## x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
japiga
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### x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Help to solve this:
If x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1. Prove: xy+yz+zx=0

Martingale
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### Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Help to solve this:
If x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1. Prove: xy+yz+zx=0
hint :

$(a+b+c)=1=1^2=(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc$

therefore

$1=1+2ab+2ac+2bc$

hence

$ab+ac+bc=0$

japiga
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### Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

And then? How to include this in the first equation x/a=y/b=z/c ?

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### Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

And then? How to include this in the first equation x/a=y/b=z/c ?

$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$

solve for x in terms of z and solve for y in terms of z

then plug those into $xy+yz+zx$

japiga
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### Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

But what is a purpose of plugging into the equation of xy + yz + zx? Actually, how to reach this equation?

Martingale
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### Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

But what is a purpose of plugging into the equation of xy + yz + zx? Actually, how to reach this equation?

Don't you want to show that xy+yz+xz=0? ....Yes!!

so start with xy+yz+xz and show that is has to be zero using the hints I provided.

japiga
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### Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

You mean that this equation x/a=y/b=z/c should be split in two equations such as x/a=y/b and y/b=z/c. So x= (az)/c and y=(bz)/c. Does it what you mean?

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### Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

... So x= (az)/c and y=(bz)/c. Does it what you mean?
yes

japiga
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### Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Sorry, but it is not so helpful!
Please give me step by step procedure.

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### Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

...So x= (az)/c and y=(bz)/c. Does it what you mean?
Sorry, but it is not so helpful!
Please give me step by step procedure.

$xy+xz+yz=\frac{a}{c}z\frac{b}{c}z+\frac{a}{c}z\cdot z+\frac{b}{c}z\cdot z=\frac{ab}{c^2}z^2+\frac{a}{c}\cdot z^2+\frac{b}{c}\cdot z^2$

...