Let D, E and F be the three points which divide internally the three sides AB, BC and CA, respectively, of a triangle ABC in the ratio of $k:(1-k)$, where $0 < k \leq \frac{1}{2}$. (1) When $k = \frac{1}{3}$, we are to find how many times greater the area of triangle ABC is than the area of triangle DEF. Since $$\triangle\mathrm{ADF} = \triangle\mathrm{BED} = \triangle\mathrm{CFE} = \frac{\mathbf{M}}{\mathbf{N}} \triangle\mathrm{ABC},$$ it follows that $$\triangle\mathrm{ABC} = \mathbf{O}\, \triangle\mathrm{DEF}.$$ (2) The area of the triangle DEF is half of the area of the triangle ABC when $$k(1-k) = \frac{\mathbf{P}}{\mathbf{Q}},$$ that is, when $$k = \frac{\mathbf{R} - \sqrt{\mathbf{S}}}{\mathbf{T}}.$$
Let D, E and F be the three points which divide internally the three sides AB, BC and CA, respectively, of a triangle ABC in the ratio of $k:(1-k)$, where $0 < k \leq \frac{1}{2}$.
(1) When $k = \frac{1}{3}$, we are to find how many times greater the area of triangle ABC is than the area of triangle DEF. Since
$$\triangle\mathrm{ADF} = \triangle\mathrm{BED} = \triangle\mathrm{CFE} = \frac{\mathbf{M}}{\mathbf{N}} \triangle\mathrm{ABC},$$
it follows that
$$\triangle\mathrm{ABC} = \mathbf{O}\, \triangle\mathrm{DEF}.$$
(2) The area of the triangle DEF is half of the area of the triangle ABC when
$$k(1-k) = \frac{\mathbf{P}}{\mathbf{Q}},$$
that is, when
$$k = \frac{\mathbf{R} - \sqrt{\mathbf{S}}}{\mathbf{T}}.$$