3sir all know dat a=b=c=0 satisfis the equation which also satisfies the given condition
This is a problem which for some strange reason is going unanswered in another forum(there's one guy who has promised to come back with a solution though):
If a,b,c≠±1, and a+b+c = abc, prove that we also have A+B+C = ABC where
A = \frac{2a}{1-a^2}; B = \frac{2b}{1-b^2}; C = \frac{2c}{1-c^2}
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6 Answers
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341See, you have to prove that whenever a+b+c = abc we also have A+B+C = ABC, not just when a=b=c=0
9this is very direct
just substitute for A,B,C and compare LHS with RHS
341its more direct to let a = tan A, b = tan B and c = tan C, where A,B, C are angles of a ...
ok, enuf said
66Its indeed strange why this sum remain unanswered given that its a straightforward one.
We have A+B+C=2\dfrac{\sum_\mathrm{cyclic}a(1-b^2)(1-c^2)}{(1-a^2)(1-b^2)(1-c^2)}
The numerator is
a+b+c - (ab2+ac2+bc2+ba2+ca2+cb2) + (ab2c2+bc2a2+ca2b2)
Using the fact that a+b+c=abc, transform the last bracket as follows:
abc(bc+ca+ab) = (a+b+c)(bc+ca+ab)=3abc + (ab2+ac2+bc2+ba2+ca2+cb2)
Hence, the numerator becomes, (a+b+c)+3abc = 4abc, from where we get
A+B+C=\dfrac{8abc}{(1-a^2)(1-b^2)(1-c^2)}=ABC
Q.E.D.
341A more straightforward way is to see that we can let a =\tan \alpha, b = \tan \beta , c = \tan \gamma with \alpha + \beta +\gamma = \pi
It follows that 2\alpha + 2\beta +2\gamma = 2\pi
so that \tan (2\alpha + 2\beta +2\gamma) = 0 \Rightarrow \tan 2\alpha + \tan 2\beta + \tan 2\gamma = \tan 2\alpha \tan 2\beta \tan 2\gamma
But \tan 2\alpha = \frac{2 \tan \alpha}{1-\tan^2 \alpha} = \frac{2a}{1-a^2} etc.
and hence the conclusion follows