Difference between revisions of "1997 AIME Problems/Problem 2"
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== Problem == | == Problem == | ||
− | The nine horizontal and nine vertical lines on an <math>8\times8</math> | + | 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 == | ||
+ | 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. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | Thus <math>\frac sr = \dfrac{204}{1296}=\dfrac{17}{108}</math>, and <math>m+n=\boxed{125}</math>. | ||
+ | |||
+ | == 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}} | |
+ | |||
+ | [[Category:Intermediate Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:09, 27 October 2022
Contents
Problem
The nine horizontal and nine vertical lines on an checkerboard form rectangles, of which are squares. The number can be written in the form where and are relatively prime positive integers. Find
Solution
To determine the two horizontal sides of a rectangle, we have to pick two of the horizontal lines of the checkerboard, or . Similarly, there are ways to pick the vertical sides, giving us rectangles.
For , there are unit squares, of the squares, and so on until of the squares. Using the sum of squares formula, that gives us .
Thus , and .
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 ways to place a x square and for a x square. This pattern can be easily generalized and we see that the number of squares is just . This can be simplified by using the well-known formula for the sum of consecutive squares to get .
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 xxxx we see they are respectively xxxx. We can quickly generalize this pattern to basically just . This gets us which is just
Now, to calculate the ratio of we divide by to get a simplified fraction of
Thus, our answer is just ~MathWhiz35
See also
1997 AIME (Problems • Answer Key • Resources) | ||
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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