TITLE: equilateral triangle; $3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.$
QUESTION [9 upvotes]: In equilateral triangle ABC of side length d, if P is an internal point with PA = a, PB = b, and PC = c, the following pleasingly symmetrical relationship holds:
$3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.$
Please prove this identity.
source: http://www.qbyte.org/puzzles/p117s.html

REPLY [4 votes]: An explanation and derivation without calculations follows.
That a relation exists can be inferred by a parameter count.  (In an equilateral triangle there are two dimensions of freedom to select a point, and three distances to the vertices.)
What the relation is, can be discovered by writing down Euler's formula for $288 V^2$ of a tetrahedron as a function of the squared edge lengths, and setting volume to zero.  (The Cayley-Menger determinant is the $n$ dimensional generalization and organization of this formula.)  
The puzzle is to explain the $S_4$ symmetry of the relation, that allows exchange of $d$ with $a$,$b$ or $c$.  The determinant is symmetric in $a,b,c$ only.  
A 60 degree rotation of the triangle around $A$ moves $P$ to $P'$ so that $APP'$ is an equilateral triangle with side $a$, whose vertices have distances $d,b,c$ to point $B$ (and $d,c,b$ to point $C$).  Hence we also have symmetries that permute $d$ with the other variables.
Conclusion: the degree $4$ polynomial relation from the Euler-Cayley-Menger determinant (which is a quadratic polynomial in the squares of $a,b,c,d$), is symmetric in all its variables.  Any such relation can be written as $A(a^2+b^2+c^2+d^2)^2 + B(a^4+b^4+c^4+d^4) = 0$.  In this problem, neither coefficient can be zero, and it is harmless to assume $A=1$.   All that is left is to pin down $B$, and by moving $P$ toward $\infty$, the relation should be consistent with $a=b=c$ and $d=0$, so that $B=-3$.
The same works in $n$ dimensions, with $(\sum a^2)^2 = (n+1)\sum a^4$.
Another way to say all this is that looking at the formula, we know it must be homogeneous, symmetric in $a,b,c$, valid for $(1,1,1,0)$, and it is not so surprising that it be of degree $4$.  The surprise is that the symmetry includes $d$, and this additional "hidden symmetry" then dictates the formula.