Question

For any two points $ M $ and $ N $ in the $ XY $-plane, let $ \overrightarrow{MN} $ denote the vector from $ M $ to $ N $, and $ \vec{0} $ denote the zero vector. Let $ P, Q $, and $ R $ be three distinct points in the $ XY $-plane. Let $ S $ be a point inside the triangle $ \Delta PQR $ such that $$ \overrightarrow{SP} + 5\overrightarrow{SQ} + 6\overrightarrow{SR} = \vec{0}. $$ Let $ E $ and $ F $ be the mid-points of the sides $ PR $ and $ QR $, respectively. Then the value of $$ \frac{\text{length of the line segment } EF}{\text{length of the line segment } ES} $$ is __________.

Answer 1

Correct Answer: 1.2

Let the position vectors of points \( P, Q, R \) be \( \vec{p}, \vec{q}, \vec{r} \) respectively, and let the position vector of point \( S \) be \( \vec{s} \).

Step 1: Use the given vector equation 

\[ \overrightarrow{SP} + 5\overrightarrow{SQ} + 6\overrightarrow{SR} = \vec{0} \]

Substitute: \[ \overrightarrow{SP} = \vec{p} - \vec{s}, \quad \overrightarrow{SQ} = \vec{q} - \vec{s}, \quad \overrightarrow{SR} = \vec{r} - \vec{s} \]

\[ (\vec{p} - \vec{s}) + 5(\vec{q} - \vec{s}) + 6(\vec{r} - \vec{s}) = \vec{0} \]

\[ \vec{p} + 5\vec{q} + 6\vec{r} - 12\vec{s} = \vec{0} \]

\[ \Rightarrow \vec{s} = \frac{1}{12}(\vec{p} + 5\vec{q} + 6\vec{r}) \]

Step 2: Find position vectors of points E and F

Since \( E \) and \( F \) are midpoints:

\[ \vec{E} = \frac{\vec{p} + \vec{r}}{2}, \quad \vec{F} = \frac{\vec{q} + \vec{r}}{2} \]

Step 3: Find vector \( \overrightarrow{EF} \)

\[ \overrightarrow{EF} = \vec{F} - \vec{E} = \frac{\vec{q} - \vec{p}}{2} \]

So, \[ |EF| = \frac{1}{2}|\vec{q} - \vec{p}| \]

Step 4: Find vector \( \overrightarrow{ES} \)

\[ \overrightarrow{ES} = \vec{s} - \vec{E} = \frac{1}{12}(\vec{p} + 5\vec{q} + 6\vec{r}) - \frac{1}{2}(\vec{p} + \vec{r}) \]

\[ = \frac{1}{12}(-5\vec{p} + 5\vec{q}) = \frac{5}{12}(\vec{q} - \vec{p}) \]

Thus, \[ |ES| = \frac{5}{12}|\vec{q} - \vec{p}| \]

Step 5: Find the required ratio

\[ \frac{EF}{ES} = \frac{\frac{1}{2}|\vec{q} - \vec{p}|}{\frac{5}{12}|\vec{q} - \vec{p}|} = \frac{1/2}{5/12} = \frac{6}{5} \]

Final Answer:

\[ \boxed{\frac{6}{5}} \]