[u]Round 5[/u]


[b]p13.[/b] Sally is at the special glasses shop, where there are many different optical lenses that distort what she sees and cause her to see things strangely. Whenever she looks at a shape through lens $A$, she sees a shape with $2$ more sides than the original (so a square would look like a hexagon). When she looks through lens $B$, she sees the shape with $3$ fewer sides (so a hexagon would look like a triangle). How many sides are in the shape that has $200$ more diagonals when looked at from lense $A$ than from lense $B$?


[b]p14.[/b] How many ways can you choose $2$ cells of a $5$ by $5$ grid such that they aren't in the same row or column?


[b]p15.[/b] If $a + \frac{1}{b} = (2015)^{-1}$ and $b +  \frac{1}{a} = (2016)^2$ then what are all the possible values of $b$?


[u]Round 6[/u]


[b]p16.[/b] In Canadian football, linebackers must wear jersey numbers from $30 -35$ while defensive linemen must wear numbers from $33 -38$ (both intervals are inclusive). If a team has $5$ linebackers and $4$ defensive linemen, how many ways can it assign jersey numbers to the $9$ players such that no two people have the same jersey number?


[b]p17.[/b] What is the maximum possible area of a right triangle with hypotenuse $8$?


[b]p18.[/b] $9$ people are to play touch football. One will be designated the quarterback, while the other eight will be divided into two (indistinct) teams of $4$. How many ways are there for this to be done?


[u]Round 7[/u]


[b]p19.[/b] Express the decimal $0.3$ in base $7$.


[b]p20.[/b] $2015$ people throw their hats in a pile. One at a time, they each take one hat out of the pile so that each has a random hat. What is the expected number of people who get their own hat?


[b]p21.[/b] What is the area of the largest possible trapezoid that can be inscribed in a semicircle of radius $4$?


[u]Round 8[/u]


[b]p22.[/b] What is the base $7$ expression of $1211_3  \cdot 1110_2  \cdot 292_{11} \cdot 20_3$ ?


[b]p23.[/b] Let $f(x)$ equal the ratio of the surface area of a sphere of radius $x$ to the volume of that same sphere. Let $g(x)$ be a quadratic polynomial in the form $x^2 + bx + c$ with $g(6) = 0$ and the minimum value of $g(x)$ equal to $c$. Express $g(x)$ as a function of $f(x)$ (e.g. in terms of $f(x)$).


[b]p24.[/b] In the country of Tahksess, the income tax code is very complicated. Citizens are taxed $40\%$ on their first $\$20, 000$ and $45\%$ on their next $\$40, 000$ and $50\%$ on their next $\$60, 000$ and so on, with each $5\%$ increase in tax rate aecting $\$20, 000$ more than the previous tax rate. The maximum tax rate, however, is $90\%$. What is the overall tax rate (percentage of money owed) on $1$ million dollars in income?


PS. You should use hide for answers. Rounds 1-4 have been posted [url=https://artofproblemsolving.com/community/c3h3157009p28696627]here [/url] and 9-12 [url=https://artofproblemsolving.com/community/c3h3158564p28715928]here[/url]. .Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. Let the people be \( P_1, P_2, P_3, \ldots, P_{2015} \), who retrieve their hats in that order.
2. Define \( x_1, x_2, x_3, \ldots, x_{2015} \) such that, for all \( i \in \{1,2,3,\ldots,2015\} \), \( x_i=0 \) if \( P_i \) does not get their hat and \( x_i=1 \) if \( P_i \) does get their hat.
3. We wish to find \( \mathbb{E}\left[\sum_{i=1}^{2015} x_i \right] \). By the Linearity of Expectation, this is \( \sum_{i=1}^{2015} \mathbb{E}[x_i] \).
4. Note that there is a \( \frac{i-1}{2015} \) probability that \( P_i \)'s hat is chosen by one of the first \( i-1 \) people and a \( \frac{2015-(i-1)}{2015} \) probability that \( P_i \)'s hat is not chosen by any of the first \( i-1 \) people.
5. If \( P_i \)'s hat is chosen by one of the first \( i-1 \) people, then there is a probability of \( 0 \) that \( P_i \) will get his own hat. If \( P_i \)'s hat is not chosen by any of the first \( i-1 \) people, then one of the remaining \( 2015-(i-1) \) hats is his, so there is a \( \frac{1}{2015-(i-1)} \) probability that he gets it.
6. Hence, the probability that \( P_i \) gets his own hat is 
   \[
   \frac{i-1}{2015} \cdot 0 + \frac{2015-(i-1)}{2015} \cdot \frac{1}{2015-(i-1)} = 0 + \frac{1}{2015} = \frac{1}{2015},
   \]
   so 
   \[
   \mathbb{E}[x_i] = \frac{1}{2015} \quad \forall \quad i \in \{1,2,3,\ldots,2015\}.
   \]
7. Therefore, the expected number of people who get their own hat is
   \[
   \sum_{i=1}^{2015} \mathbb{E}[x_i] = \sum_{i=1}^{2015} \frac{1}{2015} = 2015 \cdot \frac{1}{2015} = 1.
   \]

The final answer is \(\boxed{1}\)