Problem Of The Week # 305 - Apr 10, 2018

[Ackbach]

Here is this week's POTW:

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Solve for $x, y, z$ (in terms of $a, r, s, t$):
\begin{align*}
yz&=a(y+z)+r\\
zx&=a(z+x)+s\\
xy&=a(x+y)+t.
\end{align*}
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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

Congratulations to castor28 for his correct solution, and an honorable mention to kiwi for a mostly correct solution, to this week's POTW, which was Problem 150 in the MAA Challenges. castor28's solution follows:

[sp]Each of the equations represents a hyperbola with center $(a,a)$. If we write $x=u+a$, $y=v+a$, $z=w+a$, the equations become:
$$\begin{align*}
uv &= a^2 + t\\
uw &= a^2 + s\\
vw &= a^2 + r
\end{align*}$$
Multiplying the equations together, we get:
$$(uvw)^2 = (a^2+r)(a^2+s)(a^2+t)$$
and we obtain:
$$\begin{align*}
u &= \pm\sqrt{\frac{(a^2+s)(a^2+t)}{a^2+r}}\\
v &= \pm\sqrt{\frac{(a^2+r)(a^2+t)}{a^2+s}}\\
w &= \pm\sqrt{\frac{(a^2+r)(a^2+s)}{a^2+t}}\\
\end{align*}$$
Note that the signs are not independent. Once a sign is chosen for $u$, the signs of $v$ and $w$ are determined by the signs of $uv$ and $uw$; in general, the system has two solutions. This is also true if some unknowns are imaginary (if $(a^2+r)(a^2+s)(a^2+t)<0$).

We can then recover $x$, $y$, $z$ as $u+a$, $v+a$ and $w+a$.

If exactly one of $(a^2+r)$, $(a^2+s)$, $(a^2+t)$ is $0$, there is no solution. Indeed, if $a^2 + r = vw = 0$, then at least one of $uv$ or $uw$ must be $0$.

If at least two of $(a^2+r)$, $(a^2+s)$, $(a^2+t)$ are $0$, there are infinitely many solutions (at least two of the equations are the same). For example, if $a^2+r=a^2+s=0$, we can take $w=0$ and we are left with the equation $uv=a^2+t$, which has infinitely many solutions.[/sp]