Question

En el triángulo $ABC,$ se sabe que los ángulos $B$ y $C$ son congruentes. Los puntos $P$ y $Q$ están sobre $\overline{AC}$ y $\overline{AB},$ respectivamente, de modo que $AP = PQ = QB = BC.$ El ángulo $ACB$ es $r$ veces mayor que el ángulo $APQ,$ donde $r$ es un número real positivo. Halla el mayor entero que no excede $1000r.$

In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent. Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find the greatest integer that does not exceed $1000r.$

Respuesta

Solución:

Sea $\angle A = \alpha,$ y escala de modo que $AP = PQ = QB = BC = 1.$ En el triángulo $APQ,$ los lados iguales $AP = PQ$ dan $\angle AQP = \angle A = \alpha,$ así que $\angle APQ = 180^\circ - 2\alpha$ y, por la ley de senos, $AQ = \frac{\sin 2\alpha}{\sin \alpha} = 2\cos\alpha.$ En el triángulo $ABC,$ $\angle B = \angle C = 90^\circ - \frac{\alpha}{2},$ así que $\begin{aligned} AB &= \frac{BC \sin C}{\sin A} \\ &= \frac{\cos(\alpha/2)}{\sin\alpha} = \frac{1}{2\sin(\alpha/2)}. \end{aligned}$

Como $AQ + QB = AB,$ $\begin{aligned} 2\cos\alpha + 1 &= \frac{1}{2\sin(\alpha/2)} \\ &\quad\Longrightarrow\quad 4\sin\tfrac{\alpha}{2}\cos\alpha \\ &\quad {}+ 2\sin\tfrac{\alpha}{2} = 1. \end{aligned}$ Por la identidad de producto a suma, $4\sin\frac{\alpha}{2}\cos\alpha$ $= 2\sin\frac{3\alpha}{2}$ $- 2\sin\frac{\alpha}{2},$ así que la ecuación se reduce a $\sin\frac{3\alpha}{2} = \frac{1}{2}.$ Entonces $\alpha = 20^\circ$ o $\alpha = 100^\circ,$ pero este último hace que $AQ = 2\cos\alpha$ sea negativo, así que $\alpha = 20^\circ.$

Ahora $\angle ACB = 80^\circ$ y $\angle APQ = 140^\circ,$ así que $r = \frac{80}{140} = \frac{4}{7},$ y $\lfloor 1000r \rfloor = \left\lfloor \frac{4000}{7} \right\rfloor = 571.$

Let $\angle A = \alpha,$ and scale so $AP = PQ = QB = BC = 1.$ In triangle $APQ,$ the equal sides $AP = PQ$ give $\angle AQP = \angle A = \alpha,$ so $\angle APQ = 180^\circ - 2\alpha$ and, by the law of sines, $AQ = \frac{\sin 2\alpha}{\sin \alpha} = 2\cos\alpha.$ In triangle $ABC,$ $\angle B = \angle C = 90^\circ - \frac{\alpha}{2},$ so $\begin{aligned} AB &= \frac{BC \sin C}{\sin A} \\ &= \frac{\cos(\alpha/2)}{\sin\alpha} = \frac{1}{2\sin(\alpha/2)}. \end{aligned}$

Since $AQ + QB = AB,$ $\begin{aligned} 2\cos\alpha + 1 &= \frac{1}{2\sin(\alpha/2)} \\ &\quad\Longrightarrow\quad 4\sin\tfrac{\alpha}{2}\cos\alpha \\ &\quad {}+ 2\sin\tfrac{\alpha}{2} = 1. \end{aligned}$ By the product-to-sum identity, $4\sin\frac{\alpha}{2}\cos\alpha$ $= 2\sin\frac{3\alpha}{2}$ $- 2\sin\frac{\alpha}{2},$ so the equation collapses to $\sin\frac{3\alpha}{2} = \frac{1}{2}.$ Then $\alpha = 20^\circ$ or $\alpha = 100^\circ,$ but the latter makes $AQ = 2\cos\alpha$ negative, so $\alpha = 20^\circ.$

Now $\angle ACB = 80^\circ$ and $\angle APQ = 140^\circ,$ so $r = \frac{80}{140} = \frac{4}{7},$ and $\lfloor 1000r \rfloor = \left\lfloor \frac{4000}{7} \right\rfloor = 571.$

2000 AIME I Problema 14

Solución:

El Problema 14 en otros años