2021 AMC 10 B Problem 25

Floor Sum Method

Approximate the slope using the continuous area of a triangle, verify the exact lattice point count using the floor function, and find the next slope that captures an additional point.

Step 1: Estimate the target slope

We want exactly 300 points out of the 900 points in the grid. The line $y=mx$ bounds a triangular region. We estimate the required slope by setting the area of the triangle over the interval from the origin to the right edge of the grid equal to 300.

Step 2: Count the points for the estimated slope

To verify this slope exactly, we count the number of lattice points in our set that lie on or below the line. For each $x$, the number of valid $y$-coordinates is the floor of the line’s height. We group the terms into blocks of 3 to evaluate the sum.

Step 3: Sum the arithmetic progression

Each block of 3 consecutive $x$-values increases the sum by 6. This forms an arithmetic progression with 10 terms, starting at 3. The 10th term is 57. The total matches our target exactly.

Step 4: Determine the lower bound behavior

The line currently passes exactly through 10 grid points. If the slope decreases even slightly, all 10 of these points will simultaneously be excluded, dropping the count to 290. Thus, the slope cannot be strictly less than the estimated value.

Step 5: Find the upper bound of the interval

As the slope increases, the count remains 300 until the line captures a new grid point. The first new point captured will have the smallest ratio strictly greater than our current slope. We minimize the difference by making the cross-multiplied numerator as small as possible (which is 1) and the denominator as large as possible.

Step 6: Solve for the point and interval length

We find the largest $x$ within the grid bounds that leaves a remainder of 1 when multiplied by 2, so that the corresponding $y$ is an integer. This gives the coordinates of the next point. We then calculate the length of the interval.

Geometric Counting via Pick’s Theorem

Use Pick’s Theorem on the triangle formed by the origin and the grid boundaries to rigorously count the lattice points without explicitly summing a sequence.

Step 1: Set up the geometric region

Assuming the slope is exactly $2/3$, we analyze the triangle bounded by the x-axis, the vertical line at the right edge of the grid, and the slanted line. The continuous area of this triangle is 300.

Step 2: Count the boundary points

We count the number of lattice points strictly on the three boundary segments of the triangle. The number of intervals on a segment between $(x_1,y_1)$ and $(x_2,y_2)$ is given by the greatest common divisor of the differences in coordinates. We subtract 3 to avoid double-counting the vertices.

Step 3: Calculate the total points in the region

We apply the rearranged form of Pick’s Theorem to find the total number of lattice points (both interior and boundary) inside the triangle.

Step 4: Isolate the points that belong to the set

The problem restricts the points to those with positive $y$-coordinates. The only points in our triangle with non-positive $y$-coordinates lie on the x-axis segment. We subtract these to find the exactly valid points.

Step 5: Determine the interval and conclude

The total points are exactly 300 for this slope. As the slope increases, it captures the next point, which we find by minimizing the algebraic slope difference just as in Method 1.

Algebraic Multiple-Choice Shortcut

The algebraic structure of the interval length directly links the denominator to the answer choices, allowing us to bypass the exact point-counting arithmetic.

Step 1: Identify the algebraic structure of the interval length

Once we estimate the starting slope, the interval ends at the next fraction. To minimize the difference between the new slope and the starting slope, the cross-multiplied numerator must be 1. Thus, the length of the interval depends directly on the x-coordinate of the new point.

Step 2: Match the structure to the answer choices

Because the fraction is in simplest form, the components of the interval length relate to the denominator. We check which answer choices match this algebraic form.

Step 3: Evaluate the valid choices

Only two answer choices leave a remainder of 1 when divided by 3. We determine their corresponding x-values. To minimize the interval length, we need the largest valid $x$ within the grid boundaries.