Given this shape: diagram showing a 4000 unit wide cyan square with a 400 unit wide red square in the middle

Is it possible to divide the cyan area into 5 equal area shapes

such that:

  1. Each shape is the same
  2. Each shape has an edge touching the red square
  3. Each shape has an edge touching the outside.
  4. No diagonal lines.

Its reasonably easy to perform #2 and #3 as long as you violate #1

And I don't believe its possible to actually satisfy #1, and the very fact its contained within a square suggests #1 is not satisfiable given the presence of #2.

Though its just a bit of fun really =).

Context: Was simply devising a town plan for a guild oriented town for a minecraft world, and it became possible that we might want 5 guilds, and the fun of giving each guild a fair equal area region , in conjunction with all 5 guilds having a shared space in the middle.

Diagonals are unwanted as it makes dividing the land fairly and applying region controls onerous, as most things are rectangular, regions included.

So while you can roughly approximate a diagonal with sufficiently many rectangles, the less steps, the better.

If it is not possible

Please provide reasoning as to why not.


It would be interesting to see what sort of alternatives people can come up with, perhaps there is an optimal shape that results in all the shapes being highly similar geometrically, despite not being identical.

  • 1
    $\begingroup$ It's impossible, it is possible with 4 shapes, but not 5. For example: they need to be the same, but they have to touch the red square, so you can't divide the red squares perimeter into 5 identical pieces :) $\endgroup$
    – Gintas K
    Nov 7, 2013 at 10:59
  • $\begingroup$ @GintasK But it might be possible that 5 shapes are not symmetrically located... (I'm not sure about this) $\endgroup$
    – JiminP
    Nov 7, 2013 at 11:00
  • 4
    $\begingroup$ ok, the red squares perimeter is 1600u, so one piece should touch 320u of it. the red square has 4 corners, so at least one piece wont have that corner what makes them not the same :) $\endgroup$
    – Gintas K
    Nov 7, 2013 at 11:03
  • $\begingroup$ When I first fell upon this problem, I didn't even think it was possible to do 5 equal-area shapes of different shapes, but I've since proven myself wrong about that =) Though I came upon my knowledge by simply doing brute-force spacial geometry, I have no Idea how somebody would throw a calculator at this problem. $\endgroup$ Nov 7, 2013 at 11:26
  • 1
    $\begingroup$ @GintasK: That seems to be the right intuition, but a priori it's not entirely obvious that the "lack of corner" in the fifth shape could not be fixed by fitting it into corners made by the other pieces. $\endgroup$ Nov 7, 2013 at 11:48

2 Answers 2


What about the shape below made in geogebra? enter image description here

  • 3
    $\begingroup$ Too much free time, eh? $\endgroup$ Nov 8, 2013 at 20:22
  • 1
    $\begingroup$ @ja72 Yeah, today is my day off, and I decided to take time to this exercise. It was very interesting. $\endgroup$ Nov 8, 2013 at 20:50
  • 1
    $\begingroup$ So I am not the only one solving problems for fun. What a relief. $\endgroup$ Nov 8, 2013 at 21:00
  • $\begingroup$ That sir, is awesome =). Thats about as good as I think I could expect to get =). $\endgroup$ Nov 10, 2013 at 13:34

I don't think it possible to have all shapes identical, because the criteria of "touching the outside" and "touching the inner square" basically means there has to be at least 5 edges running from center to outside.

And because it is impossible to draw 5 edges from a square in such a way that all edges leave the square in such a way they present as being geometrically the same, the rest of the image becomes dependent on this fact.

However, if you waive the "shapes are identical rule", this is the best solution I've come up with so far.

enter image description here

enter image description here

enter image description here


The Cyan area is basically: $$ \begin{array}{cc} A_{cyan} & = & A_{outer} & - & A_{inner} \\ A_{cyan} & = & 4,000 \times 4,000 & - & 400 \times 400 \\ A_{cyan} & = & 15,840,000 \end{array} $$ So each must be

$$ \begin{array}{cc} A_{section} & = & \frac{15,840,000}{5} \\ A_{section} & = & 3,168,000 \end{array} $$

Its apparent that at least one shape may need an edge from center to edge, so you quickly find a starting rectangle of

$$ \begin{array}{cc} A_{section} & = & \frac{W_{outer}}{2} & - & \frac{W_{inner}}{2} & \times & y \\ A_{section} & = & \frac{4000}{2} & - & \frac{400}{2} & \times & y \\ A_{section} & = & 2000 & - & 200 & \times & y \\ A_{section} & = & 1800 & & & \times & y \\ y & = & \frac{3,168,000}{1800} \\ y & = & 1760 \end{array} $$

From there, it was basically a case of shape-shifting the perimeter while keeping area the same, to achieve a sort of balanced distribution, which is reasonably easy to do.

After you've created 2 shapes with that process, duplicating it was pretty simple, and you're then left with a void which will logically be the same size.

For the usecase we had, having different shapes was acceptable, and this combination of shapes was useful enough. Though somebody can probably find a combination of more similar shapes.

( You'll also see that the arrangement loosely models a pentagon, just with far less regularity )


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