2010 AMC 12A Problems/Problem 25

Problem 25

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

It should first be noted that given any quadrilateral of fixed side lengths, the angles can be manipulated so that the quadrilateral becomes cyclic.

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, sot 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 $414 + 135 + 14 + 4 + 1 = \boxed{568\ \textbf{(C)}}$ .

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2010 AMC 12A Problems/Problem 25

Problem 25

Solution