AMC12 2023 A

AMC12 2023 A · Q23

AMC12 2023 A · Q23. It mainly tests Inequalities (AM-GM etc. basic).

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

有几个正实数有序对 $(a,b)$ 满足方程 \[(1+2a)(2+2b)(2a+b) = 32ab？\]

(A) 0 0

(B) 1 1

(D) 3 3

(E) \text{an infinite number} \text{an infinite number}

Answer

Correct choice: (B)

正确答案：（B）

Solution

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

对左侧每个因式中的两项使用 AM-GM 不等式，得 \[(1+2a)(2+2b)(2a+b) \ge 8\sqrt{2a \cdot 4b \cdot 2ab}= 32ab，\] 这意味着必须取等号。因此，必须有 $1 = 2a = b$，所以唯一解为 $\boxed{\textbf{(B) }1}$。

Topics

AL.013 Inequalities (AM-GM etc. basic)