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| Some Squares Sum Square (Posted on 2012-07-30) |
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Determine all possible non zero real triplet(s) (F, G, H) that satisfy the following system of equations:
(7F)2 + (7G) 2 + (7H) 2 = (2F + 3G + 6H) 2 , and :
1/F + 1/G + 1/H = 1
Solution
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 | Comment 6 of 9 | 
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This 3D approach is the only way I can seem to get my head round this problem and confirm that only one triple exists.
Let (7F)2 + (7G)2 + (7H)2 = 49r2 , so that F2 + G2 + H2 = r2
then (2F + 3G + 6H)2 = 49r2, giving 2F + 3G + 6H = +/- 7r (r>0 say)
Considering (F, G, H) to be coordinates in 3-D space...
The first equation is of a sphere with centre at (0, 0, 0) and radius r, while
the second equation is of a plane whose normal has direction ratios (2, 3, 6).
The distance of the plane from the origin is +/- 7r/sqrt(22 + 32 + 62),
which simplifies to +/- r, so the plane can only touch the sphere at the diametrically opposite points A (2r/7, 3r/7, 6r/7) and B (-2r/7, -3r/7, -6r/7).
Substitution into the second given equation, 1/F + 1/G + 1/H = 1, then shows that only A gives a possible solution, viz r = 7/2 + 7/3 + 7/6 = 7, from which the only possible triple (F, G, H) is (2, 3, 6).
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Posted by Harry
on 2012-08-01 13:12:16 |
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