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Prove that every triangle is the orthogonal projection of some equilateral triangle.

This problem appears in a book I'm working through in the chapter on transformations in space. There is a rather boring and straightforward analytic solution which I won't detail here. However, most problems in the book have neat, constructive geometric solutions. So I expect there will be one for this problem too, but I haven't found it. Can anyone help?

(My question bears some similarity to this one, to which I've actually just added a new answer. The only difference is that in my question, the three parallel lines are non-coplanar rather than coplanar. However, I don't see how the solution to that problem would be applicable here.)

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  • $\begingroup$ An idea would be to form the cylinder based on the given triangle, and then move three points (or fix a point move the other two) on the edges of the cylinder till you obtain an equilateral triangle. I wonder if a fixed point theorem might be used, in the case when you move just two points, consider the configuration space of these two points to be a square (perhaps the latter idea is off). $\endgroup$
    – Mirko
    Dec 15, 2015 at 13:32
  • $\begingroup$ @Mirko I'm asking for a geometric proof, and-if possible-a constructive one. So anything involving a statement like "slide point $A$ along... until..." will possibly fall into the non-constructive category, depending on the details. Although some geometric proof is better than none. $\endgroup$
    – David
    Dec 15, 2015 at 14:06
  • $\begingroup$ Then, to turn the problem around, say you find a geometric construction, C, that does it. Could then one use C to solve some otherwise unsolvable problems lite trisecting an angle or doubling the cube? (I didn't think of the details and have no particular reason to believe that this would be the case... but who knows.) There are various methods to do the above two problems, assuming the construction e.g. of a certain curve is given, see Quadratrix of Hippias $\endgroup$
    – Mirko
    Dec 15, 2015 at 14:45
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    $\begingroup$ @Mirko The analytic solution only involved solving quadratic equations, so nothing like that ought to happen. $\endgroup$
    – David
    Dec 15, 2015 at 14:51
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    $\begingroup$ Here's a related (but not duplicate) question: "Can one always map a given triangle into a triangle with chosen angles by means of a parallel projection?". It's way more general than this question (the projecting triangle need not be equilateral, just "fixed"), and all of the answers are analytically derived, but seeing the analytical answers may aid the search for a geometric construction. $\endgroup$
    – Blue
    Dec 16, 2015 at 18:57

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This is not a self contained answer, and mostly has links (with a short description of each) to what I googled (which, in turn, does answer the question, I believe).

The following MathForum year 2000 page discusses the same problem and attributes the (geometric construction, as opposed to algebraic) solution to:
Simon Lhuilier (1750-1840) proved that:
The sections of an arbitrary triangular prism include all possible forms of triangles.

It refers to Section 73 of the book by Heinrich Dorrie, 100 great problems of elementary mathematics available online here, also here but not all sections.
This section discusses (with constructions) the Pohlke-Schwarz theorem (link to encyclopedia of math) which is a stronger result: for quadrilaterals instead of triangles. It states: The oblique image of a given tetrahedron can always be determined in such manner that it is similar to a given quadrilateral. As a step in the proof, Lhuilier's theorem is proven on p.305, Fig.86.

The problem that you posted is equivalent to the following problem: Given three parallel lines in space, to construct an equilateral triangle with one vertex on each line. If the three lines are in the same plane the solution is easy (using $60^\circ$ rotation), and could be found here (MSE problem), or here (UGA), or here (BrownU), or here (1915 Monthly problem 454).

Variations, in three dimensions, and when the lines were not required to be parallel could also be found in this paper by Ochonski (some of the constructions there seem pretty crowded to me, they may perhaps necessarily be complicated). More on Pohlke-Schwarz theorem here (paper by Sklenarikova and Pemova) and here (1915 paper by Emch).

The problem with three parallel lines (in a plane, or not in a plane) is stated as exercises 1 and 2 on p.37 in the book by Z. A. Melzak, Invitation to Geometry google books link.

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  • $\begingroup$ Very good answer. The statement of Lhuilier's theorem is much more satisfying and general than that of my problem. The proof of the theorem given in Dörrie's book is constructive, so it answers my question. I'll leave it open whether it's possible to make things any simpler because we want an equilateral triangle, but I'm beginning to have my doubts. $\endgroup$
    – David
    Dec 18, 2015 at 3:18
  • $\begingroup$ Thank you (I also thought the equilateral triangle case might have an easier proof, but couldn't think of anything specific). $\endgroup$
    – Mirko
    Dec 18, 2015 at 3:27

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