2007 AIME II Problems/Problem 5

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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 .

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