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Consider square of side $1.25$ can it be covered by three squares of side $1$ ?

I think it's impossible but I'm not sure how to show it.

Gerry Myerson
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Gregor
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  • It is impossible for one of square of side 1 to cover two vertices of side 1.25 so three squares can cover at most three vertices. Not sure whether that counts as proof though. – Jack Yoon May 26 '14 at 11:23
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    @JackYoon: The diagonal length of a square of side length one is $\sqrt{2} > 1.25$, so a single small square can cover two vertices of the larger square. – Michael Albanese May 26 '14 at 11:37
  • Good point. I did not consider that. – Jack Yoon May 26 '14 at 11:44
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    You should specify in your question if they are allowed to overlap and to rotate. (I guess they are, since the accepted answer does that) – o0'. May 26 '14 at 13:14
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    Makes me wonder what the minimal size would be of 3 smaller squares to fit – Ivo May 26 '14 at 13:56
  • Using a numerical solver and David's approach, I got 0.98268919 as minimum size for the small squares. – Axel Kemper May 26 '14 at 14:23
  • Generalization and proper explanation http://mathoverflow.net/questions/49674/how-many-cubes-cover-a-bigger-cube – sas May 27 '14 at 01:44

2 Answers2

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If the small squares are allowed to overlap it is possible.

I don't have any graphics software handy so this is going to be tricky.

If the big square is ABCD, place a small square EFGH such that E=A and FG passes through B. Let M be the intersection of GH and BC. Using 3-4-5 triangles you can show that BM=0.3125.

Similarly, place a small square IJKL such that I=A and JK passes through D. If N is the intersection of KL and CD then again DN=0.3125.

Since CM and CN are both less than 1, the remaining space can be covered by the third square.

Here is a really awful hand drawn image, but perhaps it's more helpful than the description. . .

enter image description here

Community
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David
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I can't compete with David's picture, but you might find Squares Covering Squares (part of Erich's Packing Center) useful. In particular it shows that Henry Dudeney found (in 1931) a covering not unlike David's which allows three unit squares to cover a square of side length $\sqrt{\frac{1+\sqrt5}{2}}\approx1.27202$.

In particular this answers Ivo Beckers' question: the minimum side length to cover a square of side 1.25 is $\sqrt{\frac{5\sqrt5+5}{8}}\approx0.982689222,$ assuming the optimality of Dudeney's construction.

hardmath
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Charles
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  • Thanks for the reference Charles - some of those coverings are truly astonishing! Small mistake: $(1+\sqrt5)/2$ is the area of the square, not the side length. My configuration is the same as Dudeney's except for the dimensions: if we calculate what the side $x$ of the large square should be to give length $CM$ equal to $1$, we get $x^4-x^2-1=0$ and hence area $x^2=(1+\sqrt5)/2$. Very satisfying to have more or less rediscovered a result of Dudeney's! – David May 27 '14 at 01:05
  • @David: Thanks for the correction, I fixed it. – Charles May 27 '14 at 01:11