# A pentagon puzzle

Consider the following pattern made of regular pentagons:

If the pattern continued, will it form a complete loop or will the pentagons overlap?

• "If the pattern continues" could mean they continue on a straight line, alternating up and down. In that case the anser is "no", they will not overlap or form a loop. – Florian F Mar 29 at 9:38
• Surely it depends on the ratio of the size (e.g. sidelength) of the pentagon to the circumference of the sphere? – smci Mar 29 at 18:30
• @smci I was about to agree with you in questioning how either answer so far could be right, then I realized the white circle is just our "viewport" not actually a sphere the pattern is on. It's misleading given a green background square on a white background already.. – TCooper Mar 29 at 23:07
• @TCooper: oh. Not a sphere. Too much soccer. – smci Mar 30 at 8:42
• Ha! i knew the answer to this cos my 3 yr old made the very shape out of magformers on my kitchen floor a week or two ago – MooN TreeS Mar 30 at 20:43

It will

form a complete loop consisting of a total of 10 pentagons.
This can be easily seen from the large regular pentagon highlighted in red:

Indeed, its sides are equal lengths by construction and the angle between the sides is that of a regular pentagon as it coincides with the angle of the given small pentagons.

This kind of generalises.

Straight forward for N=4k+1, other N require small adjustments:
N=4k-1 has the polygons point in, not out. The argument can be rescued by considering the inscribed polygon given by edge midpoints.
Even N: Chain closes after N, not 2N, steps.

• Does this work for regular heptagons and nonagons? – Anush Apr 3 at 5:39
• Yes, they are included in the second picture, right two panels. – loopy walt Apr 3 at 8:48

The shapes will

Form a complete loop

Proof

The interior angle of the pentagon is $$\frac{3\pi}{5}$$. This means that the inner angle formed between two adjacent pentagons is $$2\pi - 2\left(\frac{3\pi}{5}\right) = \frac{4\pi}{5}$$.
Hence, as we go around we are forming, in the interior, a regular polygon with regular angle $$\frac{4\pi}{5}$$, which is exactly the description of a decagon.

Image

• I suspect the same applies to any regular polygon, although I'm too busy/lazy to look for a proof right now... – r3mainer Mar 29 at 12:20
• @r3mainer Actually, using the logic above the internal angle gets smaller and smaller so we will stop getting a loop. Already the heptagon won't work but the hexagon (loops around a hexagon) and the octagon (loops around a square) do. The next shape that works is the dodecagon (loops around an equilateral triangle) and then no loops after that. Of course, we can define the loops differently, by putting more sides between adjacent joins, but I don't think that follows the natural generalisation of the problem. – hexomino Mar 29 at 13:07
• A polygon with more sides has a larger internal angle, which means you get a loop with more polygons, surely? I'm pretty sure you can arrange 14 heptagons neatly in a circle. – r3mainer Mar 29 at 14:19
• @r3mainer The angle in the polygon gets larger but the internal angle in the loop gets smaller (sorry, that's what I was talking about). Are you saying that the heptagons surround a 14-gon? Because I don't think the angles add up. – hexomino Mar 29 at 14:23
• No, of course they don't surround a regular 14-gon. I'm just suggesting they form a complete loop. – r3mainer Mar 29 at 14:27

I will try to provide as simple a solution as possible. Hopefully no pictures will be necessary.

It is well known that

the sum of the inner angles of a polygon is $$(180(n-2))^{\circ}$$, where $$n$$ is the number of sides of the polygon.

Therefore,

each of the angles of a regular pentagon is $$108^{\circ}$$. If the pentagons were to form a complete loop, the angle of the inner polygon would need to be $$360^{\circ} - 2\cdot108^{\circ} = 144^{\circ}$$ This means that the pentagons will form a complete loop if and only if the solution of the equation $$144n = 180(n-2)$$ is an integer. You can easily check that the equation is true when $$n=10$$, so the pentagons will indeed form a complete loop, namely a decagon.

• Welcome to Puzzling, and nice alternative answer! Check out our tour and help center while you're here, and feel free to poke around other puzzles or make your own. – bobble Mar 30 at 15:19
• Thank you @bobble this post was showing on the right side bar of stack overflow and I just felt curious about it. I have been taking a look into a few of them and they all seem pretty interesting to me. – Half_Bit Mar 30 at 15:26

Yes, it will form a closed loop.

And I'm going to (at least try to) proof that this generalizes to all regular polygons:

As the other answers have already noted, it's all about the angle $$α$$ between the two touching sides: So the loop won't overlap iff there is a whole number $$k$$ such that $$k*α = 2\pi$$, or $$k = \frac{2\pi}{α}$$

Let's describe the touching sides using diagonals $$d _{i}$$ as follows: Since the diagonals divide the internal angle equally, we can express the inner angle $$α$$ using $$n$$ and the diagonal index $$i$$: $$γ = i * \frac{\pi}{n}$$ $$α = \pi - (2 * γ) = \pi - \frac{2i \pi}{n} = \pi * (1 - \frac{2i}{n})$$ inserting $$α$$ to our previous equation gives: $$k = \frac{2\pi}{\pi * (1 - \frac{2i}{n})} = \frac{2}{1 - \frac{2i}{n}} = \frac{2n}{n - 2i}$$ Let's examine when the right hand side of the equation results in a whole number. The first trivial case is if the denominator becomes $$1$$: $$n - 2i = 1$$ $$i = \frac{n - 1}{2}$$ We see that for every odd $$n >= 5$$, this equation results in a whole diagonal index $$i$$, which is guaranteed to exist because a polygon with $$n$$ sides has $$n-3$$ diagonals and $$\frac{n - 1}{2} <= n - 3$$ for $$n >= 5$$. Since there is a factor of $$2$$ in the numerator, $$k$$ will also be a whole number if the denominator becomes $$2$$: $$n - 2i = 2$$ $$i = \frac{n - 2}{2}$$ This results in a whole diagonal index $$i$$ for every even $$n >= 4$$. Again, $$i$$ is guaranteed to exist because $$\frac{n - 2}{2} <= n - 3$$ for $$n >= 4$$.

So in essence, for every regular polygon with $$n >= 4$$ sides, there exists a diagonal such that a whole number of copies of the polygon touching at the sides depicted by the diagonal can be arranged in a closed loop without overlap.

### Examples

Square

$$n = 4$$, $$i = 1$$, $$α = \pi * (1 - \frac{2}{4}) = \frac{\pi}{2}$$, $$k = \frac{8}{4 - 2} = 4$$

Heptagon

$$n = 7$$, $$i = 3$$

Octagon

The octagon has two solutions, $$i = 2$$ and $$i = 3$$:

• What about a square? – Anush Apr 2 at 9:52
• @Anush added square example. – danzel Apr 2 at 10:16
• Can I check, are you saying it also works for regular heptagon and nonagon? – Anush Apr 3 at 5:38
• @Anush I added some more examples. Yes, it should work for all regular polygons. You can use the formulas to get a diagonal that works, e.g. for n=9, i=(9-1)/2=4, i.e. the fourth diagonal should work, resulting in a loop of 18 nonagons, each rotated by pi/9 around the center point. – danzel Apr 3 at 7:55
• Thank you very much! – Anush Apr 3 at 8:06

There are 180 degrees * (n-sides minus 2) in an n-sided polygon. For a pentagon, each angle is therefore 540 degrees / 5, or 108 degrees

The amount of change in angle for a pair of adjacent pentagons will be (108 + 108) - 180, or 36 degrees.

Take 360 degrees in a circle and it is evenly divisible by 36 degrees, so the completed series of pentagons will consist of 10 pentagons and they will perfectly join up.

• Hi there and welcome to Puzzling SE! This answer was already posted by Half_bit (see above), so there's no need to repost. – Voldemort's Wrath Mar 30 at 20:58
• I didn't look at the other answers. What's the fun in that? – kpkpkp Apr 5 at 23:10
• Once you solved it yourself, there was no need to repost an answer that had already been given. – Voldemort's Wrath Apr 6 at 14:56