Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution 1)
 
(7 intermediate revisions by 4 users not shown)
Line 7: Line 7:
 
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>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.
+
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>). 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>.
Line 18: Line 18:
 
<div style="text-align:center;"><math>\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,<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>.
+
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</math>.
  
 
=== Solution 3 ===
 
=== 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>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>.
 +
 +
Do you see why we simply set <math>I</math> as the answer as well? That's because every interior point, if moved down and left one (southwest direction), can have that point and the previous point create a unit square. For example, <math>(1, 1)</math> moves to <math>(0, 0)</math>, so the square of points <math>{(0, 0), (1, 0), (1, 1), (0, 1)}</math> is one example. This applies, of course, for <math>888</math> points.
  
 
=== Solution 4 ===
 
=== Solution 4 ===
Line 38: Line 40:
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 23:38, 2 January 2023

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 ($x = 1 \ldots 222$). 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. $\lfloor \frac{0}{9} \rfloor = 0, \lfloor \frac{7}{9} \rfloor = 0, \lfloor \frac{14}{9} \rfloor = 1,$ $\lfloor \frac{21}{9} \rfloor = 2, \lfloor \frac{28}{9} \rfloor = 3, \lfloor \frac{35}{9} \rfloor = 3,$ $\lfloor \frac{42}{9} \rfloor = 4, \lfloor \frac{49}{9} \rfloor = 5, \lfloor \frac{56}{9} \rfloor = 6$. 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 $888$.

Do you see why we simply set $I$ as the answer as well? That's because every interior point, if moved down and left one (southwest direction), can have that point and the previous point create a unit square. For example, $(1, 1)$ moves to $(0, 0)$, so the square of points ${(0, 0), (1, 0), (1, 1), (0, 1)}$ is one example. This applies, of course, for $888$ points.

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)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_II_Problems/Problem_5&oldid=185596"