[b]p1. [/b]Ed, Fred and George are playing on a see-saw that is slightly off center. When Ed sits on the left side and George, who weighs $100$ pounds, on the right side, they are perfectly balanced. Similarly, if Fred, who weighs $400$ pounds, sits on the left and Ed sits on the right, they are also perfectly balanced. Assuming the see-saw has negligible weight, what is the weight of Ed, in pounds?


[b]p2.[/b] How many digits does the product $2^{42}\cdot 5^{38}$ have?


[b]p3.[/b] Square $ABCD$ has equilateral triangles drawn external to each side, as pictured. If each triangle is folded upwards to meet at a point $E$, then a square pyramid can be made. If the center of square $ABCD$ is $O$, what is the measure of $\angle OEA$?
[img]https://cdn.artofproblemsolving.com/attachments/9/a/39c0096ace5b942a9d3be1eafe7aa7481fbb9f.png[/img]


[b]p4.[/b] How many solutions $(x, y)$ in the positive integers are there to $3x + 7y = 1337$ ?


[b]p5.[/b] A trapezoid with height $12$ has legs of length $20$ and $15$ and a larger base of length $42$. What are the possible lengths of the other base?


[b]p6.[/b] Let $f(x) = 6x + 7$ and $g(x) = 7x + 6$. Find the value of a such that $g^{-1}(f^{-1}(g(f(a)))) = 1$.


[b]p7.[/b] Billy and Cindy want to meet at their favorite restaurant, and they have made plans to do so sometime between $1:00$ and $2:00$ this Sunday. Unfortunately, they didn’t decide on an exact time, so they both decide to arrive at a random time between $1:00$ and $2:00$. Silly Billy is impatient, though, and if he has to wait for Cindy, he will leave after $15$ minutes. Cindy, on the other hand, will happily wait for Billy from whenever she arrives until $2:00$. What is the probability that Billy and Cindy will be able to dine together?


[b]p8.[/b] As pictured, lines are drawn from the vertices of a unit square to an opposite trisection point. If each triangle has legs with ratio $3 : 1$, what is the area of the shaded region?
[img]https://cdn.artofproblemsolving.com/attachments/e/9/35a6340018edcddfcd7e085f8f6e56686a8e07.png[/img]


[b]p9.[/b] For any positive integer $n$, let $f_1(n)$ denote the sum of the squares of the digits of $n$. For $k \ge 2$, let $f_k(n) = f_{k-1}(f_1(n))$. Then, $f_1(5) = 25$ and $f_3(5) = f_2(25) = 85$. Find $f_{2012}(15)$.


[b]p10.[/b] Given that $2012022012$ has $ 8$ distinct prime factors, find its largest prime factor.


PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To find the possible lengths of the other base of the trapezoid, we need to consider the two different orientations of the legs. We will use the Pythagorean theorem to determine the horizontal components of the legs.

1. **Case 1: Leg of length 15 slopes to the right**

   - The horizontal component of the leg of length 20:
     \[
     \sqrt{20^2 - 12^2} = \sqrt{400 - 144} = \sqrt{256} = 16
     \]
   - The horizontal component of the leg of length 15:
     \[
     \sqrt{15^2 - 12^2} = \sqrt{225 - 144} = \sqrt{81} = 9
     \]
   - The length of the other base:
     \[
     42 - 16 - 9 = 17
     \]

2. **Case 2: Leg of length 15 slopes to the left**

   - The horizontal component of the leg of length 20 remains the same:
     \[
     \sqrt{20^2 - 12^2} = 16
     \]
   - The horizontal component of the leg of length 15 remains the same:
     \[
     \sqrt{15^2 - 12^2} = 9
     \]
   - The length of the other base:
     \[
     42 - 16 + 9 = 35
     \]

Thus, the possible lengths of the other base are \(17\) and \(35\).

The final answer is \(\boxed{17, 35}\).