2021 AMC 12A Fall Problem 24

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

Answer: E

Concepts:trapezoid arithmetic sequence casework

Difficulty rating: 2520

Small hint:

With

AB = 18

the maximum, the four sides are

18, 18 - d, 18 - 2d, 18 - 3d;

place

A = (0,0),

B = (18,0),

and

D

at angle

60^\circ

Big hint:

For each assignment of the three smaller lengths to

BC, CD, DA,

impose that

\overline{CD}

is horizontal and solve for

d;

also include common difference

d = 0

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,

let

n=CD

and

\ell=BC.

Then

C=(\tfrac m2+n,\tfrac{m\sqrt3}{2}),

so

\begin{aligned} \ell^2&=m^2+(18-n)^2 \\ &\quad {}-m(18-n). \end{aligned}

Write

u_j=18-jd

for

j=1,2,3.

Substituting the permutations

123,132,213,231,312,321

of

(u_1,u_2,u_3)

for

(m,n,\ell)

gives the nonzero candidates

d=18,2,9,5,6,6,

respectively. Positivity of

u_3

requires

d<6,

leaving only

d=2

and

d=5.

The case

d=0

also gives a valid rhombus.

Thus the side sets are

\{18,16,14,12\},

\{18,13,8,3\},

and

\{18,18,18,18\}.

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.