2000 AIME I Problem 9

Question

The system of equations

\begin{aligned} &\log_{10}(2000xy) \\ &\quad {}- (\log_{10} x)(\log_{10} y) = 4 \end{aligned}

\log_{10}(2yz) - (\log_{10} y)(\log_{10} z) = 1

\log_{10}(zx) - (\log_{10} z)(\log_{10} x) = 0

has two solutions

(x_1, y_1, z_1)

and

(x_2, y_2, z_2).

Find

y_1 + y_2.

Answer

Substitute

a = \log_{10} x,

b = \log_{10} y,

c = \log_{10} z

to make each equation polynomial in

a,

b,

c

Each equation factors: the first two become

(1 - a)(1 - b)

= (1 - b)(1 - c)

= \log_{10} 2,

and the third

(1 - c)(1 - a) = 1

Let

a = \log_{10} x,

b = \log_{10} y,

c = \log_{10} z.

Using

\log_{10} 2000 = 4 - \log_{10} 5,

the first equation becomes

a + b - ab = \log_{10} 5,

which factors as

(1 - a)(1 - b)

= 1 - \log_{10} 5

= \log_{10} 2.

Similarly the second equation gives

(1 - b)(1 - c)

= 1 - (1 - \log_{10} 2)

= \log_{10} 2,

and the third gives

(1 - c)(1 - a) = 1.

Dividing the first two (note

1 - b \ne 0

) yields

1 - a = 1 - c,

and then the third equation gives

(1 - a)^2 = 1,

so

1 - a = \pm 1.

If

1 - a = 1,

then

1 - b = \log_{10} 2,

so

b = 1 - \log_{10} 2 = \log_{10} 5

and

y = 5

(indeed

(x, y, z) = (1, 5, 1)

works). If

1 - a = -1,

then

1 - b = -\log_{10} 2,

so

b = \log_{10} 20

and

y = 20

(from

(x, y, z) = (100, 20, 100)

).

Therefore

y_1 + y_2 = 5 + 20 = 25.

Problem 9 in other years