Bimedians in quadrangular pyramids.

Question

I know that in every pyramid the median connecting base with vertex is divided by the center of mass in the ratio 1:3. Addicinally, in tetrahedrons all 4 medians and 3 bimedians (joining the midpoints of two opposite edges) intersect in the same point. The question is: is there any equivalent of bimedians in quadrangular pyramids? I mean - do the line segments joining the midpoints of the base edges with the centroids of lateral faces or line segments joining the midpoints of the side edges with the vertexes in the base intersect in the center of mass? Are there any other line segments crossing the center of mass in every quadrangular pyramids?

Answer 1

The line segments joining the midpoints of the base edges with the centroids of lateral faces (such as $NF$ and $HI$ in figure below) do intersect at the same point $E$ (different from the centroid) in any quadrangular pyramid.

To see why, consider that line $HN$ is parallel to $AC$, while $IF$ is parallel to $ML$, which in turn is parallel to $AC$. Hence $NH$ and $IF$ are also parallel and points $NHFI$ are coplanar, being the vertices of a trapezoid. Thus $NF$ and $HI$ intersect at a point $E$: from $NH/AC=1/2$ and $IF/AC=1/3$ it follows $NH/IF=NE/EF=HE/EI=3/2$. The same reasoning can be carried out for the other lines joining the midpoints of the base edges with the centroids of lateral faces, which then all meet at point $E$.

In addition, notice that $NF$ lies on plane $NVL$ (where $L$ is the midpoint of $CD$), while $HI$ lies on plane $VHM$ (where $M$ is the midpoint of $AD$). Hence $E$ lies on the intersection of those planes, i.e. on the line joining vertex $V$ of the pyramid with the intersection $E'$ of the bimedians $NL$ and $HM$ of the base. But $E'$ is not, in general, the same as the centroid $G'$ of the base, as $G'$ is the intersection of the lines joining the centroids of the opposite triangles formed by the diagonals of the base.

In any case, point $E$ is not the same as the centroid $G$ of the pyramid: just notice that the distance of $E$ from the base is $(3/5)(1/3)=1/5$ of the height, while the distance of $G$ from the base is $1/4$ of the height.

On the other hand, line segments joining the midpoints of side edges with opposite base vertices (such as $SB$ and $TC$ in figure below) are, in general, skew.

In fact, line $ST$ is parallel to edge $AD$ and to the plane of the base. Hence any plane through $ST$ should intersect the plane of the base at a line parallel to $ST$ and $AD$. If edge $BC$ is not parallel to $AD$, points $BCST$ cannot then lie on the same plane.

Only if the base is a parallelogram those lines all meet at the same point $K$, for in that case $BC$ and $ST$ are parallel and, moreover, $K$ divides each line in the ratio $2:1$. Point $K$ lies on the line joining vertex $V$ with the center of the base, but it is different from centroid $G$ and from point $E$ defined above, because the distance of $K$ from the base is $(2/3)(1/2)=1/3$ of the height.

For a parallelogram base, $E'$ and $G'$ also lie in the center, hence $E$, $G$ and $K$ all lie on the same line, but at different positions.