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> checkerboard form <math>r</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>. | + | 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>. | 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>. | 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 == |
Latest revision as of 21:09, 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 x
x
x
x
we see they are respectively
x
x
x
x
. 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.