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Difference between revisions of "2007 AIME II Problems/Problem 5"

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There are <math>223 \cdot 9 = 2007</math> squares in total formed by the rectangle with edges on the x and y axes and with vertices on the intercepts of the equation, since the [[intercept]]s of the lines are <math>(223,0),\ (0,9)</math>.
 
There are <math>223 \cdot 9 = 2007</math> squares in total formed by the rectangle with edges on the x and y axes and with vertices on the intercepts of the equation, since the [[intercept]]s of the lines are <math>(223,0),\ (0,9)</math>.
  
−
Count the number of squares that the diagonal of the rectangle passes through. Since the two diagonals of a rectangle are [[congruent]], we can consider instead the diagonal <math>y = \frac{223}{9}x</math>. This passes through 8 horizontal lines (<math>\displaystyle y = 1 \ldots 8</math>) and 222 vertical lines (<math>\displaystyle x = 1 \ldots 222</math>). At every time we cross a line, we enter a new square. Since 9 and 223 are [[relatively prime]], we don’t have to worry about crossing an intersection of a horizontal and vertical line at one time. We must also account for the first square. This means that it passes through <math>222 + 8 + 1 = 231</math> squares.
+
Count the number of squares that the diagonal of the rectangle passes through. Since the two diagonals of a rectangle are [[congruent]], we can consider instead the diagonal <math>y = \frac{223}{9}x</math>. This passes through 8 horizontal lines (<math>y = 1 \ldots 8</math>) and 222 vertical lines (<math>x = 1 \ldots 222</math>). At every time we cross a line, we enter a new square. Since 9 and 223 are [[relatively prime]], we don’t have to worry about crossing an intersection of a horizontal and vertical line at one time. We must also account for the first square. This means that it passes through <math>222 + 8 + 1 = 231</math> squares.
  
 
The number of non-diagonal squares is <math>2007 - 231 = 1776</math>. Divide this in 2 to get the number of squares in one of the triangles, with the answer being <math>\frac{1776}2 = 888</math>.
 
The number of non-diagonal squares is <math>2007 - 231 = 1776</math>. Divide this in 2 to get the number of squares in one of the triangles, with the answer being <math>\frac{1776}2 = 888</math>.
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In the top row, there clearly are no squares that can be formed. In the second row, we see that the line <math>y = 8</math> gives a <math>x</math> value of <math>\frac{2007 - 8(223)}{9} = 24 \frac 79</math>, which means that <math>\lfloor 24 \frac 79\rfloor = 24</math> unit squares can fit in that row. In general, there are
 
In the top row, there clearly are no squares that can be formed. In the second row, we see that the line <math>y = 8</math> gives a <math>x</math> value of <math>\frac{2007 - 8(223)}{9} = 24 \frac 79</math>, which means that <math>\lfloor 24 \frac 79\rfloor = 24</math> unit squares can fit in that row. In general, there are
  
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<div style="text-align:center;"><math>\displaystyle\sum_{i=0}^{8} \lfloor \frac{223i}{9} \rfloor</math></div>
+
<div style="text-align:center;"><math>\sum_{i=0}^{8} \lfloor \frac{223i}{9} \rfloor</math></div>
  
−
triangles. Since <math>\lfloor \frac{223}{9} \rfloor = 24</math>, we see that there are more than <math>24(0 + 1 + \ldots + 8) = 24(\frac{8 \times 9}{2}) = 864</math> triangles. Now, count the fractional parts. <math>\lfloor \frac{0}{9} \rfloor = 0, \lfloor \frac{7}{9} \rfloor = 0, \lfloor \frac{14}{9} \rfloor = 1,</math><math> \lfloor \frac{21}{9} \rfloor = 2, \lfloor \frac{28}{9} \rfloor = 3, \lfloor \frac{35}{9} \rfloor = 3,</math><math> \lfloor \frac{42}{9} \rfloor = 4, \lfloor \frac{49}{9} \rfloor = 5, \lfloor \frac{56}{9} \rfloor = 6</math>. Adding them up, we get <math>864 + 1 + 2 + 3 + 3 + 4 + 5 + 6 = 888 \displaystyle</math>.
+
triangles. Since <math>\lfloor \frac{223}{9} \rfloor = 24</math>, we see that there are more than <math>24(0 + 1 + \ldots + 8) = 24(\frac{8 \times 9}{2}) = 864</math> triangles. Now, count the fractional parts. <math>\lfloor \frac{0}{9} \rfloor = 0, \lfloor \frac{7}{9} \rfloor = 0, \lfloor \frac{14}{9} \rfloor = 1,<cmath> \lfloor \frac{21}{9} \rfloor = 2, \lfloor \frac{28}{9} \rfloor = 3, \lfloor \frac{35}{9} \rfloor = 3,</cmath> \lfloor \frac{42}{9} \rfloor = 4, \lfloor \frac{49}{9} \rfloor = 5, \lfloor \frac{56}{9} \rfloor = 6</math>. Adding them up, we get <math>864 + 1 + 2 + 3 + 3 + 4 + 5 + 6 = 888</math>.
  
 
=== Solution 3 ===
 
=== Solution 3 ===
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From [[Pick's Theorem]], <math>\frac{2007}{2}=\frac{233}{2}-\frac{2}{2}+\frac{2I}{2}</math>. In other words, <math>2I=1776</math> and I is <math>\displaystyle 888</math>.
+
From [[Pick's Theorem]], <math>\frac{2007}{2}=\frac{233}{2}-\frac{2}{2}+\frac{2I}{2}</math>. In other words, <math>2I=1776</math> and I is <math>888</math>.
 +
 
 +
=== Solution 4 ===
 +
We know that the number of squares intersected in an <math>m\times{n}</math> rectangle is <math>m + n -\mbox{gcd}(m,n)</math>. So if we apply that here, we get that the number of intersected squares is:
 +
 
 +
<math>9 + 223 - 1 = 231</math>.
 +
 
 +
Now just subtract that from the total number of squares and divide by 2, since we want the number of squares below the line.
 +
 
 +
So,
 +
 
 +
<math>\frac{2007 - 231}{2} = \frac{1776}{2} = \fbox{888}</math>
  
 
== See also ==
 
== See also ==

Revision as of 22:48, 14 May 2008

Problem

The graph of the equation $9x+223y=2007$ is drawn on graph paper with each square representing one unit in each direction. How many of the $1$ by $1$ graph paper squares have interiors lying entirely below the graph and entirely in the first quadrant?

Solution

Solution 1

There are $223 \cdot 9 = 2007$ squares in total formed by the rectangle with edges on the x and y axes and with vertices on the intercepts of the equation, since the intercepts of the lines are $(223,0),\ (0,9)$.

Count the number of squares that the diagonal of the rectangle passes through. Since the two diagonals of a rectangle are congruent, we can consider instead the diagonal $y = \frac{223}{9}x$. This passes through 8 horizontal lines ($y = 1 \ldots 8$) and 222 vertical lines ($x = 1 \ldots 222$). At every time we cross a line, we enter a new square. Since 9 and 223 are relatively prime, we don’t have to worry about crossing an intersection of a horizontal and vertical line at one time. We must also account for the first square. This means that it passes through $222 + 8 + 1 = 231$ squares.

The number of non-diagonal squares is $2007 - 231 = 1776$. Divide this in 2 to get the number of squares in one of the triangles, with the answer being $\frac{1776}2 = 888$.

Solution 2

Count the number of each squares in each row of the triangle. The intercepts of the line are $(223,0),\ (0,9)$.

In the top row, there clearly are no squares that can be formed. In the second row, we see that the line $y = 8$ gives a $x$ value of $\frac{2007 - 8(223)}{9} = 24 \frac 79$, which means that $\lfloor 24 \frac 79\rfloor = 24$ unit squares can fit in that row. In general, there are

$\sum_{i=0}^{8} \lfloor \frac{223i}{9} \rfloor$

triangles. Since $\lfloor \frac{223}{9} \rfloor = 24$, we see that there are more than $24(0 + 1 + \ldots + 8) = 24(\frac{8 \times 9}{2}) = 864$ triangles. Now, count the fractional parts. $\lfloor \frac{0}{9} \rfloor = 0, \lfloor \frac{7}{9} \rfloor = 0, \lfloor \frac{14}{9} \rfloor = 1,<cmath> \lfloor \frac{21}{9} \rfloor = 2, \lfloor \frac{28}{9} \rfloor = 3, \lfloor \frac{35}{9} \rfloor = 3,</cmath> \lfloor \frac{42}{9} \rfloor = 4, \lfloor \frac{49}{9} \rfloor = 5, \lfloor \frac{56}{9} \rfloor = 6$. Adding them up, we get $864 + 1 + 2 + 3 + 3 + 4 + 5 + 6 = 888$.

Solution 3

From Pick's Theorem, $\frac{2007}{2}=\frac{233}{2}-\frac{2}{2}+\frac{2I}{2}$. In other words, $2I=1776$ and I is $888$.

Solution 4

We know that the number of squares intersected in an $m\times{n}$ rectangle is $m + n -\mbox{gcd}(m,n)$. So if we apply that here, we get that the number of intersected squares is:

$9 + 223 - 1 = 231$.

Now just subtract that from the total number of squares and divide by 2, since we want the number of squares below the line.

So,

$\frac{2007 - 231}{2} = \frac{1776}{2} = \fbox{888}$

See also

2007 AIME II (Problems • Answer Key • Resources)
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Problem 4 		Followed by
Problem 6
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All AIME Problems and Solutions
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