Given $$\triangle ABC$$ with $$\angle B=60^{\circ}$$ and $$\angle C=30^{\circ}$$, let $$P, Q, R$$ be points on sides $$BA, AC, CB$$ respectively such that $$BPQR$$ is an isosceles trapezium with $$PQ||BR$$ and $$BP=QR$$. Find the minimum possible value of $$\frac{2[ABC]}{[BPQR]}$$ where $$[S]$$ denotes the area of any polygon $$S$$.

Because $$\angle B = 60^\circ$$ and $$\angle C = 30^\circ$$, we have $$\angle A = 90^\circ$$. Take a right-angled Cartesian frame with $$A(0,0),\; B(1,0),\; C(0,\sqrt3)$$ so that
$$AB = 1,\; AC = \sqrt3,\; BC = 2,\; [ABC]=\dfrac{1\cdot\sqrt3}{2}=\dfrac{\sqrt3}{2}.$$

Let the required points be parameterised as
$$P(1-t,0)\;(0\lt t\le 1),$$ $$Q(0,q)\;(0\lt q\le \sqrt3),$$ $$R(s,\sqrt3(1-s))\;(0\le s\le 1).$$

Condition 1: $$PQ \parallel BR$$.
Vector $$\overrightarrow{PQ}=(-\!(1-t),\,q)$$ and $$\overrightarrow{BR}=(s-1,\,\sqrt3(1-s))$$ are parallel, so

$$\frac{1-t}{1-s}=\frac{q}{\sqrt3(1-s)} \;\;\Longrightarrow\;\; q=\sqrt3(1-t).\quad-(1)$$

Condition 2: $$BP=QR$$ (legs equal in an isosceles trapezium).
$$BP=t,$$ $$QR^2=s^2+\bigl(\sqrt3(1-s)-q\bigr)^2 =s^2+3(t-s)^2.$$

Therefore
$$t^2=s^2+3(t-s)^2 \;\;\Longrightarrow\;\; t^2-3st+2s^2=0 \;\;\Longrightarrow\;\; t=s\quad\text{or}\quad t=2s.\quad-(2)$$

Case 1: $$t=s$$ (allowed for $$0\lt s\lt1$$)

Using (1) and the shoelace formula for $$B(1,0),P,Q,R$$, the area of the quadrilateral is

$$[BPQR]=\sqrt3\,s(1-s).$$

Hence
$$\frac{2[ABC]}{[BPQR]}=\frac{\sqrt3}{\sqrt3\,s(1-s)}=\frac1{s(1-s)}.$$
The product $$s(1-s)$$ attains its maximum $$\tfrac14$$ at $$s=\tfrac12$$, giving a minimum ratio $$4$$.

Case 2: $$t=2s\;(0\lt s\le \tfrac12)$$

Again by (1) and the shoelace formula,

$$[BPQR]=\sqrt3\,s\bigl(2-3s\bigr).$$

Therefore

$$\frac{2[ABC]}{[BPQR]} =\frac{\sqrt3}{\sqrt3\,s(2-3s)} =\frac1{s(2-3s)}.$$

For $$f(s)=s(2-3s)=2s-3s^2\;(0\lt s\le\tfrac12)$$,
$$f'(s)=2-6s=0\;\Longrightarrow\;s=\frac13,$$ which gives the maximum value $$f\bigl(\tfrac13\bigr)=\tfrac13.$

Hence the minimum ratio in this case is
$$$$\frac$$1{\,\tfrac13}=3.$$

Comparing the two cases, the overall minimum of $$\dfrac{2[ABC]}{[BPQR]}$$ is $$3$$, achieved for

$$s=$$\frac$$13,\; t=$$\frac$$23,\; q=$$\sqrt$$3\!$$\left$$(1-$$\frac$$23$$\right$$)=$$\frac{\sqrt3}{3}$$.$$

Final Answer: 03