2010 AMC 12A Problems/Problem 25

Problem

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?

$\textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad \textbf{(E)}\ 2255$

Solution 1

It should first be noted that given any quadrilateral of fixed side lengths, there is exactly one way to manipulate the angles so that the quadrilateral becomes cyclic.

Proof. Given a quadrilateral $ABCD$ where all sides are fixed, we can construct the diagonal $\overline{BD}$ . When $BD$ is the minimum allowed by the triangle inequality, one of the angles $\angle DAB$ or $\angle BCD$ will be degenerate and measure $0^\circ$ , so opposite angles will sum to less than $180^\circ$ . When $BD$ is the maximum allowed, one of the angles will be degenerate and measure $180^\circ$ , so opposite angles will sum to more than $180^\circ$ . Thus, since the sum of opposite angles increases continuously as $BD$ is lengthened from the minimum to the maximum values, there is a unique value of $BD$ somewhere in the middle such that the sum of opposite angles is exactly $180^\circ$ .

Denote $a$ , $b$ , $c$ , and $d$ as the integer side lengths of the quadrilateral. Without loss of generality, let $a\ge b \ge c \ge d$ .

Since $a+b+c+d = 32$ , the Triangle Inequality implies that $a \le 15$ .

We will now split into $5$ cases.

Case $1$ : $a = b = c = d$ ( $4$ side lengths are equal)

Clearly there is only $1$ way to select the side lengths $(8,8,8,8)$ , and no matter how the sides are rearranged only $1$ unique quadrilateral can be formed.

Case $2$ : $a = b = c > d$ or $a > b = c = d$ ( $3$ side lengths are equal)

If $3$ side lengths are equal, then each of those side lengths can only be integers from $6$ to $10$ except for $8$ (because that is counted in the first case). Obviously there is still only $1$ unique quadrilateral that can be formed from one set of side lengths, resulting in a total of $4$ quadrilaterals.

Case $3$ : $a = b > c = d$ ( $2$ pairs of side lengths are equal)

$a$ and $b$ can be any integer from $9$ to $15$ , and likewise $c$ and $d$ can be any integer from $1$ to $7$ . However, a single set of side lengths can form $2$ different cyclic quadrilaterals (a rectangle and a kite), so the total number of quadrilaterals for this case is $7\cdot{2} = 14$ .

Case $4$ : $a = b > c > d$ or $a > b = c > d$ or $a > b > c = d$ ( $2$ side lengths are equal)

If the $2$ equal side lengths are each $1$ , then the other $2$ sides must each be $15$ , which we have already counted in an earlier case. If the equal side lengths are each $2$ , there is $1$ possible set of side lengths. Likewise, for side lengths of $3$ there are $2$ sets. Continuing this pattern, we find a total of $1+2+3+4+4+5+7+5+4+4+3+2+1 = 45$ sets of side lengths. (Be VERY careful when adding up the total for this case!) For each set of side lengths, there are $3$ possible quadrilaterals that can be formed, so the total number of quadrilaterals for this case is $3\cdot{45} = 135$ .

Case $5$ : $a > b > c > d$ (no side lengths are equal) Using the same counting principles starting from $a = 15$ and eventually reaching $a = 9$ , we find that the total number of possible side lengths is $69$ . There are $4!$ ways to arrange the $4$ side lengths, but there is only $1$ unique quadrilateral for $4$ rotations, so the number of quadrilaterals for each set of side lengths is $\frac{4!}{4} = 6$ . The total number of quadrilaterals is $6\cdot{69} = 414$ .

And so, the total number of quadrilaterals that can be made is $414 + 135 + 14 + 4 + 1 = \boxed{568\ \textbf{(C)}}$ .

Solution 2

As with solution $1$ we would like to note that given any quadrilateral we can change its angles to make a cyclic one.

Let $a\ge b\ge c\ge d$ be the sides of the quadrilateral.

There are $\binom{31}{3}$ ways to partition $32$ . However, some of these will not be quadrilaterals since they would have one side bigger than the sum of the other three. This occurs when $a \ge 16$ . For $a=16$ , $b+c+d=16$ . There are $\binom{15}{2}$ ways to partition $16$ . Since $a$ could be any of the four sides, we have counted $4\binom{15}{2}$ degenerate quadrilaterals. Similarly, there are $4\binom{14}{2}$ , $4\binom{13}{2} \cdots 4\binom{2}{2}$ for other values of $a$ . Thus, there are $\binom{31}{3} - 4\left(\binom{15}{2}+\binom{14}{2}+\cdots+\binom{2}{2}\right) = \binom{31}{3} - 4\binom{16}{3} = 2255$ non-degenerate partitions of $32$ by the hockey stick theorem. However, for $a\neq b \neq c \neq d$ or $a=b \neq c \neq d$ , each quadrilateral is counted $4$ times, $1$ for each rotation. Also, for $a=b\neq c=d$ , each quadrilateral is counted twice. Since there is $1$ quadrilateral for which $a=b=c=d$ , and $7$ for which $a=b\neq c=d$ , there are $2255-1-2\cdot7=2240$ quads for which $a\neq b \neq c \neq d$ or $a=b \neq c \neq d$ . Thus there are $1+7+\frac{2240}{4} = \boxed{568} = \boxed{\textbf{(C)}}$ total quadrilaterals.

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search

2010 AMC 12A Problems/Problem 25

Contents

Problem

Solution 1

Solution 2

See also