2021 Fall AMC 12A Problems/Problem 24

Problem

Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60 \textdegree$ , and $\overline{AB} \parallel \overline{CD}$ . In some order, the lengths of the four sides form an arithmetic progression, and side $\overline{AB}$ is a side of maximum length. The length of another side is $a$ . What is the sum of all possible values of $a$ ?

$\textbf{(A) } 24 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 66 \qquad \textbf{(E) } 84$

Solution

Let $E$ be a point on $\overline{AB}$ such that $BCDE$ is a parallelogram. Suppose that $BC=ED=b, CD=BE=c,$ and $DA=d,$ so $AE=18-c.$

Let $k$ be the common difference of the arithmetic progression of the side-lengths. It follows that $b,c,$ and $d$ are $18-k, 18-2k,$ and $18-3k,$ in some order.

If $k=0,$ then $ABCD$ is a rhombus with side-length $18,$ which is valid.

If $k\neq0,$ then we have six cases:

$(b,c,d)=(18-k,18-2k,18-3k)$

$(b,c,d)=(18-k,18-3k,18-2k)$

$(b,c,d)=(18-2k,18-k,18-3k)$

$(b,c,d)=(18-2k,18-3k,18-k)$

$(b,c,d)=(18-3k,18-k,18-2k)$

$(b,c,d)=(18-3k,18-2k,18-k)$

WILL COMPLETE VERY SOON. A MILLION THANKS FOR NOT EDITING THIS PAGE.

~MRENTHUSIASM

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2021 Fall AMC 12A Problems/Problem 24

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