MHB How to Create an Online Store for Your Business

[Albert1]

View attachment 2737

 

[anemone]

My solution:

Spoiler

I draw the entire rectangle $ABCD$ in the Cartesian plane, with the vertex $B$ lies in the origin, and let the point $C=(c,\,0)$ and the $\angle DBC=\theta$.

View attachment 2740

We then have the equations of $BD$ and $AE$ be $y=\tan \theta x$ and $y=\dfrac{x}{\tan \theta}+c\tan \theta$.

Solving them simultaneously we will get the coordinate points for $E$, $F$ and $G$ where:

$E=(\dfrac{c\tan^2\theta}{\sec^2 \theta},\,\dfrac{c\tan^3\theta}{\sec^2 \theta})$

$F=(\dfrac{c\tan^2\theta}{\sec^2 \theta},\,0)$

$G=(c,\,\dfrac{c\tan^3\theta}{\sec^2 \theta})$

This gives $EF^2=\dfrac{c^2\tan^6\theta}{\sec^4 \theta}$, $EG^2=\dfrac{c^2}{\sec^4 \theta}$ and $BD^2=c^2\sec^2 \theta$.

Therefore

$\begin{align*}\left[\left(\dfrac{EF}{BD}\right)^2\right]^{\dfrac{1}{3}}+\left[\left(\dfrac{EG}{BD}\right)^2\right]^{\dfrac{1}{3}}&=\left[\left(\dfrac{c^2\tan^6 \theta}{\sec^4 \theta c^2\sec^2 \theta}\right)\right]^{\dfrac{1}{3}}+\left[\left(\dfrac{\dfrac{c^2}{\sec^4 \theta}}{c^2\sec^2 \theta}\right)\right]^{\dfrac{1}{3}}\\&=\left[\dfrac{\tan^6 \theta}{\sec^6 \theta}\right]^{\dfrac{1}{3}}+\left[\dfrac{1}{\sec^6 \theta}\right]^{\dfrac{1}{3}}\\&=\dfrac{\tan^2 \theta}{\sec^2 \theta}+\dfrac{1}{\sec^2 \theta}\\&=1\end{align*}$