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

Question

This is a high school problem. It is solved by algebraic calculation (not difficult) - $\ast$.

But I have a problem in interpretation the result.

Here intersection of two surfaces $a+b+c=0,\ a^2+b^2+c^2=1$ is a circle $C$. That is, by $\ast$, $a^4+b^4+c^4$ is constant on it. What does make it so ?

Proof : Note that $S:=\{ (a,b,c)|a^4+b^4+c^4=\frac{1}{2}\}$ is a convex surface. In further, there is $x,\ y\in S$ s.t. $|x|<1<|y|$.

And I guess that $C$ is not a geodesic : Consider a norm on $\mathbb{R}^3$. For instance $\| (a,b,c) \|_1 := |a|+|b|+|c|$ In this case unit ball wrt this norm is octahedron. In further, unit balls of standard Euclidean norm, infinite norm are unit sphere $S_2$, a cube $S_3$, respectively. Clearly, intersection of $S_2$ and plane $\{ (a,b,c)| a+b+c=0\}$ is a geodesic (So is $S_3$). Here considering $S$ and $S_3$, we can assume that $S$ is like $S_3$. That is, $S$ has largest Gaussian curvature at eight points. Hence geodesic is far from these points, but $C$ is not.

I enumerate some these exercises, but how can we interpret the answer ?

[add] Consider $ \Delta':=\{ (x,y,z)\in S_2 | x,\ y,\ z\geq 0\} $, which is an equilateral geodesic triangle of side length $\frac{\pi}{2}$.

In further $\Delta \subset \Delta'$ is also an equilateral geodesic triangle of side length $\frac{\pi}{3}$, which touches mid points of sides in $\Delta'$. Clearly it is not in a plane.

Then we have a claim that if $ f (x,y,z)=(x^4,y^4 ,z^4)$, then $f(\partial \Delta )$ is in a plane.

[my answer] Consider $SO(3)$-action on $S^2(1)$. Here what is $smallest$ invariant set ? It is $S^2$. Consider a finite subgroup of $SO(3)$. For instance, a group $H$ which acts on cube whose center is origin. So smallest invariant set of $H$ is not $S^2$. It may be union of finite number of great circles, $T$. And note that $F(x,y,z)=x^4+y^4+z^4$ is invariant under $H$. Hence we can guess that $F$ is constant on $T$.

Answer 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.

Answer 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.

Answer 3

Addem, we could do like following. (I write here because it's too long in comment).

1.

$$ a+b+c=0 $$ $$ (a+b+c)^2=0 $$ $$ a^2+b^2+c^2+2ab+2bc+2ac=0 $$ $$ a^2+b^2+c^2=−2(ab+bc+ac) $$ $$ 1=−2(ab+bc+ac) $$ $$ ⇒ ab+bc+ac=−\frac{1}{2}. $$

2.

$$ (ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2$$ $$ (ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c)$$ $$ (−\frac{1}{2})^2=a^2b^2+b^2c^2+a^2c^2+0$$ $$ ⇒ a^2b^2+b^2c^2+a^2c^2=\frac{1}{4}. $$

3.

$$ a^2+b^2+c^2=1 $$ $$ (a^2+b^2 +c^2)^2=1 $$ $$ a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=1 $$ $$ a^4+b^4+c^4+2(a^2b^2+b^2c^2+a^2c^2)=1 $$ $$ a^4+b^4+c^4+2*\frac{1}{4}=1 $$ $$ ⇒ a^4+b^4+c^4=\frac{1}{2}. $$

Answer 4

I would propose a more algebraic approach. Two constraints on three variables do not seem enough to fix all three variables, hence give $a^4+b^4+c^4$. On the other hand, if $a,b,c$ are real numbers and $c=-(a+b)$, then $$ 1 = a^2+b^2+c^2 = 2(a^2+ab+b^2) $$ and $$ a^4+b^4+c^4 = 2a^4+4a^3b+6a^2 b^2+4ab^3+2b^4=2(a^2+ab+b^2)^2 = 2\left(\frac{1}{2}\right)^2=\frac{1}{2}.$$

Answer 5

Alternatively: $$(a+b+c)^2=0 \Rightarrow ab+bc+ca=-\frac12 \Rightarrow (ab)^2+(bc)^2+(ca)^2=\frac14.$$ $$(a^2+b^2+c^2)^2=1 \Rightarrow a^4+b^4+c^4=1-2((ab)^2+(bc)^2+(ca)^2)=1-2\cdot \frac14=\frac12.$$