In problem 983 (see the solution in the 3-4 issue of Volume XX), we have shown that the convex body, whose vertices are the points dividing the edges of a regular tetrahedron into 3 equal parts, has 12 body diagonals. What is the length of the diagonals if the edge of the convex body is $b$? How far are the diagonals from the center of the body? How many intersection points do the body diagonals determine?
Let the points dividing the 3b length edges of the regular tetrahedron $P Q R S=T$ into 3 equal parts - the vertices of the convex body $U$ in question - be $A, B, C, D, E, F, G, H, J, K, L, M$ as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_8548b0b544a81497a66dg-1.jpg?height=837&width=724&top_left_y=232&top_left_x=690)

From each vertex, diagonals can be drawn to the vertices dividing the opposite edge into thirds, thus the 12 body diagonals are as follows:

$A J, A K, B L, B M, C G, C H, D J, D K, E L, E M, F G, F H$.

All diagonals are of equal length because any vertex can be rotated to any other vertex by a suitable rotation, and thus any diagonal can be rotated to the position of any other diagonal. Indeed, $T$ is mapped onto itself by $120^{\circ}$ rotations about the altitudes, so any edge can be mapped to any other edge - opposite, i.e., not sharing a vertex, edges by two such rotations - and the midpoints of opposite edges can be swapped by a $180^{\circ}$ rotation about the line connecting them. These properties are inherited by $U$ because its vertices are constructed uniformly and symmetrically with respect to the midpoints of $T$'s edges.

Furthermore, $U$ has a center because, as stated above, all its vertices are equidistant from $T$'s center $O$, so $O$ is also the center of $U$. $O$ is also the centroid of $T$ due to regularity, and thus the point one quarter of the way along the altitudes from the faces. The body diagonals are equidistant from $O$ because a sphere can be circumscribed around $U$ with center $O$, the diagonals being equal chords of this sphere, and equal chords of a sphere are equidistant from the sphere's center.

$U$ has 18 edges and the diagonals of its hexagonal faces, which are $2b$ in length - a total of $4 \cdot 3=12$ - are in sets of five parallel to each other, with a common direction being the direction of one of $T$'s edges; for example, $A B\|D E\| G H\|M J\| L K$. Therefore, a plane through two non-coplanar such segments intersects $U$ in a trapezoid or possibly a rectangle. Each body diagonal passes through such a plane section. The quadrilaterals $A B K L=V_{1}$ and $D E J M=V_{2}$ are rectangles because their sides $A L$ and $D M$ are parallel to $R S$ and thus perpendicular to $P Q$.

1. Let's determine, for example, the length of the body diagonal $A K$. In $V_{1}$, $A B=b$ and $B K=2 b$, because $B K$ is one side of the equilateral triangle $B K Q$, so by the Pythagorean theorem, $A K=b \sqrt{5}$.
2. It is easily shown that the height of $T$ is $P Q \sqrt{2 / 3}=b \sqrt{6}$. Therefore, with $N$ being the centroid of the face $P Q R$, $O N=b \sqrt{6} / 4$. On the other hand, $N K=b$, so from the right triangle $O K N$, $O K=b \sqrt{22} / 4$. Now, from the isosceles triangle $A K O$, the distance of $O$ from $A K$ is $d=b \sqrt{2} / 4$.
3. To find the number of intersection points of the body diagonals, it suffices to determine, for example, the number of intersection points on the diagonal $A K$. Let's consider the other diagonals one by one alongside $A K$. $A J$ and $K D$ share a vertex with $A K$, which we do not count as an intersection. In the rectangle $V_{1}$, $B L$ intersects $A K$ at its midpoint. The plane of $V_{2}$ is parallel to the plane of $V_{1}$ and different from it, because it intersects the plane $P R S$ in the distinct lines $D M$ and $A L$. Thus, the diagonals $D J$ and $E M$ of $V_{2}$ do not intersect $A K$. (Otherwise, $A K$ and $D J$ would be skew, because the side $A B=b$ of $V_{1}$ is parallel to the side $D E=2 b$ of $V_{2}$, and thus the corresponding diagonals are not parallel; - however, this is irrelevant now.) - Now consider the body diagonals in the rectangular or trapezoidal sections passing through the edges leading to $A$ and $K$. In the trapezoidal section $A B J M$ through the edge $A B$, the diagonal $B M$ does not intersect $A K$, because the intersection line of the trapezoid with $V_{1}$, which contains $A K$, is $A B$, and $B M$ does not intersect it at $A$. In the trapezoid $A C K G$ through the edge $A C$, $C G$ intersects $A K$ at the point closer to $A$, because $A C: G K=1: 2$; however, the diagonal $C H$ in $A C J H$ does not intersect $A K$. There are no such sections through $A D$ and $K J$. Similarly, in the trapezoid $K F A H$, the diagonal $F H$ intersects $A K$ at the point closer to $K$, and in $K F D G$, the diagonal $F G$ and in $K L D E$, the diagonal $K E$ do not intersect $A K$. With this, we have examined all body diagonals: the body diagonal $A K$ is intersected internally by 3 other diagonals, each at a different point.[^0]

Since each of the 12 diagonals has 3 intersection points, and each intersection point belongs to two diagonals, the number of internal intersection points of the diagonals is $12 \cdot 3 / 2=18$. In a broader sense, including the vertices as intersection points of the diagonals, the number of common points of the diagonals is $18+12=30$.

Farkas Zoltán (Hódmezővásárhely, Bethlen G. g. II. o. t.) Gálfi László (Budapest, Fazekas M. gyak. g. II. o. t.)


[^0]:    ${ }^{2}$ L. Molnár Ferenc: On the notable points of the tetrahedron. K. M. L. 16 (1958) 1-6. and 33-38. p., specifically p. 1.