Question

En la figura, $ABCD$ es un cuadrado de lado $1.$ Los rectángulos $JKHG$ y $EBCF$ son congruentes. ¿Cuánto vale $BE$ ?

In the figure, $ABCD$ is a square of side length $1.$ The rectangles $JKHG$ and $EBCF$ are congruent. What is $BE?$

$\dfrac12(\sqrt{6} - 2)$

$\dfrac14$

$2 - \sqrt{3}$

$\dfrac{\sqrt{3}}{6}$

$1 - \dfrac{\sqrt{2}}{2}$

Solución:

Sea $x = BE = GH = CF$ y $\theta = \angle DHG = \angle AGJ$ $= \angle FKH,$ con $AD = GJ = HK = 1.$ En el triángulo rectángulo $GDH,$ $x \sin\theta = DG = 1 - \cos\theta,$ así que $x = \dfrac{1 - \cos\theta}{\sin\theta}.$

A lo largo del lado $CD,$ $\begin{gathered} 1 = CF + FH + HD \\ = x + \sin\theta + x\cos\theta. \end{gathered}$ Sustituyendo $x$ se obtiene $\begin{gathered} 1 = \dfrac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta} \\ {}+ \sin\theta \\ = \dfrac{\sin^2\theta}{\sin\theta} \\ {}+ \sin\theta \\ = 2\sin\theta. \end{gathered}$

Por lo tanto $\sin\theta = \tfrac12,$ así que $\theta = 30^\circ$ y $x = \dfrac{1 - \frac{\sqrt3}{2}}{\frac12} = 2 - \sqrt3.$

Por lo tanto, la respuesta correcta es C.

Let $x = BE = GH = CF$ and $\theta = \angle DHG = \angle AGJ$ $= \angle FKH,$ with $AD = GJ = HK = 1.$ In right triangle $GDH,$ $x \sin\theta = DG = 1 - \cos\theta,$ so $x = \dfrac{1 - \cos\theta}{\sin\theta}.$

Along side $CD,$ $\begin{gathered} 1 = CF + FH + HD \\ = x + \sin\theta + x\cos\theta. \end{gathered}$ Substituting for $x$ gives $\begin{gathered} 1 = \dfrac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta} \\ {}+ \sin\theta \\ = \dfrac{\sin^2\theta}{\sin\theta} \\ {}+ \sin\theta \\ = 2\sin\theta. \end{gathered}$

Hence $\sin\theta = \tfrac12,$ so $\theta = 30^\circ$ and $x = \dfrac{1 - \frac{\sqrt3}{2}}{\frac12} = 2 - \sqrt3.$

Thus, the correct answer is C.

2014 AMC 12B Problema 21

Solución:

El Problema 21 en otros años