olymmath

HMMT November 2012

HMMT November 2012.

General

E2

  • General
  • Answer

Let Q(x)=x2+2x+3Q(x)=x^{2}+2 x+3, and suppose that P(x)P(x) is a polynomial such that

P(Q(x))=x6+6x5+18x4+32x3+35x2+22x+8P(Q(x))=x^{6}+6 x^{5}+18 x^{4}+32 x^{3}+35 x^{2}+22 x+8

Compute P(2)P(2).

Answer, solutions and source

E3

  • General
  • Answer

ABCDA B C D is a rectangle with AB=20A B=20 and BC=3B C=3. A circle with radius 5 , centered at the midpoint of DCD C, meets the rectangle at four points: W,X,YW, X, Y, and ZZ. Find the area of quadrilateral WXYZW X Y Z.

Answer, solutions and source

E4

  • General
  • Answer

If you roll four fair 6 -sided dice, what is the probability that at least three of them will show the same value?

Answer, solutions and source

E5

  • General
  • Answer

How many ways are there to arrange three indistinguishable rooks on a 6×66 \times 6 board such that no two rooks are attacking each other? (Two rooks are attacking each other if and only if they are in the same row or the same column.)

Answer, solutions and source

E6

  • General
  • Answer

ABCDA B C D is a parallelogram satisfying AB=7,BC=2A B=7, B C=2, and ∠DAB=120∘\angle D A B=120^{\circ}. Parallelogram ECFAE C F A is contained in ABCDA B C D and is similar to it. Find the ratio of the area of ECFAE C F A to the area of ABCDA B C D.

Answer, solutions and source

E7

  • General
  • Answer

Find the number of ordered 2012-tuples of integers (x1,x2,…,x2012)\left(x_{1}, x_{2}, \ldots, x_{2012}\right), with each integer between 0 and 2011 inclusive, such that the sum x1+2x2+3x3+⋯+2012x2012x_{1}+2 x_{2}+3 x_{3}+\cdots+2012 x_{2012} is divisible by 2012 .

Answer, solutions and source

E8

  • General
  • Answer

Let nn be the 200th smallest positive real solution to the equation x−π2=tan⁡xx-\frac{\pi}{2}=\tan x. Find the greatest integer that does not exceed n2\frac{n}{2}.

Answer, solutions and source

E9

  • General
  • Answer

Consider triangle ABCA B C where BC=7,CA=8B C=7, C A=8, and AB=9.DA B=9 . D and EE are the midpoints of BCB C and CAC A, respectively, and ADA D and BEB E meet at GG. The reflection of GG across DD is G′G^{\prime}, and G′EG^{\prime} E meets CGC G at PP. Find the length PGP G.

Answer, solutions and source

E10

  • General
  • Answer

Let α\alpha and β\beta be reals. Find the least possible value of

(2cos⁡α+5sin⁡β−8)2+(2sin⁡α+5cos⁡β−15)2.(2 \cos \alpha+5 \sin \beta-8)^{2}+(2 \sin \alpha+5 \cos \beta-15)^{2} .

Answer, solutions and source

Theme

H3

  • Theme
  • Answer

Find the smallest positive integer nn such that 22⋅2⏟n>3333\underbrace{2^{2 \cdot 2}}_{n}>3^{3^{3^{3}}}. (The notation 22⋅22⏟n\underbrace{2^{2 \cdot 2^{2}}}_{n} is used to denote a power tower with n2n 2 's. For example, 222⏟n\underbrace{2^{2^{2}}}_{n} with n=4n=4 would equal 22222^{2^{2^{2}}}.)

Answer, solutions and source

H4

  • Theme
  • Answer

Find the sum of all real solutions for xx to the equation (x2+2x+3)(x2+2x+3)(x2+2x+3)=2012\left(x^{2}+2 x+3\right)^{\left(x^{2}+2 x+3\right)^{\left(x^{2}+2 x+3\right)}}=2012.

Answer, solutions and source

H5

  • Theme
  • Answer

Given any positive integer, we can write the integer in base 12 and add together the digits of its base 12 representation. We perform this operation on the number 7653217^{6^{5^{3^{2^{1}}}}} repeatedly until a single base 12 digit remains. Find this digit.

Answer, solutions and source

H6

  • Theme
  • Answer

A rectangular piece of paper with vertices ABCDA B C D is being cut by a pair of scissors. The pair of scissors starts at vertex AA, and then cuts along the angle bisector of DABD A B until it reaches another edge of the paper. One of the two resulting pieces of paper has 4 times the area of the other piece. What is the ratio of the longer side of the original paper to the shorter side?

Answer, solutions and source

H7

  • Theme
  • Answer

The game of rock-scissors is played just like rock-paper-scissors, except that neither player is allowed to play paper. You play against a poorly-designed computer program that plays rock with 50%50 \% probability and scissors with 50%50 \% probability. If you play optimally against the computer, find the probability that after 8 games you have won at least 4 .

Answer, solutions and source

H8

  • Theme
  • Answer

In the game of rock-paper-scissors-lizard-Spock, rock defeats scissors and lizard, paper defeats rock and Spock, scissors defeats paper and lizard, lizard defeats paper and Spock, and Spock defeats rock and scissors, as shown in the below diagram. As before, if two players choose the same move, then there is a draw. If three people each play a game of rock-paper-scissors-lizard-Spock at the same time by choosing one of the five moves at random, what is the probability that one player beats the other two?

Answer, solutions and source

H9

  • Theme
  • Answer

64 people are in a single elimination rock-paper-scissors tournament, which consists of a 6-round knockout bracket. Each person has a different rock-paper-scissors skill level, and in any game, the person with the higher skill level will always win. For how many players PP is it possible that PP wins the first four rounds that he plays? (A 6-round knockout bracket is a tournament which works as follows: (a) In the first round, all 64 competitors are paired into 32 groups, and the two people in each group play each other. The winners advance to the second round, and the losers are eliminated. (b) In the second round, the remaining 32 players are paired into 16 groups. Again, the winner of each group proceeds to the next round, while the loser is eliminated. (c) Each round proceeds in a similar way, eliminating half of the remaining players. After the sixth round, only one player will not have been eliminated. That player is declared the champion.)

Answer, solutions and source

H10

  • Theme
  • Answer

In a game of rock-paper-scissors with nn people, the following rules are used to determine a champion: (a) In a round, each person who has not been eliminated randomly chooses one of rock, paper, or scissors to play. (b) If at least one person plays rock, at least one person plays paper, and at least one person plays scissors, then the round is declared a tie and no one is eliminated. If everyone makes the same move, then the round is also declared a tie. (c) If exactly two moves are represented, then everyone who made the losing move is eliminated from playing in all further rounds (for example, in a game with 8 people, if 5 people play rock and 3 people play scissors, then the 3 who played scissors are eliminated). (d) The rounds continue until only one person has not been eliminated. That person is declared the champion and the game ends.

If a game begins with 4 people, what is the expected value of the number of rounds required for a champion to be determined?

Answer, solutions and source

Team

T1

  • Team
  • Answer

Find the number of integers between 1 and 200 inclusive whose distinct prime divisors sum to 16 . (For example, the sum of the distinct prime divisors of 12 is 2+3=52+3=5.)

Answer, solutions and source

T2

  • Team
  • Answer

Find the number of ordered triples of divisors (d1,d2,d3)\left(d_{1}, d_{2}, d_{3}\right) of 360 such that d1d2d3d_{1} d_{2} d_{3} is also a divisor of 360 .

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T3

  • Team
  • Answer

Find the largest integer less than 2012 all of whose divisors have at most two 1 's in their binary representations.

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T4

  • Team
  • Answer

Let π\pi be a permutation of the numbers from 2 through 2012. Find the largest possible value of log⁡2π(2)⋅log⁡3π(3)⋯log⁡2012π(2012)\log _{2} \pi(2) \cdot \log _{3} \pi(3) \cdots \log _{2012} \pi(2012).

Answer, solutions and source

T5

  • Team
  • Answer

Let π\pi be a randomly chosen permutation of the numbers from 1 through 2012 . Find the probability that π(π(2012))=2012\pi(\pi(2012))=2012.

Answer, solutions and source

T6

  • Team
  • Answer

Let π\pi be a permutation of the numbers from 1 through 2012. What is the maximum possible number of integers nn with 1≤n≤20111 \leq n \leq 2011 such that π(n)\pi(n) divides π(n+1)\pi(n+1) ?

Answer, solutions and source

T7

  • Team
  • Answer

Let A1A2…A100A_{1} A_{2} \ldots A_{100} be the vertices of a regular 100-gon. Let π\pi be a randomly chosen permutation of the numbers from 1 through 100. The segments Aπ(1)Aπ(2),Aπ(2)Aπ(3),…,Aπ(99)Aπ(100),Aπ(100)Aπ(1)A_{\pi(1)} A_{\pi(2)}, A_{\pi(2)} A_{\pi(3)}, \ldots, A_{\pi(99)} A_{\pi(100)}, A_{\pi(100)} A_{\pi(1)} are drawn. Find the expected number of pairs of line segments that intersect at a point in the interior of the 100-gon.

Answer, solutions and source

T8

  • Team
  • Answer

ABCA B C is a triangle with AB=15,BC=14A B=15, B C=14, and CA=13C A=13. The altitude from AA to BCB C is extended to meet the circumcircle of ABCA B C at DD. Find ADA D.

Answer, solutions and source

T9

  • Team
  • Answer

Triangle ABCA B C satisfies ∠B>∠C\angle B>\angle C. Let MM be the midpoint of BCB C, and let the perpendicular bisector of BCB C meet the circumcircle of △ABC\triangle A B C at a point DD such that points A,D,CA, D, C, and BB appear on the circle in that order. Given that ∠ADM=68∘\angle A D M=68^{\circ} and ∠DAC=64∘\angle D A C=64^{\circ}, find ∠B\angle B.

Answer, solutions and source

T10

  • Team
  • Answer

Triangle ABCA B C has AB=4,BC=5A B=4, B C=5, and CA=6C A=6. Points A′,B′,C′A^{\prime}, B^{\prime}, C^{\prime} are such that B′C′B^{\prime} C^{\prime} is tangent to the circumcircle of △ABC\triangle A B C at A,C′A′A, C^{\prime} A^{\prime} is tangent to the circumcircle at BB, and A′B′A^{\prime} B^{\prime} is tangent to the circumcircle at CC. Find the length B′C′B^{\prime} C^{\prime}.

Answer, solutions and source