[b]p1.[/b] Let $x_1 = 0$, $x_2 = 1/2$ and for $n >2$, let $x_n$ be the average of $x_{n-1}$ and $x_{n-2}$. Find a formula for $a_n = x_{n+1} - x_{n}$, $n = 1, 2, 3, \dots$. Justify your answer.


[b]p2.[/b] Given a triangle $ABC$. Let $h_a, h_b, h_c$ be the altitudes to its sides $a, b, c,$ respectively. Prove: $\frac{1}{h_a}+\frac{1}{h_b}>\frac{1}{h_c}$ Is it possible to construct a triangle with altitudes $7$, $11$, and $20$? Justify your answer.


[b]p3.[/b] Does there exist a polynomial $P(x)$ with integer coefficients such that $P(0) = 1$, $P(2) = 3$ and $P(4) = 9$? Justify your answer.


[b]p4.[/b] Prove that if $\cos \theta$ is rational and $n$ is an integer, then $\cos n\theta$ is rational. Let  $\alpha=\frac{1}{2010}$. Is $\cos \alpha $ rational ? Justify your answer.


[b]p5.[/b] Let function $f(x)$ be defined as $f(x) = x^2 + bx + c$, where $b, c$ are real numbers.
(A) Evaluate $f(1) -2f(5) + f(9)$ .
(B) Determine all pairs $(b, c)$ such that $|f(x)| \le 8$ for all $x$ in the interval $[1, 9]$.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. **Problem 1:**
   Given the sequence \( x_1 = 0 \), \( x_2 = \frac{1}{2} \), and for \( n > 2 \), \( x_n \) is the average of \( x_{n-1} \) and \( x_{n-2} \). We need to find a formula for \( a_n = x_{n+1} - x_n \).

   Let's start by writing the recursive formula for \( x_n \):
   \[
   x_n = \frac{x_{n-1} + x_{n-2}}{2}
   \]

   We need to find \( a_n = x_{n+1} - x_n \). Using the recursive formula, we have:
   \[
   x_{n+1} = \frac{x_n + x_{n-1}}{2}
   \]

   Therefore,
   \[
   a_n = x_{n+1} - x_n = \frac{x_n + x_{n-1}}{2} - x_n = \frac{x_{n-1} - x_n}{2}
   \]

   This can be simplified to:
   \[
   a_n = -\frac{1}{2} a_{n-1}
   \]

   We can see that \( a_n \) is a geometric sequence with the first term \( a_1 = x_2 - x_1 = \frac{1}{2} \) and common ratio \( -\frac{1}{2} \). Thus, the general formula for \( a_n \) is:
   \[
   a_n = \left( -\frac{1}{2} \right)^{n-1} \cdot \frac{1}{2}
   \]

   Simplifying, we get:
   \[
   a_n = \frac{(-1)^{n-1}}{2^n}
   \]

   Therefore, the formula for \( a_n \) is:
   \[
   \boxed{a_n = \frac{(-1)^{n-1}}{2^n}}
   \]

2. **Problem 2:**
   Given a triangle \( ABC \) with altitudes \( h_a, h_b, h_c \) to its sides \( a, b, c \) respectively, we need to prove:
   \[
   \frac{1}{h_a} + \frac{1}{h_b} > \frac{1}{h_c}
   \]

   We know that the area of the triangle \( [ABC] \) can be expressed using the altitudes:
   \[
   [ABC] = \frac{1}{2} a h_a = \frac{1}{2} b h_b = \frac{1}{2} c h_c
   \]

   Therefore, we have:
   \[
   \frac{1}{h_a} = \frac{a}{2[ABC]}, \quad \frac{1}{h_b} = \frac{b}{2[ABC]}, \quad \frac{1}{h_c} = \frac{c}{2[ABC]}
   \]

   To prove the inequality, we use the triangle inequality \( a + b > c \):
   \[
   \frac{a}{2[ABC]} + \frac{b}{2[ABC]} > \frac{c}{2[ABC]}
   \]

   Simplifying, we get:
   \[
   \frac{1}{h_a} + \frac{1}{h_b} > \frac{1}{h_c}
   \]

   Now, we need to check if it is possible to construct a triangle with altitudes 7, 11, and 20. We use the same approach:
   \[
   \frac{1}{7} + \frac{1}{11} \stackrel{?}{>} \frac{1}{20}
   \]

   Calculating the left-hand side:
   \[
   \frac{1}{7} + \frac{1}{11} = \frac{11 + 7}{77} = \frac{18}{77}
   \]

   Calculating the right-hand side:
   \[
   \frac{1}{20} = \frac{1}{20}
   \]

   Comparing the two:
   \[
   \frac{18}{77} \approx 0.2338 \quad \text{and} \quad \frac{1}{20} = 0.05
   \]

   Clearly, \( \frac{18}{77} > \frac{1}{20} \), so the inequality holds. Therefore, it is not possible to construct a triangle with altitudes 7, 11, and 20.

   \(\blacksquare\)