Problem source

The four vertices  $P_i$ ($i= 1, 2, 3, 4$) of a regular tetrahedron lie on the surface of a sphere with centre at $O$ and of radius 1.  The position vector of $P_i$ with respect to $O$ is ${\bf p}_i$ ($i= 1, 2, 3, 4$). Use the fact that ${\bf p}_1+ {\bf p}_2+{\bf p}_3+{\bf p}_4={\bf 0}\,$ to show that ${\bf p}_i \,.\, {\bf p}_j =-\frac13\,$ for $i\ne j$. 
Let $X$ be any point on the surface of the sphere, and let $XP_i$ denote the length of the line joining $X$ and $P_i$ ($i= 1, 2, 3, 4$).
\begin{questionparts}
\item
By writing  $(XP_i) ^2$ as $({\bf p}_i- {\bf x)}\,.\,({\bf p}_i- {\bf x})$, where ${\bf x}$ is the position vector of $X$ with respect to $O$, show that  
 \[
\sum_{i=1}^4(XP_i) ^2 =8\,.
\]
\item Given that $P_1$ has coordinates $(0,0,1)$ and that  the coordinates of $P_2$ are of the form $(a,0,b)$, where $a > 0$, show that  $a=2\sqrt2/3$ and $b=-1/3$, and find the coordinates of $P_3$ and $P_4$. 
\item Show that 
\[
\sum_{i=1}^4 (XP_i)^4 = 4 \sum_{i=1}^4 (1- {\bf x}\,.\,{\bf p}_i)^2\,.
\]
By letting the coordinates of $X$ be $ (x,y,z)$, show further that $\sum\limits_{i=1}^4 (XP_i)^4$ is independent of the position of $X$.
\end{questionparts}

Solution source

Note that ${\bf p}_i \cdot {\bf p}_i = 1$ and ${\bf p}_i \cdot {\bf p}_j$ are all equal when $i \neq j$ by symmetry and commutativity. 
\begin{align*}
&& 0 &= {\bf p}_i \cdot \left ( {\bf p}_1+ {\bf p}_2+{\bf p}_3+{\bf p}_4 \right) \\
&&&=  1 + \sum_{j \neq i} {\bf p}_i \cdot {\bf p}_j \\
&&&= 1 + 3 {\bf p}_i \cdot {\bf p}_j  \\
\Rightarrow && {\bf p}_i \cdot {\bf p}_j  &= -\frac13
\end{align*}

\begin{questionparts}
\item $\,$ \begin{align*}
&& (XP_i)^2 &= ({\bf p}_i- {\bf x)}\,.\,({\bf p}_i- {\bf x}) \\
&&&= {\bf p}_i \cdot {\bf p}_i - 2 {\bf p}_i \cdot {\bf x} + {\bf x} \cdot {\bf x} \\
&&&= 2 - 2  {\bf p}_i \cdot {\bf x}  \\
\Rightarrow && \sum_i (XP_i)^2  &= \sum_i \left (2 - 2  {\bf p}_i \cdot {\bf x}  \right) \\
&&&= 8 - 2 \sum_i {\bf p}_i \cdot {\bf x} \\
&&&= 8 - 2 \left (  {\bf p}_1+ {\bf p}_2+{\bf p}_3+{\bf p}_4 \right) \cdot {\bf x} \\
&&&= 8
\end{align*}

\item Notice we have $1 = \left \|\begin{pmatrix} a \\0 \\b \end{pmatrix} \right \|= a^2 + b^2$ and $-\frac13 = \begin{pmatrix} a \\0 \\b \end{pmatrix}  \cdot \begin{pmatrix} 0 \\0 \\ 1 \end{pmatrix} = b$. So $b = -1/3$ and $a = \sqrt{1-b^2} = 2\sqrt{2}/3$.

Suppose another of the vertices has coordinates $(u,v,w)$ we must have

\begin{align*}
&& 1 &= u^2+v^2+w^2 \\
&& -\frac13&=w   \\
&& -\frac13 &= \frac{2\sqrt{2}}3 u +\frac19 \\
\Rightarrow && u &= -\frac{\sqrt2}3  \\
\Rightarrow && 1 &= \frac19 + \frac29 + v^2 \\
\Rightarrow && v &= \pm \sqrt{\frac{2}{3}}
\end{align*}

So $P_3, P_4 = (-\frac{\sqrt2}3, \pm \frac{\sqrt{6}}3, -\frac13)$

\item $\,$ \begin{align*}
&& \sum_{i=1}^4 (XP_i)^4 &= \sum_i \left (2 - 2  {\bf p}_i \cdot {\bf x}  \right)^2 \\
&&&= 4 \sum_i \left (1 -   {\bf p}_i \cdot {\bf x}  \right)^2 \\
&&&= 4 \sum_i (1 - 2{\bf p}_i \cdot {\bf x} + ({\bf p}_i \cdot {\bf x})^2) \\
&&&= 16 + 4\sum_i  ({\bf p}_i \cdot {\bf x})^2 \\
&&&=16+ 4\left ( z^2+\left (\frac{2\sqrt{2}}3x-\frac13z \right)^2 +\left (-\frac{\sqrt{2}}3x-\frac{\sqrt{6}}3y-\frac13z \right)^2 +\left (-\frac{\sqrt{2}}3x+\frac{\sqrt{6}}3y-\frac13z \right)^2 \right) \\
&&&= 16+4 \left ( \frac43z^2 + \left (\frac89 + \frac29+\frac29 \right)x^2+\left (\frac69 + \frac69 \right)y^2 + 0xz + 0yz + 0zx \right) \\
&&&= 16+ 4\cdot\frac43(x^2+y^2+z^2) \\
&&&=16+\frac{16}{3}=\frac{64}{3}
\end{align*}
\end{questionparts}

Note: It may be better to view the last part of this question in terms of linear transformations.

There are two possible approaches. One is to show $T:{\bf x} \mapsto \sum_i ({\bf p}_i \cdot x) {\bf p}_i$ is $\frac43I$ (easy since it has three eigenvectors with the same eigenvalue which span $\mathbb{R}^3$ and we are interested in the value ${\bf x} \cdot T\mathbf{x} = \frac43 \lVert {\bf x} \rVert^2$.

The second is to consider $\sum_I ({\bf p}_i \cdot {\bf x})^2 = {\bf x}^TM{\bf x}$ where $M = \sum_i {\bf p}_i{\bf p}_i^T$ and note that this matrix is invariant under rotations.