It is given that , , and are nonzero real numbers such that
Find the product of all possible distinct values of .
If you think that the equality has no solution, submit 0 as your answer.
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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 .
Next, we inspect the following two cases.
Because p = a + b , r = c + a , and p = r , then we have b = c .
Therefore, a = b = c , and b a + c b + a c = 1 + 1 + 1 = 3 .
Because p = a + b , r = c + a , and p = − r , then a + b = − ( c + a ) . However, a = c , so a + b = − 2 a . It is easy to find that b = − 3 a .
Therefore, b a + c b + a c = − 3 a a + a − 3 a + a a = ( − 3 1 ) + ( − 3 ) + 1 = − 3 7 .
There are two possible values of b a + c b + a c , which are 3 and − 3 7 , and the product of these two values are − 7 .