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My problem is motivated by software like Geogebra. Suppose that we are allowed to use the following two tools only:

  1. given any two points, we can construct a line passing through both of them

  2. given any three non-collinear points, we can construct a circle passing through all of them.

I tried to use these two tools to do some simple geometric constructions, such as bisecting an angle or a line segment but failed to achieve anything. I wonder whether it is possible to do these.

Note: As mentioned in John Hughes's answer, I have to clarify my problem. The setting of the problem is just like that in the classical compass and straightedge problem, but with the compass replaced by a 'three-point-one-cirlce' instrument.

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  • $\begingroup$ I think your question is answered simply thusly: If youre given three points, how do you construct the center of said circle? Its an interesting restriction but it is grossly limited. $\endgroup$ – CogitoErgoCogitoSum Aug 19 at 4:49
  • $\begingroup$ The Poncelet-Steiner theorem states that anything you can construct with a straightedge and compass together, you can construct with a straightedge alone, provided you have a single circle with its center identified already in the plane. You have use of the straightedge and you have circles, but without centers. Your axioms prohibit the existence and constructability of a circle center, thus you dont even meet the minimum requirements of the P-S theorem to be able to perform all compass-straightedge constructions.. $\endgroup$ – CogitoErgoCogitoSum Aug 19 at 4:52
  • $\begingroup$ @CogitoErgoCogitoSum After posting this one, I managed to show that we can do constructive all figures that are constructible with compass and straightedge if we have a third tool. The third Geogebra too1 can be (a) perpendicular bisector of arbitrary line segment, (b) line perpendicular to an arbitrary line passing through an arbitrary point, (c) midpoint of arbitrary point pair, (d) bisector of arbitrary angle, or (e) line parallel to an arbitrary line passing through an arbitrary point. $\endgroup$ – CY Aries Aug 26 at 23:38
  • $\begingroup$ Yes, that makes sense. But your question explicitly states you only have the two specified tools, no third option was mentioned. $\endgroup$ – CogitoErgoCogitoSum Aug 28 at 16:13
  • $\begingroup$ I guess the problem is not solvable with only two tools. $\endgroup$ – CY Aries Aug 29 at 17:28
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At the very least, you'll need a few other "hidden" operations:

  • Given any line, can you pick two distinct points on it? Three distinct points?

  • Can you pick three noncollinear points in the plane?

  • Given a circle, can you pick points on it?

Without these, the empty geometry and the geometry consisting of one line, which contains either 1 or 2 points, are both perfectly valid, and so is the geometry consisting of just the points $(0,0), (\pm 1, 0)$, and the three lines that join them (and the circle through them, which consists of all three points!).

That latter geometry is interesting because the vertical and horizontal lines have no "angle bisector", which shows that if your axioms only allow the construction of lines and circle, constructing bisectors is impossible. (Also note that angle-measure is pretty dicey, too).

If you mean "Starting from the Euclidean plane, with its notions of angle measure and distance measure, and the collection of lines guaranteed by the axioms of Euclidean geometry, can I, using just these two construction rules, produce lots of interesting things like bisectors, etc.?" then I suspect that you can do some stuff, but not a whole lot. But you really need to clarify your question a bit.

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  • $\begingroup$ I am just thinking of replacing the 'normal' compass by a 'three-point-one-cirlce' instrument. So I would my problem starts form the Euclidean plane, with its notions of angle measure and distance measure, and the collection of lines guaranteed by the axioms of Euclidean geometry. $\endgroup$ – CY Aries Dec 9 '17 at 15:28
  • $\begingroup$ I think it comes down to this. Given the center of and two points of a circle, can construct a third point on that circle? $\endgroup$ – steven gregory Dec 9 '17 at 15:32
  • $\begingroup$ Seems as if the answer is "yes", @stevengregory: with center $C$ and points $A,B$, either $AB$ are antipodal, and we can pick a point Q not on the line $AB$, then join $C$ and $Q$ and intersect $CQ$ with the circle, OR $A$ and $B$ are not antipodal, in which case $AC$ intersects the circle in some antipodal point $A'$. I don't see how this gets you anywhere, however --- can you elaborate? I just realized that for OP, "circle" might mean "set of points" rather than "center and radius"; in that case, ignore my ravings. :) $\endgroup$ – John Hughes Dec 9 '17 at 15:34
  • $\begingroup$ You don't HAVE the circle. Just two points that are on it and its center. $\endgroup$ – steven gregory Dec 9 '17 at 15:37
  • $\begingroup$ Ah...I see what you're saying. In general, I can't see how, with these two tools alone, you can even construct any third point on the line through points $A$ and $B$, which should be even simpler. So I'm not optimistic about this toolset. $\endgroup$ – John Hughes Dec 9 '17 at 15:43

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