2023 AMC 12A Problem 23

Attempt Problem 23 of the 2023 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 12A solutions, or check the answer key.

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23.

How many ordered pairs of positive real numbers

(a,b)

satisfy the equation

\begin{gathered} (1+2a)(2+2b)(2a+b)\\ {}=32ab? \end{gathered}

0

1

2

3

an infinite number

Apply AM-GM separately to

1+2a,

2+2b,

and

2a+b

The product is at least

32ab,

so equality must hold in all three inequalities at once

By AM-GM,

1+2a\ge 2\sqrt{2a},

2+2b\ge 4\sqrt{b},

and

2a+b\ge 2\sqrt{2ab}.

Multiplying,

\begin{gathered} (1+2a)(2+2b)(2a+b)\\ {}\ge 16\sqrt{2a}\cdot\sqrt{b}\cdot\sqrt{2ab}\\ {}=32ab. \end{gathered}

Equality requires

1=2a,

2=2b,

and

2a=b

simultaneously. These give

a=\tfrac12,

b=1,

which are consistent, so there is exactly one solution.

Thus, the correct answer is B.

Problem 23 in other years