In our previous post, we have learnt about

*Similar Triangles*and the

*Right Triangle Altitude Theorem*. Today, continuing our journey in the garden of geometry, we want to find answer to the following question

Given a line segment of length $r$, by compass and straightedge, what kind of shapes can we construct?

There are two simple cases that we can see straightaway.

- We can construct a line segment whose length is a multiple of a given length, and
- We can divide a given line segment into a number of equal parts.

Here are some examples

- Given a line segment $AB=r$, by compass and straightedge, we can easily construct a line segment $AC = 3r$.

- Given a line segment $AB=r$. Construct an arbitrary ray $Ax$ and on this ray, construct $C_1$, $C_2$, $C_3$, $C_4$, $C_5$ such that $$A C_1 = C_1 C_2 = C_2 C_3 = C_3 C_4 = C_4 C_5.$$ Through $C_1$, $C_2$, $C_3$, $C_4$, construct four lines parallel to $B C_5$ which intersect with $AB$ at $D_1$, $D_2$, $D_3$, $D_4$. Then we have $$A D_1 = D_1 D_2 = D_2 D_3 = D_3 D_4 = D_4 B = \frac{r}{5}$$

- Given a line segment $AB=r$, first construct $AC = 3r$ and then divide $AC$ into 5 equal parts then we have a line segment of length $\frac{3}{5} r$.

From these simple examples we deduce the following conclusion

Given a line segment of length $r$, and $\frac{p}{q}$ is a rational number, then by compass and straightedge, we can construct a line segment of length $\frac{p}{q} r$.

Let us now look at two examples which are a bit more sophisticated: construction of a regular pentagon and construction of a regular polygon with 17 sides.

To construct a regular pentagon, first we construct a circle and pick a point $P_0$ on it. If we can construct the point $H$ then from $H$ we can construct $P_1$, and from $P_1$ we can construct other vertices of the pentagon.

Suppose that $r$ is the radius of the circle, we have the following trigonometric identity

$$\angle P_0 O P_1 = \frac{2 \pi}{5}$$ $$OH = r \cos{\frac{2 \pi}{5}} = r \frac{\sqrt{5}-1}{4} $$

So, in order to construct the pentagon, we need to construct the line segment $OH$ of length $\frac{\sqrt{5}-1}{4} r$.

Similarly, in order to construct the regular 17-gon in the picture below, we need to construct the line segment $OH = r \cos{\frac{2 \pi}{17}}$.

In 1796, at the age of 19, the mathematician Gauss made a breakthrough in solving the regular polygon construction problem. Using modulo arithmetic, Gauss calculated

$$\cos{\frac{2 \pi}{17}} = \frac{1}{16} \left( -1 + \sqrt{17} + \sqrt{34 - 2 \sqrt{17}} + 2 \sqrt{17 + 3 \sqrt{17} - \sqrt{34 - 2 \sqrt{17}} - 2 \sqrt{34 + 2 \sqrt{17}}} \right)$$

and showed that a regular 17-gon is constructible.

With the above two examples, have you spotted out any pattern yet?

Constructible numbers

Let us now define the concept of "constructible numbers"

A number $\alpha$ is called aconstructible numberif $\alpha$ is obtained from integers by addition, subtraction, multiplication, division and taking square root.

For example,

$$\frac{3}{5} \mbox{ is a constructible number,} $$

$$\cos \frac{2 \pi}{5}=\frac{\sqrt{5}-1}{4} \mbox{ is a constructible number,} $$

$$\frac{\sqrt{\sqrt{2} + 5 \sqrt{3} - 2}}{\sqrt{\frac{2}{5}} + 1} \mbox{ is a constructible number.} $$

We have the following theorem

Fundamental theorem of compass-and-straightedge construction.Given a line segment of length $r$, and $\alpha$ is aconstructible number, then by compass and straightedge, we can construct a line segment of length $\alpha \, r$.

We will prove this theorem by showing a construction algorithm.

Construction algorithm

Given three line segments of length $r$, $a r$ and $b r$. Using compass and straightedge, we can construct

- a line segment of length $(a+b)r$

- a line segment of length $(a−b)r$

- a line segment of length $(ab) r$

$AB$ is parallel to $CD$: $\frac{OD}{OC} = \frac{OA}{OB} = a \to OD = abr$ |

- a line segment of length $(a/b) r$

$AB$ is parallel to $CD$: $\frac{OD}{OC} = \frac{OA}{OB} = \frac{a}{b} \to OD = \frac{a}{b}r$ |

- a line segment of length $(\sqrt{ab}) r$

by the Right Triangle Altitude Theorem, $HC^2 = HA \times HB = ab r^2 \to HC = \sqrt{ab}r$ |

With the above construction algorithm, if $\alpha$ is a

*constructible number*then from a line segment of length $r$, we can show step by step how to construct a line segment of length $\alpha r$.

Let us stop here for now. Next time, we will learn about

*Fermat's little theorem*,

*modulo cycle*and many other exciting stuff and then we will learn how to derive Gauss trigonometric identity

$$\cos{\frac{2 \pi}{17}} = \frac{1}{16} \left( -1 + \sqrt{17} + \sqrt{34 - 2 \sqrt{17}} + 2 \sqrt{17 + 3 \sqrt{17} - \sqrt{34 - 2 \sqrt{17}} - 2 \sqrt{34 + 2 \sqrt{17}}} \right)$$

Hope to see you again then!

*Homework.*

1. Prove that $$\cos \frac{2 \pi}{5}=\frac{\sqrt{5}-1}{4}$$

2. Use Moivre formula to find a polynomial of integer coefficients such that $\cos \frac{2 \pi}{17}$ is one of its roots.

3.

*We will use the notation $(PQ)$ to denote the line $PQ$, and notation $(P, PQ)$ to denote the circle centre at $P$ and of radius $PQ$.*

**Fundamental theorem of compass-and-straightedge construction.**

Given a line segment $AB = r$. A point $M$ on the plane is said to be

**constructible**from $AB$ if there exists a sequence of points $X_1$, $X_2$, ..., $X_n$ such that

- $X_1 = A$, $X_2 = B$ and $X_n = M$
- For any $3 \leq i \leq n$, the point $X_i$ is either the intersection point of two lines $(X_{k_1} X_{k_2})$ and $(X_{k_3} X_{k_4})$, or an intersection point of the line $(X_{k_1} X_{k_2})$ with the circle $(X_{k_3}, X_{k_3} X_{k_4})$, or an intersection point of the circle $(X_{k_1}, X_{k_1} X_{k_2})$ with the circle $(X_{k_3}, X_{k_3} X_{k_4})$, for some $1\leq k_1, k_2, k_3, k_4 < i$.

Draw a Cartesian coordinate system $Axy$ where the point $B$ is on $Ax$. Each point $M$ on the plane will have a coordinate $M(x,y)$.

Prove that a point $M(x,y)$ is

*constructible*from $AB$

**if and only if**$x = \alpha_1 r$ and $y = \alpha_2 r$ where $\alpha_1$ and $\alpha_2$ are

*constructible numbers*.

4. Write about one of your favourite construction problems. Use google.com to learn more about that problem.

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