[b]p1.[/b] Two trains, $A$ and $B$, travel between cities $P$ and $Q$. On one occasion $A$ started from $P$ and $B$ from $Q$ at the same time and when they met $A$ had travelled $120$ miles more than $B$. It took $A$ four $(4)$ hours to complete the trip to $Q$ and B nine $(9)$ hours to reach $P$. Assuming each train travels at a constant speed, what is the distance from $P$ to $Q$?


[b]p2.[/b] If $a$ and $b$ are integers, $b$ odd, prove that $x^2 + 2ax + 2b = 0$ has no rational roots.


[b]p3.[/b] A diameter segment of a set of points in a plane is a segment joining two points of the set which is at least as long as any other segment joining two points of the set. Prove that any two diameter segments of a set of points in the plane must have a point in common.


[b]p4.[/b] Find all positive integers $n$ for which $\frac{n(n^2 + n + 1) (n^2 + 2n + 2)}{2n + 1}$ is an integer. Prove that the set you exhibit is complete.


[b]p5.[/b] $A, B, C, D$ are four points on a semicircle with diameter $AB = 1$. If the distances $\overline{AC}$, $\overline{BC}$, $\overline{AD}$, $\overline{BD}$ are all rational numbers, prove that $\overline{CD}$ is also rational.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. Let \( D \) be the distance between cities \( P \) and \( Q \).
2. Let \( V_A \) be the speed of train \( A \) and \( V_B \) be the speed of train \( B \).
3. Given that train \( A \) takes 4 hours to travel from \( P \) to \( Q \), we have:
   \[
   V_A = \frac{D}{4}
   \]
4. Given that train \( B \) takes 9 hours to travel from \( Q \) to \( P \), we have:
   \[
   V_B = \frac{D}{9}
   \]
5. When the trains meet, train \( A \) has traveled 120 miles more than train \( B \). Let \( t \) be the time taken for the trains to meet.
6. The distance traveled by train \( A \) when they meet is \( V_A t \) and the distance traveled by train \( B \) is \( V_B t \).
7. According to the problem, \( V_A t = V_B t + 120 \).
8. Substituting the expressions for \( V_A \) and \( V_B \):
   \[
   \frac{D}{4} t = \frac{D}{9} t + 120
   \]
9. Simplifying the equation:
   \[
   \frac{D}{4} t - \frac{D}{9} t = 120
   \]
10. Combine the terms on the left-hand side:
    \[
    \left( \frac{D}{4} - \frac{D}{9} \right) t = 120
    \]
11. Find a common denominator for the fractions:
    \[
    \left( \frac{9D - 4D}{36} \right) t = 120
    \]
    \[
    \left( \frac{5D}{36} \right) t = 120
    \]
12. Solving for \( t \):
    \[
    t = \frac{120 \cdot 36}{5D}
    \]
    \[
    t = \frac{4320}{5D}
    \]
13. Since \( t \) is the same for both trains, we can use the relationship between their speeds and times:
    \[
    t = \frac{D}{4V_A} = \frac{D}{9V_B}
    \]
14. Substituting \( V_A = \frac{D}{4} \) and \( V_B = \frac{D}{9} \):
    \[
    t = \frac{D}{4 \cdot \frac{D}{4}} = \frac{D}{9 \cdot \frac{D}{9}}
    \]
    \[
    t = 1
    \]
15. Therefore, the distance \( D \) can be found by solving:
    \[
    \frac{5D}{36} t = 120
    \]
    \[
    \frac{5D}{36} = 120
    \]
    \[
    5D = 4320
    \]
    \[
    D = \frac{4320}{5}
    \]
    \[
    D = 864
    \]

The final answer is \( \boxed{864} \) miles.