# Question on given compass and ruler construction definition wrt. angle bisection

I am trying to understand chapter 4 in John M. Howie's book called "Fields and Galois Theory" (published by Springer). In chapter 4 regarding geometric constructions he gives the following definition:

Let $$B_0$$be a set of points in the plane. There are two permitted operations on the points of $$B_0$$:

(1) (Ruler) through any points of $$B_0$$, draw a straight line;

(2) (Compass) draw a circle whose center is a point in $$B_0$$, and whose radius is the distance between two points in $$B_0$$.

Any point which is an intersection of two lines, or two circles, or a line and a circle, obtained by means of the operations (1) and (2), is said to be constructed from $$B_0$$ in one step. Denote the set of such points $$C(B_0)$$, and let $$B_1 = B_0 \cup C(B_0)$$. We can continue this process, definining $$B_n = B_{n-1} \cup C(B_{n-1})$$ ($$n=1,2,3,...)$$. A points is said to be constructible from $$B_0$$ if it belongs to $$B_n$$ for some $$n$$. A point that is constructible from $$\lbrace O,I \rbrace$$ is said to be constructible.

Phew, that was long. Notice, here the points $$O$$ and $$I$$ are the origon and unit, and we may choose any two points of our initial set to be these. Now, I have attempted a bisection of an angle $$PQR$$, which I have illustrated in the picture below. I tried to convert this bisection to the definition given above. As I see it I need atleast three points in my initial set $$B_0$$ for this. For example, I can find a bisecting line with $$B_0 = \lbrace Q, B, A \rbrace$$ ($$O$$ and $$I$$ would of course be two of these points). If I, however, try with just two, I can not. For example, suppose $$B_0 = \lbrace Q, A \rbrace$$. I can construct the red circle which has point $$B$$ on its exterior. However, to construct $$B$$ it must be an intersection point of my red circle and some other circle or line, given the definition. But to construct a line passing through $$B$$, I need atleast one more point.

1) Is the point $$C$$ in the illustration therefore not constructible, if we follow the definition given by Howie?

Howie later proves that we can not square every circle. I will not state the details on how unless you want me to, but in short, his argument is that you can not construct a square with the radius of a given circle since you can not construct such a square from $$B_0 = \lbrace O, I \rbrace$$.

2) Why is this a sufficient argument? Certainly, you could square a circle given a larger $$B_0$$ (for the trivial case, $$B_0$$ could include $$\pi$$).

I hope this was not too long to devour. Thank you very much for your patience!

• I'm not sure that I understand this, sorry. According to the definition I have provided, for a point to be constructible it must be constructible from the initial set $B_0 = \lbrace O, I \rbrace$. I do not see how we can construct the line QC from $B_0$. To me, atleast, it seems that we require one more point in our initial set $B_0$ to do so. Am I misunderstanding something? – kasp9201 Dec 18 '18 at 2:23
• $B_0=\{Q,B,A\}$ for this construction. – i. m. soloveichik Dec 18 '18 at 13:32
• Thank you very much. That lets me deduce an answer to my first question. Perhaps you can also tell me why Howie would say a point is constructible if it is constructible from $B_0 = \lbrace O, I \rbrace$? ($O$ is the origon, $I$ is the unit) He gives this criteria in form of a definition and therefore never states $\textit{why}$ he does so. – kasp9201 Dec 18 '18 at 15:47
• The constructible points are defined so that they only depend on the simplest set $B_0$. This is used later(?) to get a criterion in terms of field theory for a point to be constructible. – i. m. soloveichik Dec 19 '18 at 15:06
• Any two points would work, $B_0$ is a convenient choice. – i. m. soloveichik Dec 19 '18 at 18:29