In the figure, $\theta_1+\theta_2=\dfrac{\pi}{2}$ and $\sqrt{3}(BE)=4(AB)$. If the area of $\triangle CAB$ is $2\sqrt{3}-3$ unit$^2$, when $\dfrac{\theta_2}{\theta_1}$ is the largest, then the perimeter (in unit) of $\triangle CED$ is equal to _______.
Step-by-Step Solution
Key Concept: Let $AB=x$, $BD=y$. The constraint $\sqrt{3}\,BE=4\,AB$ and area condition give $y=\frac{4\sqrt{3}-6}{x}$. Use $\tan\theta_1\cdot\tan\theta_2=1$ (since $\theta_1+\theta_2=\pi/2$) to maximise $\theta_2/\theta_1$.
From constraints, $x=3-\sqrt{3}$ and $\theta_2=60°$. $DE=x\sqrt{3}=3\sqrt{3}-3$, $CE=2x=6-2\sqrt{3}$, $CD=x=3-\sqrt{3}$. Perimeter $=CD+DE+CE=6$.
<div class="key-concept"><strong>Key Concept:</strong> Let $AB=x$, $BD=y$. The constraint $\sqrt{3}\,BE=4\,AB$ and area condition give $y=\frac{4\sqrt{3}-6}{x}$. Use $\tan\theta_1\cdot\tan\theta_2=1$ (since $\theta_1+\theta_2=\pi/2$) to maximise $\theta_2/\theta_1$.</div>
<div class="trap-box"><strong>Trap:</strong> Maximising $\theta_2/\theta_1$ subject to $\tan\theta_1\tan\theta_2=1$ gives $x^2=12-6\sqrt{3}$, $\theta_2=60°$. Then compute $CD$, $DE$, $CE$ for perimeter.</div>
Correct Answer: 6