## Task 34/66

How many space diagonals does a rhombic dodecahedron have?
The rhombic dodecahedron consists of 12 bounding rhombi with 14 vertices and 24 edges.

1. Analytical solution: From 14 vertices, $\binom{14}{2}=91$ pairs can be formed; each pair determines a connecting line. Therefore, 14 vertices can be paired in 91 different ways. Of these 91 connections, 24 edges and 24 face diagonals (2 per rhombus) must be excluded, leaving 43 space diagonals.
2. Intuitive solution: At 6 vertices, 4 rhombi meet, and at 8 vertices, 6 rhombi meet. Each "quadruple" vertex $E_{4}$ has 8 neighboring vertices to which no space diagonals lead; the remaining 5 vertices are "distant vertices" with space diagonals. Thus, we initially identify $6 \cdot 5=30$ space diagonals.

Correspondingly, each $E_{3}$ has 6 neighboring and 7 distant vertices. Therefore, there are an additional $7 \cdot 8=56$ space diagonals, totaling 86. However, each space diagonal has clearly been counted twice, once from each vertex; thus, the rhombic dodecahedron has $\frac{86}{2}=43$ space diagonals.