2010 AMC 12A Problems/Problem 25
Contents
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?
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 where all sides are fixed (in a certain order), we can construct the diagonal . When is the minimum allowed by the triangle inequality, one of the angles or will be degenerate and measure , so opposite angles will sum to less than . When is the maximum allowed, one of the angles will be degenerate and measure , so opposite angles will sum to more than . Thus, since the sum of opposite angles increases continuously as is lengthened from the minimum to the maximum values, there is a unique value of somewhere in the middle such that the sum of opposite angles is exactly .
Denote , , , and as the integer side lengths of the quadrilateral. Without loss of generality, let .
Since , the Triangle Inequality implies that .
We will now split into cases.
Case : ( side lengths are equal)
Clearly there is only way to select the side lengths , and no matter how the sides are rearranged only unique quadrilateral can be formed.
Case : or ( side lengths are equal)
If side lengths are equal, then each of those side lengths can only be integers from to except for (because that is counted in the first case). Obviously there is still only unique quadrilateral that can be formed from one set of side lengths, resulting in a total of quadrilaterals.
Case : ( pairs of side lengths are equal)
and can be any integer from to , and likewise and can be any integer from to . However, a single set of side lengths can form different cyclic quadrilaterals (a rectangle and a kite), so the total number of quadrilaterals for this case is .
Case : or or ( side lengths are equal)
If the equal side lengths are each , then the other sides must each be , which we have already counted in an earlier case. If the equal side lengths are each , there is possible set of side lengths. Likewise, for side lengths of there are sets. Continuing this pattern, we find a total of sets of side lengths. (Be VERY careful when adding up the total for this case!) For each set of side lengths, there are possible quadrilaterals that can be formed, so the total number of quadrilaterals for this case is .
Case : (no side lengths are equal) Using the same counting principles starting from and eventually reaching , we find that the total number of possible side lengths is . There are ways to arrange the side lengths, but there is only unique quadrilateral for rotations, so the number of quadrilaterals for each set of side lengths is . The total number of quadrilaterals is .
And so, the total number of quadrilaterals that can be made is .
Solution 2
As with solution we would like to note that given any quadrilateral we can change its angles to make a cyclic one.
Let be the sides of the quadrilateral.
There are ways to partition . 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 . For , . There are ways to partition . Since could be any of the four sides, we have counted degenerate quadrilaterals. Similarly, there are , for other values of . Thus, there are non-degenerate partitions of by the hockey stick theorem. We then account for symmetry. If all sides are congruent (meaning the quadrilateral is a square), the quadrilateral will be counted once. If the quadrilateral is a rectangle (and not a square), it will be counted twice. In all other cases, it will be counted 4 times. Since there is square case, and rectangle cases, there are quadrilaterals counted 4 times. Thus there are total quadrilaterals.
See also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.