Question

The four lines drawn from the vertices of any tetrahedron to the centroid of the opposite faces meet in a point whose distance from each vertex is k times the distance from each vertex to the opposite face, where k is

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{3}{4}$
• D. $\frac{5}{4}$

Answer 1

The correct option is C $\frac{3}{4}$

Let the vertices on the tetrahedron be respectively A,B,C,D.
If E is the centroid of face BCD and G is the centroid of
Tetrahedron ABCD, then
$AG=\frac{3}{4}AE\therefore k=\frac{3}{4}$