Question

设 $x$ 为实数，且 $\sin^{10} x + \cos^{10} x = \frac{11}{36}$ 。则 $\sin^{12} x + \cos^{12} x = \frac{m}{n}$ ，其中 $m$ 与 $n$ 是互质正整数。求 $m + n$ 。

Let $x$ be a real number such that $\sin^{10} x + \cos^{10} x = \frac{11}{36}.$ Then $\sin^{12} x + \cos^{12} x = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：67

知识点：三角恒等式牛顿恒等式换元法

难度评级：2720

解答：

令 $u = \sin^2 x$ 、 $v = \cos^2 x$ ，所以 $u + v = 1$ ，并设 $p = uv$ 。展开 $(u+v)^5$ 得 $u^5 + v^5 = 1$ $- 5p(u+v)^3 + 5p^2(u+v)$ $= 1 - 5p + 5p^2$ ，所以条件变为根为 $p = \frac{1}{6}$ 与 $p = \frac{5}{6}$ 。由于 $p = \sin^2 x \cos^2 x$ $= \frac{1}{4}\sin^2 2x \le \frac{1}{4}$ ，必须有 $p = \frac{1}{6}$ 。 $\begin{aligned} 1 - 5p + 5p^2 &= \frac{11}{36} \\ &\Longrightarrow 36p^2 - 36p \\ &\quad {}+ 5 = 0, \end{aligned}$

类似地，代入 $p = \frac{1}{6}$ ，所以 $m + n = 13 + 54 = 67$ 。 $\begin{aligned} u^6 + v^6 &= (u^2 + v^2)^3 \\ &\quad {}- 3p^2(u^2+v^2) \\ &= (1 - 2p)^3 \\ &\quad {}- 3p^2(1 - 2p) \\ &= 1 - 6p + 9p^2 - 2p^3. \end{aligned}$ $\begin{aligned} u^6 + v^6 &= 1 - 1 + \frac{1}{4} - \frac{1}{108} \\ &= \frac{26}{108} = \frac{13}{54}, \end{aligned}$

Let $u = \sin^2 x$ and $v = \cos^2 x,$ so $u + v = 1,$ and set $p = uv.$ Expanding $(u+v)^5$ gives $u^5 + v^5 = 1$ $- 5p(u+v)^3 + 5p^2(u+v)$ $= 1 - 5p + 5p^2,$ so the hypothesis reads $\begin{aligned} 1 - 5p + 5p^2 &= \frac{11}{36} \\ &\Longrightarrow 36p^2 - 36p \\ &\quad {}+ 5 = 0, \end{aligned}$ with roots $p = \frac{1}{6}$ and $p = \frac{5}{6}.$ Since $p = \sin^2 x \cos^2 x$ $= \frac{1}{4}\sin^2 2x \le \frac{1}{4},$ we must have $p = \frac{1}{6}.$

Similarly $\begin{aligned} u^6 + v^6 &= (u^2 + v^2)^3 \\ &\quad {}- 3p^2(u^2+v^2) \\ &= (1 - 2p)^3 \\ &\quad {}- 3p^2(1 - 2p) \\ &= 1 - 6p + 9p^2 - 2p^3. \end{aligned}$ Substituting $p = \frac{1}{6},$ $\begin{aligned} u^6 + v^6 &= 1 - 1 + \frac{1}{4} - \frac{1}{108} \\ &= \frac{26}{108} = \frac{13}{54}, \end{aligned}$ so $m + n = 13 + 54 = 67.$

2019 AIME I 第 8 题

答案：67

难度评级：2720

解答：

其他年份的第 8 题