2021 Fall AMC 12A Problem 24

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.