2018 AMC 10A 第 20 题

Question

一个扫描码由一个

7 \times 7

方格构成，其中一些小方格涂黑，其余小方格涂白。在这个由

49

个小方格组成的网格中，两种颜色都必须至少出现一次。

如果把整个正方形绕中心逆时针旋转

90^{\circ}

的任意整数倍，或关于连接两个对角顶点的直线、连接两条对边中点的直线反射后，扫描码的外观都不改变，就称它是对称的。

一共有多少种可能的对称扫描码？

A scanning code consists of a

7 \times 7

grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of

49

squares.

A scanning code is called symmetric if its look does not change when the entire square is rotated by a multiple of

90^{\circ}

counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides.

What is the total number of possible symmetric scanning codes?

510

1022

8190

8192

65{,}534

答案：B

知识点：对称性乘法原理

难度评级：1970

小提示：

按照正方形的对称性把小方格分成若干轨道。

Classify grid squares by symmetry orbits

大提示：

每个轨道选择一种颜色，再排除两种纯色涂法。

Choose one color per orbit, excluding the two constant colorings

解答：

把行和列从

-3

编号到

3

，中心为

(0,0)

。旋转和反射可以改变坐标的符号，也可以交换两个坐标，所以

(x,y)

所在的轨道由

(|x|,|y|)

从小到大排序后得到的数对决定。这样的数对是满足

0\le u\le v\le3

的

(u,v)

，共有

4+3+2+1=10

个。

每个轨道选定一个小方格的颜色后，该轨道内其余方格的颜色都由对称性确定。

因此，在考虑两种颜色都必须出现的条件之前，共有

2^{10}

种对称涂法。

全黑和全白两种涂法不合要求，所以有效的对称扫描码共有

2^{10}-2=1022

种。因此正确答案是 B。

Number rows and columns from

-3

through

3,

with the center at

(0,0).

Rotations and reflections can change signs and exchange coordinates, so the orbit of

(x,y)

is determined by the ordered pair obtained by sorting

(|x|,|y|).

These pairs are

(u,v)

with

0\le u\le v\le3,

of which there are

4+3+2+1=10.

Once one square in each orbit is colored, symmetry forces the colors of all other squares in that orbit.

There are therefore

2^{10}

symmetric colorings before the condition about using both colors. The all-black and all-white colorings are not allowed.

The total number of valid symmetric scanning codes is

2^{10}-2=1022

. Thus, B is the correct answer.

2018 AMC 10A 第 20 题

答案：B

难度评级：1970

小提示：

大提示：

解答：

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