If a,b,c are three distinct Watch

drmath
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If a,b,c are three distinct nonzero real numbers and

a + (1 / b) .=b + (1 / c) .=c + (1 / a) .=t

for some real number t

prove that abc + t=0 .
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JohnnySPal
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(Original post by drmath)
If a,b,c are three distinct nonzero real numbers and

a + (1 / b) .=b + (1 / c) .=c + (1 / a) .=t

for some real number t

prove that abc + t=0 .
a + (1/b) = t
b + (1/c) = t
c + (1/a) = t

abc = [t-(1/b)][t-(1/c)][t-(1/a)] = [t^2 - (t/b) - (t/c) + (1/ab)][t-(1/a)]
= t^3 - (t^2/b) - (t^2/c) + (t/ab) - (t^2/a) + (t/ab) + (t/ac) - (1/a^2b)
= t^3 - t^2[(1/b) + (1/c) + (1/a)] + t[(1/ab) + (1/ab) + (1/ac)] - (1/a^2b)
= t^3 - t^2[3t-(a+b+c)] + t[...

Argh, I cannae be assed with this. I'm sure there's a very clever trick you can use that I'm not spotting.

I imgaine that by bashing away hard enough like I have you can get t[(1/ab) + (1/ab) + (1/ac)] - (1/a^2b) purely in terms of t, and then you'll get a "cubic" in t where actually all the non-linear terms cancel out with each other.
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Zhen Lin
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I would begin by noting that t is fully determined by a, b, c and that a, b, c are not entirely arbitrary - if you have two of them, the third is automatically determined simply by considering one of the three equations. I wouldn't be surprised if there is only one solution set for a, b, c - there are three variables and three equations:

\displaystyle \left\{

\begin{aligned}

a + \frac{1}{b} & = b + \frac{1}{c} \\

b + \frac{1}{c} & = c + \frac{1}{a} \\

c + \frac{1}{a} & = a + \frac{1}{b} 

\end{aligned}

\right.

Also, by considering \displaystyle \left( a + \frac{1}{b} \right) \left( b + \frac{1}{c} \right) \left( c + \frac{1}{a} \right) = t^3, and assuming that abc = -t, we find that t = \pm 1.
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DFranklin
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See also STEP 1991, FM Paper A, Q3.
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