(Moving here from a separate file for greater visibility.)

A MECHANISM TO BALANCE THE BUDGET

Suppose each member of Congress were to name a budget amount, and the median number is made law. Why the median? Because this is a number such that half the members have approved at least that number.

Of course, this number is likely to include a deficit. What to cut?

There are 435 numbers for each budget item. Let B(i,m) be mth-lowest amount chosen for budget item i; let T(m) be the sum over i of B(i,m). Let k be the highest number such that T(k) < revenue. If each budget item is enacted as B(i,k), then the budget is balanced. (Of course, if B(i,k) is higher than the median in one house or the other, it must be lowered.) This rule has the effect of temporarily requiring a supermajority for spending. Congressmen won't gain by inflating their numbers; that would just lower k. Pork-barrels would be highly vulnerable to clipping, while programs backed by a consensus would be immune.

I just thought of a new kind of packing problem, a mutant extension of the Thomson problem. In this version, each particle has coordinates in two independent spaces; in each it is confined to a sphere (of some dimension). In each space, pairs of particles repel each other with a force inversely proportional to the square of their distance in the other space.

(I was imagining a head with striped hairs, as one does, and considered making each hair’s phase anti-correlated with those of its neighbors, and what that would mean.)

On most numbered dice, opposite sides are complementary; on a cube, for example, they add to 7. As a result, if you have the skill to throw a die so that the {1,2,3} corner lands on the table, the upward face must be at least 4.

I would prefer to design dice so that, if numbers are considered as masses, the center of mass coincides with the geometric center. I think this is equivalent to saying: the sum of the numbers in any hemisphere must be equal.

You can’t do that with a tetrahedron, cube, D10 or regular dodecahedron; but I found three solutions for the octahedron, and 876 for the rhombic dodecahedron. (At least I see no obvious way to reduce that number further with symmetries.)

For a cube the best you can do is put pairs of adjacent numbers on opposite faces.
The twelve best arrangements on the regular dodecahedron all have 0 and 11 opposite each other, 1–5 around 11, and 6–10 around 0.
The two best arrangements for D10 are 0285364179 and 0582367149 (reading around the fivefold axis, alternating upper and lower faces).
(Add 1 to each number if you don’t like zero-based indexing; it doesn’t affect the math.)

I’ll update here if I come up with an approach to the icosahedron problem that won’t take thousands of years.