To feed your curiosity, today we will look at a compass and straightedge construction of the regular polygon with 15 sides and we will show that there is a connection between this construction problem with the

*measuring liquid puzzle*that we have learned from our previous post.

Measuring liquid puzzle

First, let us review the following brain teaser puzzle from our previous post.

Measuring liquid puzzle.Suppose you have a3-literjug and a5-literjug, how would you measure out exactly1 literof water?

At first glance, it seems that there are a lot of possibilities with the filling and pouring of water using the two jugs. But if we take a step back and have a careful look, we can categorize all the measuring operations into exactly 8 types:

We can see that the amount of water contained in the jugs will initially start with the value $(0,0)$, and after each step of measuring operation, this value $(a,b)$ will change into one of the following 8 possible values $$(0,b), ~~(a,0), ~~(3,b), ~~(a,5),$$ $$(0,a+b), ~~(a+b,0), ~~(a+b-5,5), ~~(3,a+b-3).$$

Therefore, we can easily prove the following fact

The amount of water (in liter) in each of the jug is always a number of the form $3 x + 5 y$ where $x$ and $y$ are two integers.

It means that if we want to measure out

**1 liter**of water then we have to write the value 1 in the form $3x+5y$. So our puzzle is reduced to the problem of solving the following linear Diophantine equation $$3x + 5y = 1.$$

This equation has infinitely many integer solutions. We can easily see that one solution is $(x=2,y=-1)$ and another one is $(x=-3,y=2)$. These two solutions give us the following two representations of the value 1: $${\bf 3} \times 2 - {\bf 5} \times 1 = 1$$ $${\bf 5} \times 2 - {\bf 3} \times 3 = 1.$$

Corresponding to these two representations, we have the following answers to our puzzle:

First solution ${\bf 3} \times 2 - {\bf 5} \times 1=1$: Filling 3-liter jug to the full twice and pouring out to 5-liter jug making it full once, we get 1 liter remaining. |

Second solution ${\bf 5} \times 2 - {\bf 3} \times 3=1$: Filling 5-liter jug to the full twice and pouring out to 3-liter jug making it full three times, we get 1 liter remaining. |

We have seen that our measuring liquid puzzle has a close connection with the Diophantine equation $$3x+5y=1.$$ Soon, we will see that this same Diophantine equation enables us to construct a regular polygon with 15 sides.

Regular polygon construction problem

Let us call a polygon with $n$ sides an

**$n$-gon**. A polygon with equal sides and equal angles is called a

**regular**polygon.

The problem of constructing regular polygons by compass and straightedge is one of a few classic problems of ancient mathematics that are easy to state but incredibly difficult to solve. For a long time, we know how to construct an equilateral triangle ($3$-gon) and a

*regular pentagon*($5$-gon). In his famous books, the

*Elements*, Euclid of ancient Greek presented a method of constructing a regular pentagon. But for the next two thousand years, no one had ever succeeded in constructing a regular $7$-gon, a regular $9$-gon, a regular $11$-gon or a regular $13$-gon. It was not until in 1796, Gauss, only at age 19, made the first breakthrough by successfully constructing a regular $17$-gon. The problem was completely solved in 1837 with the following beautiful theorem

Gauss-Wantzel's Theorem.For an odd number $n$, a regular polygon of $n$ sides can be constructed by compass and straightedgeif and only if$n = p_1 \times \dots \times p_t$, where $p_1$, ..., $p_t$ aredistinctFermat's primes.

**Fermat's primes**mentioned in Gauss' theorem? Here is the definition:

According to this definition:Fermat's prime.A prime number is called a Fermat's prime if it has the form $$F_k = 2^{2^k}+1.$$

- $k=0$, $F_0 = 2^{2^0}+1 = 2^1 + 1 = 3$ is a prime, so it is a Fermat's prime,
- $k=1$, $F_1 = 2^{2^1}+1 = 2^2 + 1 = 5$ is a prime, so it is a Fermat's prime,
- $k=2$, $F_2 = 2^{2^2}+1 = 2^4 + 1 = 17$ is a prime, so it is a Fermat's prime,
- $k=3$, $F_3 = 2^{2^3}+1 = 2^8 + 1 = 257$ is a prime, so it is a Fermat's prime,
- $k=4$, $F_4 = 2^{2^4}+1 = 2^{16} + 1 = 65537$ is a prime, so it is a Fermat's prime,
- $k=5$, $F_5 = 2^{2^5}+1 = 2^{32} + 1 = 4294967297 = 641 \times 6700417$ is a composite number.

Since $3$ and $5$ are Fermat's primes, by Gauss-Wantzel's Theorem, regular $3$-gons and regular $5$-gons are constructible. In addition, since $15=3 \times 5$, regular $15$-gons are constructible. However, there is a condition in the theorem saying that the Fermat's primes have to be

**distinct**, so $9$-gon and $25$-gon are

**not**constructible by compass and straightedge.

Construction of a regular 15-gon

As we have seen above, $15$ is a product of two

**distinct**Fermat's primes (3 and 5), so by Gauss-Wantzel's Theorem, a regular $15$-gon can be constructed by compass and straightedge. But how can we construct it?

Let us look at a picture of a regular $15$-gon. We can see that its vertices form a lot of equilateral triangles and a lot of regular pentagons.

This observation motivates us to construct first an equilateral triangle and a regular pentagon. Let us draw a circle, and pick a point $P_0$ on it. We can construct an equilateral triangle $A_0 A_1 A_2$ and a regular pentagon $B_0 B_1 B_2 B_3 B_4$ with $A_0 = B_0=P_0$. Now, even though we have not constructed a regular $15$-gon fully, but at least we have a part of it!

Can we construct other vertices of the $15$-gon? YES WE CAN!!!

Looking at the above picture, we can see that we can actually construct the whole polygon because we have already had the two sides $A_1 B_2$ and $B_3 A_2$. Using compass with radius $A_1 B_2$ (or $B_3 A_2$), from $P_0$ we can construct $P_1$, from $P_1$ we can construct $P_2$, ..., and the rest of it!

Let us now write down the steps of the construction.

- Construct a circle and pick a point $P_0$ on it.
- Construct an equilateral triangle $P_0 P_5 P_{10}$ and a regular pentagon $P_0 P_3 P_6 P_9 P_{12}$.
- Connect the sides $P_5 P_6$ and $P_9 P_{10}$.
- Draw a circle with center $P_0$ and radius equal to $P_5 P_6$, this circle intersects with the original big circle at $P_1$ and $P_{14}$.
- Draw a circle with center $P_3$ and radius equal to $P_5 P_6$, this circle intersects with the original big circle at $P_2$ and $P_{4}$.
- Other vertices $P_7$, $P_8$, $P_{11}$, $P_{13}$ can be constructed similarly.

Now we know how to construct a regular $15$-gon. Let us take a step back and see why this construction is possible. The crucial part of the construction is the fact that $A_1$ and $B_2$ are next to each other, and also $B_3$ and $A_2$. This fact can be explained by the Diophantine equation $3x+5y=1$.

Indeed, if we consider the whole circle as 15 units then each side of the equilateral triangle contains 5 units and each side of the regular pentagon contains 3 units. If we use the point $P_0$ as the origin and measure the distance along the circle, then the point $A_1$ is at 5 unit, the point $A_2$ is at 10 unit. For the regular pentagon, the point $B_1$ is at 3 unit, $B_2$ 6 unit, $B_3$ 9 unit, and $B_4$ 12 unit.

In general, the point $A_i$ is at $5i$ unit and the point $B_j$ is at $3j$ unit. The distance between $A_i$ and $B_j$ along the circle is $$A_i B_j = |5i - 3j|.$$

$A_i$ is at $5i$ unit, $B_j$ is at $3j$ unit, so the distance between $A_i$ and $B_j$ is $A_i B_j = |5i - 3j|$ |

According to this distance formula, $A_i$ is next to $B_j$ if and only if the distance between $A_i$ and $B_j$ is 1 unit, that is $$5i-3j = \pm 1$$

One of the solution is $i=2$, $j=3$, this gives us $${\bf 5} \times 2 - {\bf 3} \times 3 = 1$$

this explains why $A_2$ and $B_3$ are next to each other and $B_3 A_2$ forms a side of the 15-gon.

Another solution is $i=1$, $j=2$, $${\bf 5} \times 1 - {\bf 3} \times 2 = -1$$ and this explains why $A_1$ is next to $B_2$.

So the two sides $B_3 A_2$ and $A_1 B_2$ of the regular 15-gon correspond to the two integer solutions of the Diophantine equation $3x+5y=1$. How amazing is that!

From here, we can easily have the following generalization

Draw a circle a pick a point $P_0$. Draw a regular $p$-gon $A_0 A_1 \dots A_{p-1}$ and a regular $q$-gon $B_0 B_1 \dots B_{q-1}$ with $A_0 = B_0 = P_0$. If $p$ and $q$ are two relatively prime numbers then there must exist $A_x$ and $B_y$ such that $A_x B_y$ form a side of a regular $pq$-gon. It means that if a regular $p$-gon and a regular $q$-gon are constructible then a regular $pq$-gon is also constructible.

Let us stop here for now. In our future posts, we will look at Gauss' method of constructing regular $17$-gon. Hope to see you again then.

*Homework.*

1. Prove the generalization mentioned at the end of the post:

Let $\gcd(p,q)=1$. If a regular $p$-gon and a regular $q$-gon are constructible by compass and straightedge then a regular $pq$-gon is also constructible.

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