Problem of the Week #2

(As usual, put solutions/suggestions in comments. A solution will be posted June 5.)

The arithmetic, geometric, and harmonic mean of two positive integers are themselves three distinct positive integers. What is the smallest possible value of the arithmetic mean.

(Source: Iberoamerican Olympiad 2010, Q4. But don't look! Solve it on your own.)

Comments

  1. MaeveMay 31, 2017 at 11:40 AM

    For the geometric mean (a*b)^(1/2) to be an integer, a*b must be a perfect square. So we can try each perfect square until we find the first one that has a pair of factors with integral arithmetic and harmonic means. Note that if a = b, the means will not be distinct, so we can ignore that pair of factors. Also if a=1, the harmonic mean is 2((b^2)/(b^2 +1)). b^2 and b^2 +1 are relatively prime, so this is never an integer. So if the square is a prime p squared, it can be factored as p*p (a=b, doesn't work) or 1*p^2 (a=1, doesn't work.) So we can ignore primes.

    4: a=1, b=16: AM = 17/2. a=2, b=8: HM = 16/5.
    6: a=1, b=36: AM = 37/2. a=2, b=18: HM = 18/5. a=3, b=12: AM = 15/2. a=4, b=9: AM = 13/2.
    8: a=1, b=64: AM = 65/2. a=2, b=32: HM = 64/17. a=4, b=16: HM = 32/5.
    9: a=1, b=81: HM = 81/41. a=3, b=27: HM = 27/5.
    10: a=1, b=100: AM = 101/2. a=2, b=50: HM = 50/13. a=5, b=20: AM = 25/2.
    12: a=1, b=144: AM = 145/2. a=2, b=72: HM = 144/37. a=3, b=48: AM = 51/2. a=4, b=36: HM = 36/5. a=6,
    b=24: HM = 48/5.
    14: a=1, b=196: AM = 197/2. a=2, b=98: HM = 98/25. a=7, b=28: AM = 35/2.
    15: a=1, b=225: HM = 225/113. a=3, b=75: HM = 75/13. a=5, b=45: AM = 25. HM = 9. GM = 15.

    So a=5, b=45 is the first case that works.
    Because the arithmetic mean is always bigger than the geometric mean, any pair a,b with a geometric mean larger than 25 must also have an arithmetic mean larger than 25. So we check the rest of the squares up to 25 squared. However, we can ignore pairs of factors that sum to 50 or more, since they will have an arithmetic mean greater than or equal to 25.

    16: a=8, b=32: HM = 64/5. a=4, b=64: sum of 68. We can ignore these and all the rest of the pairs for 16.
    18: a=9, b=36: AM = 45/2. a=6, b=54: sum of 60.
    20: a=10, b=40: sum of 50.
    22: a=11, b=44: sum of 55.
    24: a=12, b=48: sum of 60.

    Since we didn't find any pairs that worked and had an arithmetic mean less than 25, and we know there can't be any after 24, we know 25 is the smallest possible arithmetic mean.

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