$$ABC$$ is an equilateral triangle with side length 6. $$P, Q, R$$ are points on the sides $$AB, BC, CA$$ respectively such that $$AP = BQ = CR = 1$$. The ratio of the area of the triangle $$ABC$$ to the area of the triangle $$PQR$$ is

The three corner triangles $$APR$$, $$BQP$$, $$CRQ$$ each have sides 1 and 5 enclosing an angle of $$60^\circ$$, so each has area $$\frac{1}{2} \times 1 \times 5 \times \sin 60^\circ = \frac{5\sqrt{3}}{4}$$. Since $$[ABC] = \frac{\sqrt{3}}{4} \times 36 = 9\sqrt{3}$$, we get $$[PQR] = 9\sqrt{3} - 3 \times \frac{5\sqrt{3}}{4} = \frac{21\sqrt{3}}{4}$$. The required ratio is $$9\sqrt{3} \colon \frac{21\sqrt{3}}{4} = 36 \colon 21 = 12 \colon 7$$.