Maths Olympiad Prep

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Combinatorics Difficulty 6.1 National olympiad Find the answer

[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 AA, she sees a shape with 22 more sides than the original (so a square would look like a hexagon). When she looks through lens BB, she sees the shape with 33 fewer sides (so a hexagon would look like a triangle). How many sides are in the shape that has 200200 more diagonals when looked at from lense AA than from lense BB?

[b]p14.[/b] How many ways can you choose 22 cells of a 55 by 55 grid such that they aren't in the same row or column?

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

[u]Round 6[/u]

[b]p16.[/b] In Canadian football, linebackers must wear jersey numbers from 303530 -35 while defensive linemen must wear numbers from 333833 -38 (both intervals are inclusive). If a team has 55 linebackers and 44 defensive linemen, how many ways can it assign jersey numbers to the 99 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 88?

[b]p18.[/b] 99 people are to play touch football. One will be designated the quarterback, while the other eight will be divided into two (indistinct) teams of 44. How many ways are there for this to be done?

[u]Round 7[/u]

[b]p19.[/b] Express the decimal 0.30.3 in base 77.

[b]p20.[/b] 20152015 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 44?

[u]Round 8[/u]

[b]p22.[/b] What is the base 77 expression of 1211311102292112031211_3 \cdot 1110_2 \cdot 292_{11} \cdot 20_3 ?

[b]p23.[/b] Let f(x)f(x) equal the ratio of the surface area of a sphere of radius xx to the volume of that same sphere. Let g(x)g(x) be a quadratic polynomial in the form x2+bx+cx^2 + bx + c with g(6)=0g(6) = 0 and the minimum value of g(x)g(x) equal to cc. Express g(x)g(x) as a function of f(x)f(x) (e.g. in terms of f(x)f(x)).

[b]p24.[/b] In the country of Tahksess, the income tax code is very complicated. Citizens are taxed 40%40\% on their first \20, 000and and 45\%ontheirnext on their next \40,00040, 000 and 50%50\% on their next \60, 000andsoon,witheach and so on, with each 5\% increase in tax rate a ecting \20,00020, 000 more than the previous tax rate. The maximum tax rate, however, is 90%90\%. What is the overall tax rate (percentage of money owed) on 11 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].

A number or a short expression. Spacing and $ signs are ignored.

Solution

1. Let the people be P1,P2,P3,,P2015 P_1, P_2, P_3, \ldots, P_{2015} , who retrieve their hats in that order.
2. Define x1,x2,x3,,x2015 x_1, x_2, x_3, \ldots, x_{2015} such that, for all i{1,2,3,,2015} i \in \{1,2,3,\ldots,2015\} , xi=0 x_i=0 if Pi P_i does not get their hat and xi=1 x_i=1 if Pi P_i does get their hat.
3. We wish to find E[i=12015xi] \mathbb{E}\left[\sum_{i=1}^{2015} x_i \right] . By the Linearity of Expectation, this is i=12015E[xi] \sum_{i=1}^{2015} \mathbb{E}[x_i] .
4. Note that there is a i12015 \frac{i-1}{2015} probability that Pi P_i 's hat is chosen by one of the first i1 i-1 people and a 2015(i1)2015 \frac{2015-(i-1)}{2015} probability that Pi P_i 's hat is not chosen by any of the first i1 i-1 people.
5. If Pi P_i 's hat is chosen by one of the first i1 i-1 people, then there is a probability of 0 0 that Pi P_i will get his own hat. If Pi P_i 's hat is not chosen by any of the first i1 i-1 people, then one of the remaining 2015(i1) 2015-(i-1) hats is his, so there is a 12015(i1) \frac{1}{2015-(i-1)} probability that he gets it.
6. Hence, the probability that Pi P_i gets his own hat is
i120150+2015(i1)201512015(i1)=0+12015=12015, \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
E[xi]=12015i{1,2,3,,2015}. \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
i=12015E[xi]=i=1201512015=201512015=1. \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 1\boxed{1}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.