2023 AMC 12A Problem 23

Below is the professionally curated solution for Problem 23 of the 2023 AMC 12A, 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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Concepts:AM-GM Inequality

Difficulty rating: 2380

23.

How many ordered pairs of positive real numbers $(a,b)$ satisfy the equation $(1+2a)(2+2b)(2a+b)=32ab?$

$0$

$1$

$2$

$3$

an infinite number

Solution:

By AM-GM, $1+2a\ge 2\sqrt{2a},$ $2+2b\ge 4\sqrt{b},$ and $2a+b\ge 2\sqrt{2ab}.$ Multiplying, $(1+2a)(2+2b)(2a+b)\ge 16\sqrt{2a}\cdot\sqrt{b}\cdot\sqrt{2ab}=32ab.$

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.

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