Equal Squares

Algebra Level 5

It is given that a a , b b , and c c are nonzero real numbers such that

( a + b ) 2 = ( b + c ) 2 = ( c + a ) 2 (a+b)^2 = (b+c)^2 = (c+a)^2

Find the product of all possible distinct values of a b + b c + c a \frac{a}{b} + \frac{b}{c} + \frac{c}{a} .

If you think that the equality has no solution, submit 0 as your answer.


The answer is -7.

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1 solution

Vincent Tandya
Jan 4, 2016

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

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

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

Remember that the values of p p , q q , and r 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 p and q q . Consequently, p = q p = q (because they have the same signs), which means a + b = b + c a + b = b + c . Obviously, a = c a = c .


Next, we inspect the following two cases.

  • p = q = r p = q = r

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

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

  • p = q = r p = q = -r

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

Therefore, a b + b c + c a = a 3 a + 3 a a + a a = ( 1 3 ) + ( 3 ) + 1 = 7 3 \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} .


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

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