[b]p1.[/b] Consider a parallelogram $ABCD$ with sides of length $a$ and $b$, where $a \ne b$. The four points of intersection of the bisectors of the interior angles of the parallelogram form a rectangle $EFGH$. A possible configuration is given below.
Show that $$\frac{Area(ABCD)}{Area(EFGH)}=\frac{2ab}{(a - b)^2}$$
[img]https://cdn.artofproblemsolving.com/attachments/e/a/afaf345f2ef7c8ecf4388918756f0b56ff20ef.png[/img]


[b]p2.[/b] A metal wire of length $4\ell$ inches (where $\ell$ is a positive integer) is used as edges to make a cardboard rectangular box with surface area $32$ square inches and volume $8$ cubic inches. Suppose that the whole wire is used.
(i) Find the dimension of the box if $\ell= 9$, i.e., find the length, the width, and the height of the box without distinguishing the different orders of the numbers. Justify your answer.
(ii) Show that it is impossible to construct such a box if $\ell = 10$.


[b]p3.[/b] A Pythagorean n-tuple is an ordered collection of counting numbers $(x_1, x_2,..., x_{n-1}, x_n)$ satisfying the equation $$x^2_1+ x^2_2+ ...+ x^2_{n-1} = x^2_{n}.$$
For example, $(3, 4, 5)$ is an ordinary Pythagorean $3$-tuple (triple) and $(1, 2, 2, 3)$ is a Pythagorean $4$-tuple.
(a) Given a Pythagorean triple $(a, b, c)$ show that the $4$-tuple $(a^2, ab, bc, c^2)$ is Pythagorean.
(b) Extending part (a) or using any other method, come up with a procedure that generates Pythagorean $5$-tuples from Pythagorean $3$- and/or $4$-tuples. Few numerical examples will not suffice. You have to find a method that will generate infinitely many such $5$-tuples.
(c) Find a procedure to generate Pythagorean $6$-tuples from Pythagorean $3$- and/or $4$- and/or $5$-tuples.

Note. You can assume without proof that there are infinitely many Pythagorean triples.



[b]p4.[/b] Consider the recursive sequence defined by $x_1 = a$, $x_2 = b$ and $$x_{n+2} =\frac{x_{n+1} + x_n - 1}{x_n - 1}, n \ge 1 .$$
We call the pair $(a, b)$ the seed for this sequence. If both $a$ and $b$ are integers, we will call it an integer seed.
(a) Start with the integer seed $(2, 2019)$ and find $x_7$.
(b) Show that there are infinitely many integer seeds for which $x_{2020} = 2020$.
(c) Show that there are no integer seeds for which $x_{2019} = 2019$.


[b]p5.[/b] Suppose there are eight people at a party. Each person has a certain amount of money. The eight people decide to play a game. Let $A_i$, for $i = 1$ to $8$, be the amount of money person $i$ has in his/her pocket at the beginning of the game. A computer picks a person at random. The chosen person is eliminated from the game and their money is put into a pot. Also magically the amount of money in the pockets of the remaining players goes up by the dollar amount in the chosen person's pocket. We continue this process and at the end of the seventh stage emerges a single person and a pot containing $M$ dollars. What is the expected value of $M$? The remaining player gets the pot and the money in his/her pocket. What is the expected value of what he/she takes home?



PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. Consider the parallelogram \(ABCD\) with sides of length \(a\) and \(b\), where \(a \ne b\). Let \(\alpha\) be the angle between sides \(a\) and \(b\). The area of the parallelogram \(ABCD\) is given by:
   \[
   \text{Area}(ABCD) = ab \sin(2\alpha)
   \]
   This follows from the formula for the area of a parallelogram, which is the product of the lengths of two adjacent sides and the sine of the included angle.

2. The four points of intersection of the bisectors of the interior angles of the parallelogram form a rectangle \(EFGH\). Let us denote the sides of this rectangle as \(EF\) and \(EH\).

3. Since \(HF \parallel CD\) and \(HF = EG = a - b\), we can use trigonometric relationships to find the lengths of \(EF\) and \(EH\):
   \[
   EF = HF \sin(\alpha) = (a - b) \sin(\alpha)
   \]
   \[
   EH = HF \cos(\alpha) = (a - b) \cos(\alpha)
   \]

4. The area of the rectangle \(EFGH\) is given by:
   \[
   \text{Area}(EFGH) = EF \cdot EH = (a - b) \sin(\alpha) \cdot (a - b) \cos(\alpha) = (a - b)^2 \sin(\alpha) \cos(\alpha)
   \]

5. To find the ratio of the areas, we divide the area of the parallelogram \(ABCD\) by the area of the rectangle \(EFGH\):
   \[
   \frac{\text{Area}(ABCD)}{\text{Area}(EFGH)} = \frac{ab \sin(2\alpha)}{(a - b)^2 \sin(\alpha) \cos(\alpha)}
   \]

6. Using the double-angle identity for sine, \(\sin(2\alpha) = 2 \sin(\alpha) \cos(\alpha)\), we can simplify the expression:
   \[
   \frac{\text{Area}(ABCD)}{\text{Area}(EFGH)} = \frac{ab \cdot 2 \sin(\alpha) \cos(\alpha)}{(a - b)^2 \sin(\alpha) \cos(\alpha)} = \frac{2ab}{(a - b)^2}
   \]

Thus, we have shown that:
\[
\frac{\text{Area}(ABCD)}{\text{Area}(EFGH)} = \frac{2ab}{(a - b)^2}
\]

The final answer is \( \boxed{ \frac{2ab}{(a - b)^2} } \).