2010 AMC 12A Problem 25

Question

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

Answer: C

Concepts:Burnside’s Lemma stars and bars cyclic quadrilateral

Difficulty rating: 2520

Small hint:

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

15

Big hint:

Count ordered quadruples summing to

32,

then quotient by cyclic rotation using Burnside’s lemma

Solution:

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.