IMO 1964 · Problem 6

In tetrahedron \(ABCD\), vertex \(D\) is connected with \(D_0\) the centroid of \(\Delta ABC\). Lines parallel to \(DD_0\) are drawn through \(A, B\) and \(C\). These lines intersect the planes \(BCD, CAD\) and \(ABD\) in points \(A_1, B_1\) and \(C_1\), respectively. Prove that the volume of \(ABCD\) is one third the volume of \(A_1B_1C_1D_0\). Is the result true if point \(D_0\) is selected anywhere within \(\Delta ABC\)?

Official text of the problem. Official problem paper (PDF)

Solution

We prove that the equality holds for every interior point \(D_0\) of \(\triangle ABC\).

Take \(D\) as the origin, and write the position vectors of \(A,B,C,D_0\) as \(\mathbf a,\mathbf b,\mathbf c,\mathbf p\), respectively. Since the tetrahedron is nondegenerate, \(\mathbf a,\mathbf b,\mathbf c\) are linearly independent. Since \(D_0\) is inside \(\triangle ABC\), its barycentric coordinates satisfy \[ \mathbf p=\alpha\mathbf a+\beta\mathbf b+\gamma\mathbf c, \qquad \alpha,\beta,\gamma>0, \qquad \alpha+\beta+\gamma=1. \]

The line through \(A\) parallel to \(DD_0\) consists of the points with position vectors \[ \mathbf a+t\mathbf p =(1+t\alpha)\mathbf a+t\beta\mathbf b+t\gamma\mathbf c. \] The plane \(BCD\) is the span of \(\mathbf b,\mathbf c\). Consequently, the line meets this plane precisely when \(t=-1/\alpha\). Thus the position vectors of the three intersections are \[ \mathbf a_1=\mathbf a-\frac{\mathbf p}{\alpha}, \qquad \mathbf b_1=\mathbf b-\frac{\mathbf p}{\beta}, \qquad \mathbf c_1=\mathbf c-\frac{\mathbf p}{\gamma}. \]

Put \(K=\det(\mathbf a,\mathbf b,\mathbf c)\ne0\). The original tetrahedron has volume \(|K|/6\). For the new tetrahedron, use the three edge vectors from \(D_0\): \[ \mathbf u=\mathbf a-\left(1+\frac1\alpha\right)\mathbf p,\quad \mathbf v=\mathbf b-\left(1+\frac1\beta\right)\mathbf p,\quad \mathbf w=\mathbf c-\left(1+\frac1\gamma\right)\mathbf p. \] By multilinearity, all determinant terms containing \(\mathbf p\) twice vanish. Also, \[ \det(\mathbf p,\mathbf b,\mathbf c)=\alpha K,\qquad \det(\mathbf a,\mathbf p,\mathbf c)=\beta K,\qquad \det(\mathbf a,\mathbf b,\mathbf p)=\gamma K. \] Therefore \[ \begin{aligned} \det(\mathbf u,\mathbf v,\mathbf w) &=K-\left(1+\frac1\alpha\right)\alpha K -\left(1+\frac1\beta\right)\beta K -\left(1+\frac1\gamma\right)\gamma K\\ &=\bigl(1-(\alpha+\beta+\gamma)-3\bigr)K\\ &=-3K. \end{aligned} \] It follows that \[ \operatorname{Vol}(A_1B_1C_1D_0) =\frac{|-3K|}{6} =3\operatorname{Vol}(ABCD). \] In particular, this proves the required assertion for the centroid, where \(\alpha=\beta=\gamma=1/3\), and answers the final question affirmatively for every interior choice.

If “within” includes the boundary, the construction is undefined there. Indeed, if \(D_0\in BC\), then \(DD_0\) lies in the plane \(BCD\); its parallel through \(A\), which is outside that plane, never meets \(BCD\). The same argument applies to the other sides, including their endpoints. Thus the asserted equality holds for every interior point, while boundary points do not yield all three required intersections.

The solution was checked against John Scholes' IMO solution archive (Kalva).