Dan asked ... > Hmm, what about other rectangles also with area pi^2/6 ? > For example the pi/sqrt(6) x pi/sqrt(6) square. and I replied ... No, you need one edge of the rectangle to be at least 1.5, if the squares are placed othogonally. It looks even worse if they're rotated 45 degrees. But pi/2 x pi/3 looks like a possibility. This was too optimistic: With orthogonal placement, any packable rectangle must have a larger dimension >= 1 + 1/3 +1/4 = 1.58333... or smaller dimension >= 1/2 + 1/3 + 1/4 = 1.08333... But pi/2 x pi/3 is 1.5707... x 1.047..., which fails both criteria. It doesn't look like rotating any of the squares helps, but I haven't done the arithmetic to check this. Perhaps things go the other way: Maybe only the 1 x pi^2/6 rectangle works. Rich
Until we can pack the unit square with 1/n x 1/(n+1) rectangles, I would not be hopeful of packing a pi^2/6 rectangle with 1/n x 1/n squares.
This is an old problem of Leo Moser. Here's one of the most recent advances in it: MR1620857 (99e:52028) Paulhus, Marc M.(3-CALG) An algorithm for packing squares. (English. English summary) J. Combin. Theory Ser. A 82 (1998), no. 2, 147--157. 52C20 (05B45) References: 0 Reference Citations: 3 Review Citations: 0 The paper under review studies the problem of packing sets of squares (or rectangles) into rectangles. More exactly, the following three (open) problems are dealt with in the paper. (1) Find the smallest $\varepsilon>0$ such that the (infinite) set of squares with sides of lengths $1/n, n=2,3,4,\cdots$, can be packed in a rectangle of area $\pi^2/6-1+\varepsilon$. This problem appears in \ref[H. T. Croft, K. J. Falconer and R. K. Guy, Unsolved problems in geometry, Springer, New York, 1991; MR1107516 (92c:52001); correction; MR1316393 (95k:52001)]. The first results concerning it are due to A. Meir and L. Moser \ref[J. Combinatorial Theory 5 (1968), 126--134; MR0229142 (37 \#4716)]. (2) Find the smallest $\varepsilon>0$ such that the (infinite) set of squares with sides of lengths $1/(2n+1), n=1,2,3,\cdots$, can be packed in a rectangle of area $\pi^2/8-1+\varepsilon$. (3) Find the smallest $\varepsilon>0$ such that the (infinite) set of rectangles of widths $1/(n+1)$ and lengths $1/n, n=1,2,3,\cdots$, can be packed in a square of side $1+\varepsilon$ (the first results on this problem also are due to Meir and Moser [op. cit.]). In the paper under review an algorithm is presented and it is shown how the best known results can be improved (roughly speaking by a factor of at least $10^6$) using the algorithm. Also, it is proved that if $n\geq (1+l)/ wl$ and $n\geq (1+2l)/2wl$, then the squares of (1) and (2), respectively, from $n$ to infinity can be packed in a rectangle of width $w$ and length $l$. Reviewed by Béla Uhrin R. On Thu, 19 May 2005, Richard Schroeppel wrote:
Dan asked ...
Hmm, what about other rectangles also with area pi^2/6 ? For example the pi/sqrt(6) x pi/sqrt(6) square.
and I replied ... No, you need one edge of the rectangle to be at least 1.5, if the squares are placed othogonally. It looks even worse if they're rotated 45 degrees. But pi/2 x pi/3 looks like a possibility.
This was too optimistic: With orthogonal placement, any packable rectangle must have a larger dimension >= 1 + 1/3 +1/4 = 1.58333... or smaller dimension >= 1/2 + 1/3 + 1/4 = 1.08333... But pi/2 x pi/3 is 1.5707... x 1.047..., which fails both criteria. It doesn't look like rotating any of the squares helps, but I haven't done the arithmetic to check this.
Perhaps things go the other way: Maybe only the 1 x pi^2/6 rectangle works.
Rich
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