In the figure, $$\theta_1 + \theta_2 = \frac{\pi}{2}$$ and $$\sqrt{3}BE = 4AB$$. If the area of $$\triangle CAB$$ is $$2\sqrt{3} - 3$$ unit$$^2$$, when $$\frac{\theta_2}{\theta_1}$$ is the largest, then the perimeter (in unit) of $$\triangle CED$$ is equal to _______.

From the figure:

• $$CD$$ is parallel to $$AB$$ (both are perpendicular to $$BE$$). Thus, $$CD = AB$$.
• In $$\triangle CAB$$: $$AC = AB \tan \theta_1$$.
• Area $$(\triangle CAB) = \frac{1}{2} \cdot AB \cdot AC = \frac{1}{2} (AB)^2 \tan \theta_1 = 2\sqrt{3} - 3$$.
• In $$\triangle CED$$: $$ED = CD \tan \theta_2 = AB \tan \theta_2$$.
• We are given $$\theta_1 + \theta_2 = \frac{\pi}{2}$$, which implies $$\theta_2 = \frac{\pi}{2} - \theta_1$$.
• Therefore, $$\tan \theta_2 = \cot \theta_1$$.

You are given that $$\sqrt{3}BE = 4AB$$, which means $$BE = \frac{4}{\sqrt{3}}AB$$.

From the geometry, $$BE = BD + DE$$.

Since $$BD = AC$$ (opposite sides of rectangle $$ACDB$$), we have:

$$BE = AB \tan \theta_1 + AB \cot \theta_1$$

$$\frac{4}{\sqrt{3}}AB = AB (\tan \theta_1 + \cot \theta_1)$$

$$\frac{4}{\sqrt{3}} = \frac{\sin \theta_1}{\cos \theta_1} + \frac{\cos \theta_1}{\sin \theta_1} = \frac{\sin^2 \theta_1 + \cos^2 \theta_1}{\sin \theta_1 \cos \theta_1} = \frac{1}{\sin \theta_1 \cos \theta_1}$$

$$\sin(2\theta_1) = 2 \sin \theta_1 \cos \theta_1 = 2 \left( \frac{\sqrt{3}}{4} \right) = \frac{\sqrt{3}}{2}$$

This gives two possible values for $$2\theta_1$$: $$60^\circ$$ or $$120^\circ$$.

• If $$2\theta_1 = 60^\circ \implies \theta_1 = 30^\circ$$ and $$\theta_2 = 60^\circ$$.
• If $$2\theta_1 = 120^\circ \implies \theta_1 = 60^\circ$$ and $$\theta_2 = 30^\circ$$.

The problem states $$\frac{\theta_2}{\theta_1}$$ is largest, so we choose $$\theta_1 = 30^\circ$$ and $$\theta_2 = 60^\circ$$.

Using the area of $$\triangle CAB$$:

$$\frac{1}{2} (AB)^2 \tan 30^\circ = 2\sqrt{3} - 3$$

$$\frac{1}{2} (AB)^2 \left( \frac{1}{\sqrt{3}} \right) = 2\sqrt{3} - 3$$

$$(AB)^2 = 2\sqrt{3}(2\sqrt{3} - 3) = 4(3) - 6\sqrt{3} = 12 - 6\sqrt{3}$$

Using the identity $$12 - 6\sqrt{3} = (3 - \sqrt{3})^2$$:

$$AB = 3 - \sqrt{3}$$

In $$\triangle CED$$ (a $$30^\circ-60^\circ-90^\circ$$ triangle with $$\theta_2 = 60^\circ$$):

• $$CD = AB = 3 - \sqrt{3}$$
• $$ED = CD \tan 60^\circ = (3 - \sqrt{3})\sqrt{3} = 3\sqrt{3} - 3$$
• $$CE = \frac{CD}{\cos 60^\circ} = 2(3 - \sqrt{3}) = 6 - 2\sqrt{3}$$

Perimeter $$= CD + ED + CE$$:

$$(3 - \sqrt{3}) + (3\sqrt{3} - 3) + (6 - 2\sqrt{3})$$

$$3 - 3 + 6 - \sqrt{3} + 3\sqrt{3} - 2\sqrt{3}$$

$$6 + 0\sqrt{3} = 6$$

Final Answer: 6