Math Is Fun Forum

I have had a problem with this perplexing problem: The triangle ACB is an isosceles triangle.

AC = AP = PQ = QB
So I though that since ACB is isosceles, we can conclude that 180-6x+2x=3x. 180-4x=3x 180=7x, x = 180/7.
Confirmation needed

Today I took a intriguing mathematical quiz(cannot disclose: sorry!)
Anyhow I feel like it devastated me, and I have two questions that were perplexing, namely..
1) Andy and Barry are running around a circular track that is 400 ft. Andy and Barry start at the same place, though Barry pulls up, travelling at 125% the speed of Andy. How many laps of the track has he traversed once Barry surpasses Andy?

2) There are 216 sprinters participating in a race. Unfortunately, the track only holds 6 people at a time. Once a sprinter wins the race, the others are eliminated. This process is reiterated until a final champion wins. How many races are held to find the champion?

I have been bombarded with a particularly perplexing logic problem which states as follows.
Middle, Rear and Front are standing in a line. A eccentric millionaire challenges them to a game. Choosing arbitrarily from 3 white hats and 2 black hats, the millionaire puts a hat on each of the three men. The men cannot see their own hats. The millionaire then asks each of the three the color of their hats. Rear, who can see two other hats, declines to give an answer(he cannot determine his hat's colour.). Middle, who can see only Front's hat, also declines to give an answer(he cannot determine his hat's colour.). Finally, Front, who cannot see any hats, states the color of his own hat.
How?
If somebody decides to link me to the MIF "Three Hats Puzzle", keep in mind we have THREE white hats and TWO black ones.
Now I disprove that scenarios with Front having a black hat could happen in these conditions(Either Middle or Rear would be able to clearly define the colour of their hat.), and thus Front has a white hat. Wikipedia concurs.
However, I have noticed that I end with a convoluted, case by case solution. Can anyone provide me with a simpler solution(not case by case..)?

A string is wrapper around the Earth's equator(i.e, the circumference)  and the two ends of the string just touch. Now suppose that another string is tied to the original string so that it is 100 ft longer. If this new string is placed around the equator and pulled tight so it is suspended over the earth, how high will the string be above the ground?
a. An atom high
b. A bowling ball high
c. A Mack truck high
d. Exactly 100 ft.
I believe it is C.
Explanation:
Let CE = Earth's circumference and let CS = String's new circumference
Let RE= Radius of the Earth and let RS = Radius of the string
CE = 2πRE
and CR+100 = 2πRS.
We want to find the difference between the two radiuses(aka height.)
C/2π = RE
C+100/2π= RS
thus...
RS - RE = (C+100-C)/2π.
RS - RE = 100/2π = 50/π
50/pi ≈ 17.
Therefore Mack truck, aka C, is the solution.
I would LaTeX the aforementioned equations but I'm in a hurry.

has been given to me as homework to be done. It asks for me to give a proper explanation why as well. Optimal path refers to the path that would minimize the length of the voracious ant's travel. I am thoroughly befuddled by the "it includes the floor and ceiling" hunt as the optimal path does not seem to go through as presupposed.