2018 AIME I Problem 7

Count triangles with all vertices on one hexagon separately from those using both; each hexagon alone contributes $8$ isosceles triangles

For two vertices on one base, the apex on the other base must lie above the perpendicular bisector of the pair, or a lateral side $\sqrt{d^2 + 4}$ must equal the base chord $2$

The chords of a unit regular hexagon have lengths $1,$ $\sqrt{3},$ and $2.$ Among the $\binom{6}{3} = 20$ triangles in one hexagon, $6$ have sides $1, 1, \sqrt{3}$ and $2$ are equilateral with side $\sqrt{3};$ the other $12,$ with sides $1, \sqrt{3}, 2,$ are scalene. So each base contributes $8$ isosceles triangles, for $16$ in all.

Otherwise two vertices lie on one base ($2$ choices of that base) and one on the other. A vertex of the top base at horizontal distance $d$ from a bottom vertex is at distance $\sqrt{d^2 + 4} \ge 2$ from it. If the bottom pair is adjacent (chord $1$): the perpendicular bisector of a hexagon edge passes through no vertices, and no slant side can equal $1,$ so there are no isosceles triangles. If the pair has one vertex between them (chord $\sqrt{3},$ $6$ pairs): the top vertices above that middle vertex and above the opposite vertex are equidistant from the pair, giving $6 \cdot 2 = 12.$ If the pair is diametrically opposite (chord $2,$ $3$ pairs): no vertex lies above the perpendicular bisector, but the top vertex directly above either endpoint gives a slant side $\sqrt{0 + 4} = 2$ equal to the chord, giving $3 \cdot 2 = 6.$

The total is $16 + 2\,(12 + 6) = 52.$