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We are given a rectangle \( ABCD \) with center \( O \) and the ratio of side lengths \( \frac{AB}{AD} = k \). A point is chosen randomly from the interior of the rectangle, and we are tasked with finding the probability that the point is closer to \( O \) than to any of the vertices \( A \), \( B \), \( C \), or \( D \).

### Step-by-Step Approach

1. **Geometrical Setup**:
   - Let the dimensions of the rectangle be \( AB = w \) (width) and \( AD = h \) (height), such that \( \frac{w}{h} = k \).
   - The center \( O \) is equidistant from all four sides of the rectangle.
   - We need to determine the region inside the rectangle where any point is closer to \( O \) than to any of the vertices.

2. **Region of Interest**:
   - The boundary where a point is equidistant to \( O \) and a vertex forms a parabola. For each vertex, the set of points closer to \( O \) than to that vertex is inside this parabola.
   - The required region is the intersection of the areas inside these parabolas for all four vertices.
   - The resulting region is known as the **medial region**, which will be symmetric about \( O \).

3. **Probability Calculation**:
   - The probability is the ratio of the area of the medial region (where points are closer to \( O \) than the vertices) to the area of the entire rectangle.

   - The exact area of this medial region for rectangles is non-trivial to compute directly, but in many similar problems, the solution depends on the ratio of the side lengths \( k = \frac{w}{h} \).
   
   - For a square (i.e., \( k = 1 \)), it is known that the medial region occupies approximately \( 1/2 \) of the area of the square. For rectangles with \( k \neq 1 \), the exact ratio will depend on \( k \), but the answer will generally be less than \( 1/2 \) as \( k \) deviates from 1.

### Conclusion:

- The probability depends on the ratio \( k = \frac{AB}{AD} \), and for general rectangles, this requires complex geometric analysis.
  
- For the specific case where \( k = 1 \) (the square), the probability is approximately \( \frac{1}{2} \).
  
Would you like more details on how the medial region is calculated or specific examples with other values of \( k \)? 

### Follow-up Questions:
1. How is the medial region derived geometrically for a rectangle?
2. How does the probability change as \( k \) increases or decreases from 1?
3. What happens if the rectangle is long and thin (i.e., \( k \gg 1 \))?
4. Can this problem be generalized for higher dimensions (e.g., cubes)?
5. How is this problem solved for different shapes like triangles or circles?

**Tip**: Problems involving distances between points and geometric shapes often involve loci (like parabolas) and can be solved using symmetry and area ratios.

Rectangle ABCD has center O and AB/AD = k. A point is randomly chosen from the interior of rectangle ABCD. What is the probability that it is closer to O than to any of the four vertices?

Explore the probability that a randomly chosen point inside a rectangle is closer to its center than to any of its vertices. This problem involves concepts of geometry, symmetry, and the properties of loci within geometric shapes. Suitable for advanced high school students in Grades 11-12.

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