Let `G_(1), G(2) and G_(3)` be the centroid of the triangular faces OBC, OCA and OAB of a tetrahedron OABC. If `V_(1)` denotes the volume of tetrahedron OABC and `V_(2)` that of the parallelepiped with `OG_(1), OG_(2) and OG_(3)` as three concurrent edges, then the value of `(4V_(1))/(V_2)` is (where O is the origin

Answer

To solve the problem, we need to find the value of \( \frac{4V_1}{V_2} \), where \( V_1 \) is the volume of tetrahedron \( OABC \) and \( V_2 \) is the volume of the parallelepiped formed by the centroids \( G_1, G_2, \) and \( G_3 \) of the triangular faces \( OBC, OCA, \) and \( OAB \).

### Step 1: Calculate the Volume of Tetrahedron \( OABC \)

The volume \( V_1 \) of tetrahedron \( OABC \) can be calculated using the formula:

\[ V_1 = \frac{1}{6} \left| \text{det}(\vec{OA}, \vec{OB}, \vec{OC}) \right| \]

Where \( \vec{OA}, \vec{OB}, \vec{OC} \) are the position vectors of points \( A, B, C \) from the origin \( O \). Let’s denote these position vectors as:

- \( \vec{OA} = \vec{a} \)
- \( \vec{OB} = \vec{b} \)
- \( \vec{OC} = \vec{c} \)

Thus, we have:

\[ V_1 = \frac{1}{6} \left| \text{det}(\vec{a}, \vec{b}, \vec{c}) \right| \]

### Step 2: Calculate the Centroids \( G_1, G_2, G_3 \)

The centroids of the triangular faces are given by:

- \( G_1 \) for triangle \( OBC \):
\[ \vec{OG_1} = \frac{\vec{OB} + \vec{OC}}{3} = \frac{\vec{b} + \vec{c}}{3} \]

- \( G_2 \) for triangle \( OCA \):
\[ \vec{OG_2} = \frac{\vec{OC} + \vec{OA}}{3} = \frac{\vec{c} + \vec{a}}{3} \]

- \( G_3 \) for triangle \( OAB \):
\[ \vec{OG_3} = \frac{\vec{OA} + \vec{OB}}{3} = \frac{\vec{a} + \vec{b}}{3} \]

### Step 3: Calculate the Volume of Parallelepiped \( V_2 \)

The volume \( V_2 \) of the parallelepiped formed by vectors \( \vec{OG_1}, \vec{OG_2}, \vec{OG_3} \) is given by:

\[ V_2 = \left| \text{det}(\vec{OG_1}, \vec{OG_2}, \vec{OG_3}) \right| \]

Substituting the expressions for \( \vec{OG_1}, \vec{OG_2}, \vec{OG_3} \):

\[ V_2 = \left| \text{det}\left(\frac{\vec{b} + \vec{c}}{3}, \frac{\vec{c} + \vec{a}}{3}, \frac{\vec{a} + \vec{b}}{3}\right) \right| \]

Factoring out \( \frac{1}{3} \):

\[ V_2 = \frac{1}{27} \left| \text{det}(\vec{b} + \vec{c}, \vec{c} + \vec{a}, \vec{a} + \vec{b}) \right| \]

Using the property of determinants, we can express:

\[ \text{det}(\vec{b} + \vec{c}, \vec{c} + \vec{a}, \vec{a} + \vec{b}) = 2 \cdot \text{det}(\vec{a}, \vec{b}, \vec{c}) \]

Thus, we have:

\[ V_2 = \frac{2}{27} \left| \text{det}(\vec{a}, \vec{b}, \vec{c}) \right| \]

### Step 4: Calculate \( \frac{4V_1}{V_2} \)

Now substituting \( V_1 \) and \( V_2 \):

\[ V_1 = \frac{1}{6} \left| \text{det}(\vec{a}, \vec{b}, \vec{c}) \right| \]

\[ V_2 = \frac{2}{27} \left| \text{det}(\vec{a}, \vec{b}, \vec{c}) \right| \]

Now, we can compute \( \frac{4V_1}{V_2} \):

\[ \frac{4V_1}{V_2} = \frac{4 \cdot \frac{1}{6} \left| \text{det}(\vec{a}, \vec{b}, \vec{c}) \right|}{\frac{2}{27} \left| \text{det}(\vec{a}, \vec{b}, \vec{c}) \right|} \]

This simplifies to:

\[ \frac{4V_1}{V_2} = \frac{4 \cdot 27}{6 \cdot 2} = \frac{108}{12} = 9 \]

### Final Answer

Thus, the value of \( \frac{4V_1}{V_2} \) is:

\[ \boxed{9} \]