2005 AMC 12B Problem 23

Question

Let

S

be the set of ordered triples

(x, y, z)

of real numbers for which

\log_{10}(x + y) = z

and

\log_{10}(x^2 + y^2) = z + 1.

There are real numbers

a

and

b

such that for all ordered triples

(x, y, z)

in

S

we have

x^3 + y^3 = a \cdot 10^{3z} + b \cdot 10^{2z}.

What is the value of

a + b?

\dfrac{15}{2}

\dfrac{29}{2}

15

\dfrac{39}{2}

24

Rewrite the conditions as

x + y = 10^z

and

x^2 + y^2 = 10 \cdot 10^z

Use

x^3 + y^3 = (x+y)^3 - 3xy(x+y)

with

xy = \tfrac12\left[(x+y)^2 - (x^2+y^2)\right]

The conditions give

x + y = 10^z

and

x^2 + y^2 = 10 \cdot 10^z.

Then

\begin{aligned} &2xy = (x+y)^2 - (x^2+y^2) \\ &= 10^{2z} - 10 \cdot 10^z, \end{aligned}

so

xy = \dfrac12\left(10^{2z} - 10 \cdot 10^z\right).

Using

x^3 + y^3

= (x+y)^3 - 3xy(x+y),

\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 = -\dfrac12

and

b = 15,

giving

a + b = \dfrac{29}{2}.

These coefficients are determined: setting

z=0

gives real solutions

x,y=\dfrac{1\pm\sqrt{19}}{2},

and their equation forces this same value of

a+b.

Thus, the correct answer is B.

Problem 23 in other years