Too many monics

Algebra Level 5

Monic quadratic polynomials P ( x ) P(x) and Q ( x ) Q(x) have the property that P ( Q ( x ) ) P(Q(x)) has zeros at x = 23 , 21 , 17 , x = -23, -21, -17, and 15 -15 , and Q ( P ( x ) ) Q(P(x)) has zeros at x = 59 , 57 , 51 , x = -59, -57, -51, and 49 -49 . What is the sum of the minimum values of P ( x ) P(x) and Q ( x ) Q(x) ?


The answer is -100.

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1 solution

Revanth Gumpu
Aug 19, 2015

Since both P ( Q ( x ) ) P(Q(x)) and Q ( P ( x ) ) Q(P(x)) have four distinct real zeros, both P ( x ) P(x) and Q ( x ) Q(x) must have two distinct real zeros. Hence, we can write P ( x ) = ( x h 1 ) 2 k 1 2 P(x) = (x - h_1)^2 - k_{1}^2 and Q ( x ) = ( x h 2 ) 2 k 2 2 Q(x) = (x - h_{2})^2 - k_{2}^2 .

The zeros of P ( x ) P(x) are h 1 ± k 1 h_{1} \pm k_{1} , so the zeros of P ( Q ( x ) ) P(Q(x)) are the roots of Q ( x ) = h 1 ± k 1 Q(x) = h_{1} \pm k_{1} . Since the function Q ( x ) = ( x h 2 ) 2 k 2 2 Q(x) = (x - h_{2})^2 - k_{2}^2 is symmetric around x = h 2 x = h_{2} , the roots of Q ( x ) = h 1 + k 1 Q(x) = h_{1} + k_{1} must be symmetric around x = h 2 x = h_{2} , and the roots of Q ( x ) = h 1 k 1 Q(x) = h_{1} - k_{1} must also be symmetric around x = h 2 x = h_{2} . The values 23 , 21 , 17 -23, -21, -17 , and 15 -15 are symmetric around 19 -19 , so h 2 = 19 h_{2} = -19 . Then Q ( x ) = ( x + 19 ) 2 k 2 2 Q(x) = (x + 19)^2 - k_{2}^2 .

Since Q ( x ) Q(x) is monic, Q ( x ) Q(x) is decreasing on the interval ( , 19 ] (-\infty,-19] and increasing on the interval [ 19 , ) [-19,\infty) . Therefore, the roots of Q ( x ) = h 1 k 1 Q(x) = h_{1} - k_{1} are 21 -21 and 17 -17 , and the roots of Q ( x ) = h 1 + k 1 Q(x) = h_{1} + k_{1} are 23 -23 and 15 -15 . Hence, Q ( 17 ) = h 1 k 1 Q(-17) = h_{1} - k_{1} and Q ( 15 ) = h 1 + k 1 Q(-15) = h_{1} + k_{1} , so Q ( 15 ) Q ( 17 ) = 2 k 1 Q(-15) - Q(-17) = 2k_{1} . But Q ( 15 ) Q ( 17 ) = [ ( 15 + 19 ) 2 k 2 2 ] [ ( 17 + 19 ) 2 k 2 2 ] = 16 4 = 12 Q(-15) - Q(-17) = [(-15 + 19)^2 - k_{2}^2]- [(-17 + 19)^2 - k_{2}^2] = 16 - 4 = 12 , so 2 k 1 = 12 2k_{1} = 12 , which means k 1 = 6 k_{1} = 6 .

Similarly, the zeros of Q ( x ) Q(x) are h 2 ± k 2 h_2 \pm k_{2} , so the zeros of Q ( P ( x ) ) Q(P(x)) are the roots of P ( x ) = h 2 ± k 2 P(x) = h_{2} \pm k_{2} . Since the function P ( x ) = ( x h 1 ) 2 k 1 2 P(x) = (x - h_{1})^2 - k_{1}^2 is symmetric around x = h 1 x = h_{1} , the roots of P ( x ) = h 2 + k 2 P(x) = h_{2} + k_{2} must be symmetric around x = h 1 x = h_{1} , and the roots of P ( x ) = h 2 k 2 P(x) = h_{2} - k_{2} must also be symmetric around x = h 1 x = h_{1} . The values 59 , 57 , 51 , -59, -57, -51, and 49 -49 are symmetric around 54 -54 , so h 1 = 54 h_{1} = -54 . Then P ( x ) = ( x + 54 ) 2 k 1 2 P(x) = (x + 54)^2 - k_{1}^2 .

Since P ( x ) P(x) is monic, P ( x ) P(x) is decreasing on the interval ( , 54 ] (-\infty,-54] and increasing on the interval [ 54 , ) [-54,\infty) . Therefore, the roots of P ( x ) = h 2 k 2 P(x) = h_{2} - k_{2} are 57 -57 and 51 -51 , and the roots of P ( x ) = h 2 + k 2 P(x) = h_{2} + k_{2} are 59 -59 and 49 -49 . Hence, P ( 51 ) = h 2 k 2 P(-51) = h_{2} - k_{2} and P ( 49 ) = h 2 + k 2 P(-49) = h_{2} + k_{2} , so P ( 49 ) P ( 51 ) = 2 k 2 P(-49) - P(-51) = 2k_{2} . But P ( 49 ) P ( 51 ) = [ ( 49 + 54 ) 2 k 1 2 ] [ ( 51 + 54 ) 2 k 1 2 ] = 25 9 = 16 P(-49) - P(-51) = [(-49 + 54)^2 - k_1^2] - [(-51 + 54)^2 - k_1^2] = 25 - 9 = 16 , so 2 k 2 = 16 2k_{2} = 16 , which means k 2 = 8 k_{2} = 8 .

The minimum value of P ( x ) = ( x h 1 ) 2 k 1 2 P(x) = (x - h_{1})^2 - k_{1}^2 is k 1 2 -k_{1}^2 , and the minimum value of Q ( x ) = ( x h 2 ) 2 k 2 2 Q(x) = (x - h_{2})^2 - k_{2}^2 is k 2 2 -k_{2}^2 , so the sum of their minimum values is k 1 2 k 2 2 = 36 64 = 100 -k_{1}^2 - k_{2}^2 = -36 - 64 = \boxed{-100} .

Moderator note:

Great approach! Converting it into the parabola form makes it easier to manipulate and understand these polynomials in the context of the problem.

We only have 4 unknowns (the coefficients), but are given 8 constants (the zeroes). Does this mean that we can reduce the number of roots, and still be able to solve the problem?

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