Inspired by Muhammad Rasel Parvej

Geometry Level 5

What is the minimum side length of a square which can contain 5 non-overlapping unit squares?

Give your answer to 3 decimal places.


Inspiration .


The answer is 2.707.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Calvin Lin Staff
Dec 29, 2016

[This solution is incomplete.]

If the unit squares had to be parallel to the sides of the larger square, then the answer would be 3. See this solution for why that is the case.

The difficulty of this problem comes about from being able to tilt the squares . While it is clear that the large square has area at least 5 hence side length at least 2.23, and that a large square of side length 3 can fit 9 unit squares hence gives us an upper bound, it's not immediately apparent what the actual minimum is.

The generalized problem is still open, though specific cases have been worked on. Proving the minimum is extremely hard, even for small values. See this survey of known results .

Here is how we can pack 5 unit squares into a square of length 2 + 2 2 2 + \frac{ \sqrt{2}}{2} .

There is a relatively simple proof. Can you find it?


Hint: Here is a proof that the minimum side length of a square which can fit 3 squares is 2.

We will show that a square of side length t < 2 t < 2 allows only 1 unit square.
Orientate the square T T with the lower left corner at the origin, meaning that the square has vertices ( 0 , 0 ) , ( t , 0 ) , ( t , t ) , ( 0 , t ) (0,0), (t, 0), (t, t), (0, t) .
Consider the point ( t / 2 , t / 2 ) (t/2, t/2) . Since it is distance t / 2 < 1 t/2 < 1 away from any sides of T, thus any unit square in T T will have to contain this point.
Since all unit squares must contain this point, hence we can place at most 1 non-overlapping unit square in T.
Conversely, it is clear how we can fit 3 unit squares into a square of slide length 2.


How can we extend this argument to the case of 5 squares?


Here is an outline of the proof that no square of side length t < 2 + 2 2 t < 2 + \frac{ \sqrt{2} } { 2 } will allow 5 squares.

Set L = 2 + 2 2 L = 2 + \frac{ \sqrt{2} } { 2 } .
Orientate the square T T with the lower left corner at the origin, meaning that the square has vertices ( 0 , 0 ) , ( t , 0 ) , ( t , t ) , ( 0 , t ) (0,0), (t, 0), (t, t), (0, t) .
Consider the 4 points ( t L , t L ) , ( t L , ( L 1 ) t L ) , ( ( L 1 ) t L , t L ) , ( ( L 1 ) t L , ( L 1 ) t L ) ( \frac{t}{L}, \frac{t}{L} ), ( \frac{ t}{L}, \frac{ (L-1)t} { L } ), ( \frac{ (L-1)t} { L } , \frac{ t}{L}), ( \frac{ (L-1)t} { L } , \frac{ (L-1)t} { L } ) .

(Claim, which is harder than I initially expected) Any unit square in T T must contain at least one of these 4 points in the interior.
Hence, we can place at most 4 non-overlapping unit squares in T.

Note: For t = L t = L , these 4 points correspond to the touching points of the 5 square setup above, which is why it makes sense to consider these 4 points.

Consider packing 4 unit squares and 1 unit circle inside a larger square. Since a side length of 3 is not optimal, then the unit circle has to be at the center of the larger square. To complete this proof, this graphic is provided, showing locus of vertices as the corner squares are rotated, which shows that the unit circle is locked in place when the corner squares are aligned with the sides of the large square

Still, this proof isn't complete.

Michael Mendrin - 4 years, 5 months ago

Log in to reply

Playing devil's advocate, why must the corner squares be flush against the sides? Why can't we tilt them slightly?

Note: In all of the known best packings, the corner squares are indeed flush.

Calvin Lin Staff - 4 years, 5 months ago

Log in to reply

The green lines are loci of the vertices of the squares when they are tilted...all the way around. So, from this, we can determine that the circle is "locked" right where it is, i.e., cannot budge from where it is now. However, that doesn't necessarily preclude the existence of other "locked" configurations. I am not seeing a "relatively simple proof" here just yet.

Michael Mendrin - 4 years, 5 months ago

Log in to reply

@Michael Mendrin K, added a hint by working out a smaller case.

Calvin Lin Staff - 4 years, 5 months ago

Log in to reply

@Calvin Lin Hm, the proof isn't as simple as I thought it would be. There's a claim which requires some work.

Calvin Lin Staff - 4 years, 5 months ago

I've just visually guessed this orientation of the unit squares. It's just my intuition was so strong about its optimality. Then this link verified my guess.

Muhammad Rasel Parvej - 4 years, 5 months ago

Log in to reply

Can you add a complete proof that this is indeed the minimum? Thanks!

Calvin Lin Staff - 4 years, 5 months ago

Log in to reply

In the link, there is a proof. I'll try to understand it and then to post a solution.

Muhammad Rasel Parvej - 4 years, 5 months ago

The assumption of (four-fold) symmetry if a huge assumption. There isn't a strong reason why they should be symmetrical, if at all. E.g. with the 3 square case, the best setup is in a 2 × 2 2 \times 2 square, where 2 squares are packed nicely and 1 square could be placed anywhere.

Calvin Lin Staff - 4 years, 5 months ago

Log in to reply

I am talking only for FIVE squares. Not a general case. I can not give any proof, but in this case, both minimum and maximum must have 4-fold symmetry. I am weak in proof and theory of numbers.

Niranjan Khanderia - 4 years, 5 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...