Mathematics : Cyclic Equations

(Initially written on 2/1/2011, revised 13/8/2025)

Let us look at a few equations with cyclic properties. You might think there are tricks in it, but 

Problem 1. Find all values of $x$ so that there exists real numbers $a,b,c$ satisfying $x = \frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{a+b}$.

It is an ancient HKCEE question but it looks more like Olympiad questions...anyway. 3 equations and 4 unknowns, how can we solve that?

Notice that if $(x,a,b,c)$ is a solution then so as $(x,ka,kb,kc)$ for any multiple $k$. For that we can fix $a = 1$ -- by eliminating the parameter we now have one unknown less. The equation reduces to $1+c = b^2+bc$ and $b^2+b = c^2+c$.

Notice that the second equation is symmetric. Or in other words it factorizes to $(b-c)(b+c+1) = 0$, so $b = c$ or $b+c=-1$. The first case gives $x = 1/2$ and the second gives $x = 1/2$ or $x = -1$.

Here is another beautiful approach to reach $x = 1/2$ by rearrangement:

$b+c = a/x, a+c = b/x, a+b = c/x$

Summing gives $2(a+b+c) = (a+b+c)/x$ so $x = 1/2$.

But wait, where is the $x=-1$ solution then? We have to check the case $a+b+c = 0$. In that case, $a/x = b+c = -a$, so $x = -1$.

How is that a public exam question???

Problem 2. (Prelim 09') Find all values of positive $t$ for there exist $u,v,w,x$ all distinct and real that satisfy

$t = u + \frac{1}{v} = v + \frac{1}{w} = w + \frac{1}{x} = x + \frac{1}{u}$.

Bad news is we can't de-parametrize because solutions aren't linear. What can we do now? Well, we parametrize instead:

$x = t-\frac{1}{u} = \frac{ut-1}{u}$

$w = t-\frac{1}{x} = \frac{ut^2-t-u}{ut-1}$

$v = t-\frac{1}{w} = \frac{ut^3-t^2-2ut+1}{ut^2-t-x}$

Putting back into $t = u+\frac{1}{v}$ and some factorization we reached the equation $t(t^2-2)(ut^2-u^2-1) = 0$. That says, either $t = 0, \pm \sqrt{2}$ or all 4 distinct variables got to satisfy the equation $\alpha ^2 - t^2\alpha + 1=0$ which is impossible. Therefore $t = \sqrt{2}$.

(Can you figure out possible solutions of $u,v,x,w$?)

It turns out that cyclic equations isn't really all about cyclicity, and each one has their own way to be tackled. Here are a few more similar problems with distinct solutions:

1) Find all values of $x$ so that there exists real numbers $a,b,c$ satisfying $x = \frac{ab}{ac+bc} = \frac{bc}{ac+ab} = \frac{ac}{ab+bc}$.

2) (Modified past IMO) Find all values of $x$ so that there exists real numbers $a,b,c,d$ satisfying $x = \frac{a}{b+c+d} = \frac{b}{c+d+a} = \frac{c}{d+a+b} = \frac{d}{a+b+c}$.

3) Find all values of $x$ so that there exists real numbers $a,b,c$ satisfying $x = \frac{a^2}{b^2+c^2} = \frac{b^2}{a^2+c^2} = \frac{c^2}{a^2+b^2}$.

4) (MO, 10') Solve for integer solutions: $x+y+z = 3, x^3+y^3+z^3 = 3$.