MHB The Value of |xyz| Given x, y, and z

[anemone]

Let $$x,\;y,$$ and $$z$$ be distinct non-zero real numbers such that $$x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}$$.

What is the value of $$|xyz|$$?

 

[MarkFL]

My solution:

Spoiler

Let's begin with:

$$x+\frac{1}{y}=y+\frac{1}{z}$$

Solving this for $x$, we obtain:

$$x=\frac{y^2z+y-z}{yz}$$

Next, taking:

$$y+\frac{1}{z}=z+\frac{1}{x}$$

and solving this for $x$, we get:

$$x=\frac{z}{yz+1-z^2}$$

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

$$\frac{y^2z+y-z}{yz}=\frac{z}{yz+1-z^2}$$

Simplifying, we get:

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

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

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

Using the first factor, we obtain:

$$z=\frac{1}{1-y}$$

and so we find:

$$x=\frac{y-1}{y}$$

Hence:

$$|xyz|=\left|\frac{y-1}{y}\cdot y\cdot\frac{1}{1-y} \right|=|-1|=1$$

The same result is obtained from the other factor.

 

[anemone]

My solution:

Spoiler

Let's begin with:

$$x+\frac{1}{y}=y+\frac{1}{z}$$

Solving this for $x$, we obtain:

$$x=\frac{y^2z+y-z}{yz}$$

Next, taking:

$$y+\frac{1}{z}=z+\frac{1}{x}$$

and solving this for $x$, we get:

$$x=\frac{z}{yz+1-z^2}$$

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

$$\frac{y^2z+y-z}{yz}=\frac{z}{yz+1-z^2}$$

Simplifying, we get:

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

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

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

Using the first factor, we obtain:

$$z=\frac{1}{1-y}$$

and so we find:

$$x=\frac{y-1}{y}$$

Hence:

$$|xyz|=\left|\frac{y-1}{y}\cdot y\cdot\frac{1}{1-y} \right|=|-1|=1$$

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

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

$$\therefore \frac{y^2z+y-z}{yz}=\frac{z}{yz+1-z^2}$$

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]

...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)

 

[anemone]

Ouch! Dan was right...(Giggle)

Oh Dan...is he your buddy?(Tongueout)

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

Hahaha...this cracks me up!

 

[kaliprasad]

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