Today we will start a series of posts on

**compass-and-straightedge construction**. Believe it or not, there are a few construction problems that sound simple but it had required more than two thousand years to settle! The most famous ones are the problem of

**regular polygon**construction and the problem of

**angle trisection**. These problems were known in ancient times but it was not until the late 18th-19th centuries that mathematicians could finally solve them using the most modern tools of mathematics.

In this first post, we will learn about some

*basic compass-and-straightedge constructions*. These constructions are assumed as obvious and can be employed in more complex construction problems.

Basic constructions

**Construction of the perpendicular bisector of a line segment**

Given a line segment $AB$. To construct theperpendicular bisectorof $AB$, we follow the steps below:

- Construct two circles of
equal radiuswith centres at $A$ and $B$ so that they intersect at two points.

- The line connecting the two intersection points is the perpendicular bisector of $AB$.

**Construction of the midpoint of a line segment**

Given a line segment $AB$. To construct themidpointof the line segment $AB$, we follow the steps below:

- Construct the perpendicular bisector of $AB$.

- The perpendicular bisector meets $AB$ at the midpoint $M$ of $AB$.

**Through a point, construct a line perpendicular to a given line**

Given a line $\ell$ and a point $A$. To construct a line passing though $A$ and perpendicular to $\ell$, we follow the steps below:

- Construct a circle with centre $A$ so that it intersects with the line $\ell$ at two points $B$ and $C$.

- Construct the perpendicular bisector of $BC$, this is the line passing through $A$ and perpendicular to $\ell$.

**Through a point, construct a line parallel to a given line**

Given a line $\ell$ and a point $A$. To construct a line passing though $A$ and parallel to $\ell$, we follow the steps below:

- Construct the line $t$ passing through $A$ and perpendicular to $\ell$.

- Construct the line $u$ passing through $A$ and perpendicular to $t$, the line $u$ is the line that passes through $A$ and is parallel to $\ell$.

**Construction of the angle bisector**

Given an angle $\angle xOy$, to construct theangle bisectorwe follow the steps below:

- Construct a circle with centre $O$ that intersects $Ox$ and $Oy$ at $A$ and $B$, respectively.

- Construct the perpendicular bisector of $AB$, this is the required angle bisector of $\angle xOy$.

**Construction of an equal angle to a given angle**

Given an angle $\angle xOy$ and a ray $A \ell$, to construct an angle equal to $\angle xOy$ that has $A \ell$ as one side, we follow the steps below:

- Take a point $B$ on the ray $A \ell$.

- Construct the circle with centre $O$ and radius equal to $AB$, this circle meets $Ox$ and $Oy$ at $D$ and $C$, respectively.

- Construct the circle with centre $A$ and radius $AB$, and construct the circle with centre $B$ and radius $CD$, these two circles intersect at $E$ and $F$.

- The two angles $\angle EA \ell$ and $\angle FA \ell$ are both equal to $\angle xOy$.

**Through a point, construct a tangent line to a circle**

Given a circle with centre $O$ and a point $A$ lying outside of the circle, to construct a line passing though $A$ and tangent to the circle $(O)$, we follow the steps below:

- Construct the midpoint $B$ of $OA$.

- Construct the circle with centre $B$ and radius equal to $AB$, this circle intersects with the circle $(O)$ at two points $C$ and $D$.

- The two lines $AC$ and $AD$ are the two required tangent lines to the circle $(O)$.

Some construction problems

**Example 1.**Given a line segment $AB$. Use compass and straightedge to divide this line segment into five equal parts.

**Solution:**

- Through $A$ construct a ray, and on this ray, use compass to subsequently construct the points $C_1$, $C_2$, $C_3$, $C_4$, $C_5$ so that $AC_1 = C_1C_2=C_2C_3=C_3C_4=C_4C_5$.
- Connect $BC_5$.
- Construct the four lines passing through $C_1$, $C_2$, $C_3$, $C_4$, respectively, and parallel to $BC_5$ which meet $AB$ at $D_1$, $D_2$, $D_3$, $D_4$.
- We have $AD_1 = D_1D_2=D_2D_3=D_3D_4=D_4B$.

**Example 2.**Given an angle $xOy$ and a point $M$. Use compass and straightedge to construct a point $A$ on $Ox$ and a point $B$ on $Oy$ so that $M$ is the midpoint of $AB$.

**Solution:**

- Draw the line $OM$ and use compass to construct the point $N$ on $OM$ such that $OM=MN$.
- Construct a line passing through $N$ and parallel to $Oy$ which meets $Ox$ at $A$.
- Construct a line passing through $N$ and parallel to $Ox$ which meets $Oy$ at $B$.
- $OANB$ is a parallelogram so the midpoint $M$ of the diagonal $ON$ is also the midpoint of the diagonal $AB$.

**Example 3.**Given a triangle $ABC$. Use compass and straightedge to construct a square $PQRS$ so that the vertex $Q$ lies on the side $AB$ of the triangle, the vertex $R$ lies on the side $AC$, and the two vertices $P$, $S$ lie on the side $BC$.

**Solution:**

- Take a point $U$ on $AB$.
- Construct the line $UV$ perpendicular to $BC$.
- Use compass to construct the point $F$ on the ray $VC$ so that $VF=VU$.
- Construct the square $UVFE$.
- Construct the intersection point $R$ of the two lines $BE$ and $AC$.
- Construct the line $RS$ perpendicular to $BC$.
- Use compass to construct the point $P$ on the ray $SB$ so that $SP=SR$.
- Construct the square $PQRS$.

Let us stop here for now. We will continue this topic of compass-and-straightedge construction in the next post. Hope to see you again there.

*Homework.*

1. Give compass-and-straightedge constructions of an

*equilateral triangle*, a

*square*, a

*regular hexagon*(6 sides), and a

*regular octagon*(8 sides).

2. Given two circles, use compass and straightedge to construct all

**common tangents**to these two circles.

3. Given two line segments with lengths $a$ and $b$, respectively, use compass and straightedge give a construction of a line segment of length $\sqrt{ab}$.

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

and use this formula to derive a compass-and-straightedge construction of a regular pentagon.

5. Given a quadrilateral $ABCD$ and four points $M_1$, $M_2$, $M_3$, $M_4$, in this order, on $AB$. Using compass and straightedge, show how to construct four points $N_1$, $N_2$, $N_3$, $N_4$ on $CD$ so that the four line segments $M_1 N_1$, $M_2 N_2$, $M_3 N_3$ and $M_4 N_4$ divide the quadrilateral into 5 small pieces of

**equal area**.

6. Given two points $A$ and $B$, only use compass (

**), show how to construct four points $D_1$, $D_2$, $D_3$, $D_4$ on the line segment $AB$ so that they divide the line segment into five equal parts.**

*straightedge is not allowed*
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