IMO 1964 - Problem 6

Problem: Given a tetrahedron ABCD, let D1 be the centroid ò the triangle ABC and let A1,B1,C1 be the intersection points of the lines parallel to DD1 and passing through the points A,B,C with the opposite faces of the tetrahedron. Prove that the volume of the tetrahedron ABCD is one-third the volume of the tetrahedron A1B1C1D1. Does the result remain true if the point D1 is replaced with any point inside the triangle ABC?
Solution: We shall prove that the statement is valid in the general case, for an arbitrary point D1 inside $\triangle{ABC}$. Since D1 belongs to the plane ABC, there are real numbers a,b,c such that $(a+b+c)\vec{DD_1}=a\vec{DA}+b\vec{DB}+c\vec{DC}$. Since $AA_\parallel DD_1$, it holds that $\vec{AA_1}=k\vec{DD_1}$ for some $k\in \mathbb{R}$. Now it is easy to get $\vec{DA_1}=-\frac{b\vec{DA}+c\vec{DC}}{a},\vec{DB_1}=-\frac{a\vec{DA}+c\vec{DC}}{b}$, and $\vec{DC_1}=-\frac{a\vec{DA}+b\vec{DB}}{c}$. This implies: $\vec{D_1A_1}=-\frac{a^2\vec{DA}+b(a+2b+c)\vec{DB}+c(a+b+2c)\vec{DC}}{a(a+b+c)},\vec{D_1B_1}=-\frac{a(2a+b+c)\vec{DA}+b^2\vec{DB}+c(a+b+2c)\vec{DC}}{b(a+b+c)}$, and $\vec{D_1C_1}=-\frac{a(2a+b+c)\vec{DA}+b(a+2b+c)\vec{DB}+c^2\vec{DC}}{c(a+b+c)}$.
By using $6V_{D_1A_1B_1C_1}=\big |[\vec{D_1A_1},\vec{D_1B_1},\vec{D_1C_1}]\big |$ and $6V_{DABC}=\big |[\vec{DA},\vec{DB},\vec{DC}]\big |$ we get
$V_{D_1A_1B_1C_1}=\frac{S}{6abc(a+b+c)^3}=3V_{DABC}$, where $S=\begin{align*}\begin{vmatrix} a^2&b(a+2b+c)&c(a+b+2c)\\ a(2a+b+c)&b^2&c(a+b+2c)\\ a(2a+b+c)& b(a+2b+c)&c^2\end{vmatrix}\end{align*}$.