considers the following proof. Now I'd like you to show that the proof is no longer true if we place ourselves in a two-dimensional space just as in the proof, but where the space is bounded by two intervals X and Z [min_x ; max_x ] and [min_z ; max_z] the proof is no longer valid. You can assume that the point p_k is not collinear (not on the same straight line) with p_i and p_r, but it is in the same direction as this straight line, p_k is behind p_r, it is very, very close to the straight line that passes through p_k and p_r and the boundaries of the space are small enough that there is no intersection. Generate a python code to visualize it. You can draw, in the python code, the straight line connecting p_r to p_i, and the one connecting p_r to p_k. I intuitively believe that the smaller the angle between these two lines, the less likely they are to intersect. Proof : 
   - **Definitions**: The regions \( R(a, b) \) are defined as half-planes consisting of all points closer to \( a \) than to \( b \), i.e., \( R(a, b) = \{ x \in \mathbb{R}^2 : |x - a| \leq |x - b| \} \).
   - **Goal**: Prove that for any point \( p_k \) not on the line through \( p_i \) and \( p_r \), the regions \( R(p_k, p_i) \) and \( R(p_i, p_r) \) have a non-empty intersection, i.e., \( R(p_k, p_i) \cap R(p_i, p_r) \neq \emptyset \).

**Placing \( p_i \) at the Origin**:
   - Without loss of generality, you assume that \( p_i \) is at the origin \((0, 0)\). This is valid since any rigid transformation that moves \( p_i \) to the origin will preserve the distances between points, and thus the structure of the bisectors and regions.

**Bisectors**:
   - You correctly identify that the bisectors are key. The bisector between \( p_i \) and \( p_r \) is the set of points equidistant from \( p_i \) and \( p_r \), which forms a line, as shown:
   \[
   \sqrt{x^2 + y^2} = \sqrt{(x - x_r)^2 + (y - y_r)^2}
   \]
   After squaring and simplifying, you derive the equation of the bisector \( B_{i,r} \):
   \[
   2x x_r + 2y y_r = x_r^2 + y_r^2
   \]
   Similarly, you find the equation for the bisector between \( p_i \) and \( p_k \), \( B_{k,i} \), which is:
   \[
   2x x_k + 2y y_k = x_k^2 + y_k^2
   \]

**Intersection of the Half-Planes**:
   - You rightly conclude that since \( p_k \) is not on the line through \( p_i \) and \( p_r \), the bisectors \( B_{i,r} \) and \( B_{k,i} \) cannot be parallel—they must intersect at a single point.
   - As a result, the half-planes \( R(p_i, p_r) \) and \( R(p_k, p_i) \) form a convex wedge, implying that their intersection is non-empty. This step is crucial, as the convexity of the regions ensures that any intersection between the bisectors leads to a non-empty overlap of the half-planes.

**Convexity and Non-Empty Intersection**:
   - The argument here is correct: the regions are convex, and the intersection of convex sets is also convex, so the intersection must contain points (unless the regions are disjoint, which you argue only happens if \( p_k \) is on the line through \( p_i \) and \( p_r \)).

**Collinear Case**:
   - In the case where \( p_k \) lies on the line through \( p_i \) and \( p_r \), the bisectors \( B_{i,r} \) and \( B_{k,i} \) may coincide or be parallel, and the regions may not overlap. This correctly addresses the exceptional case where the intersection could be empty, aligning with the hypothesis that \( p_k \) must not be on the line through \( p_i \) and \( p_r \) for the intersection to be guaranteed non-empty.