Recently I asked a question about Mohr-Mascheroni theorem. I read the paper "A short elementary proof of the Mohr-Mascheroni Theorem" by Norbert Hungerbuhler but was unsatisfied with it.

In Construction 1 the author constructs intersection points of a circle with center $M$ and line non-passing a center of the circle given by two points.

The Construction 1 is as follows:

If the straight line $g$ is given by the points $P_1$ and $P_2$ we reflect the center $M$ of given circle $K$ with respect to $g$. (It is done by means of two circles one with center in $P_1$ and second with center in $P_2$ through $M$) Then we find the two points of intersection $\{X,Y\}=K\cap g$ as the point of intersection of $K$ and the reflected circle $K'$.

enter image description here

As the user @Aretino indicated in comment on the recent question Euclid's compass could only draw a circle given its center AND a point of the circle, which was my point when I asked the recent question about an reference of alternative proof.

So my question is: How to construct $\{X,Y\}$ as above with "Euclid's compass"?

I can't understand if construction of reflected circle is carried out in a correct way in the construction mentioned above, we have a center $M'$ of reflected circle, but haven't a point constructed to draw it through.

May be I misunderstood a Mohr-Mascheroni theorem, may be it is not about an "Euclide's compoass"?


I agree that the construction is unclear, though it is easily fixed. We just need to reflect any point on circle $K$ through $g$. For example:

  1. The point $Q$ is already constructed as the intersection of $K$ with the circle with centre $P_1$ through $M$.
  2. Reflect $Q$ through $g$ to $Q'$ using the already given point reflection construction. (Circles with centres $P_1$ and $P_2$ through $Q$, not shown in diagram.)
  3. Construct $K'$ with centre $M'$ through $Q'$.

Image of amended construction

Alternatively, it is already known that a circle with a given radius can be reconstructed with a different centre using only a collapsing compass. See Wikipedia.

  • $\begingroup$ Oh, I see, a reflection of Q does what was needed. Thank you wery much. $\endgroup$ – Evgeny Kuznetsov Sep 27 '18 at 14:47

Just follow the numbered steps in order. I think every step is legitimate.

enter image description here


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