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Compass-and-straightedge construction


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 the perpendicular bisector of $AB$, we follow the steps below:
  • Construct two circles of equal radius with 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 the midpoint of 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 the angle bisector we 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 (straightedge is not allowed), 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.






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