Difference between revisions of "1996 AIME Problems/Problem 7"

Latest revision as of 17:25, 31 August 2024

Problem

Two squares of a $7\times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible?

Solution 1 (Generalized)

There are ${49 \choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards.

$[asy] pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,3)*unitsquare,rgb(1,1,.4));fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(3,4)*unitsquare,rgb(.8,.8,.5));fill(shift(1,4)*unitsquare,rgb(.8,.8,.5)); fill(shift(2,3)*unitsquare,rgb(.8,.8,.5));fill(shift(2,1)*unitsquare,rgb(.8,.8,.5)); fill(shift(3,2)*unitsquare,rgb(.8,.8,.5));fill(shift(5,2)*unitsquare,rgb(.8,.8,.5)); D(arc(O,1,280,350),EndArrow(4)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,1,10,80),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,1,190,260),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,1,100,170),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]$

$[asy]pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(2,1)*unitsquare,rgb(1,1,.4)); fill(shift(1,4)*unitsquare,rgb(.8,.8,.5)); fill(shift(5,2)*unitsquare,rgb(.8,.8,.5)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]$

For most pairs, there will be
three other equivalent boards.

For those symmetric about the center,
there is only one other.

Note that a pair of yellow squares will only yield $2$ distinct boards upon rotation iff the yellow squares are rotationally symmetric about the center square; there are $\frac{49-1}{2}=24$ such pairs. There are then ${49 \choose 2}-24$ pairs that yield $4$ distinct boards upon rotation; in other words, for each of the ${49 \choose 2}-24$ pairs, there are three other pairs that yield an equivalent board.

Thus, the number of inequivalent boards is $\frac{{49 \choose 2} - 24}{4} + \frac{24}{2} = \boxed{300}$ . For a $(2n+1) \times (2n+1)$ board, this argument generalizes to $n(n+1)(2n^2+2n+1)$ inequivalent configurations.

Solution 2 (Casework)

There are 4 cases: 1. The center square is occupied, in which there are $12$ cases. 2. The center square isn't occupied and the two squares that are opposite to each other with respect to the center square, in which there are $12$ cases. 3. The center square isn't occupied and the two squares can rotate to each other with a $90^{\circ}$ rotation with each other and with respect to the center square, in which case there are $12$ cases. 4. And finally, the two squares can't rotate to each other with respect to the center square in which case there are $\dbinom{12}{2} \cdot \frac{16}{4} = 264$ cases. Add up all the values for each case to get $\boxed{300}$ as your answer.

~First

Solution 3 (Burnside’s Lemma)

Consider the group $G=\mathbb{Z}/4\mathbb{Z}$ with addition acting on set $X$ , where $X$ is the set of color schemes on the checkerboard. We now apply Burnside’s Lemma. Since $\mathbb{Z}/4\mathbb{Z}=\langle r \mid r^4=1\rangle$ , we only need to consider 4 cases for $g$ an element of $\mathbb{Z}/4\mathbb{Z}$ .

1. $g=1$ . There are clearly ${49 \choose 2}$ cases.

2. $g=r$ . Since this is a $90$ degree rotation, there are no fixed points under this transformation.

3. $g=r^2$ . There are $24$ cases (which can be manually checked).

4. $g=r^3$ . There are no cases since this is just Case $2$ but with a $-90$ degree rotation.

Therefore, by Burnside’s Lemma there are $\frac{1176+24+0+0}{|G|}=\frac{1200}{4}=\boxed{300}$ inequivalent configurations (also known as orbits in group-theoretic terms).

@@ Line 51: / Line 51: @@
 ~First
+== Solution 3 (Burnside’s Lemma)==
+Consider the group <math>G=\mathbb{Z}/4\mathbb{Z}</math> with addition acting on set <math>X</math>, where <math>X</math> is the set of color schemes on the checkerboard. We now apply Burnside’s Lemma. Since <math>\mathbb{Z}/4\mathbb{Z}=\langle r \mid r^4=1\rangle</math>, we only need to consider 4 cases for <math>g</math> an element of <math>\mathbb{Z}/4\mathbb{Z}</math>.
+. <math>g=1</math>. There are clearly <math>{49 \choose 2}</math> cases.
+. <math>g=r</math>. Since this is a <math>90</math> degree rotation, there are no fixed points under this transformation.
+. <math>g=r^2</math>. There are <math>24</math> cases (which can be manually checked).
+. <math>g=r^3</math>. There are no cases since this is just Case <math>2</math> but with a <math>-90</math> degree rotation.
+Therefore, by Burnside’s Lemma there are <math>\frac{1176+24+0+0}{|G|}=\frac{1200}{4}=\boxed{300}</math> inequivalent configurations (also known as orbits in group-theoretic terms).
 == See also ==

1996 AIME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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