Divide a line segment by compass

Using compass and straightedge, we can construct the midpoint of a line segment, and we can easily divide a line segment into, say three equal parts. The question is, is it possible to do these constructions with just the compass.

The answer is "yes"! Indeed, it is possible to construct the midpoint of a line segment, and it is possible divide a line segment into a number of equal parts by using the compass alone. How amazing is that?! Today, we will look at these constructions!



Basic compass-and-straightedge constructions

First, let us recall the basic compass-and-straightedge constructions. In our previous post, we learn about these constructions. These constructions are assumed as obvious and can be employed in more complex construction problems:

• Construction of the perpendicular bisector of a line segment and the midpoint

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

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

• Construction of the angle bisector

• Construction of an equal angle to a given angle

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

You can read more about these basic compass-and-straightedge constructions here.

A simple compass-and-straightedge construction problem is how to divide a line segment into a number of equal parts. For instance, suppose we are given a line segment $AB$, using compass and straightedge, how can we divide $AB$ into three equal parts?

The solution is quite trivial: through A construct an arbitrary ray, and on this ray, use compass to subsequently construct arbitrary points $C_1$, $C_2$, $C_3$ so that $AC_1=C_1C_2=C_2C_3$. Connect $BC_3$. Then through $C_1$ and $C_2$, respectively, construct the two lines parallel to $BC_3$ which meet $AB$ at $D_1$ and $D_2$. Finally, we obtain $AD_1 = D_1D_2=D_2B$.



We now move on to consider constructions that use compass but not the straightedge.

Construction by compass alone

There is a special theorem in mathematics called the Mohr-Mascheroni Theorem. This theorem states that any geometric construction which can be done with compass and straightedge can also be done with compass alone. That means that the use of straightedge is actually not necessary.

Mohr-Mascheroni Theorem. Any point can be constructed by compass and straightedge can be constructed by compass alone.

By Mohr-Mascheroni Theorem, it is then possible to use compass alone to multiply and divide a line segment.


Multiply a line segment by compass alone

Problem. Given two points A and B, only using compass, construct a point C on AB so that $$AC = AB \times 3.$$

If we construct two equal circles with radius AB and center at A and B, respectively, then they will intersect at two points X and Y. We obtain two equilateral triangles ABX and ABY. If we keep doing like this then we can construct a lattice of equilateral triangles. Thus, we can multiply a line segment easily.

Divide a line segment by compass alone

Problem. Given two points A and B, only using compass, construct a point D on AB so that $$AD = AB / 3.$$

We can see that the two points C and D are kind of inverse of one another. On one hand, we have $$AC = AB \times 3$$ and on the other hand, we have $$AD = AB / 3$$

This is called the inversion operation. We will learn more about this inversion operation in the future. 

Because of this inversion, we have $$AC \times AD = AB^2$$

In geometry, inversion equation like the above is often derived from similar triangles.

If P is an arbitrary point such that AP = AB then the two triangles ADP and APC are similar triangles. This is because the two triangles share the same angle A and they have equal side ratios $$\frac{AD}{AP} = \frac{AP}{AC}.$$

Since P is quite arbitrary, we will choose a convenient position for P. We will choose P so that the two triangles ADP and APC become isosceles triangles. That is $$PA = PD, ~~~~~~ CP = CA$$

Since $AP = AB$ and $CP = CA$, the point P is an intersection point of the circle centered at A with radius AB and the circle centered at C with radius CA. It means that we can construct this point P by compass alone.

Furthermore, since PD = PA = AB, the point D lies on the circle centered at P with radius equal to AB. By symmetry, we can construct the point D. The construction is as follows:

• Construct the circle centered at A with radius AB and the circle centered at C with radius CA. The two circles intersect at P and Q.

• Construct the circle centered at P with radius equal to AB and the circle centered at Q with radius equal to AB. The two circles intersect at A and D.

So by using compass alone we indeed can construct the point D to divide the line segment into three equal parts. Let us stop here for now. Hope to see you again next time.

Homework.

1. Given three points A, B, and C, only using compass, construct the circumcircle of the triangle ABC.

2. Given three points A, B, and C, only using compass, construct the incircle of the triangle ABC.

3. Go to google.com and search about the geometric inversion operation.