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Difference between revisions of "1997 AIME Problems/Problem 2"

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== Problem ==
 
== Problem ==
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The nine horizontal and nine vertical lines on an <math>8\times8</math> checkerboard form <math>r</math> rectangles, of which <math>s</math> are squares.  The number <math>s/r</math> can be written in the form <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m + n.</math>
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The nine horizontal and nine vertical lines on an <math>8\times8</math> checkerboard form <math>r</math> [[rectangle]]s, of which <math>s</math> are [[square]]s.  The number <math>s/r</math> can be written in the form <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m + n.</math>
  
 
== Solution ==
 
== Solution ==
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For r, we can choose two out of 9 lines, and 2 out of nine lines again, to get <math>r=(\binom{9}{2})^2=36^2=1296</math>
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To determine the two horizontal sides of a rectangle, we have to pick two of the horizontal lines of the checkerboard, or <math>{9\choose 2} = 36</math>. Similarly, there are <math>{9\choose 2}</math> ways to pick the vertical sides, giving us <math>r = 1296</math> rectangles.
  
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For s, there are 8^2 unit squares, 7^2 2*2 squares, .... 1^1 8*8 squares. That gives us <math>s=1^2+2^2+\cdots+8^2=\dfrac{8*9*17}{6}=12*17=204</math>
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For <math>s</math>, there are <math>8^2</math> [[unit square]]s, <math>7^2</math> of the <math>2\times2</math> squares, and so on until <math>1^2</math> of the <math>8\times 8</math> squares. Using the sum of squares formula, that gives us <math>s=1^2+2^2+\cdots+8^2=\dfrac{(8)(8+1)(2\cdot8+1)}{6}=12*17=204</math>.
  
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<math>\dfrac{204}{1296}=\dfrac{17}{108}</math>
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Thus <math>\frac sr = \dfrac{204}{1296}=\dfrac{17}{108}</math>, and <math>m+n=\boxed{125}</math>.
  
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<math>m+n=125</math>
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== 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 <math>8^2</math> ways to place a <math>1</math> x <math>1</math> square and <math>7^2</math> for a <math>2</math> x <math>2</math> square. This pattern can be easily generalized and we see that the number of squares is just <math>\sum^8_{i=1}{i^2}</math>. This can be simplified by using the well-known formula for the sum of consecutive squares <math>\frac{n(n+1)(2n+1)}{6}</math> to get <math>204</math>.
 +
 
 +
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 <math>1</math>x<math>1, 1</math>x<math>2 , 2</math>x<math>1, 2</math>x<math>2,...,</math> we see they are respectively <math>8</math>x<math>8, 8</math>x<math>7, 7</math>x<math>8, 7</math>x<math>7, ...</math>. We can quickly generalize this pattern to basically just <math>{\sum^8_{i=1}{i}}\cdot{\sum^8_{i=1}{i}}</math>. This gets us <math>{(\frac{9\cdot8}{2})}^2,</math> which is just <math>1296.</math>
 +
 
 +
Now, to calculate the ratio of <math>s/r,</math> we divide <math>204</math> by <math>1296</math> to get a simplified fraction of <math>\frac{17}{108}.</math>
 +
 
 +
Thus, our answer is just <math>s + r = 17+108 = \boxed{125}</math>
 +
~MathWhiz35
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1997|num-b=1|num-a=3}}
 
{{AIME box|year=1997|num-b=1|num-a=3}}
 +
 +
[[Category:Intermediate Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 20:09, 27 October 2022

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 ${9\choose 2} = 36$. Similarly, 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 (Problems • Answer Key • Resources)
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Problem 1 		Followed by
Problem 3
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All AIME Problems and Solutions

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