AMC10 2011 B

AMC10 2011 B · Q24

AMC10 2011 B · Q24. It mainly tests Coordinate geometry, Counting in geometry (lattice points).

A lattice point in an xy-coordinate system is any point (x, y) where both x and y are integers. The graph of y = mx + 2 passes through no lattice point with 0 < x ≤100 for all m such that $\frac{1}{2}$ < m < a. What is the maximum possible value of a ?

在 $xy$ 坐标系中，晶格点是任意点 $(x, y)$ 其中 $x$ 和 $y$ 均为整数。对于所有满足 $\frac{1}{2} < m < a$ 的 $m$，直线 $y = mx + 2$ 在 $0 < x \leq 100$ 时不经过任何晶格点。$a$ 的最大可能值是多少？

(A) $\frac{51}{101}$ $\frac{51}{101}$

(B) $\frac{50}{99}$ $\frac{50}{99}$

(D) $\frac{52}{101}$ $\frac{52}{101}$

(E) $\frac{13}{25}$ $\frac{13}{25}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): For $0<x\le 100$, the nearest lattice point directly above the line $y=\frac12x+2$ is $(x,\frac12x+3)$ if $x$ is even and $(x,\frac12x+\frac52)$ if $x$ is odd. The slope of the line that contains this point and $(0,2)$ is $\frac12+\frac1x$ if $x$ is even and $\frac12+\frac1{2x}$ if $x$ is odd. The minimum value of the slope is $\frac{51}{100}$ if $x$ is even and $\frac{50}{99}$ if $x$ is odd. Therefore the line $y=mx+2$ contains no lattice point with $0<x\le 100$ for $\frac12<m<\frac{50}{99}$.

答案（B）：对于 $0<x\le 100$，在直线 $y=\frac12x+2$ 正上方最近的格点，当 $x$ 为偶数时是 $(x,\frac12x+3)$，当 $x$ 为奇数时是 $(x,\frac12x+\frac52)$。经过该点与 $(0,2)$ 的直线斜率：若 $x$ 为偶数则为 $\frac12+\frac1x$，若 $x$ 为奇数则为 $\frac12+\frac1{2x}$。斜率的最小值：$x$ 为偶数时为 $\frac{51}{100}$，$x$ 为奇数时为 $\frac{50}{99}$。因此，当 $\frac12<m<\frac{50}{99}$ 时，直线 $y=mx+2$ 在 $0<x\le 100$ 范围内不包含任何格点。

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