2010 AMC 12A 第 18 题

Question

一条

16

步路径要从

(-4,-4)

走到

(4,4)

，每一步都使

x

坐标或

y

坐标增加

1

。有多少条这样的路径在每一步都位于正方形

-2\le x\le2

、

-2\le y\le2

的外部或边界上？

A

16

-step path is to go from

(-4,-4)

to

(4,4)

with each step increasing either the

x

-coordinate or the

y

-coordinate by

1.

How many such paths stay outside or on the boundary of the square

-2\le x\le2,

-2\le y\le2

at each step?

92

144

1568

1698

12{,}800

答案：D

知识点：格路组合对称性

难度评级：1880

小提示：

每一步都使

x+y

增加

1

，所以路径与直线

x+y=0

恰好相交于一个格点。

Each step increases

x+y

by

1,

so the path meets the line

x+y=0

at exactly one lattice point

大提示：

为避开正方形内部，该点只能是

(\pm2,\mp2),(\pm3,\mp3)

，或

(\pm4,\mp4)

；分别计数。

To avoid the square’s interior, that point is

(\pm2,\mp2),(\pm3,\mp3),

or

(\pm4,\mp4);

count paths through each

解答：

每一步都使

x+y

增加

1

，而坐标和从

-8

变到

8

，所以每条路径恰好经过一个满足

x+y=0

的格点。

为了不进入开正方形内部，该点

(t,-t)

必须满足

|t|\ge2

，所以它是

(\pm2,\mp2),(\pm3,\mp3),(\pm4,\mp4)

之一。

由对称性，只考虑三点

(-4,4),(-3,3),(-2,2)

，然后将结果加倍。从

(-4,-4)

到

(-(4-j),4-j)

的路径数为

\binom{8}{j}

，继续到

(4,4)

的路径数也是

\binom{8}{j}

。

因此总数为

\begin{gathered} 2\left(\binom80^2+\binom81^2+\binom82^2\right) \\ =2(1+64+784) \\ =1698 \end{gathered}\text{。}

所以正确答案是 D。

Every step increases

x+y

by

1,

which runs from

-8

to

8,

so each path passes through exactly one lattice point with

x+y=0.

To stay out of the open square, that point

(t,-t)

must have

|t|\ge2,

so it is one of

(\pm2,\mp2),(\pm3,\mp3),(\pm4,\mp4).

By symmetry consider the three points

(-4,4),(-3,3),(-2,2)

and double. The number of paths from

(-4,-4)

to

(-(4-j),4-j)

is

\binom{8}{j},

and the number continuing on to

(4,4)

is also

\binom{8}{j}.

Therefore the total is

\begin{gathered} 2\left(\binom80^2+\binom81^2+\binom82^2\right) \\ =2(1+64+784) \\ =1698. \end{gathered}

Thus, D is the correct answer.

2010 AMC 12A 第 18 题

答案：D

难度评级：1880

小提示：

大提示：

解答：

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