Question 52: Let the positive integer $\mathrm{n} \geq 3$, find the maximum number of diagonals that can be selected from a regular $\mathrm{n}$-sided polygon such that any two of the selected diagonals either do not intersect internally or are perpendicular to each other.
Question 52:
Answer: When $n$ is odd, the maximum number of diagonals that can be taken is $n-3$; when $n$ is even, the maximum number of diagonals that can be taken is $n-2$.

First, we use mathematical induction to prove: for a convex $\mathrm{n}$-gon, the maximum number of diagonals that can be selected such that no two diagonals intersect inside the polygon is $\mathrm{n}-3$.
When $n=3, 4$, the proposition is obviously true.
If the proposition holds for convex polygons with fewer than $\mathrm{n}$ sides, in a convex $\mathrm{n}$-gon, select one diagonal. This diagonal divides the original polygon into two smaller convex polygons, one with $\mathrm{m}$ sides and the other with $\mathrm{n}-\mathrm{m}+2$ sides. The other diagonals that can be selected are the diagonals of these two smaller polygons, and they do not intersect inside their respective polygons. Therefore, the total number of diagonals that can be selected is no more than: $1+(m-3)+(n-m+2-3)=n-3$.
Furthermore, in a convex $\mathrm{n}$-gon, $\mathrm{n}-3$ diagonals drawn from one point satisfy the condition.
The proposition is proved.
Let the regular $n$-gon be $A_{1} A_{2} \ldots A_{n} \circ$ Suppose diagonals $A_{i} A_{j}(i<j)$ and $A_{s} A_{t}(i<s<t)$ intersect and are perpendicular inside the regular $n$-gon, then $s<j$, and $(s+t)-(i+j)=\frac{n}{2}$, thus, $n$ is even.

When $\mathrm{n}$ is odd, any two intersecting diagonals are not perpendicular, so the selected diagonals do not intersect inside the polygon. Therefore, according to the above proposition, the maximum number of diagonals that can be selected is $\mathrm{n}-3$.

When $\mathrm{n}$ is even, if the selected diagonals do not intersect inside the polygon, then the maximum number of diagonals that can be selected is also $\mathrm{n}-3$. If two diagonals that intersect inside the polygon are selected, consider the subset $\mathrm{U}$ of diagonals that are parallel to one of these two diagonals and intersect with other diagonals. Suppose $\mathrm{U}$ contains $\mathrm{u}$ diagonals and has $\mathrm{v}$ different endpoints.

Consider the longest diagonal in set $U$ in one direction, then the other diagonals drawn from the endpoints of this diagonal are not in $U$. Otherwise, select a diagonal that intersects with this diagonal, it is parallel to the longest diagonal, and is in $U$, but it is longer than the longest diagonal, which is a contradiction.
Similarly, the longest diagonal in set $\mathrm{U}$ in the other direction also has this property.
In set $U$, excluding these two diagonals and their four endpoints, there are $u-2$ diagonals and $v-4$ different endpoints left. Since each endpoint can belong to at most two diagonals, we have
$$
2(u-2) \leq 2(v-4) \Rightarrow u \leq v-2
$$

Suppose the endpoints in set $U$ divide the vertices of the polygon into $A_{1}, A_{2}, \ldots, A_{v}$ parts, where $A_{j}$ part consists of $a_{i}$ consecutive adjacent non-$U$ vertices and the vertices on both sides of $U$. Diagonals between different parts cannot be selected, otherwise, they would intersect with the diagonals in $U$. Furthermore, the vertices in the same part must belong to a semicircle arc, and no two perpendicular intersecting diagonals can be selected from its vertices. Therefore, the maximum number of diagonals that can be selected from $A_{i}$ part is: $1+\left(a_{i}+2-3\right)=a_{i}$.
Hence, the maximum number of diagonals that can be selected is:
$$
\sum_{i=1}^{v} a_{i}+u=n-v+u \leq n-2
$$

On the other hand, for any vertex $\mathrm{A}$ of a regular $\mathrm{n}$-gon, let $\mathrm{A}^{\prime}$ be the antipodal point of $\mathrm{A}$. Select all diagonals drawn from point $\mathrm{A}$ and diagonal $d$, where $d$ is the diagonal connecting the two vertices adjacent to $A^{\prime}$. At this time, only one diagonal $A A^{\prime}$ intersects with $\mathrm{d}$, and they are perpendicular. These diagonals total $\mathrm{n}-2$.
In summary, when $n$ is even, the maximum number of diagonals that can be selected is $n-2$.