A pentagon puzzle

Question

Consider the following pattern made of regular pentagons:

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

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

The answer is

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 $i = 3$:

Answer 5

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.