2010 AMC 12A 第 25 题

Question

如果一个四边形可以通过旋转和平移得到另一个四边形，则认为这两个四边形相同。边长为整数且周长等于

32

的不同凸圆内接四边形有多少个？

Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to

32?

560

564

568

1498

2255

答案：C

知识点：伯恩赛德引理隔板法圆内接四边形

难度评级：2520

小提示：

给定一组按圆周顺序排列的边长后，凸圆内接四边形唯一；每条边至多为

15

。

A convex cyclic quadrilateral with given side lengths in a given cyclic order is unique; each side is at most

15

大提示：

先数和为

32

的有序四元组，再用 Burnside 引理按循环旋转取商。

Count ordered quadruples summing to

32,

then quotient by cyclic rotation using Burnside’s lemma

解答：

一个凸圆内接四边形由其按圆周顺序排列的边长唯一确定，且存在的条件是最大边长小于其他三边之和。周长为

32

时，每条边至多为

15

。

先数正整数有序四元组

(a,b,c,d)

，满足

a+b+c+d=32

且每项至多为

15

。若没有上界限制，有

\binom{31}{3}=4495

个；去掉某一项至少为

16

的情况，需减去

4\binom{16}{3}=2240

，剩下

2255

个。

四边形的旋转对应于

(a,b,c,d)

的循环置换。由 Burnside 引理，不同四边形的数量为

\frac{1}{4}\big(2255+f_1+f_2+f_3\big)\text{，}

其中

f_i

表示旋转

i

个位置后不变的四元组数。

旋转一步或三步只固定

(8,8,8,8)

，所以

f_1=f_3=1

。旋转两步固定形如

(a,b,a,b)

且

a+b=16

、

1\le a,b\le15

的四元组，所以

f_2=15

。

因此数量为

\begin{aligned} \frac{1}{4}(2255+1+15+1) &= \frac{2272}{4} \\ &= 568\text{。} \end{aligned}

所以正确答案是 C。

A convex cyclic quadrilateral is determined up to rotation and translation by its cyclic sequence of side lengths, and it exists exactly when the largest side is less than the sum of the others. With perimeter

32,

this means each side is at most

15.

First count ordered quadruples

(a,b,c,d)

of positive integers with

a+b+c+d=32

and each entry at most

15.

Without the upper bound there are

\binom{31}{3}=4495;

removing those with some entry at least

16

subtracts

4\binom{16}{3}=2240,

leaving

2255.

Rotations of the quadrilateral correspond to cyclic permutations of

(a,b,c,d).

By Burnside’s lemma the number of distinct quadrilaterals is

\frac{1}{4}\big(2255+f_1+f_2+f_3\big),

where

f_i

counts quadruples fixed by rotating

i

positions.

A one- or three-step rotation fixes only

(8,8,8,8),

so

f_1=f_3=1.

A two-step rotation fixes

(a,b,a,b)

with

a+b=16

and

1\le a,b\le15,

giving

f_2=15.

Hence the count is

\begin{aligned} \frac{1}{4}(2255+1+15+1) &= \frac{2272}{4} \\ &= 568. \end{aligned}

Thus, C is the correct answer.

2010 AMC 12A 第 25 题

答案：C

难度评级：2520

小提示：

大提示：

解答：

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