5
$\begingroup$

I have a square cloth with side length $x$ cm, and I am going to cut it into at least $n$ squares with side length $1$ cm for my customer, and also you cannot cut the cloth to thinner pieces (reminded by @risky mysteries). I cannot glue any bits of cloth together. What is the minimum value of $x$?


Problem inspired from a math test problem in my school (about a month ago).


Thanks to @Jaap Scherphuis, I now know this is an unsolved problem. So of course I still haven't solved it. You can use a computer!

$\endgroup$
  • $\begingroup$ Is folding valid? $\endgroup$ – risky mysteries Jul 28 at 4:04
  • 1
    $\begingroup$ @riskymysteries Yes, but you can't say a $1$ by $1.1$ rectangle is a $1$ by $1$ square with folding. You need to cut it. $\endgroup$ – Culver Kwan Jul 28 at 4:09
  • 1
    $\begingroup$ Just to make this clear, so is this $x$ should be defined in term of $n$ a.k.a $x = f(n)$? $\endgroup$ – athin Jul 28 at 5:07
  • 10
    $\begingroup$ Isn't this an open problem in mathematics? en.wikipedia.org/wiki/Square_packing_in_a_square $\endgroup$ – Jaap Scherphuis Jul 28 at 5:51
  • 3
    $\begingroup$ This isn't clear: the problem is inspired from a math test problem, but has not been solved by the OP. So there is no definitive answer? This should be closed IMO. $\endgroup$ – Earlien Jul 28 at 11:14
1
$\begingroup$

$\lceil\sqrt{n}\;\rceil$

(the ceiling of the square root of n)

Because

If you need n squares which are 1cm wide, you essentially require a piece of cloth with an area of , i.e. a square of length √n.
But given you can't glue pieces together we take the ceiling of √n to ensure we have enough pieces whose length are 1cm

| improve this answer | |
$\endgroup$
0
$\begingroup$

Is the answer

1

Reason:

The cloth can be cut into 2 pieces like this

Image

| improve this answer | |
$\endgroup$
  • $\begingroup$ No, you cannot cut it to thinner pieces. I will add that. $\endgroup$ – Culver Kwan Jul 28 at 3:29
  • $\begingroup$ @CulverKwan Make sure to acknowledge the edit! $\endgroup$ – risky mysteries Jul 28 at 3:30
  • 1
    $\begingroup$ Ok, I have acknowledge it! $\endgroup$ – Culver Kwan Jul 28 at 3:35
0
$\begingroup$

Clearly, $x \geq \sqrt n$, otherwise your original square would have less area than the n smaller squares.

In most cases, you will need $x = \lceil \sqrt n \rceil$, as that will be the smallest square that allows you to cut the necessary pieces.

However, for $n=5$,

You can make do with $x = 2\sqrt2$:
Square with diagonal smaller squares
This wastes 3 units, which is less than would be wasted for a size 3 square.

This does not continue to work for higher $n$. For example, for $x=3\sqrt2$, you could fit 13 squares in a similar pattern, but the slightly smaller $x = 4$ would have room for 16.

So, in conclusion,

you need $x = \lceil \sqrt n \rceil$, except for $n=5$, where $x=2\sqrt 2$ will work.

| improve this answer | |
$\endgroup$
-2
$\begingroup$

In both the square, area will be the same.

suppose big square has area, A = x*x

all small square total area will be A = n * 1 *1

so, x*x = n * 1 * 1

x = √n, for all x>=n

| improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ How can you cut for example 2 whole squares from a piece of cloth of size $\sqrt{2}\times\sqrt{2}$? $\endgroup$ – Jaap Scherphuis Jul 28 at 6:52
  • $\begingroup$ i think solution will work for all x>=n $\endgroup$ – Amit Huda Jul 29 at 8:41
  • $\begingroup$ You already said that $x=\sqrt{n}$, so how can it be that $x\ge n$? $\endgroup$ – Jaap Scherphuis Jul 29 at 8:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.