If $P$ is the intersection of the altitudes of a tetrahedron $ABCD$ and $r$ is the circumradius, then $PA^2+PB^2+PC^2+PD^2=4r^2$ Prove that, if $P$ is the intersection of the altitudes of a tetrahedron $ABCD$ and $r$ is the circumradius, then $$PA^2+PB^2+PC^2+PD^2=4r^2$$

An orthocentric tetrahedron is by definition any tetrahedron for which the altitudes meet at a point, and that point is called the orthocenter of the tetrahedron. I use a fact relating the orthocenter, the circumcenter, and the centroid which holds for any orthocentric tetrahedron. Reference:
http://en.wikipedia.org/wiki/Orthocentric_tetrahedron
Let $P$ be the orthocenter, $K$ be the centroid $(A+B+C+D)/4$, and $O$ be the circumcenter. The relation between these points is that $P$ is the symmetric point to $O$ with respect to $K$, i.e. that $K$ is the midpoint of segment $PO$. The squared length of any segment $XY$ between points $X,Y$ in 3-space is the dot product $(X-Y)\cdot(X-Y)$, viewing the points $X,Y$ as vectors based at the origin. We will write $(X-Y)^2$ for this dot product, so that the relation between $P,K,O$ mentioned above implies that
$$[1]\ \ \ \ (P-K)^2-(O-K)^2=0.$$
Now the squared lengths $PA^2$ etc. are $(P-A)^2$ etc. in our dot product notation. The relation
$$PA^2+PB^2+PC^2+PD^2-4r^2=0$$ becomes in our notation, since $r^2=(O-A)^2=...=(O-D)^2,$
$$[2]\ \ \ \ (P-A)^2+(P-B)^2+(P-C)^2+(P-D)^2\\ -(O-A)^2-(O-B)^2-(O-C)^2-(O-D)^2=0.$$
When the dot products are all multiplied out, it turns out that the expression on the left of [2] is (surprisingly, at least to me) exactly 4 times the expression on the left of [1]. So since [1] is a known fact about orthocentric tetrahedra, it follows that [2] also holds, finishing the proof.
