A point taken on each median of a triangle divides the median in the ratio 1 : 3, reckoning from the vertex. Then the ratio of the area of the triangle with vertices at these points to that of the original triangle is

Options

A. `5:13`
B. `25:64`
C. `13:32`
D. none of these

Answer

To solve the problem, we need to find the ratio of the area of triangle EFG (formed by points on the medians of triangle ABC) to the area of triangle ABC.

### Step-by-Step Solution:

1. **Define the Coordinates of Triangle ABC:**
Let the coordinates of the vertices of triangle ABC be:
- A(0, 0)
- B(4m, 0)
- C(4p, 4q)

2. **Find the Midpoints of the Medians:**
The midpoints of the sides of triangle ABC (which are the points where the medians intersect) are:
- M1 (midpoint of BC) = ((4m + 4p)/2, (0 + 4q)/2) = (2m + 2p, 2q)
- M2 (midpoint of AC) = ((0 + 4p)/2, (0 + 4q)/2) = (2p, 2q)
- M3 (midpoint of AB) = ((0 + 4m)/2, (0 + 0)/2) = (2m, 0)

3. **Determine the Points on the Medians:**
Each median is divided in the ratio 1:3 from the vertex:
- Point E on AM1:
Using the section formula, the coordinates of E are:
\[ E = \left(\frac{1 \cdot (2m + 2p) + 3 \cdot 0}{1 + 3}, \frac{1 \cdot 2q + 3 \cdot 0}{1 + 3}\right) = \left(\frac{2m + 2p}{4}, \frac{2q}{4}\right) = \left(\frac{m + p}{2}, \frac{q}{2}\right) \]
- Point F on BM2:
\[ F = \left(\frac{1 \cdot (2p) + 3 \cdot (4m)}{1 + 3}, \frac{1 \cdot 2q + 3 \cdot 0}{1 + 3}\right) = \left(\frac{2p + 12m}{4}, \frac{2q}{4}\right) = \left(\frac{p + 6m}{2}, \frac{q}{2}\right) \]
- Point G on CM3:
\[ G = \left(\frac{1 \cdot (2m) + 3 \cdot (4p)}{1 + 3}, \frac{1 \cdot 0 + 3 \cdot (4q)}{1 + 3}\right) = \left(\frac{2m + 12p}{4}, \frac{12q}{4}\right) = \left(\frac{m + 6p}{2}, 3q\right) \]

4. **Calculate the Area of Triangle ABC:**
The area of triangle ABC can be calculated using the determinant formula:
\[ \text{Area}_{ABC} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Plugging in the coordinates:
\[ \text{Area}_{ABC} = \frac{1}{2} \left| 0(0 - 4q) + 4m(4q - 0) + 4p(0 - 0) \right| = \frac{1}{2} \left| 16mq \right| = 8mq \]

5. **Calculate the Area of Triangle EFG:**
Using the same determinant formula for triangle EFG:
\[ \text{Area}_{EFG} = \frac{1}{2} \left| \frac{m + p}{2} \left(\frac{q}{2} - 3q\right) + \frac{p + 6m}{2} \left(3q - \frac{q}{2}\right) + \frac{m + 6p}{2} \left(\frac{q}{2} - \frac{q}{2}\right) \right| \]
After calculating the determinant, we find:
\[ \text{Area}_{EFG} = \frac{25mq}{8} \]

6. **Find the Ratio of Areas:**
Now, we find the ratio of the areas:
\[ \text{Ratio} = \frac{\text{Area}_{EFG}}{\text{Area}_{ABC}} = \frac{\frac{25mq}{8}}{8mq} = \frac{25}{64} \]

### Final Answer:
The ratio of the area of triangle EFG to that of triangle ABC is \( \frac{25}{64} \).