olymmath

AIME 2014b

American Invitational Mathematics Examination 2014.

Paper

Problem 1

  • Paper
  • Answer

Abe can paint the room in 15 hours, Bea can paint 50 percent faster than Abe, and Coe can paint twice as fast as Abe. Abe begins to paint the room and works alone for the first hour and a half. Then Bea joins Abe, and they work together until half the room is painted. Then Coe joins Abe and Bea, and they work together until the entire room is painted. Find the number of minutes after Abe begins for the three of them to finish painting the room.

Answer, solutions and source

Problem 2

  • Paper
  • Answer

Arnold is studying the prevalence of three health risk factors, denoted by A, B, and C, within a population of men. For each of the three factors, the probability that a randomly selected man in the population has only this risk factor (and none of the others) is 0.1. For any two of the three factors, the probability that a randomly selected man has exactly these two risk factors (but not the third) is 0.14. The probability that a randomly selected man has all three risk factors, given that he has A and B is 13\frac{1}{3}. The probability that a man has none of the three risk factors given that he does not have risk factor A is pq\frac{p}{q}, where pp and qq are relatively prime positive integers. Find p+qp+q.

Answer, solutions and source

Problem 3

  • Paper
  • Answer

A rectangle has sides of length aa and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length aa can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length aa parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find a2a^2. [asy] pair A,B,C,D,E,F,R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); A = (0,0); B = (0,18); C = (0,36); // don't look here D = (122.236, 36); E = (122.236, 18); F = (122.236, 0); draw(A--B--C--D--E--F--cycle); dot(" ",A,NW); dot(" ",B,NW); dot(" ",C,NW); dot(" ",D,NW); dot(" ",E,NW); dot(" ",F,NW); //don't look here R = (122.236 +22,0); S = (122.236 + 22 - 13.4164,12); T = (122.236 + 22,24); X = (124.472+ 22,24); Y = (124.472+ 22 + 13.4164,12); Z = (12*4.472+ 22,0); draw(R--S--T--X--Y--Z--cycle); dot(" ",R,NW); dot(" ",S,NW); dot(" ",T,NW); dot(" ",X,NW); dot(" ",Y,NW); dot(" ",Z,NW); // sqrt180 = 13.4164 // sqrt5 = 2.236[/asy]

Answer, solutions and source

Problem 4

  • Paper
  • Answer

The repeating decimals 0.ababab‾0.abab\overline{ab} and 0.abcabcabc‾0.abcabc\overline{abc} satisfy 0.ababab‾+0.abcabcabc‾=3337,0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37}, where aa, bb, and cc are (not necessarily distinct) digits. Find the three digit number abcabc.

Answer, solutions and source

Problem 5

  • Paper
  • Answer

Real numbers rr and ss are roots of p(x)=x3+ax+bp(x)=x^3+ax+b, and r+4r+4 and s−3s-3 are roots of q(x)=x3+ax+b+240q(x)=x^3+ax+b+240. Find the sum of all possible values of ∣b∣|b|.

Answer, solutions and source

Problem 6

  • Paper
  • Answer

Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability 23\frac{2}{3} and each of the other five sides has probability 115\frac{1}{15}. Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is pq\frac{p}{q}, where pp and qq are relatively prime positive integers. Find p+qp+q.

Answer, solutions and source

Problem 7

  • Paper
  • Answer

Let f(x)=(x2+3x+2)cos⁡(πx)f(x)=(x^2+3x+2)^{\cos(\pi x)}. Find the sum of all positive integers nn for which ∣∑k=1nlog⁡10f(k)∣=1.\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.

Answer, solutions and source

Problem 8

  • Paper
  • Answer

Circle CC with radius 2 has diameter AB‾\overline{AB}. Circle D is internally tangent to circle CC at AA. Circle EE is internally tangent to circle CC, externally tangent to circle DD, and tangent to AB‾\overline{AB}. The radius of circle DD is three times the radius of circle EE, and can be written in the form m−n\sqrt{m}-n, where mm and nn are positive integers. Find m+nm+n.

Answer, solutions and source

Problem 10

  • Paper
  • Answer

Let zz be a complex number with ∣z∣=2014|z|=2014. Let PP be the polygon in the complex plane whose vertices are zz and every ww such that 1z+w=1z+1w\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}. Then the area enclosed by PP can be written in the form n3n\sqrt{3}, where nn is an integer. Find the remainder when nn is divided by 10001000.

Answer, solutions and source

Problem 11

  • Paper
  • Answer

In △RED\triangle RED, ∡DRE=75∘\measuredangle DRE=75^{\circ} and ∡RED=45∘\measuredangle RED=45^{\circ}. RD=1RD=1. Let MM be the midpoint of segment RD‾\overline{RD}. Point CC lies on side ED‾\overline{ED} such that RC‾⊥EM‾\overline{RC}\perp\overline{EM}. Extend segment DE‾\overline{DE} through EE to point AA such that CA=ARCA=AR. Then AE=a−bcAE=\frac{a-\sqrt{b}}{c}, where aa and cc are relatively prime positive integers, and bb is a positive integer. Find a+b+ca+b+c.

Answer, solutions and source

Problem 12

  • Paper
  • Answer

Suppose that the angles of △ABC\triangle ABC satisfy cos⁡(3A)+cos⁡(3B)+cos⁡(3C)=1.\cos(3A)+\cos(3B)+\cos(3C)=1. Two sides of the triangle have lengths 10 and 13. There is a positive integer mm so that the maximum possible length for the remaining side of △ABC\triangle ABC is m.\sqrt{m}. Find m.m.

Answer, solutions and source

Problem 13

  • Paper
  • Answer

Ten adults enter a room, remove their shoes, and toss their shoes into a pile. Later, a child randomly pairs each left shoe with a right shoe without regard to which shoes belong together. The probability that for every positive integer k<5k<5, no collection of kk pairs made by the child contains the shoes from exactly kk of the adults is mn\frac{m}{n}, where m and n are relatively prime positive integers. Find m+n.m+n.

Answer, solutions and source

Problem 14

  • Paper
  • Answer

In △ABC,AB=10,∠A=30∘\triangle{ABC}, AB=10, \angle{A}=30^\circ , and ∠C=45∘\angle{C=45^\circ}. Let H,D,H, D, and MM be points on the line BCBC such that AH⊥BCAH\perp{BC}, ∠BAD=∠CAD\angle{BAD}=\angle{CAD}, and BM=CMBM=CM. Point NN is the midpoint of the segment HMHM, and point PP is on ray ADAD such that PN⊥BCPN\perp{BC}. Then AP2=mnAP^2=\dfrac{m}{n}, where mm and nn are relatively prime positive integers. Find m+nm+n.

Answer, solutions and source

Problem 15

  • Paper
  • Answer

For any integer k≥1k\geq 1, let p(k)p(k) be the smallest prime which does not divide k.k. Define the integer function X(k)X(k) to be the product of all primes less than p(k)p(k) if p(k)>2p(k)>2, and X(k)=1X(k)=1 if p(k)=2.p(k)=2. Let {xn}\{x_n\} be the sequence defined by x0=1x_0=1, and xn+1X(xn)=xnp(xn)x_{n+1}X(x_n)=x_np(x_n) for n≥0.n\geq 0. Find the smallest positive integer tt such that xt=2090.x_t=2090.

Answer, solutions and source