Problem 1587

1. Define the vertices of the rectangle and the points in the plane:
Let $W, X, Y, Z$ be the vertices of the rectangle inscribed in the circle of radius 1. Let $A_1, A_2, \ldots, A_{2002}$ be arbitrary points in the plane.

2. Sum of distances from a point to the vertices of the rectangle:
Consider a point $A$ in the plane. The sum of the distances from $A$ to the vertices of the rectangle is given by:
$$WA + XA + YA + ZA$$

3. Use the properties of the rectangle and the circle:
Since the rectangle is inscribed in the circle, the diagonals of the rectangle are equal to the diameter of the circle, which is 2 (since the radius is 1). Let $d_1$ and $d_2$ be the lengths of the diagonals of the rectangle. Then:
$$d_1 = d_2 = 2$$

4. Apply the triangle inequality:
For any point $A$ in the plane, the sum of the distances from $A$ to the vertices of the rectangle can be split into pairs of opposite vertices:
$$WA + YA \geq WY \quad \text{and} \quad XA + ZA \geq XZ$$
Since $WY = XZ = 2$, we have:
$$WA + YA + XA + ZA \geq 4$$

5. **Sum over all points $A_1, A_2, \ldots, A_{2002}$:**
Summing the above inequality over all points $A_1, A_2, \ldots, A_{2002}$, we get:
$$\sum_{k=1}^{2002} (WA_k + YA_k + XA_k + ZA_k) \geq 4 \times 2002 = 8008$$

6. Select the largest three terms:
Since the sum of the distances from all four vertices to all points is at least 8008, the sum of the distances from the three vertices with the largest sums will be at least:
$$\frac{3}{4} \times 8008 = 6006$$

Therefore, there exist three vertices $M, N, P$ of the rectangle such that:
$$\sum_{k=1}^{2002} MA_k + \sum_{k=1}^{2002} NA_k + \sum_{k=1}^{2002} PA_k \geq 6006$$

$\blacksquare$