This looks ugly (but I assure you it isn't)

Algebra Level 5

For real numbers A , a , b , A, a, b, and c c , minimize

( A a ) ( A b ) ( c a ) ( c b ) + ( A a ) ( A c ) ( b a ) ( b c ) + ( A b ) ( A c ) ( a b ) ( a c ) \frac { (A-a)(A-b) }{ (c-a)(c-b) } +\frac { (A-a)(A-c) }{ (b-a)(b-c) } +\frac { (A-b)(A-c) }{ (a-b)(a-c) }


The answer is 1.

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1 solution

Dylan Pentland
Mar 26, 2015

I apologize for making the question somewhat deceiving: the equation is actually a very fancy way of saying 1 1 ! However, if I asked you to find the constant value for it, it would be trivial to find it because you could just plug in some values and guess 1 1 .

If we fix a , b , c a, b, c then the expression becomes a quadratic with a variable of A A . Note that at A = a , A = b , A = c A=a, A=b, A=c two of the large terms become zero because they contain A a A-a , A b A-b , or A c A-c on top and the last term has a matching numerator and denominator. Thus, at these values the expression is 1 1 .

Three points define a quadratic equation! The only one that fits is that for any A A the quadratic yields 1 1 , therefore the expression simplifies to 1 1 .

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