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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.

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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 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 at 11:37
    
Good point. I did not consider that. –  Jack Yoon May 26 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) –  Lohoris May 26 at 13:14
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Makes me wonder what the minimal size would be of 3 smaller squares to fit –  Ivo Beckers May 26 at 13:56

2 Answers 2

up vote 17 down vote accepted

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

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If anyone feels like editing in a better image please feel free. . . –  David May 26 at 11:59
    
I actually really like this image and it conveys the meaning concisely. –  githubphagocyte May 27 at 0:56

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.

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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 at 1:05
    
@David: Thanks for the correction, I fixed it. –  Charles May 27 at 1:11

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