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

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Let <math>a\ge b\ge c\ge d</math> be the sides of the quadrilateral.
 
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.
+
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, if all sides are distinct or there is exactly one pair of congruent sides, each quadrilateral is counted <math>4</math> times, <math>1</math> for each rotation. Also, when opposite sides are equal (but not adjacent sides), each quadrilateral is counted twice. Since there is <math>1</math> quadrilateral for which all sides are distinct, and <math>7</math> for which there is exactly one pair of congruent sides, there are <math>2255-1-2\cdot7=2240</math> quadrilaterals counted 4 times. Thus there are <math>1+7+\frac{2240}{4} = \boxed{568} = \boxed{\textbf{(C)}}</math> total quadrilaterals.
  
 
== See also ==
 
== See also ==

Revision as of 11:30, 4 August 2020

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 (in a certain order), 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, if all sides are distinct or there is exactly one pair of congruent sides, each quadrilateral is counted $4$ times, $1$ for each rotation. Also, when opposite sides are equal (but not adjacent sides), each quadrilateral is counted twice. Since there is $1$ quadrilateral for which all sides are distinct, and $7$ for which there is exactly one pair of congruent sides, there are $2255-1-2\cdot7=2240$ quadrilaterals counted 4 times. Thus there are $1+7+\frac{2240}{4} = \boxed{568} = \boxed{\textbf{(C)}}$ total quadrilaterals.

See also

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