5. If on each face of a cube we draw one of the two diagonals of that face, we find that some of the drawn diagonals touch - they emanate from a common vertex. Let the number of pairs of touching diagonals be denoted by $N$, where a single diagonal can also appear in multiple pairs. Determine the maximum and minimum possible value of the number $N$.

## 52nd Mathematical Competition for High School Students in Slovenia

Optional Competition, April 2, 2008

## Solutions to Problems

Each problem is worth 7 points. All mathematically and logically correct solutions are equally valid. When evaluating each problem, reasonably consider the provided scoring guide. A competitor should not receive more than 3 points for a particular problem if the partial solution does not clearly lead to the final solution of the problem.
IV/5. In the first image, $N=12$, and in the second, $N=4$. Let's prove that these are the maximum and minimum possible values of $N$.
![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-18.jpg?height=428&width=1034&top_left_y=340&top_left_x=518)

There are 8 vertices of the cube. There are 6 drawn diagonals, and thus 12 endpoints of the diagonals. From a vertex of the cube, 0, 1, 2, or 3 diagonals can emanate. Therefore, each vertex of the cube can contribute 0, 1, or 3 different pairs of touching diagonals.

Suppose we can achieve $N \leq 3$. If 3 diagonals emanate from a vertex, then from the remaining vertices, at most one diagonal can emanate. This would mean that the number of endpoints of the diagonals is at most $1 \cdot 3 + 7 \cdot 1 = 10$, which is not possible. Therefore, in no vertex do three diagonals touch, and we can have at most 3 vertices where two diagonals touch. From each of the remaining vertices, at most one diagonal can emanate. Thus, the total number of endpoints of the diagonals is at most $3 \cdot 2 + 5 \cdot 1 = 11$. This is a contradiction. Therefore, for any selection of allowed diagonals, $N \geq 4$.

The total number of different pairs of diagonals is $\frac{6 \cdot 5}{2} = 15$. Diagonals lying on opposite faces of the cube cannot touch. Therefore, there are at most $15 - 3 = 12$ different pairs of touching diagonals. Therefore, for any selection of allowed diagonals, $N \leq 12$.

Diagram or listing of diagonals with coordinates for $N=4 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

Diagram or listing of diagonals with coordinates for $N=12 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ Conclusion that from each vertex, 0 to 3 diagonals can emanate, and this means 0, 1, or 3 pairs 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-18.jpg?height=46&width=1625&top_left_y=1662&top_left_x=221)

Conclusion that it is not possible to achieve $N \geq 13$.................................................. 2 points


[^0]:    Conclusion that $K^{\prime}, L^{\prime}, K, L$ are concyclic 1 point

    Conclusion $\frac{|A B|}{|A C|}=\frac{|B L|}{|C L|}$ 1 point

    Conclusion $\frac{|A D|}{|A C|}=\frac{|D K|}{|C K|}$ 1 point

    ![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-15.jpg?height=68&width=1633&top_left_y=2056&top_left_x=220)
    Conclusion that triangles $D B C$ and $K L C$ are similar ........................ 1 point

    ![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-15.jpg?height=48&width=1633&top_left_y=2169&top_left_x=220)

    ![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-15.jpg?height=49&width=1633&top_left_y=2214&top_left_x=220)
    (If the concyclicity of $K^{\prime}, L^{\prime}, K, L$ is proven and it is shown that the concyclicity of $B, D, L^{\prime}, K^{\prime}$ is equivalent to the statement that $K L$ is parallel to $B D$, but the ratios are not used, award 2 points.)