AMC8 2005 8, Problem 3

What is the minimum number of small squares that must be colored black so that a line of symmetry lies on the diagonal $\overline{BD}$ of square $ABCD$? [asy] defaultpen(linewidth(1)); for ( int x = 0; x &lt; 5; ++x ) {     draw((0,x)--(4,x));     draw((x,0)--(x,4)); }  fill((1,0)--(2,0)--(2,1)--(1,1)--cycle); fill((0,3)--(1,3)--(1,4)--(0,4)--cycle); fill((2,3)--(4,3)--(4,4)--(2,4)--cycle); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle); label("$A$", (0, 4), NW); label("$B$", (4, 4), NE); label("$C$", (4, 0), SE); label("$D$", (0, 0), SW); [/asy]

Solution: Rotating square $ABCD$ counterclockwise $45^\circ$ so that the line of symmetry $BD$ is a vertical line makes it easier to see that $\fbox{4}$ squares need to be colored to match its corresponding square.

Answer: 4