MathDB

5

Part of 2018 Harvard-MIT Mathematics Tournament

Problems(7)

2018 Combinatorics #5

Source:

2/12/2018
A bag contains nine blue marbles, ten ugly marbles, and one special marble. Ryan picks marbles randomly from this bag with replacement until he draws the special marble. He notices that none of the marbles he drew were ugly. Given this information, what is the expected value of the number of total marbles he drew?
2018 Algebra / NT #5

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2/12/2018
Let {ω1,ω2,⋯ ,ω100}\{\omega_1,\omega_2,\cdots,\omega_{100}\} be the roots of x101−1x−1\frac{x^{101}-1}{x-1} (in some order). Consider the set S={ω11,ω22,ω33,⋯ ,ω100100}.S=\{\omega_1^1,\omega_2^2,\omega_3^3,\cdots,\omega_{100}^{100}\}. Let MM be the maximum possible number of unique values in S,S, and let NN be the minimum possible number of unique values in S.S. Find M−N.M-N.
2018 Geometry #5

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2/12/2018
In the quadrilateral MAREMARE inscribed in a unit circle ω,\omega, AMAM is a diameter of ω,\omega, and EE lies on the angle bisector of ∠RAM.\angle RAM. Given that triangles RAMRAM and REMREM have the same area, find the area of quadrilateral MARE.MARE.
geometry
2018 Team #5

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2/12/2018
Is it possible for the projection of the set of points (x,y,z)(x, y, z) with 0≤x,y,z≤10 \leq x, y, z \leq 1 onto some two-dimensional plane to be a simple convex pentagon?
2018 General #5

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11/12/2018
Compute the smallest positive integer nn for which 100+n+100−n\sqrt{100+\sqrt{n}}+\sqrt{100-\sqrt{n}} is an integer.
algebra
2018 Theme #5

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11/13/2018
Lil Wayne, the rain god, determines the weather. If Lil Wayne makes it rain on any given day, the probability that he makes it rain the next day is 75%75\%. If Lil Wayne doesn't make it rain on one day, the probability that he makes it rain the next day is 25%25\%. He decides not to make it rain today. Find the smallest positive integer nn such that the probability that Lil Wayne makes it rain nn days from today is greater than 49.9%49.9\%.
probability
2018 November Team #5

Source:

11/13/2018
Find the sum of all positive integers nn such that 1+2+⋯+n1+2+\cdots+n divides 15[(n+1)2+(n+2)2+⋯+(2n)2].15\left[(n+1)^2+(n+2)^2+\cdots+(2n)^2\right].