Show that one can construct (with ruler and compasses) a length equal to the altitude from A of the tetrahedron ABCD, given the lengths of all the sides. [So for each pair of vertices, one is given a pair of points in the plane the appropriate distance apart.]
Solution
Let the altitude from A be AH with H in the plane BCD. The plane normal to BC through A also contains H. Suppose it meets BC at X. Then HX and AX are both perpendicular to BC.
Since we have the side lengths we can construct a cardboard cutout of the tetrahedron: the base BCD and the face BCA next to it, also the face CDA' (and the face BDA", although we do not need it. If we folded along the lines BC, CD and BD, then A, A' and A" would become coincident and we would get the tetrahedron.) We have just shown that in the plane AH is a straight line perpendicular to BC (and meeting it at X). So we draw this line and also the line through A' perpendicular to CD, giving H as their point of intersection. Thus we have AH and HX and we know that in the tetrahedron AH is perpendicular to HX. So draw a circle diameter AX and take a circle center H radius HX meeting the circle at K. Then AK is the required length.
© John Scholes
jscholes@kalva.demon.co.uk
24 Aug 2002