An Algebraic Lemma with Geometric Consequences

Given six quantities $x,y,z,x',y',z'$ such that the four sums $x+y+z,$ $x+y'+z',$ $x'+y+z',$ and $x'+y'+z,$ are equal. Then $x=x',$ $y=y',$ and $z=z'.$

The proof is immediate. Subtract in turn from the first sum the other three to obtain three equations:

$\begin{align} (y-y')&+(z-z')=0\\ (x-x')&+(z-z')=0\\ (x-x')&+(y-y')=0. \end{align}$

Summing up gives $(x-x')+(y-y')+(z-z')=0.$ Subtract now from this each of the three equations: $x-x'=0,$ $y-y'=0,$ $z-z'=0,$ as required.

This simple lemma has several interesting applications.

Any tetrahedron with isoperimetric faces, i.e., faces of the same perimeter, is isosceles.

For a proof, let $a,a',$ $b,b',$ $c,c'$ denoted the pairs of opposite edges of the tetrahedron.

The four faces of the tetrahedron have perimeters $a+b+c,$ $a+b'+c',$ $a'+b+c',$ and $a'+b'+c$ that are equal. By direct application of the lemma, $a=a',$ $b=b',$ $c=c',$ as required.

The accompanying figure shows the lines drawn in each face, the tetrahedron being opened out and developed on the plane. It is obvious that the small triangles which share a tetrahedral edge are congruent and the angles of the theorem may then be marked as in the figure. The claim is then an immediate consequence of the lemma.

Theorem

If the faces of a tetrahedron have equal areas, the tetrahedron is isosceles.

Let us represent the areas of the small triangles in the figure by the letters $a, b, c; a', b', c':$

Then, by the lemma, if the faces of the tetrahedron are equiareal $a=a',$ $b=b',$ $c=c'$. Let us denote the tangential distances from the vertices to the inscribed sphere by $A, B, C, D.$ The equality of areas of small triangles gives three equations, a typical example of which is

$\displaystyle\frac{1}{2}A\cdot B\sin\beta =\frac{1}{2}C\cdot D\sin\beta',$

where $\beta$ and $\beta'$ are the Bang angles which were shown to be equal: $\beta =\beta'.$ It follows that $A\cdot B=C\cdot D.$ Similarly, we obtain $A\cdot C=B\cdot D$ and $B\cdot C=A\cdot D.$

Taking the product of the first two equations $A^{2}BC=BCD^{2}$ we see that $A=D.$ The product of the last two equations - $ABC^{2}=ABD^{2}$ - gives $C=D$. The product of the first and the third which is $AB^{2}C=ACD^{2}$ delivers $B=D$ such that all four quantities are equal: $A=B=C=D.$

The claim that the tetrahedron is isosceles follows readily.

1. B. H. Brown, Theorem of Bang. Isosceles Tetrahedra, The American Mathematical Monthly, Vol. 33, No. 4 (Apr., 1926), pp. 224-228