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I do not know the original source of this problem; I have seen it a few places. The solution that I have seen is not pretty in the sense that it lacks symmetry. Can you find an elegant solution? Of course, elegance is subjective.

User Stiv asked if the answer must be a single region. I will allow answers that consist of multiple regions.

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    $\begingroup$ Does it have to be a single region? Because two separate rectangles springs to mind... $\endgroup$
    – Stiv
    Commented Oct 1, 2022 at 5:23
  • $\begingroup$ @Stiv I will allow multiple regions. $\endgroup$ Commented Oct 1, 2022 at 5:41
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    $\begingroup$ Does this puzzle "invite speculative answers"? Yes. That is its intention. Should it remain open? I say yes because it is interesting, and I'm totally fine that there are many "correct" answers. The answers posted so far are all great! However, there are other types of puzzles that "invite speculative answers" that should be closed -- overly-vague riddles come to mind. $\endgroup$
    – JLee
    Commented Oct 1, 2022 at 9:08
  • $\begingroup$ A one-by-four rectangle takes only 10 unit lines. What am I missing? $\endgroup$
    – keshlam
    Commented Oct 1, 2022 at 18:11
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    $\begingroup$ If you want to keep it open, edit it to define the problem better. For example, as posed I can make my 10-stick solution into a 12-stick one by stacking the last two on top of sticks in the outline, or just pointing them off into nowhere so they don't enclose any space... Or just by dropping them across the middle of the box... or using them to offset part of the box, which I would argue meets not only the letter but the intent of the problem as posed. If there's a good puzzle here, it can be posed more clearly. $\endgroup$
    – keshlam
    Commented Oct 2, 2022 at 15:38

6 Answers 6

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Here's a very simple solution:

Make a parallelogram with base 4 and height 1, as shown:
image of parallelogram
Here, you can see that the distance from $A$ to $C$ is 2 units (since the black circle has radius 2, and $\overline{AC}$ is just another radius). Similarly, $\overline{BD}$ has length 2; so you can place 4 matchsticks along the top and bottom, and 2 on the two diagonals.

(This, of course, generalizes to any area from 0 to 8 - you just have to change the angles of the parallelogram.)

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  • $\begingroup$ Cool! And simple! $\endgroup$ Commented Oct 1, 2022 at 6:04
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The funny thing about this problem is that once you understand the sticks don't have to follow a square grid, you see that you can enclose any surface up to 9 and some units by just making any loop and adjusting the space between the matches, flattening the figure, until it encloses the right surface.

I will try for "elegant".

![paralellogram with sides 1 and 5

According to Pythagoras the diagonals are 5 units.
This gives a perimeter of 2x5 + 2x1 = 12.
The suface is the base times the height, 1x4 = 4.

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  • $\begingroup$ This is a nice solution. $\endgroup$ Commented Oct 1, 2022 at 9:16
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Using 12 matches to obtain an area of 4 units:


enter image description here

Four regions but squarely square.

Edit: another distinct solution with a single enclosed area:

Based on the 3:4:5 Pythagorean triangle, and very easy to construct.
It would normally have perimeter = 12, area = 6 units.
But moving 3 matches at the right-angle reduces the area by 2 units.
So it now has perimeter = 12, area = 4 units.

enter image description here

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  • $\begingroup$ A nice simple answer. $\endgroup$ Commented Oct 1, 2022 at 9:14
  • $\begingroup$ I like it. One might wonder whether you twist the rules, but you definitely have enclosed 4 unit squares. $\endgroup$
    – Florian F
    Commented Oct 1, 2022 at 15:47
  • $\begingroup$ @FlorianF thanks. I realise that in effect I was 'hiding' four matches but the rules say "answers that consist of multiple regions". I had looked at a Pythagorean 3:4:5 possibility too, but your answer got it nicely. $\endgroup$ Commented Oct 1, 2022 at 21:33
  • $\begingroup$ If you use the four internal matches to create a single square inside, would the whole figure be enclosing only three unit squares? $\endgroup$
    – Muqo
    Commented Oct 2, 2022 at 0:32
  • $\begingroup$ @Muqo a shape with a hole in it? There's an idea. Not that but I have added another (fully legal) solution to this answer. $\endgroup$ Commented Oct 2, 2022 at 19:20
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One area, some symmetry, and easy to create from a square:

enter image description here

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  • $\begingroup$ @ParclyTaxel If you fold one part inwards, you have to fold another part outwards to stay at area 4. $\endgroup$
    – Retudin
    Commented Oct 1, 2022 at 8:12
  • $\begingroup$ Another nice answer. $\endgroup$ Commented Oct 1, 2022 at 9:13
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If the matchsticks do not need to form one single enclosed region, and instead the areas of multiple enclosed regions can be combined to sum to an area of 4 square units, there are two more trivial solutions:

Two trivial solutions

In fact, any arrangement where you can form four unit squares from sixteen matches, and then combine them in a way so as to remove four of them will provide a valid answer.

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  • $\begingroup$ Another nice answer. $\endgroup$ Commented Oct 1, 2022 at 6:11
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Perhaps the most symmetric solution:

a six-pointed star à la Twilight Sparkle.

The inner vertices form a regular hexagon with side length $2s$, where $s=0.4107231313\dots$ solves $6s(s\sqrt3+\sqrt{1-s^2})=4$.

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  • $\begingroup$ Nice solution & diagram $\endgroup$ Commented Oct 1, 2022 at 9:24
  • $\begingroup$ Considering that you answered this question, why did you vote to close the question? $\endgroup$ Commented Oct 2, 2022 at 1:08
  • $\begingroup$ One reason for closing a question is when it has too many solutions. And this one does have many solutions. I myself did not vote to close it though because it still gets people stumped because they are looking for a solution based on squares. $\endgroup$
    – Florian F
    Commented Oct 2, 2022 at 9:11
  • $\begingroup$ @FlorianF Before posting I knew there were at least two solutions. I hope people had fun trying to solve this puzzle. I think some solvers would be happy to find even one solution while other solvers would want to find as many solutions as possible. $\endgroup$ Commented Oct 3, 2022 at 4:27

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