Today was the 5th Kettering Olympiad - and here are the problems, which are very good intermediate problems.

1. Find all real $x$ so that $(1+x^2)(1+x^4)=4x^3$

2. Mark and John play a game. They have $100$ pebbles on a table. They take turns taking at least one at at most eight pebbles away. The person to claim the last pebble wins. Mark goes first. Can you find a way for Mark to always win? What about John?

3. Prove that

$\sin x + \sin 3x + \sin 5x + ... + \sin 11 x = (1-\cos 12 x)/(2 \sin x)$

4. Mark has $7$ pieces of paper. He takes some of them and splits each into $7$ pieces of paper. He repeats this process some number of times. He then tells John he has $2000$ pieces of paper. John tells him he is wrong. Why is John right?

5. In a triangle $ABC$, the altitude, angle bisector, and median split angle $A$ into four equal angles. Find the angles of $ABC.$

6. There are $100$ cities. There exist airlines connecting pairs of cities. 

a) Find the minimal number of airlines such that with at most $k$ plane changes, one can go from any city to any other city.

b) Given that there are $4852$ airlines, show that, given any schematic, one can go from any city to any other city.
1. We start with the given equation:
   \[
   (1 + x^2)(1 + x^4) = 4x^3
   \]
   Expanding the left-hand side, we get:
   \[
   1 + x^2 + x^4 + x^6 = 4x^3
   \]
   Rearranging the terms, we have:
   \[
   x^6 + x^4 + x^2 + 1 = 4x^3
   \]
   Dividing both sides by \(x^3\) (assuming \(x \neq 0\)), we get:
   \[
   \frac{x^6}{x^3} + \frac{x^4}{x^3} + \frac{x^2}{x^3} + \frac{1}{x^3} = 4
   \]
   Simplifying, we obtain:
   \[
   x^3 + x + \frac{1}{x} + \frac{1}{x^3} = 4
   \]
   Let \(y = x + \frac{1}{x}\). Then:
   \[
   y^3 = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}
   \]
   Substituting \(y\) into the equation, we get:
   \[
   y^3 - 3y = x^3 + \frac{1}{x^3} + x + \frac{1}{x}
   \]
   Therefore:
   \[
   y^3 - 3y + y = 4
   \]
   Simplifying, we have:
   \[
   y^3 - 2y - 4 = 0
   \]
   We solve this cubic equation by guessing and checking. Trying \(y = 2\):
   \[
   2^3 - 2(2) - 4 = 8 - 4 - 4 = 0
   \]
   So, \(y = 2\) is a root. Factoring out \(y - 2\), we get:
   \[
   (y - 2)(y^2 + 2y + 2) = 0
   \]
   Solving \(y^2 + 2y + 2 = 0\) using the quadratic formula:
   \[
   y = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2} = -1 \pm i
   \]
   Since these are imaginary roots, the only real solution is \(y = 2\). Thus:
   \[
   x + \frac{1}{x} = 2
   \]
   Multiplying both sides by \(x\), we get:
   \[
   x^2 - 2x + 1 = 0
   \]
   Factoring, we have:
   \[
   (x - 1)^2 = 0
   \]
   Therefore, \(x = 1\) is the only solution.

The final answer is \( \boxed{ x = 1 } \).