Question

Para $\pi \le \theta \lt 2\pi,$ sea $\begin{aligned} P &= \frac{1}{2}\cos\theta - \frac{1}{4}\sin 2\theta \\ &\quad {}- \frac{1}{8}\cos 3\theta + \frac{1}{16}\sin 4\theta \\ &\quad {}+ \frac{1}{32}\cos 5\theta - \frac{1}{64}\sin 6\theta \\ &\quad {}- \frac{1}{128}\cos 7\theta + \ldots \end{aligned}$ y $\begin{aligned} Q &= 1 - \frac{1}{2}\sin\theta - \frac{1}{4}\cos 2\theta \\ &\quad {}+ \frac{1}{8}\sin 3\theta + \frac{1}{16}\cos 4\theta \\ &\quad {}- \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta \\ &\quad {}+ \frac{1}{128}\sin 7\theta + \ldots \end{aligned}$ de modo que $\frac{P}{Q} = \frac{2\sqrt{2}}{7}.$ Entonces $\sin\theta = -\frac{m}{n}$ donde $m$ y $n$ son enteros positivos primos entre sí. Halla $m + n.$

For $\pi \le \theta \lt 2\pi,$ let $\begin{aligned} P &= \frac{1}{2}\cos\theta - \frac{1}{4}\sin 2\theta \\ &\quad {}- \frac{1}{8}\cos 3\theta + \frac{1}{16}\sin 4\theta \\ &\quad {}+ \frac{1}{32}\cos 5\theta - \frac{1}{64}\sin 6\theta \\ &\quad {}- \frac{1}{128}\cos 7\theta + \ldots \end{aligned}$ and $\begin{aligned} Q &= 1 - \frac{1}{2}\sin\theta - \frac{1}{4}\cos 2\theta \\ &\quad {}+ \frac{1}{8}\sin 3\theta + \frac{1}{16}\cos 4\theta \\ &\quad {}- \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta \\ &\quad {}+ \frac{1}{128}\sin 7\theta + \ldots \end{aligned}$ so that $\frac{P}{Q} = \frac{2\sqrt{2}}{7}.$ Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Respuesta

Solución:

Los signos y la alternancia entre senos y cosenos sugieren potencias de $i:$ en efecto $\begin{aligned} Q + iP &= 1 + \frac{1}{2}ie^{i\theta} + \frac{1}{4}i^2e^{2i\theta} \\ &\quad {}+ \frac{1}{8}i^3e^{3i\theta} + \cdots \\ &= \frac{1}{1 - \frac{ie^{i\theta}}{2}} = \frac{2}{2 - ie^{i\theta}}. \end{aligned}$ Como $2 - ie^{i\theta} = (2 + \sin\theta) - i\cos\theta,$ multiplicando por el conjugado se obtiene $\begin{aligned} &Q + iP \\ &= \frac{2(2 + \sin\theta) + 2i\cos\theta}{5 + 4\sin\theta}, \end{aligned}$ así que $\frac{P}{Q} = \frac{\cos\theta}{2 + \sin\theta}.$

Poniendo $\frac{\cos\theta}{2 + \sin\theta} = \frac{2\sqrt{2}}{7}$ y elevando al cuadrado, $49(1 - \sin^2\theta) = 8(2 + \sin\theta)^2,$ que se reordena como $\begin{aligned} &57\sin^2\theta + 32\sin\theta - 17 \\ &= (3\sin\theta - 1)(19\sin\theta + 17) \\ &= 0. \end{aligned}$

Como $\pi \le \theta \lt 2\pi$ obliga a $\sin\theta \le 0,$ obtenemos $\sin\theta = -\frac{17}{19}$ (y entonces $\cos\theta = \frac{6\sqrt{2}}{19} \gt 0,$ consistente con la razón positiva). Por lo tanto $m + n = 17 + 19 = 36.$

The signs and the alternation between sines and cosines suggest powers of $i:$ indeed $\begin{aligned} Q + iP &= 1 + \frac{1}{2}ie^{i\theta} + \frac{1}{4}i^2e^{2i\theta} \\ &\quad {}+ \frac{1}{8}i^3e^{3i\theta} + \cdots \\ &= \frac{1}{1 - \frac{ie^{i\theta}}{2}} = \frac{2}{2 - ie^{i\theta}}. \end{aligned}$ Since $2 - ie^{i\theta} = (2 + \sin\theta) - i\cos\theta,$ multiplying by the conjugate gives $\begin{aligned} &Q + iP \\ &= \frac{2(2 + \sin\theta) + 2i\cos\theta}{5 + 4\sin\theta}, \end{aligned}$ so $\frac{P}{Q} = \frac{\cos\theta}{2 + \sin\theta}.$

Setting $\frac{\cos\theta}{2 + \sin\theta} = \frac{2\sqrt{2}}{7}$ and squaring, $49(1 - \sin^2\theta) = 8(2 + \sin\theta)^2,$ which rearranges to $\begin{aligned} &57\sin^2\theta + 32\sin\theta - 17 \\ &= (3\sin\theta - 1)(19\sin\theta + 17) \\ &= 0. \end{aligned}$

Since $\pi \le \theta \lt 2\pi$ forces $\sin\theta \le 0,$ we get $\sin\theta = -\frac{17}{19}$ (and then $\cos\theta = \frac{6\sqrt{2}}{19} \gt 0,$ consistent with the positive ratio). Thus $m + n = 17 + 19 = 36.$

2013 AIME I Problema 14

Solución:

El Problema 14 en otros años