Billy and Bobby are located at points $A$ and $B$, respectively. They each walk directly toward the other point at a constant rate; once the opposite point is reached, they immediately turn around and walk back at the same rate. The first time they meet, they are located 3 units from point $A$; the second time they meet, they are located 10 units from point $B$. Find all possible values for the distance between $A$ and $B$.

[i]Proposed by Isabella Grabski[/i]
1. Let the distance between points \( A \) and \( B \) be \( x \).

2. The first encounter happens before either Billy or Bobby has turned around. At this point, Billy has walked 3 units, and Bobby has walked \( x - 3 \) units. Since they meet at this point, the ratio of their speeds is given by:
   \[
   \frac{\text{Billy's speed}}{\text{Bobby's speed}} = \frac{3}{x-3}
   \]

3. The second encounter happens when they are 10 units away from point \( B \). We need to consider the possible scenarios for this encounter:
   - (a) Only Billy has turned around.
   - (b) Only Bobby has turned around.
   - (c) Both have turned around.

4. Scenario (a) is impossible because \( x > 10 \) and Bobby, who walks faster, will turn around first.

5. For scenario (b), at the time of the second meeting:
   - Bobby has walked \( x \) units to reach \( A \) and \( x - 10 \) units back towards \( B \).
   - Billy has walked \( x - 10 \) units towards \( B \).

   Since their speeds are constant, the ratio of their distances traveled must be the same as before:
   \[
   \frac{3}{x-3} = \frac{x-10}{2x-10}
   \]

6. Cross-multiplying to clear the fractions:
   \[
   3(2x-10) = (x-3)(x-10)
   \]
   \[
   6x - 30 = x^2 - 13x + 30
   \]
   \[
   0 = x^2 - 19x + 60
   \]

7. Solving the quadratic equation:
   \[
   x^2 - 19x + 60 = 0
   \]
   \[
   (x-4)(x-15) = 0
   \]

8. Since \( x > 10 \), the viable solution is:
   \[
   x = 15
   \]

9. For scenario (c), using the same strategy:
   \[
   \frac{3}{x-3} = \frac{x+10}{2x-10}
   \]
   \[
   3(2x-10) = (x-3)(x+10)
   \]
   \[
   6x - 30 = x^2 + 7x - 30
   \]
   \[
   0 = x^2 + x
   \]
   \[
   x(x+1) = 0
   \]

10. This has no meaningful solutions since \( x \) must be positive.

Therefore, the only solution is \( \boxed{15} \).