1997 AIME Problems/Problem 2

Problem

The nine horizontal and nine vertical lines on an $8\times8$ checkerboard form $r$ rectangles, of which $s$ are squares. The number $s/r$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution

To determine the two horizontal sides of a rectangle, we have to pick two of the horizontal lines of the checkerboard, or ${9\choose 2} = 36$. Similarily, there are ${9\choose 2}$ ways to pick the vertical sides, giving us $r = 1296$ rectangles.

For $s$, there are $8^2$ unit squares, $7^2$ of the $2\times2$ squares, and so on until $1^2$ of the $8\times 8$ squares. Using the sum of squares formula, that gives us $s=1^2+2^2+\cdots+8^2=\dfrac{(8)(8+1)(2\cdot8+1)}{6}=12*17=204$.

Thus $\frac sr = \dfrac{204}{1296}=\dfrac{17}{108}$, and $m+n=\boxed{125}$.

Solution

First, to find the number of squares, we can look case by case by the side length of the possible squares on the checkerboard. We see that there are $8^2$ ways to place a $1$ x $1$ square and $7^2$ for a $2$ x $2$ square. This pattern can be easily generalized and we see that the number of squares is just $\sum^8_{i=1}{i^2}$. This can be simplified by using the well-known formula for the sum of consecutive squares $\frac{n(n+1)(2n+1)}{6}$ to get $204$.

Then, to find the number of rectangles, first note that a square falls under the definition of a rectangle. We can break up the rectangles into cases for the length x width. As we note down the cases for $1$x$1, 1$x$2 , 2$x$1, 2$x$2,...,$ we see they are respectively $8$x$8, 8$x$7, 7$x$8, 7$x$7, ...$. We can quickly generalize this pattern to basically just ${\sum^8_{i=1}{i}}\cdot{\sum^8_{i=1}{i}}$. This gets us ${(\frac{9\cdot8}{2})}^2,$ which is just $1296.$

Now, to calculate the ratio of $s/r,$ we divide $204$ by $1296$ to get a simplified fraction of $\frac{17}{108}.$

Thus, our answer is just $s + r = 17+108 = \boxed{125}$ ~MathWhiz35

See also

1997 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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