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By moving the concept of geometric construction into three dimensions, could one trace the 3D wireframe of any of the five platonic solids using only a compass and straightedge?

If not, what additional tools would be required?

I imagine the construction taking place in a "void" of sorts, without the luxury of a preexisting plane. No one $xy$, $xz$ or $yz$ plane is visualized.

Rules copied from TheNullHypodermic:

Draw a line between any two distinct points.

Draw a circle with one point as the center, and any other point on its circumference.

Draw an arbitrary point on a line or a circle, or off it.

Draw the point at the intersection of two lines (if they intersect).

Draw the point (or two) at the intersection of two circles (if they intersect).

Draw the point (or two) at the intersection of a line and a circle (if they intersect).

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    $\begingroup$ Some would argue, that not only that this can be done, but that in The Elements Euclid actually did so. $\endgroup$ – Lord Shark the Unknown May 30 '17 at 17:59
  • $\begingroup$ In 2D constructions, I can select an arbitrary point and swing a circle around it. If I have a known radius I can make that circle match in size. What are the equivalents in your 3D universe? Please make the question clear. $\endgroup$ – Ross Millikan May 30 '17 at 18:25
  • $\begingroup$ @RossMillikan Sorry, you say you can make the circle match in size. I do not understand what the circle is "matching" in size. $\endgroup$ – Ola May 30 '17 at 18:43
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    $\begingroup$ This might interest you: thenullhypodermic.blogspot.com/2012/01/… $\endgroup$ – David K May 30 '17 at 18:59
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    $\begingroup$ @Ola By "matching in size", he means that if you have a segment of some desired length, and a center point then you can make a circle with that center point and a radius that "matches the length" of that segment. $\endgroup$ – Arthur May 30 '17 at 19:08
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You have to define how you use the compass and straightedge to define points that are not in the original plane. You can construct segments of the proper lengths to have the coordinates of the corners of the Platonic solids. How you translate those coordinates into a point in 3D is not clear to me. How do you get the $z$ axis given the $xy$ plane?

Arthur suggested we consider the compass to make spheres and the straightedge to be able to make planes. To construct the regular tetrahedron then you just construct an equilateral triangle in the plane and swing spheres from each corner with radius equal to the side of the triangle. The point where the three spheres intersect is the top point. Constructing a cube is easy as well. Construct a square, then swing spheres of size $\sqrt 2$ times the side from to diagonally opposite corners plus one of side $\sqrt 3$ times the side from the other two. The intersections get two of the top corners of the cube. The others can also be done because the corners are all at points that can be expressed with addition, subtraction, multiplication, division, and square roots.

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    $\begingroup$ If this is meant as an answer, and not a comment, perhaps you could at the very least suggest some way to solve the problems you raise. Such as "if your compass draws spheres instead of circles, then yes, you can construct the Platonic solids" $\endgroup$ – Arthur May 30 '17 at 18:08
  • $\begingroup$ @Arthur: that is a good way of translating compasses and straightedges into 3D that I hadn't thought about. I will suggest constructions using them. $\endgroup$ – Ross Millikan May 30 '17 at 18:10
  • $\begingroup$ I imagine the construction taking place in a "void" of sorts. No one $xy$, $xz$ or $yz$ plane can be isolated.The best analogy i have is tracing one-dimensional curves in thin air with a 3D-pen. $\endgroup$ – Ola May 30 '17 at 18:17
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    $\begingroup$ Another possibility is to do your geometric construction on the plane (regarded as a sheet of paper), and then fold it to get your 3D shape. $\endgroup$ – Robert Israel May 30 '17 at 18:17
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    $\begingroup$ @Ola: You need to define what the compass and straightedge can do, then. $\endgroup$ – Ross Millikan May 30 '17 at 18:23

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