Square Business

Geometry Level 5

The largest square which can be covered completely with four unit squares is a 2 × 2 2 \times 2 square, as shown on the left.

Without cutting or bending the unit squares, what is the side length (to 3 decimal places) of the largest square that can be covered completely with three unit squares?

Disclaimer: The image on the right is not necessarily the right configuration to obtain the largest square.


The answer is 1.272.

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1 solution

Michael Mendrin
Jan 26, 2017

Any complete covering of a larger square by 3 unit squares is going to partition the perimeter of the larger square into at least 3 segments. Suppose one of them covers one segment. The maximum segment any one square can cover has a length of 2 2 , which occurs when one of the unit squares shares one same vertex with the larger square and is aligned with it---as the graphic shows. Then it's a matter of how much more the other two unit squares can cover.

If S S is the side length of the larger square, and point P P has the coordinates ( x , y ) (x,y) , then if we let s = y x s=\dfrac{y}{x} , point Q Q has the coordinates

Q = ( S s 2 + 1 , s S s 2 + 1 ) Q= \left( \dfrac{S}{ {s}^{2}+1 }, \; s \dfrac{S}{ {s}^{2}+1 } \right)

then we have the following system of equations to solve for x , y , s , S x, y, s, S

x 2 + x 2 = 1 {x}^{2} + {x}^{2}=1
s = y x s=\dfrac{y}{x}
1 s ( S x ) + y = S 1 -\dfrac{1}{s} (S-x)+y=S-1
( S S s s + 1 ) 2 + ( s S s s + 1 ) 2 = 1 {\left( S-\dfrac{S}{{s}^{s}+1}\right) }^{2}+{ \left( s\dfrac{S}{{s}^{s}+1} \right) }^{2}=1


which yields the result

S = 1 + 5 2 = 1.27202... S=\sqrt{ \dfrac{ 1+\sqrt{5} } {2} }=1.27202...

Interestingly the area of this largest square is equal to the golden ratio.

Maria Kozlowska - 4 years, 4 months ago

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Yeah, like π \pi , the golden ratio keeps popping up everywhere. I did notice that.

Michael Mendrin - 4 years, 4 months ago

Wow! What a great problem... Thanks for posting!

Geoff Pilling - 4 years, 4 months ago

I don't think the argument about "perimeter covering" is rigorous.

Note that there is another way to cover this square of side length S S :

Calvin Lin Staff - 4 years, 4 months ago

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The argument may not be rigorous (I have a habit of doing a lot of hand-waving), but the maximum perimeter any one square can cover is still 2 2 --if the larger square is sufficiently larger than the unit square. Whether or not that necessarily leads to the largest possible covered square is another question!

Now I'll have to find out just how large that covered square can be made with your suggested figure. Let me go put that on my schedule.

Michael Mendrin - 4 years, 4 months ago

Wow, close! . This one works out to a maximum square side length of 1.268623... 1.268623...

This one got ugly pretty fast.

Michael Mendrin - 4 years, 4 months ago

Great problem! Here's a followup to it. More Square Business .

Sharky Kesa - 4 years, 4 months ago

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