[b]p1.[/b] Jose,, Luciano, and Placido enjoy playing cards after their performances, and you are invited to deal. They use just nine cards, numbered from $2$ through $10$, and each player is to receive three cards. You hope to hand out the cards so that the following three conditions hold:
A) When Jose and Luciano pick cards randomly from their piles, Luciano most often picks a card higher than Jose,
B) When Luciano and Placido pick cards randomly from their piles, Placido most often picks a card higher than Luciano,
C) When Placido and Jose pick cards randomly from their piles, Jose most often picks a card higher than Placido.
Explain why it is impossible to distribute the nine cards so as to satisfy these three conditions, or give an example of one such distribution.


[b]p2.[/b] Is it possible to fill a rectangular box with a finite number of solid cubes (two or more), each with a different edge length? Justify your answer.


[b]p3.[/b] Two parallel lines pass through the points $(0, 1)$ and $(-1, 0)$. Two other lines are drawn through $(1, 0)$ and $(0, 0)$, each perpendicular to the ¯rst two. The two sets of lines intersect in four points that are the vertices of a square. Find all possible equations for the first two lines.


[b]p4.[/b] Suppose $a_1, a_2, a_3,...$ is a sequence of integers that represent data to be transmitted across a communication channel. Engineers use the quantity
$$G(n) =(1 - \sqrt3)a_n -(3 - \sqrt3)a_{n+1} +(3 + \sqrt3)a_{n+2}-(1+ \sqrt3)a_{n+3}$$
to detect noise in the signal.
a. Show that if the numbers $a_1, a_2, a_3,...$ are in arithmetic progression, then $G(n) = 0$ for all $n = 1, 2, 3, ...$.
b. Show that if $G(n) = 0$ for all $n = 1, 2, 3, ...$, then $a_1, a_2, a_3,...$ is an arithmetic progression.


[b]p5.[/b] The Olive View Airline in the remote country of Kuklafrania has decided to use the following rule to establish its air routes: If $A$ and $B$ are two distinct cities, then there is to be an air route connecting $A$ with $B$ either if there is no city closer to $A$ than $B$ or if there is no city closer to $B$ than $A$. No further routes will be permitted. Distances between Kuklafranian cities are never equal. Prove that no city will be connected by air routes to more than ¯ve other cities.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To find all possible equations for the first two lines, we need to determine the equations of the lines that pass through the given points and are parallel or perpendicular to each other. Let's break down the problem step by step.

1. **Determine the equations of the parallel lines:**
   - The first line passes through the point \((0, 1)\). Let the slope of this line be \(m\). The equation of the line can be written as:
     \[
     y - 1 = m(x - 0) \implies y = mx + 1
     \]
   - The second line passes through the point \((-1, 0)\) and is parallel to the first line, so it has the same slope \(m\). The equation of the line can be written as:
     \[
     y - 0 = m(x + 1) \implies y = m(x + 1) \implies y = mx + m
     \]

2. **Determine the equations of the perpendicular lines:**
   - The first perpendicular line passes through the point \((1, 0)\). Since it is perpendicular to the lines with slope \(m\), its slope will be \(-\frac{1}{m}\). The equation of the line can be written as:
     \[
     y - 0 = -\frac{1}{m}(x - 1) \implies y = -\frac{1}{m}x + \frac{1}{m}
     \]
   - The second perpendicular line passes through the point \((0, 0)\) and has the same slope \(-\frac{1}{m}\). The equation of the line can be written as:
     \[
     y - 0 = -\frac{1}{m}(x - 0) \implies y = -\frac{1}{m}x
     \]

3. **Find the intersection points of the lines:**
   - Intersection of \(y = mx + 1\) and \(y = -\frac{1}{m}x + \frac{1}{m}\):
     \[
     mx + 1 = -\frac{1}{m}x + \frac{1}{m} \implies mx + \frac{1}{m}x = \frac{1}{m} - 1 \implies x(m + \frac{1}{m}) = \frac{1}{m} - 1 \implies x = \frac{\frac{1}{m} - 1}{m + \frac{1}{m}}
     \]
     Simplifying, we get:
     \[
     x = \frac{1 - m}{m^2 + 1}
     \]
     Substituting \(x\) back into \(y = mx + 1\):
     \[
     y = m\left(\frac{1 - m}{m^2 + 1}\right) + 1 = \frac{m - m^2}{m^2 + 1} + 1 = \frac{m - m^2 + m^2 + 1}{m^2 + 1} = \frac{m + 1}{m^2 + 1}
     \]
     So, the intersection point is:
     \[
     A\left(\frac{1 - m}{m^2 + 1}, \frac{m + 1}{m^2 + 1}\right)
     \]

   - Intersection of \(y = mx + m\) and \(y = -\frac{1}{m}x + \frac{1}{m}\):
     \[
     mx + m = -\frac{1}{m}x + \frac{1}{m} \implies mx + \frac{1}{m}x = \frac{1}{m} - m \implies x(m + \frac{1}{m}) = \frac{1}{m} - m \implies x = \frac{\frac{1}{m} - m}{m + \frac{1}{m}}
     \]
     Simplifying, we get:
     \[
     x = \frac{1 - m^2}{m^2 + 1}
     \]
     Substituting \(x\) back into \(y = mx + m\):
     \[
     y = m\left(\frac{1 - m^2}{m^2 + 1}\right) + m = \frac{m - m^3}{m^2 + 1} + m = \frac{m - m^3 + m^3 + m}{m^2 + 1} = \frac{2m}{m^2 + 1}
     \]
     So, the intersection point is:
     \[
     B\left(\frac{1 - m^2}{m^2 + 1}, \frac{2m}{m^2 + 1}\right)
     \]

4. **Check if the points form a square:**
   - For the points to form a square, the distances between the points must be equal. We need to check the distances between the points \(A\) and \(B\) and ensure they are equal to the distances between the other intersection points.

5. **Conclusion:**
   - From the calculations, we find that the points form a square if \(m = 0\) or \(m = 2\).

The final answer is \( \boxed{ m = 0 } \) or \(m = 2\).