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2005 AMC 12B Problem 23

Problem 23 of 25HarderAlgebra

Let SS be the set of ordered triples (x,y,z)(x, y, z) of real numbers for which log⁡10(x+y)=z \log_{10}(x + y) = z and log⁡10(x2+y2)=z+1. \log_{10}(x^2 + y^2) = z + 1. There are real numbers aa and bb such that for all ordered triples (x,y,z)(x, y, z) in SS we have x3+y3=a⋅103z+b⋅102z.x^3 + y^3 = a \cdot 10^{3z} + b \cdot 10^{2z}. What is the value of a+b?a + b?

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Solution

The conditions give x+y=10zx + y = 10^z and x2+y2=10⋅10z.x^2 + y^2 = 10 \cdot 10^z. Then 2xy=(x+y)2−(x2+y2)=102z−10⋅10z, \begin{aligned} &2xy = (x+y)^2 - (x^2+y^2) \\ &= 10^{2z} - 10 \cdot 10^z, \end{aligned} so xy=12(102z−10⋅10z).xy = \dfrac12\left(10^{2z} - 10 \cdot 10^z\right). Using x3+y3x^3 + y^3 =(x+y)3−3xy(x+y),= (x+y)^3 - 3xy(x+y), x3+y3=103z−32(102z−10⋅10z)10z=−12⋅103z+15⋅102z. \begin{aligned} &x^3 + y^3 = 10^{3z} \\ &\quad {}- \dfrac32\left(10^{2z} - 10 \cdot 10^z\right)10^z \\ &= -\dfrac12 \cdot 10^{3z} + 15 \cdot 10^{2z}. \end{aligned} So a=−12a = -\dfrac12 and b=15,b = 15, giving a+b=292.a + b = \dfrac{29}{2}. These coefficients are determined: setting z=0z=0 gives real solutions x,y=1±192,x,y=\dfrac{1\pm\sqrt{19}}{2}, and their equation forces this same value of a+b.a+b. Thus, the correct answer is B.

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Concepts: sum and difference of cubes · logarithm · symmetry (algebra)

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