Question

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

$24$

$42$

$60$

$66$

$84$

Solution:

Since $AB = 18$ is the largest, the four sides are $18, 18 - d, 18 - 2d, 18 - 3d.$ Placing $A = (0,0),$ $B = (18,0),$ and $D = \left(\tfrac{m}{2}, \tfrac{m\sqrt3}{2}\right)$ with $m = DA,$ the base $\overline{CD}$ is horizontal, giving $C$ and hence a length condition on $BC.$

Solving over the assignments yields two genuine trapezoids: sides $\{18, 16, 14, 12\}$ (with $d = 2$ ) and sides $\{18, 13, 8, 3\}$ (with $d = 5$ ). The degenerate $d = 0$ case is the rhombus with all sides $18.$

The possible values of a non- $AB$ side length are $\{3, 8, 12, 13, 14, 16, 18\},$ whose sum is $84.$

Thus, the correct answer is E.

2021 AMC 12A Fall Problem 24

Solution:

Problem 24 in Other Years