A slightly altered 2000 AMC 12 #20

There's like many trash and trivial and way too personal posts on my blog recently, so thought we should do something hardcore here.

Consider this problem:

Solve the system of equations: $x+\dfrac 1y=-1$, $y+\dfrac 1z=\dfrac12$, and $z+\dfrac 1x=2$.

Some of y'all may recognize this as the 2000 AMC 12 Problem 20.  But there's two differences here: 1) the numbers are changed (which matters BIG TIME!) and 2) we need to find $x$, $y$, and $z$, while AMC12/20 asks us to find the product $xyz$.

A naive way

So we find that the first and second equations allows us to relate all the other equations to $y$.  Now, the third equation relates the other two variables, which means we can represent everything in terms of $y$ by first writing the first and second equations as $x$ and $z$ in terms of $y$, substituting into the third equation, and then solving for $y$.  From there, we can get $x$ and $z$.

We note that we can reduce things to a single variable $y$ if we represent $x$ and $z$ in terms of $y$ using the first and second equations, and then substitute that into the third equation to solve for $y$.  This gets us $x=-1-\frac1y$ and $\frac1z=\frac12-y$ so $z=\frac2{1-2y}$.  Substituting both into the third equation, we have $$\frac2{1-2y}-\frac{y}{y+1}=2$$ (in which $\frac1x=-\frac{y}{y+1}$.)

Multiplying through, we get $2(y+1)-y(1-2y)=2(1-2y)(y+1)$, which is $y+2+2y^2=-4y^2-2y+2$, which is $3y+6y^2=0$.  The solutions are either $y=-\frac12$ or $y=0$, and since $y$ appears in denominators the latter is obviously not acceptable.  So $y=-\frac12$.

Now, we can substitute this $y$ into the first and second equations in this problem to solve for $x$ and $z$.  You should get $(1,-\frac12,1)$.

An elegant observation

Now, note that the $3$ equations all have $x$, $\frac1x$, $y$, $\frac1y$, $z$ and $\frac1z$ on the left.  This means that if we multiply the expressions on the left, a lot of things can cancel out.

And indeed, the parts that are left over is pretty nice.  If we actually multiply everything through and reorder the terms slightly, we note that

$$\begin{aligned}\left(x+\frac1y\right)\left(y+\frac1z\right)\left(z+\frac1x\right)&=xyz+xy\frac1x+x\frac1zz+x\frac1z\frac1x+\frac1yyz+\frac1yy\frac1x+\frac1y\frac1zz+\frac1y\frac1z\frac1x\\&=xyz+y+x+\frac1z+z+\frac1x+\frac1y+\frac1{xyz}\\&=xyz+\frac1{xyz}+\left(x+\frac1y\right)+\left(y+\frac1z\right)+\left(z+\frac1x\right)\end{aligned}$$

....which is simply... magical.  Multiplying these three left-sides gives something that includes the sum of the three terms; and since we already know these three terms, let's substitute in the values!  We have $-1\cdot \frac12\cdot 2=xyz+\frac1{xyz}+(-1)+\frac12+2$, and rearranging this is $xyz+\frac1{xyz}=-\frac52$; multiplying through by $xyz$, we have a quadratic in $xyz$ which we solve to get either $xyz=-2$ or $xyz=-\frac12$.

But, how the heck do we get $x$, $y$, and $z$ from $xyz$?  Well, I thought about this for about half an hour, but nothing really came to me.  I then had to turn to Google Gemini, and it turns out there may be one approach, namely to do this:

$$\left(x+\frac1y\right)\left(y+\frac1z\right)z=xyz+x+z+\frac1y=xyz+\left(x+\frac1y\right)+z$$

which means that if we substitute values for $\left(x+\frac1y\right)$ and $\left(y+\frac1z\right)$ and rearrange, we get $\frac32z=1-xyz$.  From here, we can plug in the two possible values of $xyz$ and get either $z=1$ from $xyz=-\frac12$ or $z=2$ from $xyz=-2$.  Now we plug the $z$'s into the second and third equations to solve for $x$ and $y$; note that, if we plug $z=2$ into the third equation we get $\frac1x=0$ which has No Roots (pun intended).  But if we plug $z=1$ into the first and second equations, we again get $(1,-\frac12,1)$.

I suspect that this is motivated by the fact as long as we get a single value, we can substitute into the two different equations, and by noticing that to isolate a single value, we have to break symmetry somewhere, so we multiply things by dropping one term in one expression.

Useless magic

Notice how we had to find another way to do things in order to extract the actual $x$, $y$, and $z$ values?  Notice we have to do all the work twice because one of the $xyz$ values is actually invalid?  Notice how the second writeup is simply longer?  Ever seen someone sweat so much about a single problem?

Now for the million-dollar question: Ever noticed how Rapunzel's hair seems to glow when she's singing, but how the glow does absolutely zero good in the movie Tangled (please Disney don't trash the live action like you've done with previous movies)?  Looking at you, Flynn Rider.  You're a thief.  Go steal some candles and pencils.  Light the candles up in a cave, grab a pencil, and get Rapunzel to chain you to a chair until you just bash your math problems in the most naive way possible.  At least that shows Raps you're hard working and therefore trustworthy.

Well yeah.  That's sort of the point.  The second approach is noticably more complicated, even though it has a certain elegance.

But still, one can instantly demolish one very specific AMC 12 problem with this "elegant" approach.

That AMC Problem

So notice how we made one elegant observation, but then had to bend backwards to try to use another observation to easily extract the $x$, $y$, and $z$ values to finish up our problem?  Well, this exact form was not neccessary on an AMC problem.  This is what it was:

If $x,y,$ and $z$ are positive numbers satisfying

$$x + \frac{1}{y} = 4,\qquad y + \frac{1}{z} = 1, \qquad \text{and} \qquad z + \frac{1}{x} = \frac{7}{3}$$

Then what is the value of $xyz$ ?

$$\text {(A)}\ \frac{2}{3} \qquad \text {(B)}\ 1 \qquad \text {(C)}\ \frac{4}{3} \qquad \text {(D)}\ 2 \qquad \text {(E)}\ \frac{7}{3}$$

Now, the same elegant observation about $xyz$ may be used, and we will get a quadratic to $xyz$.  But hold on, it may have 2 solutions, how do we know which one is the valid one that doesn't require $\frac1x=0$ or $\frac1y=0$?

Well, that problem doesn't have this issue, because they designed the numbers so that the quadratic only have one root, and AMC 12's are guaranteed to have an answer, so there's no need for verifying things or finding the specific $x$'s, $y$'s, and $z$'s.  I won't bother repeating myself, so here's the solution: https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_20

And so, I'm both impressed and baffled by the fact that such a technique would be much more complicated if one had just asked a slightly different question with slightly different numbers...

Conclusion

This is, ultimately, another life lesson I've had to learn.  There's beauty and all, but you gotta survive.  You can't waste your time on trivial things like this, and miss out on all the more complicated math that is actually beautiful.  You can hike through endless miles of barren landscapes to explore because you like the way the sunset touches the horizon, but why not take a car ride and hike at a more national park instead?

While Rapunzel's hair is beautiful, she did have to cut it at the end to actually improve her situations.  And sometimes, we do have to cut our magical long hair for things to work out, like in this case...