# Largest and smallest dodecagon with sides $1, 2, 3, \dots,12$

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?

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

The smallest area I can find:

This has an area = $$40$$

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:

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 $$6:8:10$$.

• 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 – El-Guest Jun 25 at 22:43
• 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 – El-Guest Jun 25 at 22:45
• Ah! I think I can do the same the other other side too. Hang on... that will take 5 minutes or so. – Weather Vane Jun 25 at 22:45
• @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). – Weather Vane Jun 25 at 22:58
• @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. – Weather Vane Jun 25 at 23:20

The largest area I have found:

Area = 378, as indicated

• Hah! you posted this while I was busy editing the similar solution! Well done. – Weather Vane Jun 26 at 10:15

Minimizing area, I present the "snake".

Should be smaller than the others found so far.

Area = 37 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.