$ a+b+c=0,\ a^2+b^2+c^2=1$ implies $ a^4+b^4+c^4=\frac{1}{2}$

Solution 1:

HINT:

From the identity $$(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$$

the equality $a^4+b^4+c^4=1/2$ is a consequence of $a+b+c=0$ and $a^2+b^2+c^2=1$. One can also look at gradients and see a geometric reason from the equality.

Note that conversely $a^4+b^4+c^4=1/2$ and $a+b+c=0$ implies $(a^2+b^2+c^2)^2=1$ and so (reals!) $a^2+b^2+c^2=1$. So indeed the plane cuts the surface along that circle.

One can cook up different surfaces of equation $$P(a,b,c)(a^2+b^2+c^2-1)+Q(a,b,c)(a+b+c)=0$$ that will contain the same circle.

If one looks at the surface $a^4+b^4+c^4=1/2$, it has the shape of a rounded cube. How is it that the plane $a+b+c=0$ cuts it along a circle?

Recall that a cube $\max(|a|,|b|,|c|)=1$ is cut by this plane along a regular hexagon with vertices the permutations of $(-1,0,1)$. Now the permutations of $(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$ lie on both of the surfaces $a^2+b^2+c^2=1$ and $a^4+b^4+c^4=\frac{1}{2}$. That should make it more intuitive.

Note that each of the $4$ planes $a\pm b\pm c = 0$ cuts our surface along a circle, as the identity tells us.

Solution 2:

The geometric approach that you have outlined...

The intersection of $a+b+c = 0$ and $a^2 + b^2 + c^2 = 1$

A plane cutting a sphere, producing a circle.

$a^4 + b^4 + c^4 = \frac 12$ is a convex surface that resembles a cube with rounded edges and corners.

As it happens every point on the circle $C$ lies on this "cubeoid." But, it is still not obvious that this would be the case.

You could parameterize the circle:

$\begin {bmatrix} \frac {\sqrt 3}{3} & \frac {\sqrt 2}{2} &\frac {\sqrt 6}{6}\\ \frac {\sqrt 3}{3} & -\frac {\sqrt 2}{2} &\frac {\sqrt 6}{6}\\ \frac {\sqrt 3}{3} & 0 &-\frac {\sqrt 6}{3}\end{bmatrix}\begin {bmatrix} 0\\\cos\theta\\\sin\theta\end{bmatrix}= \begin {bmatrix} \frac {\sqrt 2}{ 2}\cos\theta + \frac {\sqrt {6}}{6}\sin\theta\\-\frac {\sqrt 2}{ 2}\cos\theta + \frac {\sqrt {6}}{6}\sin\theta\\-\frac {\sqrt 6}{3}\sin\theta\end{bmatrix}$

Would give you a parameterization of the circle.

And $a^4 +b^4 + c^4$ indeed does equal $\frac 12$

I don't think that this is easier than the more algebraic approaches...But it does work.