least value of $2(PA^2+PB^2+PC^2+PO^2)$

Question

Let $OABC$ be regular tetrahedron and $P$ be any point in a space . If the edge length of tetrahedron is $1$ unit.then least value of $2\big(PA^2+PB^2+PC^2+PO^2\bigg)$

Try: assuming position vector of $O,A,B,C$ be $\vec{o},\vec{a},\vec{b},\vec{c}$ and point $P$ is $\vec{p}$

Then $$(\vec{a}-\vec{p})^2+(\vec{b}-\vec{p})^2+(\vec{c}-\vec{p})^2+(\vec{p})^2$$

How i find least value, help me thanks

Answer 1

Expand all summands a la $(x-y)^2=x^2-2xy+y^2$ to obtain $$ \vec a^2+\vec b^2+\vec c^2-2(\vec a+\vec b+\vec c)\vec p+4\vec p^2,$$ which is $$ \left(2\vec p-\frac{\vec a+\vec b+\vec c}2\right)^2 + \text{something}$$

Answer 2

$PA^2+PB^2+PC^2+PO^2$ is the moment of inertia of $\{A,B,C,O\}$ with respect to $P$, which by the parallel axis theorem is minimized iff $P$ lies at the centroid $G$ of $ABCO$. If we embed $ABCO$ in $\mathbb{R}^4$ as follows: $$ A=\frac{(1,0,0,0)}{\sqrt{2}},\quad B=\frac{(0,1,0,0)}{\sqrt{2}},\quad C=\frac{(0,0,1,0)}{\sqrt{2}},\quad O=\frac{(0,0,0,1)}{\sqrt{2}}$$ then $G$ lies at $\frac{(1,1,1,1)}{4\sqrt{2}}$ and $GA^2+GB^2+GC^2+GO^2=4\cdot\frac{3^2+1^2+1^2+1^2}{32}=\frac{3}{2}$, so the wanted minimum equals $\color{red}{3}$.