2023 AMC 12A 第 23 题

Question

有多少个正实数有序对

(a,b)

满足方程

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

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

无穷多个

an infinite number

分别对

1+2a

、

2+2b

和

2a+b

使用 AM-GM

Apply AM-GM separately to

1+2a,

2+2b,

and

2a+b

乘积至少为

32ab

，所以等号必须在三个不等式中同时成立

The product is at least

32ab,

so equality must hold in all three inequalities at once

由 AM-GM，

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

，

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

，且

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

。相乘得

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

等号要求

1=2a

、

2=2b

和

2a=b

同时成立。这给出

a=\tfrac12

、

b=1

，且三者相容，所以恰有一个解。

所以正确答案是 B。

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.

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