Given a point $P$ in regular tetrahedron $T$, is there a triangle with vertices on the boundary of $T$ whose center of mass is $P$?

Question

Here's a "simple'' problem in basic geometry.

Given is a regular tetrahedron $T$.

Consider the Barycentric coordinate system on $T$. Given is also a point $P\in T$ in the interior of $T$ such that no two barycentric coordinates of $P$ are equal. ($P$ is not part of any half-plane of symmetry of $T$.)

Does there exist a triangle $\Delta\subset T$ with both of the following properties?

1. The vertices of $\Delta$ are in the boundary of $T$
2. $P$ is the center of mass of $\Delta$

I think that there's a unique such triangle $\Delta$. Furthermore, I feel that this should (somehow!!) generalize to simplexes of greater dimension. [I'm pretty sure it holds for a triangle $T$ and a line segment $\Delta$.]

Intelligenti pauca shows below that there's more than one such triangle. Great!

Here's a more complicated version, which I hope has a unique solution (due to fewer degrees of freedom).

Does there exists a plane $\pi\subset\mathbb R^3$ such that $P$ is the centroid of $\pi\cap T$? (If so, then $P\in\pi$. $\pi\cap T$ is a triangle $\Delta\subset T$ such that all vertices of $\Delta$ are part of an edge of $T$.)

All help is much appreciated, Jürgen

Answer 1

Here's a geometric construction of the desired triangle (see figure below). As it was predictable (six free coordinates with three equations) we have three degrees of freedom left, hence the triangle is far from being unique.

Let $ABCD$ be the given tetrahedron and $P$ a given point inside it. We want to construct a triangle $QRS$ having $P$ as centroid and its vertices on faces $ABD$, $ABC$ and $ACD$ (we can choose any three faces at will). Start by choosing an arbitrary point on one of the faces, for instance point $S$ on $ACD$. Join $SP$ and extend it to $M$, such that $PS=2PM$. Point $M$ is then the midpoint of side $QR$ of the triangle.

Construct then $E$ on $AB$ as the perpendicular projection of $M$; from it draw line $EJ\perp AB$ on face $ABD$. Choose $J$ such that $MJ\perp EJ$ and extend $EJ$ to $L$ so that $EJ=JL$. Finally, from $L$ construct a line parallel to $AB$: any point $Q$ on that line can be chosen as the second vertex of the triangle, while the third vertex $R$ is the reflection of $Q$ about $M$.

Here's the construction made with GeoGebra

The construction works because $R$ lies on face $ABC$. In fact we can construct on $ABC$ point $K$ such that $EKMJ$ is a parallelogram, and point $N$ such that $K$ is the midpoint of $EN$. Point $N$ is then the reflection of $L$ about $M$ and line $NR$ is parallel to $LQ$, lying thus on $ABC$.

The construction fails if points $M$ or $R$ lie outside the tetrahedron, but we can avoid that with a different choice of points $S$ and $Q$, or changing the face where $S$ lies.