Assumption:

• Let $S$ be a circle with center $O$ and radius 5.
• Let $A$ be a point inside the circle $S$ and $B$ be a point on the boundary of $S$, i.e., $\|B - O\| = 5$.
• Let $d(A, B)$ denote the distance between points $A$ and $B$.
• The interval $(0, 5)$ represents the possible values of $c$.

Want to show:

• For every $c \in (0, 5)$, there exists a point $A$ inside $S$ such that the distance between $A$ and any point $B$ on the boundary of $S$ is strictly greater than $c$.

Proof:

1. Position of Point $A$:

   • Since $A$ is inside the circle, the distance from $A$ to the center $O$ is less than 5, i.e., $\|A - O\| = r$ where $0 \leq r < 5$.
   • Let $B$ be any point on the boundary of the circle $S$. Since $B$ is on the boundary, the distance from $B$ to $O$ is exactly 5, i.e., $\|B - O\| = 5$.

2. Distance Formula between $A$ and $B$:

   • The distance between $A$ and $B$ can be derived using the law of cosines. Let $\theta$ be the angle between the vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$. $d(A, B) = \sqrt{r^2 + 5^2 - 2 \cdot r \cdot 5 \cdot \cos(\theta)}.$ Since $r < 5$, the minimum value of $d(A, B)$ occurs when $\theta = 0$ (i.e., when $A$ and $B$ lie along the same radius): $d_{\text{min}}(A, B) = 5 - r.$ The maximum distance occurs when $\theta = \pi$ (i.e., $A$ and $B$ are directly opposite each other along a diameter): $d_{\text{max}}(A, B) = 5 + r.$

3. Conditions for $d(A, B) > c$:

   • We want to show that for any $c \in (0, 5)$, there exists an $A$ such that $d(A, B) > c$ for all $B$ on the boundary of $S$.
   • The minimum distance $d_{\text{min}}(A, B) = 5 - r$. For $d(A, B) > c$ for all $B$, it must hold that: $5 - r > c \quad \Rightarrow \quad r < 5 - c.$ This implies that as long as $A$ is positioned such that its distance from the center $O$ is less than $5 - c$, the distance between $A$ and any point $B$ on the boundary will be greater than $c$.

4. Existence of such a point $A$:

   • Since $c \in (0, 5)$, we can always find $r \in [0, 5 - c)$. In this range, the point $A$ exists inside the circle $S$ with $\|A - O\| = r$, ensuring that the distance between $A$ and every point $B$ on the boundary of $S$ satisfies $d(A, B) > c$.

Conclusion:

For every $c \in (0, 5)$, there exists a point $A$ inside the circle $S$ such that the distance between $A$ and every point $B$ on the boundary of $S$ is greater than $c$. Thus, the statement is proved.

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Let me know if you need further details or explanations!

5 Related Questions:

1. How does the law of cosines apply to calculating distances between points on a circle?
2. How would the proof change if the circle had a different radius?
3. What happens if point $A$ lies on the boundary of the circle?
4. How can this proof be adapted for a higher-dimensional sphere?
5. What other geometric properties are derived from a circle’s boundary?

Tip:

Always sketch diagrams for geometric proofs. They help clarify the relationships between points, angles, and distances.