Difference between revisions of "2007 AIME II Problems/Problem 5"
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
| − | Count the number of each squares in each row of the triangle. The [[intercept]]s of the [[line]] are <math>(223,0),\ (9 | + | 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>. |
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| + | 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 = 233</math> squares. | ||
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| + | 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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| + | === Solution 2 === | ||
| + | Count the number of each squares in each row of the triangle. The [[intercept]]s of the [[line]] are <math>(223,0),\ (0,9)</math>. | ||
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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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,</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>. | ||
| − | === Solution | + | === Solution 3 === |
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>\displaystyle 888</math>. | ||
Revision as of 08:40, 5 April 2007
Problem
The graph of the equation 
is drawn on graph paper with each square representing one unit in each direction. How many of the 
by graph paper squares have interiors lying entirely below the graph and entirely in the first quadrant?
Solution
Solution 1
There are 
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 
.
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 
. This passes through 8 horizontal lines (
) and 222 vertical lines (
). 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 
squares.
The number of non-diagonal squares is 
. Divide this in 2 to get the number of squares in one of the triangles, with the answer being 
.
Solution 2
Count the number of each squares in each row of the triangle. The intercepts of the line are .
In the top row, there clearly are no squares that can be formed. In the second row, we see that the line 
gives a 
value of 
, which means that 
unit squares can fit in that row. In general, there are
triangles. Since 
, we see that there are more than 
triangles. Now, count the fractional parts. 
. Adding them up, we get 
.
Solution 3
From Pick's Theorem, 
. In other words, 
and I is 
.
See also
| 2007 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
