Let $SABC$ be a tetrahedron with the lenght of the side equal to $a>0$. We construct $PA$ perpendicular to the plane $(ABC)$ such that $P$ and $S$ to be in different half-space (see picture below). I have to find the lenght of the segment $PS$ and the maximum of $$ \left| TP-TS \right|, $$ where $T$ is in the plane $(ABC)$.
I have no idea how to get the lenght of segment $PS$. For the second part of the question, I think that the maximum value I think that
- $PT$ should be as large as possible such that
- $TS$ to have the smallest possibe lenght.
So the maximum value of $PT$ is when $T$ is $B$ or $C$. So, I think that $T$ should be between $B$ and $C$. For a fixed point $T$ between $B$ and $C$, the minimum values of $ST$ is the shortest path from $S$ to egde $BC$. Hence the minimum values is the height of the face $SBC$. So $T$ should be the midpoint of the segment $BC$ and the maximum values is $PT+TS$. Am I right? Thank you!
To find $PS$ you can draw a right triangle with hypotenuse $PS$ and legs $AH$ and $PA+SH$, where $H$ is the projection of $S$ on plane $ABC$.
The difference $|TP-TS|$ is constant when $T$ lies on the surface of a hyperboloid obtained by rotating about $PS$ a hyperbola having $P$ and $S$ as foci. If $T$ is on plane $ABC$ then $|TP-TS|$ gets its maximum value when the plane is tangent to one of those hyperboloids and $T$ is the tangency point.
It is clear by symmetry that $T$ must lie on line $AH$, so we can consider only plane $PASH$: in that plane $T$ is the tangency point between a hyperbola with foci $P$, $S$ and line $AH$. But the tangent of that hyperbola is the bisector of angle $\angle PTS$, that is: $\angle STA=\angle PTA$. Then one can find the position of $T$, because triangles $TSH$ and $TPA$ are similar:
$$ TA:AP=TH:SH. $$
Setting $TA=ax$ this can be rewritten as:
$$ x:1=\left(x-{1\over\sqrt3}\right):\sqrt{2\over3}, $$
which gives: $x=\sqrt2+\sqrt3$. Finally, apply Pythagoras' theorem to get $PT-ST$.