Equal Squares

Note that $(a+b)^2 = (b+c)^2 = (c+a)^2$ doesn't necessarily mean $a+b = b+c = c+a$ , because for any real number $x$ , we have $(x)^2 = (-x)^2$ .

The correct equation is, instead, $|a+b| = |b+c| = |c+a|$ (with $|x|$ denotes the absolute value of $x$ )

Let $a+b = p$ , $b+c = q$ , and $c+a = r$ . Thus, $|p| = |q| = |r|$ .

Remember that the values of $p$ , $q$ , and $r$ can either have a negative sign or not.

By Pigeonhole Principle , there must be at least two out of those three values that have a same sign (in this case, there are three pigeons to be placed in two pigeonholes, so there is a pigeonhole containing at least two pigeons).

Without loss of generality, let those two values be $p$ and $q$ . Consequently, $p = q$ (because they have the same signs), which means $a + b = b + c$ . Obviously, $a = c$ .

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Next, we inspect the following two cases.

• $p = q = r$

Because $p = a + b$ , $r = c + a$ , and $p = r$ , then we have $b = c$ .

Therefore, $a = b = c$ , and $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 1 + 1 + 1 = 3$ .

• $p = q = -r$

Because $p = a + b$ , $r = c + a$ , and $p = -r$ , then $a + b = - (c+a)$ . However, $a = c$ , so $a + b = -2a$ . It is easy to find that $b = -3a$ .

Therefore, $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = \frac{a}{-3a} + \frac{-3a}{a} + \frac{a}{a} = (-\frac{1}{3}) + (-3) + 1 = -\frac{7}{3}$ .

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There are two possible values of $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ , which are $3$ and $-\frac{7}{3}$ , and the product of these two values are $\boxed{-7}$ .