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I recently came across this in my textbook:

Any three vertices of a cube determine a right triangle. Is this a true statment?

My initial thought was that is was, but the answers say otherwise. I cannot, however, come up with a counterexample.

What three vertices of a cube don't determine a right triangle?

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You are standing at a corner. Consider the three corners closest to you.

These three corners determine a triangle, all of whose sides are face diagonals. So all these sides have equal length, the triangle is equilateral.

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  • $\begingroup$ That is brilliant. It seems that there still should be right angles where lines from perpendicular planes meet, though, but that can't be... $\endgroup$ – John Dec 17 '12 at 17:19
  • $\begingroup$ You are right about where three perpendicular planes meet, but this triangle kind of slices across. All triangles without a right angle arise in this way. There are some slightly weird ones. For example, let $A$ be a vertex, let $B$ be a neighbouring vertex, and let $C$ be the corner furthest from $B$. Then $\triangle ABC$ has sides $1,\sqrt{2},\sqrt{3}$ so by the converse of the Pythagorean Theorem it has a right angle at $A$. $\endgroup$ – André Nicolas Dec 17 '12 at 17:31
  • $\begingroup$ Ah, I see! Thanks! $\endgroup$ – John Dec 17 '12 at 17:44
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Hint: Draw the cube so that it looks like a hexagon.

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  • $\begingroup$ If you're referring to a drawing of a cube with the illusion of 3 dimensions, I have tried this. Otherwise, I haven't the slightest idea how to make a cube look like a hexagon. >.> $\endgroup$ – John Dec 17 '12 at 17:15
  • $\begingroup$ Then draw it like you did before, and then find a non-right triangle on this projection. $\endgroup$ – dtldarek Dec 17 '12 at 17:42
  • $\begingroup$ I have looked, and had others look. This question is my "I give up; where is it?" :\ $\endgroup$ – John Dec 17 '12 at 17:45
  • $\begingroup$ Connect every second vertex of this hexagon. What happens? $\endgroup$ – dtldarek Dec 17 '12 at 17:49
  • $\begingroup$ Ah, there it is! That is an interesting way to look at it! Thanks! $\endgroup$ – John Dec 17 '12 at 17:52
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Three that involve no edges from the cube. If the vertices are the standard ones for the unit cube, then, $(0,0,0)$, $(0,1,1)$, and $(1,1,0)$ will do.

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