Question

设 $R$ 为一个正方形区域， $n \ge 4$ 为整数。若从 $R$ 内部一点 $X$ 发出 $n$ 条射线，可以把 $R$ 分成 $n$ 个面积相等的三角形，则称 $X$ 为 $n$ -射线分割点。有多少个点是 $100$ -射线分割点但不是 $60$ -射线分割点？

Let $R$ be a square region and $n \ge 4$ an integer. A point $X$ in the interior of $R$ is called $n$ -ray partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are $100$ -ray partitional but not $60$ -ray partitional?

$1500$

$1560$

$2320$

$2480$

$2500$

答案：C

知识点：面积格点补集计数

难度评级：2460

小提示：

$n$ -射线分割点形成网格：当 $n = 2m$ ，它们是内部点 $\left(\tfrac{i}{m}, \tfrac{j}{m}\right)$ ，构成 $(m-1) \times (m-1)$ 个点

The $n$ -ray partitional points form a grid: for $n = 2m,$ they are the interior points $\left(\tfrac{i}{m}, \tfrac{j}{m}\right),$ an $(m-1) \times (m-1)$ array

大提示：

一个点同时是 $100$ -射线和 $60$ -射线分割点，当且仅当它是 $20$ -射线分割点

A point is both $100$ - and $60$ -ray partitional exactly when it is $20$ -ray partitional

解答：

将正方形缩放为 $[0,1]^2$ ，并令 $X=(x,y)$ 。射线必须包括通向四个顶点的射线。每个小三角形的面积都是 $\frac{1}{n}$ 。以底边各段为底的三角形面积之和为 $\frac{y}{2}$ ，所以这样的三角形有 $\frac{ny}{2}$ 个。同理，沿上、左、右三边的个数分别是 $\frac{n(1-y)}{2}$ ， $\frac{nx}{2}$ ，和 $\frac{n(1-x)}{2}$ 。

这四个数都必须是正整数。因此 $n$ 是偶数，并且 $\begin{gathered} X=\left(\dfrac{2i}{n},\dfrac{2j}{n}\right), \\ 1\le i,j\le\dfrac n2-1 \end{gathered}\text{。}$ 反过来，按上述数量把每条边等分，并将分点连接到 $X$ ，就会得到 $n$ 个等面积三角形。因此这些恰好是分割点。

当 $n=100$ 时，这些点为 $(\frac{i}{50},\frac{j}{50})$ ，其中 $1\le i,j\le49$ ，共 $49^2=2401$ 个。这样的点同时也是 $60$ 射线分割点，当且仅当对某些整数 $c,d$ ，有 $\frac{i}{50}=\frac{c}{30}$ 且 $\frac{j}{50}=\frac{d}{30}$ 。因此 $i$ 和 $j$ 都必须是 $5$ 的倍数。每个坐标有 $9$ 种选择，所以重合的点有 $9^2=81$ 个。

所以所求数量为 $2401 - 81 = 2320$ 。

因此，正确答案是 C。

Scale the square to $[0,1]^2$ and write $X=(x,y).$ The rays must include those through the four vertices. Every small triangle has area $\frac{1}{n}.$ The triangles whose bases partition the bottom side together have area $\frac{y}{2},$ so their number is $\frac{ny}{2}.$ Similarly, the numbers along the top, left, and right sides are $\frac{n(1-y)}{2},$ $\frac{nx}{2},$ and $\frac{n(1-x)}{2}.$

These four numbers must be positive integers. Hence $n$ is even and $\begin{gathered} X=\left(\dfrac{2i}{n},\dfrac{2j}{n}\right), \\ 1\le i,j\le\dfrac n2-1. \end{gathered}$ Conversely, partitioning each side into the indicated number of equal segments and joining the division points to $X$ produces $n$ equal-area triangles. Thus these are exactly the partitional points.

For $n=100,$ the points are $(\frac{i}{50},\frac{j}{50})$ with $1\le i,j\le49,$ giving $49^2=2401.$ Such a point is also $60$ -ray partitional exactly when $\frac{i}{50}=\frac{c}{30}$ and $\frac{j}{50}=\frac{d}{30}$ for integers $c,d.$ Thus $i$ and $j$ must both be multiples of $5.$ There are $9$ choices for each, so the overlap has $9^2=81$ points.

So the count is $2401 - 81 = 2320.$

Thus, the correct answer is C.

2011 AMC 12A 第 22 题

答案：C

难度评级：2460

小提示：

大提示：

解答：

其他年份的第 22 题