Math and Logic Puzzles: Redux

[BTD6_maker]

Turbo the Snail is back. This time, there are n circles on the plane. No three circles meet at a point. Turbo begins her journey on one of the circles. She initially moves clockwise along this circle. She moves along a circle until she reaches an intersection. Then she starts moving along the other circle and also changes direction (from clockwise to anticlockwise and vice versa). Prove that, if Turbo's path entirely covers all the circles, n is odd.

[BTD6_maker]

None of the circles are tangent to each other.

[BTD6_maker]

Here's another problem (although you can still try my earlier one).

Let n be a positive integer. In the Cartesian plane, there is a butterfly on each lattice point with non-negative coordinates (and there are no other butterflies). The neighbourhood of a bitterfly is the (2n+1)*(2n+1) square centred on the butterfly. A butterfly is lonely, comfortable, or crowded, respectively, if the number of butterflies in its neighbourhood is less than, equal to, or greater than half the number of lattice points in its neighbourhood (excluding the butterfly itself) respectively.

Every minute, all lonely butterflies fly away.

a: Prove that after some point there are no lonely butterflies left.
b: Find the number of comfortable butterflies left at this point.

[BTD6_maker]

Part of the difficulty in this problem is that there is no easy way to make tetrahedral numbers into nth powers. I have a proof that, for any n >= 3, there are only finitely many triangular numbers that are nth powers, but I haven't yet got it to work for tetrahedral numbers. This is an interesting problem in itself, so you can try this problem (as well as my two previous combinatorics problems).

For geometry's geometry problem, let O be the centre of W and let it have radius R. Let one of the tangents from A touch W at C, and let theta be the angle AOC. Then OA' = OC cos theta and OC = OA cos theta so OA' OA = OC^2 = R^2. Likewise, OB' OB = R^2. Also, O, A, A' are collinear, and so are O, B, B'. Since OA' OA = OB' OB, we have AA'B'B is cyclic.