IMO 1972 · Problem 6

Given four distinct parallel planes, prove that there exists a regular tetrahedron with a vertex on each plane.

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

Solution

Choose Cartesian coordinates so that the given planes have equations \[ z=h_1,\qquad z=h_2,\qquad z=h_3,\qquad z=h_4, \] where the \(h_i\) are pairwise distinct. Put \[ m=\frac{h_1+h_2+h_3+h_4}{4}, \qquad t_i=h_i-m. \] Thus \(\sum_{i=1}^4t_i=0\).

Consider the four vectors \[ v_1=(1,1,1),\quad v_2=(1,-1,-1),\quad v_3=(-1,1,-1),\quad v_4=(-1,-1,1). \] Their endpoints form a regular tetrahedron: every pair is at distance \(2\sqrt2\), and they are not coplanar because \[ \det(v_2-v_1,\ v_3-v_1,\ v_4-v_1)=-16\ne0. \] Moreover, \[ v_i\cdot v_i=3,\qquad v_i\cdot v_j=-1\quad(i\ne j). \]

Define \[ w=\frac14\sum_{j=1}^4t_jv_j. \] For each \(i\), the preceding identities give \[ w\cdot v_i =\frac14\left(3t_i-\sum_{j\ne i}t_j\right) =t_i. \] Since the \(h_i\) are distinct, the \(t_i\) are not all zero; hence \(w\ne0\). Write \(r=\lVert w\rVert>0\).

Choose a rotation \(R\) taking the unit vector \(w/r\) to the vertical unit vector \(e_3=(0,0,1)\). Rotations preserve scalar products, so \[ e_3\cdot Rv_i =R(w/r)\cdot Rv_i =\frac{w\cdot v_i}{r} =\frac{t_i}{r}. \] Now define four points by their position vectors \[ p_i=rRv_i+me_3\qquad(i=1,2,3,4). \] The vertical coordinate of \(p_i\) is \[ e_3\cdot p_i=t_i+m=h_i, \] so \(p_i\) lies on the \(i\)-th prescribed plane.

Finally, these four points are obtained from the original nondegenerate regular tetrahedron by a rotation, a dilation of positive factor \(r\), and a translation. They therefore form a nondegenerate regular tetrahedron, with one vertex on each given plane. \(\square\)

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