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

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<math>\textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad \textbf{(E)}\ 2255</math>
 
<math>\textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad \textbf{(E)}\ 2255</math>
  
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== Solution ==
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== Solution 1 ==
 
It should first be noted that given any quadrilateral of fixed side lengths, the angles can be manipulated so that the quadrilateral becomes cyclic.
 
It should first be noted that given any quadrilateral of fixed side lengths, the angles can be manipulated so that the quadrilateral becomes cyclic.
  
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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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==Solution 2==
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As with solution <math>1</math> we would  like to note that given any quadrilateral we can change its angles to make a cyclic one.
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 +
Let <math>a\ge b\ge c\ge d</math> be the sides of the quadrilateral.
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There are <math>\binom{31}{3}</math> ways to partition <math>32</math>. 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 <math>a \ge 16</math>.  For <math>a=16</math>, <math>b+c+d=16</math>. There are <math>\binom{15}{2}</math> ways to partition <math>16</math>. Since <math>a</math> could be any of the four sides, we have counted <math>4\binom{15}{2}</math> degenerate quadrilaterals. Similarly, there are <math>4\binom{14}{2}</math>, <math>4\binom{13}{2} \cdots 4\binom{2}{2}</math> for other values of <math>a</math>.  Thus, there are <math>\binom{31}{3} - 4\left(\binom{15}{2}+\binom{14}{2}+\cdots+\binom{2}{2}\right) = \binom{31}{3} - 4\binom{16}{3} = 2255</math> non-degenerate partitions of <math>32</math> by the hockey stick theorem.  However, for <math>a\neq b \neq c \neq d</math> or <math>a=b \neq c \neq d</math>, each quadrilateral is counted <math>4</math> times, <math>1</math> for each rotation. Also, for <math>a=b\neq c=d</math>, each quadrilateral is counted twice. Since there is <math>1</math> quadrilateral for which <math>a=b=c=d</math>, and <math>7</math> for which <math>a=b\neq  c=d</math>, there are <math>2255-1-2\cdot7=2240</math> quads for which <math>a\neq b \neq c \neq d</math> or <math>a=b \neq c \neq d</math>. Thus there are <math>1+7+\frac{2240}{4} = \boxed{568} = \boxed{\textbf{(C)}}</math> total quadrilaterals.
  
 
== See also ==
 
== See also ==

Revision as of 09:56, 5 February 2012

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, 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, 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.

See also

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