2024 AIME II Problem 4

Question

Let

x,

y,

and

z

be positive real numbers that satisfy the following system of equations:

\log_2\left(\frac{x}{yz}\right) = \frac{1}{2}

\log_2\left(\frac{y}{xz}\right) = \frac{1}{3}

\log_2\left(\frac{z}{xy}\right) = \frac{1}{4}

Then the value of

\left|\log_2(x^4 y^3 z^2)\right|

is

\frac{m}{n}

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

Set

a = \log_2 x,

b = \log_2 y,

c = \log_2 z

to turn the system into three linear equations

Add all three equations to find

a + b + c,

then combine it with each equation to solve for

a,

b,

c

one at a time

Let

a = \log_2 x,

b = \log_2 y,

c = \log_2 z.

The equations become

a - b - c = \frac{1}{2},

b - a - c = \frac{1}{3},

c - a - b = \frac{1}{4}.

Adding all three gives

-(a + b + c) = \frac{13}{12}.

Since

a - b - c = 2a - (a + b + c),

we get

2a = \frac{1}{2} - \frac{13}{12} = -\frac{7}{12},

so

a = -\frac{7}{24},

and similarly

b = -\frac{3}{8}

and

c = -\frac{5}{12}.

Therefore

4a + 3b + 2c = -\frac{28}{24}

- \frac{27}{24}

- \frac{20}{24} = -\frac{75}{24}

= -\frac{25}{8},

so

\left|\log_2(x^4 y^3 z^2)\right| = \frac{25}{8}

and

m + n = 25 + 8 = 33.

Problem 4 in other years