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By famous Mohr-Mascheroni theorem

Every geometric construction that can be carried out by compass and straightedge can be done with the compass only (without a straightedge).

To say in short, to prove the theorem, we have to prove that the three following constructions can be done with only compass:

  1. Points of intersection of two circles given by center and one of the points for each circle
  2. Points of intersection of a circle (given by center and one of its points) and a straight line (given by two points).
  3. Point of intersection of two straight lines each of them given by two points.

I was reading "A short elementery proof of the Mohr-Mascheroni Theorem" by Norbert Hungerbuhler.

But it seems to me that the autor uses transport of the measure by the compass.

I suspect that we can avoid the usage of transport of measure by compass in the proof of Mohr-Mascheroni theorem. That is I do believe that every point constructible by collapsing compass and a straightedge can be constructed by means of collapsing compass only. But unfortunately I still find myself unable to do that.

P.S. It seems to me that despite the comments below, the construction in the Problem 4 of the book of Kostovskiy mentioned in the answer by @saltandpepper uses the measure tramsport as well [constructing the circles $(O,a)$, $(C, OE)$, $(D, OE)$].

Evgeny Kuznetsov
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  • This may answer your question about "measurement by means of compasses": https://math.stackexchange.com/questions/2682519/euclidean-proposition-8-of-book-i/2682526#2682526 – Ethan Bolker Mar 27 '18 at 17:42
  • @Ethan Bolker I can not see how does your comment help if we have nothing but collapsing compass (i.e we have no ruler) – Evgeny Kuznetsov Nov 21 '21 at 22:41
  • The answer I point to only that a collapsing compass can do anything a rigid one can - in particular, transport a measure on a line. It says nothing about rulers. (You should say "straightedge" not "ruler" since rulers have markings. Straightedges just join points.) – Ethan Bolker Nov 21 '21 at 22:41
  • Ok, sorry about the inconsistent terminology. Ofcourse I mean the straightedge. – Evgeny Kuznetsov Nov 21 '21 at 22:44
  • But anyway the answer you are pointing to says that collapsing compass and straightedge is equivalent to rigid one and straightedge. But what I seek to understand is that if they are equivalent without the straightedge. – Evgeny Kuznetsov Nov 21 '21 at 22:58

2 Answers2

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Why do you say that they require measurements by compass?

Is it because of the reflections? Observe that you can construct the reflection of $M$ with respect to the line passing through $P_1$ and $P_2$, by drawing the circle with center $P_1$ and radius $P_1M$ and drawing the circle with center $P_2$ and radius $P_2M$ you get two intersection points $M$ and $M'$.

Besides constructing reflections (which they claim but I didn't see proven) I didn't see any other step that can be seen as using measurements.

The midpoint construction is explicitly there on page $2$. Did I miss some other instance?

Doubling is also OK. Take into account that necessarily one should be allowed to, given two points, place the needles of the compass on each of them, otherwise we wouldn't be able to draw the circle with center on one of them and passing through the other.

This Mir book also has it: Geometrical Constructions Using Compasses Only by A. N. Kostovskii. The link has the translation in Spanish.

The first and second constructions in Kostovskii's book are precisely the reflection of a point, and multiplying a segment by an integer number.

saltandpepper
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  • Well, Euclid's compass could only draw a circle given its center AND a point of the circle. Using a compass to transport a measure was not allowed and Euclid gave a specific construction for that. – Intelligenti pauca Mar 27 '18 at 17:44
  • @Aretino There is no transportation used in the construction. At least in the book, let me double check the OP's paper. – saltandpepper Mar 27 '18 at 17:47
  • @Aretino It looks like they do assume it possible in construction 4 in the OP's paper, without proving it. But well, maybe they skipped as they skipped the reflection. – saltandpepper Mar 27 '18 at 17:52
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    How do they construct circle $K'$ in Construction 1? – Intelligenti pauca Mar 27 '18 at 19:41
  • @EvgenyKuznetsov It is true that the paper leaves some things to the reader. But it is not hard to work around that problem. We already know how to duplicate a segment just with compass and no length transportation (right?). Then we can get $B'$ on $BM$ such that $BB'$ is double of $BM$. Draw the circle with center $B$ and passing through $B'$. Then do the same with $AM$. The intersections of those two circles gives you such a point $M_1$. – saltandpepper Mar 27 '18 at 23:19
  • @EvgenyKuznetsov In case you prefer a reference instead of completing the argument in the paper you linked: The book I linked has an alternative construction if the intersection of a circle and a line passing through its center, in Problem 6, Figure 8. The construction is based on dividing arcs of a circle in half, which is done in Problem 4. – saltandpepper Mar 27 '18 at 23:27
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The obstacle that bound me can be elininated by the elegant equivalence between two kinds of compasses a "rigid" one and a collapsing "divider".

Suppose we want to build a circle with center $A$ and the radius $BC$. Then we do the following: first draw blue circles, then red ones, then resulting green one; as it is shown below: [source: Wikipedia] (https://i.stack.imgur.com/aA1CA.png) Source of the answer is the underrated comment on my other old quaestion: https://math.stackexchange.com/a/2933016/239005

Evgeny Kuznetsov
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