2019 AIME I 第 8 题

Set $u = \sin^2 x$ and $v = \cos^2 x,$ so $u + v = 1;$ every symmetric power sum is a polynomial in $p = uv$

$u^5 + v^5 = 1 - 5p + 5p^2;$ solve for $p$ (only one root satisfies $p \le \frac{1}{4}$) and plug into $u^6 + v^6 = 1 - 6p + 9p^2 - 2p^3$

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 $m + n = 13 + 54 = 67.$