15
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Of all dodecagons laying in the cartesian plane, all of whose vertices are lattice points, and whose sides are of length $1, 2, 3, \dots,$ and $12$ in some order, which two have the largest and smallest area?

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  • $\begingroup$ Are degenerate dodecagons (with some of their internal angles being 180°) allowed? $\endgroup$ – trolley813 Jun 25 at 16:37
  • $\begingroup$ Also, what about self-intersecting shapes? $\endgroup$ – user3294068 Jun 25 at 16:41
  • $\begingroup$ No, degenerate dodecagons are not allowed, nor self-intersecting shapes. $\endgroup$ – Bernardo Recamán Santos Jun 25 at 16:54
  • $\begingroup$ Or, is it allowed that some internal angles be zero degrees? $\endgroup$ – user3294068 Jun 25 at 17:05
  • $\begingroup$ No sir, internal angles cannot be 0. $\endgroup$ – Bernardo Recamán Santos Jun 25 at 17:07
14
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The smallest area I can find:

This has an area = $40$

enter image description here

The area of the angled section can be seen to be $ (6 \times 8 / 2) - (4 \times 3 / 2) - (3 \times 2) = 24 - 6 - 6 = 12$
The remainder can be counted with an area of $28$
$ 12 + 28 = 40$

I don't know this is the smallest – it is my smallest.
I found another solution without any angles, area $41$.

The largest area I can find (another edit):

This has an area = $378$

This was a lot more difficult than finding a smallest area.
The improved solution was found by looking for an enclosing rectangle or square that would maximise the area, comprised of the available dimensions. I found the possibilities
$26 \times 16$ (as used in an earlier post)
$25 \times 17$
$21 \times 21$
So I continued with that last one.

I then juggled around the 6 smallest dimensions (apart from $5$ which I wanted on a corner) to find the least area which would be lost by using rectangles as cut-outs, and I found that the smallest area which would be lost is $33$ from those rectangular cutouts.

Along with that are two mitres at the other corners, losing another $24 + 6 = 30$ area. $441 - 33 - 30 = 378$

Finally I juggled around these parts and the four remaining lengths to obtain this:

enter image description here

In detail, $(21 \times 21) - (6 \times 8 / 2)- (3 \times 4 / 2) - (3 \times 4) - (1 \times 7) - (2 \times 6) - (1 \times 2) = $
$ 441 - 24 - 6 - 12 - 7 - 12 - 2 = 378 $
I am fairly sure this is the largest possible - but I may be wrong.

The 5 and the 10 lengths are the only ones that can go diagonally.
They are the hypotenuse of Pythagorean triples $3:4:5$ and $6:8:10$.

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    $\begingroup$ In your biggest, What if you flip your six and seven around? ie. coming down from 9, you make the seven go to the right, then 1 down, then 6 to the left to connect to the 10? That should get you another 14 to make 221 $\endgroup$ – El-Guest Jun 25 at 22:43
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    $\begingroup$ Same thing with your 4,3,2 — connect the five down and left with a 3 going left, then 2 down, then 4 to the right to connect with the 11 — gives you another 14 I think for 235 total $\endgroup$ – El-Guest Jun 25 at 22:45
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    $\begingroup$ Ah! I think I can do the same the other other side too. Hang on... that will take 5 minutes or so. $\endgroup$ – Weather Vane Jun 25 at 22:45
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    $\begingroup$ @El-Guest thanks for that: it gained 6+7 on the right and 6+8 on the left. But it leaves me thinking there should be yet a larger area to be found, as my original approach was wrong (apart from the angles). $\endgroup$ – Weather Vane Jun 25 at 22:58
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    $\begingroup$ @El-Guest following your input I revised the answer completely. This is a bit larger that OP's ordered example, so there could still be room for improvement, e.g. a shape that is more square in its overall dimensions. $\endgroup$ – Weather Vane Jun 25 at 23:20
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The largest area I have found:

Area = 378, as indicated
enter image description here

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    $\begingroup$ Hah! you posted this while I was busy editing the similar solution! Well done. $\endgroup$ – Weather Vane Jun 26 at 10:15
8
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Minimizing area, I present the "snake".

Should be smaller than the others found so far.

Area = 37 the snake polygon The triangular part is a 3x4 triangle for an area of 6, minus two squares at the right angle for a total of 4. The rest are 33 squares.

Generalization of solution:

After the parts 2,3 and 5 are used for the head, and 1 for the end, all the others can can be divided into pairs of a and a+2, and one pair of b and b+1. These pairs can all be steered to either direction, so the tail can be made to not collide with itself. This works for n-gons, where n is divisible by 4.

Daniel Mathias used this generalization for his hexadecagon answer

An alternative snake with the same area:

Area = 37 alternative snake

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    $\begingroup$ Very nice. I was convinced that WV's solution was optimal, and did not put much effort into finding a better minimum area. $\endgroup$ – Daniel Mathias Jun 26 at 19:45
  • $\begingroup$ Vey good. I didn't think mine was optimal but I've been looking in a different direction. $\endgroup$ – Weather Vane Jun 26 at 23:00

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