The volume of a triangular pyramid is 1. Find the volume of the pyramid with vertices at the points of intersection of the medians of the given pyramid.

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Let $A 1, B 1, C 1$ and $D 1$ be the points of intersection of the medians of the faces $B C D, A C D, A B D$ and $A B C$ of the triangular pyramid $A B C D$. Then the segments $A A 1, B B 1, C C 1$ and $D D 1$ (medians of the tetrahedron) intersect at one point (denote it by $M$) and are divided by it in the ratio 3:1, counting from the vertices of the pyramid. Therefore, under a homothety relative to the point $M$ with a coefficient of $-\frac{1}{-3}$, the point $A$ goes to the point $A 1$, the point $B$ to the point $B 1$, $C$ to $C 1$, and $D$ to $D 1$. Thus, the tetrahedron $A 1 B 1 C 1 D 1$ is similar to the tetrahedron $A B C D$ with a coefficient of $\frac{1}{\frac{1}{3}}$. Therefore,

| Problem 110396 Topics: | $[$ [Volume helps solve the problem.] | Difficulty: 3  Grades: 10,11 |
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$$
V_{A} 1 B 1 C 1 D 1=\left(\frac{1}{3}\right) 3 \cdot V_{A B C D}=\frac{1}{27}
$$

## Answer

$\frac{1}{27}$