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 $$PA^2+PB^2+PC^2+PD^2=4r^2$$
 A: 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.
A: This is a supplement to coffeemath's answer. It is a pity to let the amazing facts metioned there leave as a mystery.
Let $\vec{x}_1, \vec{x}_2, \vec{x}_3, \vec{x}_4$ be the vertices of a non-degenerate tetrahedron. WOLOG, we will pick the circumcenter of the tetrahedron as the origin.
We have $|\vec{x}_1| = |\vec{x}_2| = |\vec{x}_3| = |\vec{x}_4| = r$, the circumradius.
Let us also assume the four altitudes intersect at a single point $\vec{p}$. 
The vector $\vec{x}_1 - \vec{p}$ is in the direction of the normal vector for the affine plane $\langle x_2x_3x_4\rangle$ generated by the 3 vectors
$\vec{x}_2$, $\vec{x}_3$ and $\vec{x}_4$. This means it will be perpendicular to every vector that lives in the plane. In particular, the vector $\vec{x}_3 - \vec{x}_4$, i.e.
$$(\vec{x}_1 - \vec{p}) \cdot (\vec{x}_3 - \vec{x}_4) = 0\tag{*1a}$$
By a similar argument to $\vec{x}_2 - \vec{p}$ and the affine plane $\langle x_4x_3x_1\rangle$, we get:
$$(\vec{x}_2 - \vec{p}) \cdot (\vec{x}_3 - \vec{x}_4) = 0\tag{*1b}$$
Now 
$$(*1a) - (*1b) \implies (\vec{x}_1 - \vec{x}_2) \cdot (\vec{x}_3 - \vec{x}_4) = 0$$
i.e. the pair of opposite edges $\langle x_1x_2 \rangle$ and $\langle x_3x_4\rangle$ are perpendicular to each other. By a similar argument, the other two pairs of opposite edges,
$\{ \langle x_1x_3\rangle$, $\langle x_2x_4\rangle \}$ and $\{ \langle x_1x_4\rangle$, $\langle x_2x_3\rangle \}$ are perpendicular among themselves. This means the tetrahedron
is an orthocentric tetrahedron as described in coffeemath's answer.
Next,
$$(*1a) + (*1b) \implies (\frac{\vec{x}_1 + \vec{x}_2}{2} - \vec{p}) \cdot (\vec{x}_3 - \vec{x}_4) = 0\tag{*2a}$$
Notice
$$(\vec{x}_3 + \vec{x}_4) \cdot (\vec{x}_3 - \vec{x}_4) = |\vec{x}_3|^2 - |\vec{x}_4|^2 = r^2 - r^2 = 0\tag{*2b}$$
$(*2a) + \frac12 (*2b)$ then gives us:
$$(2 \vec{c} - \vec{p}) \cdot (\vec{x}_3 - \vec{x}_4) = 0$$
where $\vec{c} = \frac{\vec{x}_1 + \vec{x}_2 + \vec{x}_3 + \vec{x}_4}{4}$ is the centroid of the tetrahedron. By a similar argument, we have:
$$(2 \vec{c} - \vec{p}) \cdot (\vec{x}_1 - \vec{x}_4) = 
  (2 \vec{c} - \vec{p}) \cdot (\vec{x}_2 - \vec{x}_4) =
  (2 \vec{c} - \vec{p}) \cdot (\vec{x}_3 - \vec{x}_4) = 0$$
For a non-degenerate tetrahedron, the 3 vectors $\vec{x}_1 - \vec{x}_4, \vec{x}_2 - \vec{x}_4, \vec{x}_3 - \vec{x}_4$ froms a basis for the vector space $\mathbb{R}^3$.
As a result, we get
$$2\vec{c} - \vec{p} = \vec{0} \iff \vec{p} = 2 \vec{c} \iff  \vec{c} = \frac12 ( \vec{p} + \vec{0} )$$
This means the centroid $\vec{c}$ is located at the mid-point between the orthocenter $\vec{p}$ and the circumcenter $\vec{0}$.
Finally, we have:
$$\sum_{i=1}^4 |\vec{x_i} - \vec{p}|^2 
=\sum_{i=1}^4 |\vec{x}_i|^2 - 2 ( \sum_{i=1}^4 \vec{x}_i ) \cdot \vec{p} + 4|\vec{p}|^2
=4 r^2 + 4 \vec{p}\cdot(\vec{p} - 2\vec{c}) = 4 r^2$$
A: The problem splits into two logically independent theorems.


*

*For any point $P$ and any other $n$ points $A,B,C,D,(E,F,G,...)$ on a sphere with center $Q$ and radius $r$, the equation $PA^2 + PB^2 + \dots= nr^2$ is equivalent to $PM^2=QM^2$ where $M$ is the center of mass of the $n$ points.  Proof:   the moments of inertia of the $n$ points are the same relative to $P$ and to $Q$.  This is a general algebraic fact about sets of points on spheres, in Euclidean space of any dimension.  For the problem at hand, it means we should prove $|PM|=|QM|$.

*In an orthocentric tetrahedron, $M$ is the midpoint of $PQ$ (the centroid is half-way between the orthocenter and the circumcenter).  Starting from the first fact about points on spheres, and looking for a reason why this problem is correct, you could guess that $PM=QM$ holds with the points collinear, based on analogy with the Euler line in a triangle.  A proof can be found in Nathan Althshiller-Court's article Notes On the Orthcentric Tetrahedron from 1934, which is available in several online forms.
For a general tetrahedron take $P$ to be the Monge point.  Then the formula and both parts of the theorem are true, and $P$ coincides with the orthocenter when one exists. 
