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Given a circle with diameter AB, how can we determine the center of the circle with a straightedge (we cannot measure lengths, cannot measure angles, or draw parallel lines,... We can only draw straight lines)?

I received this problem as homework from my professor. In fact, we were finding a way to apply a trapezoid theorem(!) to the problem in which we have to construct a line through any point C on a circle perpendicular to its diameter AB with only a straightedge. That problem was indeed doable. The professor asked me, as an advance exercise, to try using the theorem to find the center of a circle with A and B being two given points on the circle and segment AB being the diameter.

(!) Given a trapezoid, the straight line joining the point of intersection of its diagonals and the point of intersection of its non-parallel sides bisects each of the parallel sides.

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  • $\begingroup$ Can the straightedge take any shape, so long as every segment is straight? Can the straight edge be a straight corner? $\endgroup$
    – user110391
    Commented Jul 19, 2022 at 2:11
  • $\begingroup$ See "Mathematical Snapshots" by Hugo Steinhaus. The same idea applies to the closely related problem above (the OP's question). $\endgroup$
    – Wlod AA
    Commented Jul 19, 2022 at 11:27
  • $\begingroup$ Does "straightedge" mean "idealized straightedge" as described in the Wikipedia article as including: Always assumed to be without graduations or marks, or the ability to mark? If so, please include that specifically in the question, as non-mathematicians (like me) are unclear about that, and since this has hit HNQ, I'm likely not the only non-mathematician to read this question. $\endgroup$
    – manassehkatz-Moving 2 Codidact
    Commented Jul 19, 2022 at 14:15
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    $\begingroup$ I am voting to reopen. It's clear from the answers that this is an interesting question. $\endgroup$
    – Steven Landsburg
    Commented Jul 20, 2022 at 12:20

4 Answers 4

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I do not think that it is possible.

Consider a projective map $f$ which preserves the circle; maps the center $O$ to a point $P\ne O$; and maps some point $Q$ to $O$. The line $QO$ is mapped to the line $OP$. So, there is a rotation $g$ with respect to $O$ such that $g(f)$ maps the line $QO$ to itself, although $O$ not to itself. It yields that given a circle and a line through $O$, you can not find the center by using straightedge only, since such construction is projective-equivariant.

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    $\begingroup$ Explicitly, if the diameter is $y=0$ and the circle is $x^2+y^2=z^2$, rewrite the latter as $y^2 = z^2-x^2 = (z-x)(z+x)$, and use projective linear transformations $T_c$ that fix $y$ and take $(z-x,z+x)$ to $(c^{-1}(z-x). c(z+x))$ for some nonzero $c$. $$ $$ The existence of such $T_c$ is not surprising: the projective linear group ${\rm PGL}_3({\bf R})$ has dimension $3^2-1 = 8$, while conics and lines vary over spaces of dimension $5$ and $2$ respectively, so we expect that a typical conic and line will have a stabilizer of dimension $8-(5+2) = 1$, which is what we found. $\endgroup$
    – Noam D. Elkies
    Commented Jul 18, 2022 at 14:31
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    $\begingroup$ @NoamD.Elkies strangely your comment comes up without the little upvote arrow (either in mobile or laptop). Oh well, +1 anyway! $\endgroup$
    – Yaakov Baruch
    Commented Jul 19, 2022 at 9:55
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    $\begingroup$ @YaakovBaruch I think it's just that there are three lines of whitespace in Noam's comment, so that what looks like two comments is in fact one. $\endgroup$
    – James Martin
    Commented Jul 19, 2022 at 10:02
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Edit: The following answer was written assuming that the ambiguous phrase “given a circle with diameter $AB$” in the question is to be interpreted as “given a circle having diameter $AB$” rather than “given a circle and given one of its diameters $AB$”. Assuming the latter interpretation, the argument can be modified by considering the isometries of the hyperbolic plane (represented through its Klein model) which fix two ideal points, where the circle is seen as the set of ideal points.

It is a classical theorem due to Jacob Steiner and David Hilbert that it is not possible to construct the center of a circle with straightedge alone: the classical argument is essentially that stated by Fedor Petrov in his answer; see Cut the Knot for an explanation, and Shen - Hilbert's error? for a criticism of the way this argument is (sometimes carelessly) formulated.

Of related interest is the Poncelet–Steiner theorem which states essentially that given a single circle and its center anywhere in the plane, any compass-and-straightedge construction can be performed with straightedge only.

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    $\begingroup$ The OP has however a circle and a diameter, so it's not clear that this applies. $\endgroup$
    – godelian
    Commented Jul 18, 2022 at 15:49
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Please refer to Chapter 18 of "The Ruler in Geometrical Constructions" by Smogorzhevskii. There the author provides a provides a proof of sorts that in general where you have two(!) circles, you can't construct their midpoints using only a ruler. If you can't do it in general with two, you certainly can't do it with a single given circle and only a ruler.

There are cases where you can construct the midpoints of two circles, and one of these is two given intersecting circles. See chapter 19 of the same book.

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    $\begingroup$ You are not given just a circle, you are given a circle and one of its diameters. $\endgroup$
    – Emil Jeřábek
    Commented Jul 19, 2022 at 9:28
  • $\begingroup$ @EmilJeřábek True, but that doesn't really matter, does it? As the straight edge as posited by the OP can't measure length. So, there is no way to transfer that information by construction under the constraints of the given problem by the OP. $\endgroup$
    – nanitous
    Commented Jul 20, 2022 at 8:38
  • $\begingroup$ @Peter Mortensen, thanks for the edit. Learning every day as a non-native speaker. $\endgroup$
    – nanitous
    Commented Jul 20, 2022 at 8:39
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    $\begingroup$ I don't see how measuring lengths is relevant. You are given two more points to work with, so in principle the problem may be easier. I don't see a way of a priori ruling out that it helps. $\endgroup$
    – Emil Jeřábek
    Commented Jul 20, 2022 at 9:51
  • $\begingroup$ @EmilJeřábek Ah, wait. I now realise that the diameter is given as a segment. Thank you for pointing that out. But Poncelet-Steiner constructions are still not possible $\endgroup$
    – nanitous
    Commented Jul 21, 2022 at 11:01
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Coming at this from a total non-mathematician (found this on HNQ). So no fancy proofs (last did any of those in the mid-1980s).

  • Do I start with the diagram (circle with one line already marked)? (Not clear, but some comments seem to indicate that is the case.)

  • Can I mark things ("only draw straight lines" - but can I draw some short straight lines on the straightedge?)?

If both are true then:

  • Put the straightedge across AB
  • Mark A and B on the straightedge
  • Move the straightedge around (angle doesn't matter) and put the A mark on the edge of the circle (we'll call this spot C)
  • Rotate the straightedge until the B mark is on the opposite side of the circle (we'll call this spot D)
  • Draw a line CD
  • The intersection of AB and CD is the center

Or am I vastly oversimplifying or cheating or somehow missing the point? (And if any of those are going on, I will delete this answer.)

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    $\begingroup$ There's no way to find "the opposite side of the circle" of a point. $\endgroup$
    – Sam Hopkins
    Commented Jul 19, 2022 at 13:55
  • $\begingroup$ If you mark A and B on a straightedge and put A on one edge of the circle (call that C), the B marking will be outside the circle except if you have the straightedge positioned on another diameter. So you rotate around C until the edge of the circle is at the other marking on the straightedge. No measurement (AB = 17.3cm or whatever) needed because as you move it you can see if the distance between the marking and the edge of the circle is getting bigger or smaller - the point where it gets larger whether you move to the right or left is where you are at another diameter. $\endgroup$
    – manassehkatz-Moving 2 Codidact
    Commented Jul 19, 2022 at 14:01
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    $\begingroup$ The word straightedge seems to imply a physical object that we can mark things on. Depends really on what the situation is. It seems like it could go both ways. $\endgroup$
    – IsawU
    Commented Jul 19, 2022 at 14:07
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    $\begingroup$ @manassehkatz-Moving2Codidact: It is known among mathematicians (but you mentioned that you are not one) that straightedges (or "rulers") have no markings. In the meantime, the OP has clarified the question, which means that your post, sadly, no longer addresses the question. $\endgroup$
    – Alex M.
    Commented Jul 19, 2022 at 15:00
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    $\begingroup$ Also, if you can record lengths on the straightedge and then hold one end of the length at a point and use the other end to find the opposite point of a circle, you're basically using it as a compass.... $\endgroup$
    – Kevin Casto
    Commented Jul 19, 2022 at 15:25

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