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.
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.