Difference between revisions of "2007 AIME II Problems/Problem 5"
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=== 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>. | ||
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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. | 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. | ||
Revision as of 20:57, 9 March 2017
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?
Contents
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 
.
Do you see why we simply set 
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, 
moves to 
, so the square of points 
is one example. This applies, of course, for points.
Solution 4
We know that the number of squares intersected in an 
rectangle is 
. So if we apply that here, we get that the number of intersected squares is:
.
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,
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. 
