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Find |xyz|

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anemone

MHB POTW Director
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Feb 14, 2012
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Let \(\displaystyle x,\;y,\) and \(\displaystyle z\) be distinct non-zero real numbers such that \(\displaystyle x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}\).

What is the value of \(\displaystyle |xyz|\)?
 

MarkFL

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Feb 24, 2012
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My solution:

Let's begin with:

\(\displaystyle x+\frac{1}{y}=y+\frac{1}{z}\)

Solving this for $x$, we obtain:

\(\displaystyle x=\frac{y^2z+y-z}{yz}\)

Next, taking:

\(\displaystyle y+\frac{1}{z}=z+\frac{1}{x}\)

and solving this for $x$, we get:

\(\displaystyle x=\frac{z}{yz+1-z^2}\)

Hence, we may equate the two expressions for $x$ to get:

\(\displaystyle \frac{y^2z+y-z}{yz}=\frac{z}{yz+1-z^2}\)

Simplifying, we get:

\(\displaystyle (y-z)\left(yz-z+1 \right)\left(yz+z+1 \right)=0\)

Since $y$ and $z$ must be distinct, we are left with:

\(\displaystyle \left(yz-z+1 \right)\left(yz+z+1 \right)=0\)

Using the first factor, we obtain:

\(\displaystyle z=\frac{1}{1-y}\)

and so we find:

\(\displaystyle x=\frac{y-1}{y}\)

Hence:

\(\displaystyle |xyz|=\left|\frac{y-1}{y}\cdot y\cdot\frac{1}{1-y} \right|=|-1|=1\)

The same result is obtained from the other factor.
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,680
My solution:

Let's begin with:

\(\displaystyle x+\frac{1}{y}=y+\frac{1}{z}\)

Solving this for $x$, we obtain:

\(\displaystyle x=\frac{y^2z+y-z}{yz}\)

Next, taking:

\(\displaystyle y+\frac{1}{z}=z+\frac{1}{x}\)

and solving this for $x$, we get:

\(\displaystyle x=\frac{z}{yz+1-z^2}\)

Hence, we may equate the two expressions for $x$ to get:

\(\displaystyle \frac{y^2z+y-z}{yz}=\frac{z}{yz+1-z^2}\)

Simplifying, we get:

\(\displaystyle (y-z)\left(yz-z+1 \right)\left(yz+z+1 \right)=0\)

Since $y$ and $z$ must be distinct, we are left with:

\(\displaystyle \left(yz-z+1 \right)\left(yz+z+1 \right)=0\)

Using the first factor, we obtain:

\(\displaystyle z=\frac{1}{1-y}\)

and so we find:

\(\displaystyle x=\frac{y-1}{y}\)

Hence:

\(\displaystyle |xyz|=\left|\frac{y-1}{y}\cdot y\cdot\frac{1}{1-y} \right|=|-1|=1\)

The same result is obtained from the other factor.
Thanks for participating again in my challenge problem, MarkFL and I think to expand and verify this

\(\displaystyle (y-z)\left(yz-z+1 \right)\left(yz+z+1 \right)=0\)

\(\displaystyle \therefore \frac{y^2z+y-z}{yz}=\frac{z}{yz+1-z^2}\)

to be true is easy, but to contract it to become a product of three factors...that is much more difficult, and so I'll deduct 2 marks from you for this...hehehe...
 

MarkFL

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Feb 24, 2012
13,775
...so I'll deduct 2 marks from you for this...hehehe...
Ouch! Dan was right...(Giggle)

I figured I could leave the drudgery of the details to the reader...you know, like a good textbook. (Wasntme)
 
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anemone

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Feb 14, 2012
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kaliprasad

Well-known member
Mar 31, 2013
1,309
x+1/y=y+1/z=z+1/x. ...(1)

this question would not have been asked has the solution not been unique

now putting 1/a for z , 1/b for x and 1/c for y we get

1/b + c = 1/c + a = 1/a + b

the above equation is same as (1) with c for x b for y and a for z and hence

|xyz| = |abc| as the value is unique

= | 1/z 1/x 1/y| = | 1/xyz|

so |xyz| = 1