Can all four heights be equally long in a not regular tetrahedron? Give an example that they can or prove that it is impossible. (First published on Facebook/Mathematics)
YES it is possible for all four heights to be equally long in a not regular tetrahedron.
If alla the faces are congruent then they have equal area and so the corresponding heights are equal (since the volume V = Bh/3 where B is any face area and h the corresponding height). The problem is then, can four congruent, not equilateral triangles fit together to form a tetrahedron? – They can, and they need not even be isosceles; any acute-angled triangles will do. To see this look at this picture:
Three congruent, acute-angled triangles are placed in a plane as shown. They form together a larger triangle, similar to the smaller ones. The broken lines will, if extended be heights in the large triangle. Thus they coincide at the one point H. Then, if the triangles surronding the one in the middle are folded upwards their outer corners will, as seen from above, all reach the point H. But their distances to the other two corners are equal for adjacent triangles so they are also at the same altitude and the triangle sides will coincide.
Trying the same with right-angled or obtuse-angled triangles will rather obviously not work. Another way to see that acute angles are necessary is to look at a formula for the volume of the tetrahedron. – Let the sides of each small triangle be a, b, c. Then, if vectors with lengths a, b, c from one corner of the tetrahedron along three edges will point at the other three vertices. The volume is then given by the scalar-cross product, V = (1/6)( a · (b × c)), which can be written as a determinant with row or column vectors a, b, c. (Scalar multiplikation *, cross product x, underscore _ denotes vector.) The square V² is then essentially the determinant of the matrix squared,
36V² = = det [ [a·a , a·b , a·c ] , [b·a , b·b , b·c ] , [c·a , c·b , c·c ] ].
Using the cosine theorem this can be expressed in the side lengths a, b, c:
72V² = (a² + c² – b²)(b² + c² – a²)(a² + b² – c²)
(see also http://en.wikipedia.org/wiki/Tetrahedron).
From this either all three factors on the right hand side are positive or one is positive and two negative. But by the cosine theorem, a² + b² – c² > 0 means that the angle opposite to c is acute, and a² + b² – c² < 0 that it is obtuse, and there cannot be two obtuse angles in a tringle, so all three factors are positive. So the triangles are acute-angled. Of course, they can also not be right-angled since that would make one factor zero.
Using Herons formula and V = Bh/3 we can also find a nice expression for the height:
h² = 2(a² + b² – c²)(b² + c² – a²)(c² + a² – b²)/
((a + b + c)·(a + b – c)·(b + c – a)·(c + a – b))
By the above the numerator is positive so the denominator must also be positive. This means simply that the triangle inequalities are fulfilled so that the edges with lengths a, b, c actually do form triangles!
This answers the question and more. However, as often happens new questions pop up. – Is it possible for the heights of the tetrahedron to be equal without the faces being congruent?