(In reply to
Solution by Harry)
Having now read the hint from KS, I’m wondering if this is what was needed:
The given equation can be rearranged to
10(2G – 3F)2 – 18(2G – 3F)(H – 3F) + 13(H – 3F)2 = 0 (1)
Clearly, a solution is 3F = 2G = H, which gives triples of the form (f, 3f/2, 3f).
Substitution into the equation 1/F + 1/G + 1/H = 1 then gives
f = 1 + 2/3 + 1/3 = 2, so the triple is (2, 3, 6).
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No other triple is possible. Writing (1) as 10a2 – 18ab + 13b2 = 0 and solving
this quadratic for a/b would involve the discriminant (-18)2 – 4(10)(13),
which is negative and would deny real values for at least one of a and b and
therefore at least one of F, G and H.
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Posted by Harry
on 2012-08-01 14:06:37 |
re: Solution (spoiler)
