Question

Sean $A, B, C$ los ángulos de un triángulo con $A$ y $C$ agudos y $B$ mayor que un ángulo recto, que satisfacen $\begin{aligned} &\cos^2 A + \cos^2 B \\ &\quad {}+ 2 \sin A \sin B \cos C = \frac{15}{8} \end{aligned}$ y $\begin{aligned} &\cos^2 B + \cos^2 C \\ &\quad {}+ 2 \sin B \sin C \cos A = \frac{14}{9}. \end{aligned}$ Existen enteros positivos $p,$ $q,$ $r,$ y $s$ para los cuales $\begin{aligned} &\cos^2 C + \cos^2 A \\ &\quad {}+ 2 \sin C \sin A \cos B \\ &= \frac{p - q\sqrt{r}}{s}, \end{aligned}$ donde $p + q$ y $s$ son primos entre sí y $r$ no es divisible por el cuadrado de ningún primo. Halla $p + q + r + s.$

Let $A, B, C$ be angles of a triangle with $A$ and $C$ acute and $B$ greater than a right angle satisfying $\begin{aligned} &\cos^2 A + \cos^2 B \\ &\quad {}+ 2 \sin A \sin B \cos C = \frac{15}{8} \end{aligned}$ and $\begin{aligned} &\cos^2 B + \cos^2 C \\ &\quad {}+ 2 \sin B \sin C \cos A = \frac{14}{9}. \end{aligned}$ There are positive integers $p,$ $q,$ $r,$ and $s$ for which $\begin{aligned} &\cos^2 C + \cos^2 A \\ &\quad {}+ 2 \sin C \sin A \cos B \\ &= \frac{p - q\sqrt{r}}{s}, \end{aligned}$ where $p + q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p + q + r + s.$

Respuesta

Solución:

Reemplazando cada $\cos^2$ por $1 - \sin^2,$ la primera ecuación se convierte en $\sin^2 A + \sin^2 B$ $- 2 \sin A \sin B \cos C = \frac{1}{8}.$ Por la ley de senos, $\sin A = \frac{a}{2R}$ y así sucesivamente, así que el lado izquierdo es igual a $\begin{aligned} \frac{a^2 + b^2 - 2ab\cos C}{4R^2} &= \frac{c^2}{4R^2} \\ &= \sin^2 C \end{aligned}$ por la ley de cosenos. De aquí $\sin^2 C = 2 - \frac{15}{8} = \frac{1}{8}.$ El mismo argumento convierte la segunda ecuación en $\sin^2 A = 2 - \frac{14}{9} = \frac{4}{9},$ y muestra que la expresión pedida es igual a $2 - \sin^2 B.$

Como $A$ y $C$ son agudos, $\cos A = \frac{\sqrt{5}}{3}$ y $\cos C = \frac{\sqrt{14}}{4},$ con $\sin A = \frac{2}{3}$ y $\sin C = \frac{\sqrt{2}}{4}.$ Entonces $\begin{aligned} \sin B &= \sin(A + C) \\ &= \frac{2}{3} \cdot \frac{\sqrt{14}}{4} + \frac{\sqrt{5}}{3} \cdot \frac{\sqrt{2}}{4} \\ &= \frac{2\sqrt{14} + \sqrt{10}}{12}, \end{aligned}$ así que $\sin^2 B = \frac{66 + 8\sqrt{35}}{144} = \frac{33 + 4\sqrt{35}}{72}.$

Por lo tanto $2 - \sin^2 B = \frac{111 - 4\sqrt{35}}{72},$ y $p + q + r + s$ $= 111 + 4 + 35 + 72$ $= 222.$

Replacing each $\cos^2$ by $1 - \sin^2,$ the first equation becomes $\sin^2 A + \sin^2 B$ $- 2 \sin A \sin B \cos C = \frac{1}{8}.$ By the law of sines, $\sin A = \frac{a}{2R}$ and so on, so the left side equals $\begin{aligned} \frac{a^2 + b^2 - 2ab\cos C}{4R^2} &= \frac{c^2}{4R^2} \\ &= \sin^2 C \end{aligned}$ by the law of cosines. Hence $\sin^2 C = 2 - \frac{15}{8} = \frac{1}{8}.$ The same argument turns the second equation into $\sin^2 A = 2 - \frac{14}{9} = \frac{4}{9},$ and shows the requested expression equals $2 - \sin^2 B.$

Since $A$ and $C$ are acute, $\cos A = \frac{\sqrt{5}}{3}$ and $\cos C = \frac{\sqrt{14}}{4},$ with $\sin A = \frac{2}{3}$ and $\sin C = \frac{\sqrt{2}}{4}.$ Then $\begin{aligned} \sin B &= \sin(A + C) \\ &= \frac{2}{3} \cdot \frac{\sqrt{14}}{4} + \frac{\sqrt{5}}{3} \cdot \frac{\sqrt{2}}{4} \\ &= \frac{2\sqrt{14} + \sqrt{10}}{12}, \end{aligned}$ so $\sin^2 B = \frac{66 + 8\sqrt{35}}{144} = \frac{33 + 4\sqrt{35}}{72}.$

Therefore $2 - \sin^2 B = \frac{111 - 4\sqrt{35}}{72},$ and $p + q + r + s$ $= 111 + 4 + 35 + 72$ $= 222.$

2013 AIME II Problema 15

Solución:

El Problema 15 en otros años