Problem 1645

1. Understanding the Problem:
We are given six pairwise non-intersecting circles, each with a radius of at least 1. We need to prove that any circle intersecting all six of these circles must have a radius of at least 1.

2. Notation and Setup:
Let the centers of the six given circles be $O_1, O_2, \ldots, O_6$ and their radii be $r_1, r_2, \ldots, r_6$ respectively. We know that $r_i \geq 1$ for all $i$. Let $O$ be the center and $r$ be the radius of the circle that intersects all six given circles.

3. Geometric Considerations:
Since the circle centered at $O$ intersects all six circles, the distance from $O$ to each $O_i$ must be at most $r + r_i$. That is,
$$OO_i \leq r + r_i \quad \text{for all } i.$$

4. Pairwise Non-Intersecting Property:
Given that the six circles are pairwise non-intersecting, the distance between any two centers $O_i$ and $O_j$ must be at least $r_i + r_j$. That is,
$$O_iO_j \geq r_i + r_j \quad \text{for all } i \neq j.$$

5. Angle Consideration:
Consider the angles formed by the lines connecting $O$ to $O_i$ and $O_j$. Since there are six circles, by the pigeonhole principle, there must be at least one pair of circles such that the angle $\angle O_iOO_j$ is at most $60^\circ$.

6. Distance Inequality:
For such a pair $(O_i, O_j)$ with $\angle O_iOO_j \leq 60^\circ$, we can use the Law of Cosines in $\triangle OO_iO_j$:
$$O_iO_j^2 = OO_i^2 + OO_j^2 - 2 \cdot OO_i \cdot OO_j \cdot \cos(\angle O_iOO_j).$$
Since $\cos(60^\circ) = \frac{1}{2}$, we have:
$$O_iO_j^2 = OO_i^2 + OO_j^2 - OO_i \cdot OO_j.$$

7. Bounding the Distances:
Using the inequalities $OO_i \leq r + r_i$ and $OO_j \leq r + r_j$, we get:
$$O_iO_j \leq (r + r_i) + (r + r_j) = 2r + r_i + r_j.$$
However, from the pairwise non-intersecting property, we also have:
$$O_iO_j \geq r_i + r_j.$$

8. Combining Inequalities:
Combining the above inequalities, we get:
$$r_i + r_j \leq 2r + r_i + r_j.$$
Simplifying this, we find:
$$0 \leq 2r.$$
This is always true, but it does not provide a lower bound for $r$. We need a stricter inequality.

9. Stricter Bound:
Consider the case where $O_iO_j > r_i + r_j$. Then:
$$r + r_i \geq OO_i \geq O_iO_j - OO_j \geq (r_i + r_j) - (r + r_j) = r_i - r.$$
This implies:
$$r \geq r_i - r.$$
Since $r_i \geq 1$, we have:
$$r \geq 1 - r.$$
Solving for $r$, we get:
$$2r \geq 1 \implies r \geq \frac{1}{2}.$$

10. Final Bound:
To ensure that $r$ is at least 1, we need to consider the worst-case scenario where $r_i = 1$. In this case, the circle centered at $O$ must intersect all six circles, and the minimum radius $r$ must be at least 1 to ensure intersection with all six circles.

$\blacksquare$