2018 AIME I Problem 8

In an equiangular hexagon $AB + BC = DE + EF$ and $FA + AB = CD + DE;$ find $EF$ and $FA$

Opposite sides are parallel, at distance $\frac{\sqrt{3}}{2}$ times the sum of the two sides connecting them; the narrowest of the three strips limits the circle

All interior angles are $120^\circ,$ so opposite sides are parallel. Attaching equilateral triangles to two opposite sides produces a parallelogram, which forces $AB + BC = DE + EF$ and $FA + AB = CD + DE.$ Hence $EF = 2$ and $FA = 16.$

Walking from one side to the opposite side along the two connecting sides shows that the distance between a pair of opposite sides is $\frac{\sqrt{3}}{2}$ times the sum of those two connecting sides: the strips have widths $\frac{\sqrt{3}}{2}(BC + CD) = 9\sqrt{3}$ between $AB$ and $DE,$ $\frac{\sqrt{3}}{2}(CD + DE) = 11\sqrt{3}$ between $BC$ and $EF,$ and $\frac{\sqrt{3}}{2}(DE + EF) = 7\sqrt{3}$ between $CD$ and $FA.$ Any circle inside the hexagon fits in the narrowest strip, so $d \le 7\sqrt{3}.$

A circle of diameter $7\sqrt{3}$ tangent to lines $CD$ and $FA$ can be centered so that it also touches $DE$ exactly and has distances $6\sqrt{3},$ $5\sqrt{3},$ and $\frac{11\sqrt{3}}{2}$ from lines $EF,$ $BC,$ and $AB,$ all more than its radius $\frac{7\sqrt{3}}{2},$ so it fits inside the hexagon. Therefore $d = 7\sqrt{3}$ and $d^2 = 147.$