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A square has a side length of 5 units. Is it possible to cover this square with three squares each with a side length of 4 units?

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yes

see picture red is 5x5 the other three are 4x4 enter image description here few measurements on the green square: the triangle outside the red square to the left has sides $4,3,5$, therefore the triangle outside the red to the top has sides $1,0.75,1.25$ the crucial thing being $1.25 \ge 5-4$

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    $\begingroup$ Your diagram looks stretched; none of the squares look square. $\endgroup$ – user3294068 Aug 3 at 16:42
  • $\begingroup$ @user3294068 If you rotate the screen about 55 degrees to one side, then they all do. $\endgroup$ – Penguino Aug 6 at 1:58
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This is a known problem.

The answer is “yes”.

Because the golden ratio $\varphi=\frac{\sqrt{5}+1}{2}=1.618\dots$ is bigger than $\left(\frac 54\right)^2=1.44$

The following cover of a square of area $\varphi$ by three unit squares was found by Henry E. Dudeney in 1931.

enter image description here

Erich Friedman had a page “Squares Covering Squares” with coverings of the largest known square by $n$ unit squares.

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I think the answer is

no.

Explanation:

  • First place one of the $4\times4$ squares within the $5\times5$ square. The remaining area (9 units) should be as compact as possible, so let's shift the $4\times4$ square right up to one corner, leaving an L-shape remaining. (I'm not sure how to prove rigorously that this is optimal.)
  • Now we need to place the other two squares so as to cover that L-shape. Surely the best way to do this is

    diagonally, so that the longest possible part of the $4\times4$ square is along the length of the L on each side.

It seems this should be enough to cover the $5\times5$ square, since

$4\sqrt{2}>5$,

but actually it's not that easy,

because the total $4\sqrt{2}$ length only covers zero width when the square is placed diagonally.

Let's do a quick bit of calculation:

calculation

Placed diagonally inside the $4\times4$ square for maximum length, the longest width-$1$ box that can fit is of length $4\sqrt{2}-1=4.66$.

Even allowing for the fact that the two additional $4\times4$ squares will meet at the corner of the L-shape, there's no way we can cover all of the two open edges of the $5\times5$ square, since our coverage of them will only go as far as $4.66$ along before the coverage starts to slope away from the edge.

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    $\begingroup$ Good effort undone by one little inaccuracy ("Surely the best way to do this is "), still +1 $\endgroup$ – Paul Panzer Aug 2 at 23:10
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    $\begingroup$ Ouch, I feel for Rand. Keep answering puzzles! And welcome Paul, I see you answered a lot recently. $\endgroup$ – justhalf Aug 3 at 13:20
  • $\begingroup$ Certainly not "rigorous" either, but definitely seems likely that you need to go to a corner because otherwise you will have to cover areas on both sides of the smaller square. With only 3 squares to work with (2 remaining besides the first one not on a corner in this scenario), that seems really hard to do. $\endgroup$ – Chipster Aug 3 at 23:31
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You need to cover the four corners, so one of the smaller squares covers two corners of the bigger one and the other two smalls cover one corner each.

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  • $\begingroup$ How can you cover two corners of the bigger square using one of the smaller squares? $\endgroup$ – Rand al'Thor Aug 3 at 19:41
  • $\begingroup$ Turn it 45 degrees and make each of the two adjacent top edges coincident with one of the top vertices of the bigger square. $\endgroup$ – AND Aug 3 at 19:44

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