AMC12 2011 B

AMC12 2011 B · Q19

AMC12 2011 B · Q19. It mainly tests Counting in geometry (lattice points), Inequalities with integers (floor/ceiling basics).

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 $0 < x \leq 100$ for all $m$ such that $\frac{1}{2} < m < a$. What is the maximum possible value of $a$?

在 $xy$ 坐标系中，晶格点是指 $(x, y)$ 中 $x$ 和 $y$ 都为整数的点。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) $51/101$ $51/101$

(B) $50/99$ $50/99$

(D) $52/101$ $52/101$

(E) $13/25$ $13/25$

Answer

Correct choice: (B)

正确答案：（B）

Solution

It is very easy to see that the $+2$ in the graph does not impact whether it passes through the lattice. We need to make sure that $m$ cannot be in the form of $\frac{a}{b}$ for $1\le b\le 100$. Otherwise, the graph $y= mx$ passes through the lattice point at $x = b$. We only need to worry about $\frac{a}{b}$ very close to $\frac{1}{2}$, $\frac{n+1}{2n+1}$, $\frac{n+1}{2n}$ will be the only case we need to worry about and we want the minimum of those, clearly for $1\le b\le 100$, the smallest is $\frac{50}{99}$, so answer is $\boxed{\frac{50}{99} \textbf{(B)}}$ (In other words we are trying to find the smallest $m>\frac{1}{2}$ such that $b\le 100$.)

很容易看出，直线中的 $+2$ 并不影响它是否经过晶格点。我们需要确保 $m$ 不能写成 $\frac{a}{b}$ 的形式，其中 $1\le b\le 100$。否则，直线 $y= mx$ 会在 $x = b$ 处经过晶格点。我们只需要担心那些非常接近 $\frac{1}{2}$ 的 $\frac{a}{b}$，其中 $\frac{n+1}{2n+1}$、$\frac{n+1}{2n}$ 是唯一需要考虑的情况，并且我们要取其中的最小值。显然在 $1\le b\le 100$ 时，最小的是 $\frac{50}{99}$，所以答案是 $\boxed{\frac{50}{99} \textbf{(B)}}$（换句话说，我们是在寻找满足 $b\le 100$ 的最小的 $m>\frac{1}{2}$）。

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