(In reply to
Almost solution by Jer)
I'm getting stuck in the same spot. But while working on this problem, I imagined an easier version where we are working with a hexagonal base - taking just points A1,A3,A5,A7,A9,A11 to form the base.
Then OA1A5 is exactly 1/6 of the hexagon, seen by drawing OA1, OA5, OA9, A1A5, A5,A9, and A1A9 partitioning the hexagon into six congruent triangles. And then some simple math on the 30-30-120 triangle OA1A5 gets everything, including length A1A5 to work out.
But I can't easily extend that to the dodecagon. In addition to the six triangles I drew for the hexagon, there's also six more 15-15-150 triangles like A1A2A3 that make the ratio of areas more difficult than the simple 1/6 in the hexagon version.
re: Almost solution
