equilateral triangle; $3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.$

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

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.

A Cayley-Menger determinant gives the volume of "tetrahedron" $PABC$:

$$288\; V^2 = \left|\begin{array}{ccccc} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & d^2 & d^2 & a^2 \\ 1 & d^2 & 0 & d^2 & b^2 \\ 1 & d^2 & d^2 & 0 & c^2 \\ 1 & a^2 & b^2 & c^2 & 0 \end{array}\right| = d^2 \left( \left( a^2+b^2+c^2+d^2 \right)^2-3\left(a^4+b^4+c^4+d^4\right) \right)$$

Since the tetrahedron is flat, its volume is zero.

Here's a brute-force approach:

Abbreviating the measures of angles surrounding point $P$ thusly, $$\alpha := |\angle BPC| \qquad \beta := |\angle CPA| \qquad \gamma := |\angle APB|$$ we have $$\cos\gamma = \cos\left(2\pi - \alpha - \beta\right) = \cos\left(\alpha+\beta\right) = \cos\alpha \cos \beta - \sin\alpha \sin\beta$$ so that $$\left( \cos\alpha \cos\beta - \cos\gamma \right)^2 = \left(1-\cos^2\alpha\right)\left(1-\cos^2\beta\right)$$

Now, substitute-in these relations from the Law of Cosines $$\cos\alpha = \frac{b^2+c^2-d^2}{2bc} \qquad \cos\beta = \frac{c^2+a^2-d^2}{2ca} \qquad \cos\gamma = \frac{a^2+b^2-d^2}{2ab}$$ and do a bit of algebra.

Here's an example in planar euclidean geometry. Consider an equilateral triangle of side $a$ and a general point in the plane distant $b$, $c$, and $d$ from the respective vertices. Then

$$3(a^4+b^4+c^4+d^4)=(a^2+b^2+c^2+d^2)^2.$$

This is an an awful slog to get by planar trigonometry. Even harder to do by trig in three dimensions is the corresponding result for the regular tetrahedron. However, it's easy to get the $(n−1)$-dimensional result for a regular $(n−1)$-dimensional simplex of side $d_0$, with vertex distances $d_1 ,..., d_n$ :

$$n(d^4_0+\cdots+d^4_n)=(d^2_0+\cdots+d^2_n)^2.$$

You can do this by embedding the euclidean $(n−1)$-dimensional space as the hyperplane of points $(x_1,...,x_n)$ in euclidean $n$-space such that $x_1+\cdots+x_n=d_0/\sqrt2.$ The vertices of the simplex can then be represented as the points $(d_0/\sqrt2)(1,0,...,0), ... , (d_0/\sqrt2)(0,...,0,1)$ in the hyperplane, and the result drops out in a few lines.

Edit: Following the prompting of "zyx", here are the details. Let $\bf x$ be a general point in the hyperplane such that $|\mathbf x-a\mathbf e_i|=d_i\;(i=1,...,n),$ where $a=d_0/\sqrt2$. Then, writing $s=a^2+x_1^2+\cdots+x_n^2,$ we have $$d_i^2=\sum_{j=1}^n(x_j-\delta _{ij}a)^2=s-2ax_i\quad(i=1,...,n).$$Since $x_1+\cdots+x_n=a,$ it follows that $d_0^2+\cdots+d_n^2=ns$ and also $$d_0^4+\cdots+d_n^4=4a^4+\sum_{i=1}^n(s^2-4sax_i+4a^2x_i^2)=ns^2,$$giving the required result.

• @zyx: Sorry, I don't see what you are getting at. The formula works for $n=2$ (my $n$) as well as for any other $n$. By the way, it isn't obvious to me how to obtain the (true) quartic result that you claim for general $n$ from the Cayley--Menger determinant corresponding to $n$ dimensions, which is a polynomial of total degree $2n$. – John Bentin Aug 15 '13 at 20:59
• For the relation in dimension $n$ I did not mean to suggest that it is read off from the Cayley Menger determinant, but that a quartic relation symmetric in all variables has to be the one written down. – zyx Aug 16 '13 at 4:41
• @zyx: I have given the details in my edit above. Regarding your derivation, I can see that, if there is a quartic identity with the appropriate symmetries, then it is the one given. What isn't obvious to me, though, is why any identity satisfied by the $n+1$ variables has to be quartic. – John Bentin Aug 16 '13 at 10:52