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

(solution)
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== Solution ==
 
== Solution ==
−
There are <math>{49 \choose 2}</math> possible ways to select two squares to be painted green. There are four possible ways to rotate each board. Given an arbitrary pair of green squares, these four rotations will either yield two or four equivalent but distinct boards.  
+
There are <math>{49 \choose 2}</math> 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.  
  
 
<center><table><tr><td>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<asy>
 
<center><table><tr><td>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<asy>
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  for(int j=0;j<7;++j)
 
  for(int j=0;j<7;++j)
 
   D(shift(i,j)*unitsquare);
 
   D(shift(i,j)*unitsquare);
−
fill(shift(4,5)*unitsquare,rgb(.4,.8,.4));
+
fill(shift(4,5)*unitsquare,rgb(1,1,.4));
−
fill(shift(2,1)*unitsquare,rgb(.4,.8,.4));
+
fill(shift(2,1)*unitsquare,rgb(1,1,.4));
−
fill(shift(1,4)*unitsquare,rgb(.7,1,.7));
+
fill(shift(1,4)*unitsquare,rgb(.4,.4,.7));
−
fill(shift(5,2)*unitsquare,rgb(.7,1,.7));
+
fill(shift(5,2)*unitsquare,rgb(.4,.4,.7));
  
 
D(arc(O,5^.5,-20,50),EndArrow(4));
 
D(arc(O,5^.5,-20,50),EndArrow(4));
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</asy></td></tr><tr><td><font style="font-size:85%">For most pairs, there will be <br /> three other equivalent boards.</font></td><td><font style="font-size:85%">For those symmetric about the center, <br /> there is only one other.</font></td></tr></table></center>
 
</asy></td></tr><tr><td><font style="font-size:85%">For most pairs, there will be <br /> three other equivalent boards.</font></td><td><font style="font-size:85%">For those symmetric about the center, <br /> there is only one other.</font></td></tr></table></center>
  
−
Note that a pair of green squares will only yield <math>2</math> distinct boards upon rotation [[iff]] the green squares are rotationally symmetric about the center square; there are <math>\frac{49-1}{2}=24</math> such pairs. There are then <math>{49 \choose 2}-24</math> pairs that yield <math>4</math> distinct boards upon rotation; in other words, for each of the <math>{49 \choose 2}-24</math> pairs, there are three other pairs that yield an equivalent board.
+
Note that a pair of yellow squares will only yield <math>2</math> distinct boards upon rotation [[iff]] the yellow squares are rotationally symmetric about the center square; there are <math>\frac{49-1}{2}=24</math> such pairs. There are then <math>{49 \choose 2}-24</math> pairs that yield <math>4</math> distinct boards upon rotation; in other words, for each of the <math>{49 \choose 2}-24</math> pairs, there are three other pairs that yield an equivalent board.
  
 
Thus, the number of inequivalent boards is <math>\frac{{49 \choose 2} - 24}{4} + \frac{24}{2} = \boxed{300}</math>. For a <math>(2n+1) \times (2n+1)</math> board, this argument generalizes to <math>n(n+1)(2n^2+2n+1)</math> inequivalent configurations.  
 
Thus, the number of inequivalent boards is <math>\frac{{49 \choose 2} - 24}{4} + \frac{24}{2} = \boxed{300}</math>. For a <math>(2n+1) \times (2n+1)</math> board, this argument generalizes to <math>n(n+1)(2n^2+2n+1)</math> inequivalent configurations.  

Revision as of 23:12, 12 March 2011

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

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(.4,.8,.4));fill(shift(4,5)*unitsquare,rgb(.4,.8,.4)); fill(shift(3,4)*unitsquare,rgb(.7,1,.7));fill(shift(1,4)*unitsquare,rgb(.7,1,.7)); fill(shift(2,3)*unitsquare,rgb(.7,1,.7));fill(shift(2,1)*unitsquare,rgb(.7,1,.7)); fill(shift(3,2)*unitsquare,rgb(.7,1,.7));fill(shift(5,2)*unitsquare,rgb(.7,1,.7)); 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(.4,.4,.7)); fill(shift(5,2)*unitsquare,rgb(.4,.4,.7)); 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.

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
All AIME Problems and Solutions
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