The 34th question: A set of $n$ points $P_{1}, P_{2}, \ldots, P_{n}$ in the plane, denoted as $D$, satisfies that no three points are collinear. A line segment is drawn between any two points, and the lengths of all these line segments are distinct. In a triangle, the side that is neither the longest nor the shortest is called the "middle side" of the triangle; if all three sides of a triangle are "middle sides" (which is necessarily the middle side of this triangle), then this triangle is called a "middle side triangle" in the set $D$. A line $l$ that does not pass through any point $P_{i} (1 \leq i \leq n)$ divides the set $D$ into two subsets $D_{1}$ and $D_{2}$. If, regardless of how these $n$ points are distributed and regardless of how $l$ is chosen, there always exists a subset $D_{k} (k \in \{1,2\})$ such that $D_{k}$ contains a middle side triangle, find the minimum value of $n$.

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The 34th question: A set of $n$ points $P_{1}, P_{2}, \ldots, P_{n}$ in the plane, denoted as $D$, satisfies that no three points are collinear. A line segment is drawn between any two points, and the lengths of all these line segments are distinct. In a triangle, the side that is neither the longest nor the shortest is called the "middle side" of the triangle; if all three sides of a triangle are "middle sides" (which is necessarily the middle side of this triangle), then this triangle is called a "middle side triangle" in the set $D$. A line $l$ that does not pass through any point $P_{i} (1 \leq i \leq n)$ divides the set $D$ into two subsets $D_{1}$ and $D_{2}$. If, regardless of how these $n$ points are distributed and regardless of how $l$ is chosen, there always exists a subset $D_{k} (k \in \{1,2\})$ such that $D_{k}$ contains a middle side triangle, find the minimum value of $n$.
Question 34:
Solution: First, we prove a lemma: Given any 6 points, with all pairwise distances being distinct, there must be a median-edge triangle.

Proof of the lemma: If an edge is the median edge of a triangle, then color it red; otherwise, color it blue. According to Ramsey's theorem, in a two-colored \( \mathrm{K}_{6} \), there must exist a monochromatic triangle. If there is a blue triangle, then it must contain a median edge, which should be colored red, leading to a contradiction. Therefore, there must be a red triangle, which is a median-edge triangle. The lemma is proved! Now, let's return to the original problem.
When \( n \geq 11 \), regardless of how \( l \) is chosen, one of \( D_{1} \) or \( D_{2} \) must contain at least 6 points. According to the lemma, there must be a median-edge triangle in that subset.
Next, we prove that when \( n = 10 \), there exist a point set \( D \) and a line \( l \) such that neither \( D_{1} \) nor \( D_{2} \) contains a median-edge triangle. Consider the two subsets \( D_{1} \) and \( D_{2} \) on either side of the line \( l \), each containing 5 points, and they are distributed identically.
Assume the 5 points in \( D_{1} \) are \( P_{1}, P_{2}, P_{3}, P_{4}, P_{5} \), arranged in a counterclockwise order on a circle. Let the arc lengths \( \widehat{P_{1}P_{2}}: \widehat{P_{2}P_{3}}: \widehat{P_{3}P_{4}}: \overline{P_{4}P_{5}}: \widehat{P_{5}P_{1}} = 2: 12: 6: 15: 5 \). Then, the pairwise distances between \( P_{1}, P_{2}, P_{3}, P_{4}, P_{5} \) are all distinct, and exactly \( P_{2}P_{3}, P_{3}P_{1}, P_{1}P_{5}, P_{5}P_{4}, P_{4}P_{2} \) are median edges, and there is no median-edge triangle. Similarly, there is no median-edge triangle in \( D_{2} \).
In summary, the minimum value of \( n \) is 11.
Note: If \( n \) points \( P_{1}, P_{2}, \ldots, P_{n} \) in the plane, no three of which are collinear, form a set \( D \), and any \( m \) lines in the plane, none of which pass through any point in \( D \), divide \( D \) into several subsets, if regardless of how the \( n \) points are distributed and how the \( m \) lines are chosen, there must be a subset containing a median-edge triangle, then the minimum value of \( n \) is \( \frac{5m(m+1)}{2} + 6 \). This is because \( m \) lines can divide the plane into at most \( 1 + \frac{m(m+1)}{2} \) regions, and thus can divide \( D \) into at most \( 1 + \frac{m(m+1)}{2} \) subsets.