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Erich Friedman has collected solutions and notes:

All of these are probably optimal, except for possibly n=20.

But by adding the domino areas, one gets:

$$A=\frac{20(20+1)(2\cdot20+1)}3=5740$$

So the side length must be at least $\lceil\sqrt A\rceil=76$, as given. Is it really that simple or have I misunderstood something?

What (better) lower and upper bound can be given for the general case of $n$ consecutive dominoes?

Edit: Ah, it does not say anywhere that the sides of the dominoes have to be parallel to a side of the square. So I have no proof that the solution is optimal. Got it.

hardmath
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Ivan
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1 Answers1

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what? There's a gap in the numbers (and there are clearly some white spaces in the drawing)

Prelude> sum [ 2*k^2 |k<-[1..20] ]
5740
Prelude> 76^2
5776

I fail to see why the problem is interesting.

I like much better the first part of Problem D5 in Croft/Falconer/Guy: Unsolved Problems in Geometry: pack rectangles $(1/n, 1/(n+1))$ into the unit square. (This can only work if there are no gaps, because the sum of the areas is 1.)

d8d0d65b3f7cf42
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