1)--cycle,gray); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle,linewidth(1)); draw((8,0)--(8,3),linewidth(1)); draw((6,1)--(9,1),linewidth(1)); draw((7,0)--(7,3),linewidth(1)); draw((6,2)--(9,2),linewidth(1)); fill((14,1)--(15,1)--(15,3)--(14,3)--cycle,gray); draw((12,0)--(15,0)--(15,3)--(12,3)--cycle,linewidth(1)); draw((14,0)--(14,3),linewidth(1)); draw((12,1)--(15,1),linewidth(1)); draw((13,0)--(13,3),linewidth(1)); draw((12,2)--(15,2),linewidth(1)); fill((18,1)--(19,1)--(19,3)--(18,3)--cycle,gray); draw((18,0)--(21,0)--(21,3)--(18,3)--cycle,linewidth(1)); draw((20,0)--(20,3),linewidth(1)); draw((18,1)--(21,1),linewidth(1)); draw((19,0)--(19,3),linewidth(1)); draw((18,2)--(21,2),linewidth(1)); [/asy]$

$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18$

## Solution

We break this into cases.

Case 1: At least one square is a vertex: WLOG, suppose one of them is in the upper-left corner. Then, consider the diagonal through that square. The two squares on that diagonal could be the second square, or the second square is on one side of the diagonal.

The square is reflectionally symmetric about this diagonal, so we only consider the squares on one side, giving another three possibilities.

In this case, there are $2+3=5$ distinct squares.

Case 2: At least one square is on an edge, but no square is on a vertex: There are clearly two edge-edge combinations and one edge-center combination, so this case has $3$ squares.

In total, there are $3+5=8$ distinct squares $\rightarrow \boxed{\text{C}}$