# Cover a square with three smaller squares

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?

yes

see picture red is 5x5 the other three are 4x4 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 $$1.25 \ge 5-4$$

• Your diagram looks stretched; none of the squares look square. Aug 3, 2020 at 16:42
• @user3294068 If you rotate the screen about 55 degrees to one side, then they all do. Aug 6, 2020 at 1:58

This is a known problem.

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.

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

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:

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.

• Good effort undone by one little inaccuracy ("Surely the best way to do this is "), still +1 Aug 2, 2020 at 23:10
• Ouch, I feel for Rand. Keep answering puzzles! And welcome Paul, I see you answered a lot recently. Aug 3, 2020 at 13:20
• 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. Aug 3, 2020 at 23:31

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.

• How can you cover two corners of the bigger square using one of the smaller squares? Aug 3, 2020 at 19:41
• Turn it 45 degrees and make each of the two adjacent top edges coincident with one of the top vertices of the bigger square.
– AND
Aug 3, 2020 at 19:44