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I'm currently reading Antoine Chambert-Loirs "A field guide to algebra". The very start of the book is dedicated to the topic of Construction with ruler and compass.

The most import definition used:

Let $\Sigma$ be a set of points in the plane $R^2$. One says that a point $P$ is constructible with ruler and compass from $\Sigma$ if there is an integer $n$ and a sequence of points $(P_1,...,P_n)$ with $P_n = P$ and such that for any $i \in {1,...,n}$, denoting $\Sigma_i := \Sigma \cup \big\{ P_1,...,P_{i−1} \big\} $, one of the following holds:

  • there are four points $A, B, C$ and $D \in \Sigma_i$ such that $P_i$ is the intersection point of the two nonparallel lines $(AB)$ and $(CD)$;

  • there are four points $A, B, C$, and $D \in \Sigma_i$ such that $P_i$ is one of the (at most) two intersection points of the line $(AB)$ and the circle with center $C$ and radius $CD$;

  • there are four points $O, M, O$ and $M \in \Sigma_i$ such that $P_i$ is one of the (at most) two intersection points of the distinct circles with, respectively, center $O$ and radius $OM$, and center $O$ radius $OM$ .

Note: The definition above does not include the creation of a new point via the intersection between the circle of radius $AB$ around a (from $A, B$ distinct) point $C$ and a line/different circle.

The author asks the reader to explain, how this type of creation of new points is possible, with just the three given moves.

At this point I'm stuck and have no clue how to construct these kind of points. Any ideas?

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  • $\begingroup$ I think the author is asking you how to set the compass and strike a circle about the center $C$. $\endgroup$ – Lubin Mar 8 '17 at 23:35
  • $\begingroup$ @Lubin Yes, I thought my question was clear enough, sorry. I have no idea how to do this, though. $\endgroup$ – user326377 Mar 8 '17 at 23:36
  • $\begingroup$ @Herickson You gave the definition for one point to be constructible, but are asking for a proof that a circle is constructible. Unless you mean to solve it by proving that each of the infinitely many points on the circle is constructible, you'll need some additional definitions and/or assumptions. $\endgroup$ – dxiv Mar 9 '17 at 0:03
  • $\begingroup$ @dxiv Sorry, I'll edit my answer. What I mean is to prove, that if you can construct a point via the intersection of an circle of radius $AB$ around a distinct center $C$ with a line $(D, E)$ - same with a second circle instead of a line, you can also construct this point with the three rules above. $\endgroup$ – user326377 Mar 9 '17 at 0:09
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For short, the allowed moves are

  1. Intersecting two lines through two pairs of given points;
  2. Intersecting a line through two given points with the circle through a third point with a known center;
  3. Intersecting two circles with known centers through two given points.

We may use two moves of type-3 to draw the midpoint and the perpendicular bisector of a segment, hence four moves of type-3 and one move of type-1 are enough to find the circumcenter and circumcircle of a triangle. Moreover, we may find the symmetric of $C$ with respect to the midpoint of $AB$ with two moves of type-3, one move of type-1 and one move of type-2:

  • We consider the circles centered at $A$ and $B$ with radius $AB$: their radical axis meet $AB$ at its midpoint $M$;
  • The $CM$ line and the circle with center $M$ and radius $CM$ meet at $C$ and at the symmetric of $C$ with respect to $M$, say $D$.

$ADBC$ is a parallelogram, hence the circle with center $A$ and radius $AD$ solves the problem.

enter image description here

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  • $\begingroup$ That's what I was looking for. Thank you! $\endgroup$ – user326377 Mar 9 '17 at 2:02
  • $\begingroup$ @Herickson: you're welcome. $\endgroup$ – Jack D'Aurizio Mar 9 '17 at 2:02

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