A piece of paper has the precise shape of a triangle ABC with AB=2, AC=3, BC=4.

Show how you can construct a tetrahedron by folding three lines.

*** No geometric instrumental aid like straightedge, compass, ruler etc. is permissible.

Although only three folds are needed for the base edges of the tetrahedron,
I’m hoping that more folds are allowed in total. I seem to need three
extra folds, each applied to the original starting shape as follows.

a.         Fold so that B coincides with C. Use the fold to identify the
            mid-point, M, of BC.
b.         Fold so that the edges CA and CB are collinear and call the
            crease CD (D on AB).
c.          Fold to form a crease through M while aligning the two parts
            of crease CD. Call this new crease ME (E on AC).
d.         Find the mid-point, N, of AE by folding A on to E.
e.         Fold along BN,
f.          Fold along MN.

The folds along BN, MN and ME can now be used as the edges of the
required tetrahedron ABMN, with A, E and B, C being coincident pairs.

Proof:
CM = BM = 2 (from a).
Triangle CEM isosceles (from b & c), so CE = CM =2 therefore CE = BA.
EN = AN (from d).
These three underlined statements prove that the edge segments match.

Posted by Harry on 2014-05-06 17:40:26