Interesting tetrahedron problem with right dihedral angles A tetrahedron WYXZ, which all sides are acute triangles, has right dihedral angles at WY and XZ. Is there a way to prove that the orthocenters of all faces are on one plane?
The way I tried to solve it was by connecting pairs of orthocenters on opposite sides with lines and proving that they intersect each other but I was unsuccessful.
 A: If $ABCD$ are the vertices of the tetrahedron, we can assume without loss of generality that $A$ and $D$ lie on the $x$-axis, $B$ lies on the $z$-axis and $C$ on $xy$ plane:
$$
A=(a,0,0),\quad B=(0,0,b),\quad C=(c,d,0),\quad D=(t,0,0),
$$
where $a$, $b$, $c$, $d$ are free parameters, while
$$
t=\frac{b^2 \left(a c-c^2-d^2\right)}{a b^2+a d^2-b^2 c}
$$
to ensure planes $ABC$ and $BCD$ to be perpendicular (this expression for $t$ can be found from $(B-D)\times(C-D)\cdot(B-A)\times(C-A)=0$).
The orthocenter $O$ of a generic triangle $PQR$ can be found using:
$$
O={\alpha P+\beta Q+\gamma R\over\delta},
$$
where:
$$
\def\dt#1#2{#1\!\cdot\! #2}
\alpha=(\dt PR)^2 + (\dt PQ)^2 - (\dt QR)^2 - 
2 (\dt PR) (\dt PQ) + (Q^2-R^2) (\dt PR - \dt PQ) + 
(Q^2+R^2)(\dt QR) - Q^2 R^2,
$$
$$
\beta=(\dt QP)^2 + (\dt QR)^2 - (\dt RP)^2 - 
2 (\dt QP) (\dt QR) + (R^2-P^2) (\dt QP - \dt QR) + 
(R^2+P^2)(\dt RP) - R^2 P^2,
$$
$$
\gamma=(\dt RQ)^2 + (\dt RP)^2 - (\dt PQ)^2 - 
2 (\dt RQ) (\dt RP) + (P^2-Q^2) (\dt RQ - \dt RP) + 
(P^2+Q^2)(\dt PQ) - P^2 Q^2,
$$
$$
\begin{align}
\delta=&
(\dt PQ)^2 + (\dt QR)^2 + (\dt RP)^2 - 2 (\dt PQ) (\dt QR) - 2 (\dt QR) (\dt RP) - 
 2 (\dt RP) (\dt PQ) +\cr 
&2 P^2 (\dt QR) + 2 Q^2 (\dt RP) + 
 2 R^2 (\dt PQ) - P^2 Q^2 - Q^2 R^2 - R^2 P^2.
\end{align}
$$
Inserting in this formula the coordinates of the vertices as given above, we can find the coordinates of the orthocenters $O_{ABC}$, $O_{BCD}$, $O_{ACD}$, $O_{ABD}$, of the faces. To check they belong to the same plane one can show that they form a parallelogram, because one finds the midpoint of $O_{ABC}O_{BCD}$ is the same as the midpoint of $O_{ACD}O_{ABD}$:
$$
{O_{ABC}+O_{BCD}\over2}={O_{ACD}+O_{ABD}\over2}=
\left(
{c\over2},\
{(a - c) (b^2 + a c) d\over 2a (b^2 + d^2)-2b^2 c},\ 
{a b (a c - c^2 - d^2)\over 2a (b^2 + d^2)-2b^2 c}
\right).
$$
One can also prove that the diagonals of that parallelogram have the same length, hence the four orthocenters are the vertices of a rectangle.
