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 $AE=18-c,$ as shown below.

DIAGRAM WILL BE READY VERY VERY SOON ...

We apply the Law of Cosines to $\triangle ADE:$ $\begin{align*} AD^2 + AE^2 - 2\cdot AD\cdot AE\cdot\cos 60^\circ &= DE^2 \\ d^2 + (18-c)^2 - d(18-c) &= b^2 \\ (18-c)^2 - d(18-c) &= b^2 - d^2 \\ (18-c)(18-c-d) &= (b+d)(b-d). \hspace{15mm}(\bigstar) \end{align*}$ 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. Note that $0\leq k<6.$

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
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.

Page

Toolbox

Search

2021 Fall AMC 12A Problems/Problem 24

Problem

Solution

See Also