# 2023 AMC 12A Problems/Problem 23

## Problem

How many ordered pairs of positive real numbers $(a,b)$ satisfy the equation $$(1+2a)(2+2b)(2a+b) = 32ab?$$ $\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{an infinite number}$

## Solution 1: AM-GM Inequality

Using AM-GM on the two terms in each factor on the left, we get $$(1+2a)(2+2b)(2a+b) \ge 8\sqrt{2a \cdot 4b \cdot 2ab}= 32ab,$$ meaning the equality condition must be satisfied. This means $1 = 2a = b$, so we only have $\boxed{1}$ solution.

## Solution 2: Sum Of Squares

Equation $(1+2a)(2+2b)(2a+b)=32ab$ is equivalent to $$b(2a-1)^2+2a(b-1)^2+(2a-b)^2=0,$$ where $a$, $b>0$. Therefore $2a-1=b-1=2a-b=0$, so $(a,b)=\left(\tfrac12,1\right)$. Hence the answer is $\boxed{\textbf{(B) }1}$.

~r00tsOfUnity

## Video Solution

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 