Question

方程 $2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1$ 有三个实根。已知它们的和为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数，求 $m + n$ 。

The equation $2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1$ has three real roots. Given that their sum is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

答案

答案：113

知识点：指数换元法韦达定理对数

难度评级：2500

解答：

令 $y = 2^{111x}$ 。则 $2^{333x-2} = \frac{y^3}{4}$ ， $2^{111x+2} = 4y$ ， $2^{222x+1} = 2y^2$ ，所以方程变为 $\frac{y^3}{4} + 4y = 2y^2 + 1$ ，即 $y^3 - 8y^2 + 16y - 4 = 0.$ 因为三个根 $x_1, x_2, x_3$ 都是实数，且 $y = 2^{111x}$ 严格递增，所以它们对应于这个三次方程的三个正实根 $y_1, y_2, y_3$ 。

每个 $x_i = \frac{1}{111}\log_2 y_i$ ，所以 $\begin{aligned} x_1 + x_2 + x_3 &= \frac{1}{111}\log_2(y_1 y_2 y_3) \\ &= \frac{1}{111}\log_2 4 \\ &= \frac{2}{111}, \end{aligned}$ 这里用了韦达定理求根的乘积。因此 $m + n = 2 + 111 = 113$ 。

Let $y = 2^{111x}.$ Then $2^{333x-2} = \frac{y^3}{4},$ $2^{111x+2} = 4y,$ and $2^{222x+1} = 2y^2,$ so the equation becomes $\frac{y^3}{4} + 4y = 2y^2 + 1,$ that is, $y^3 - 8y^2 + 16y - 4 = 0.$ Since the three roots $x_1, x_2, x_3$ are real and $y = 2^{111x}$ is strictly increasing, they correspond to three positive real roots $y_1, y_2, y_3$ of the cubic.

Each $x_i = \frac{1}{111}\log_2 y_i,$ so $\begin{aligned} x_1 + x_2 + x_3 &= \frac{1}{111}\log_2(y_1 y_2 y_3) \\ &= \frac{1}{111}\log_2 4 \\ &= \frac{2}{111}, \end{aligned}$ using Vieta's formulas for the product of the roots. Then $m + n = 2 + 111 = 113.$

2005 AIME I 第 8 题

答案：113

难度评级：2500

解答：

其他年份的第 8 题