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Difference between revisions of "2010 AMC 12A Problems/Problem 25"

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== Problem 25 ==
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== 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?
 
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?
  
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And so, the total number of quadrilaterals that can be made is <math>414 + 135 + 14 + 4 + 1 = \boxed{568\ \textbf{(C)}}</math>.
 
And so, the total number of quadrilaterals that can be made is <math>414 + 135 + 14 + 4 + 1 = \boxed{568\ \textbf{(C)}}</math>.
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== See also ==
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{{AMC12 box|year=2010|num-b=24|after=Last Problem|ab=A}}
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 22:36, 25 February 2010

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

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

See also

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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All AMC 12 Problems and Solutions
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