[b]p1.[/b] Al usually arrives at the train station on the commuter train at $6:00$, where his wife Jane meets him and drives him home. Today Al caught the early train and arrived at $5:00$. Rather than waiting for Jane, he decided to jog along the route he knew Jane would take and hail her when he saw her. As a result, Al and Jane arrived home $12$ minutes earlier than usual. If Al was jogging at a constant speed of $5$ miles per hour, and Jane always drives at the constant speed that would put her at the station at $6:00$, what was her speed, in miles per hour?


[b]p2.[/b] In the figure, points $M$ and $N$ are the respective midpoints of the sides $AB$ and $CD$ of quadrilateral $ABCD$. Diagonal $AC$ meets segment $MN$ at $P$, which is the midpoint of $MN$, and $AP$ is twice as long as $PC$. The area of triangle $ABC$ is $6$ square feet.
(a) Find, with proof, the area of triangle $AMP$.
(b) Find, with proof, the area of triangle $CNP$.
(c) Find, with proof, the area of quadrilateral $ABCD$.
[img]https://cdn.artofproblemsolving.com/attachments/a/c/4bdcd8390bae26bc90fc7eae398ace06900a67.png[/img]


[b]p3.[/b] (a) Show that there is a triangle whose angles have measure $\tan^{-1}1$, $\tan^{-1}2$ and $\tan^{-1}3$.
(b) Find all values of $k$ for which there is a triangle whose angles have measure $\tan^{-1}\left(\frac12 \right)$, $\tan^{-1}\left(\frac12 +k\right)$, and $\tan^{-1}\left(\frac12 +2k\right)$


[b]p4.[/b] (a) Find $19$ consecutive integers whose sum is as close to $1000$ as possible.
(b) Find the longest possible sequence of consecutive odd integers whose sum is exactly $1000$, and prove that your sequence is the longest.


[b]p5.[/b] Let $AB$ and $CD$ be chords of a circle which meet at a point $X$ inside the circle.
(a) Suppose that $\frac{AX}{BX}=\frac{CX}{DX}$. Prove that $|AB|=|CD|$.
(b) Suppose that $\frac{AX}{BX}>\frac{CX}{DX}>1$. Prove that $|AB|>|CD|$.
($|PQ|$ means the length of the segment $PQ$.)


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
### Part (a)

1. Let \( m \) be the middle term of the sequence of 19 consecutive integers. The sequence can be written as:
   \[
   m-9, m-8, \ldots, m-1, m, m+1, \ldots, m+8, m+9
   \]

2. The sum \( S \) of these 19 consecutive integers is:
   \[
   S = 19m
   \]

3. We want this sum to be as close to 1000 as possible. Therefore, we solve the inequality:
   \[
   1000 - 19 \leq 19m \leq 1000 + 19
   \]

4. Simplifying the inequality:
   \[
   981 \leq 19m \leq 1019
   \]

5. Dividing by 19:
   \[
   51.63 \leq m \leq 53.63
   \]

6. Since \( m \) must be an integer, the closest integer value is \( m = 53 \).

7. Therefore, the 19 consecutive integers are:
   \[
   44, 45, 46, \ldots, 52, 53, 54, \ldots, 60, 61, 62
   \]

The final answer is \( \boxed{ -497, -495, -493, \ldots, 497, 499 } \).