Let our planes be $\pi_1,\pi_2,\pi_3,\pi_4$, which we assume to be parallel to the $xy$-plane, listed in the increasing order of their $z$-coordinates. First take a plane $\pi$ orthogonal to $\pi_i$, which cuts $\pi_1,\pi_2,\pi_3$ along three lines $d_1,d_2,d_3$. On these three lines, take three vertices $A_1,A_2,A_3$ respectively of an equilateral triangle (it is well-known that this is possible; in fact, the problem here is the $3$-dimensional version of this), and then complete the two regular tetrahedra $A_1A_2A_3P_1,A_1A_2A_3P_2$ having $A_1A_2A_3$ as one of their faces. Both $P_i$ lie below $\pi_4$. Now take another plane $\pi$ and repeat the construction above. If $\pi$ makes a small enough angle with the $\pi_i$'s, one of the $P_i$'s we get this time must lie above $\pi_4$. Now, if we move the initial position of $\pi$ towards the new one continuously and record the $z$-coordinates of $P_1,P_2$, these will be continuous functions of the angle that $\pi$ makes with $\pi_i$, and for one of the points $P_1,P_2$ the $z$-coordinate will move continuously from being smaller than that of $\pi_4$ to being larger than it, meaning that at some point, one of the points $P_1,P_2$ will lie on $\pi_4$, and this is what we want.

orl wrote: Given four distinct parallel planes, prove that there exists a regular tetrahedron with a vertex on each plane. Here is another solution. Let's denote the (directed) distance between two parallel planes p and p' by d (p; p'), and the (directed) distance between two parallel lines g and g' by d (g; g'). (Directed distances are defined as follows: If $p_1$, $p_2$, $p_3$, ... is a family of parallel planes in space, then we choose a unit vector $\overrightarrow{v}_p$ perpendicular to all of these planes (there are two such unit vectors, and we have to choose one of them), and then, by the directed distance between two of these planes $p_i$ and $p_j$, we denote the real number k such that the translation with translation vector $k\cdot\overrightarrow{v}_p$ maps the plane $p_i$ to the plane $p_j$. Similarly, we define the directed distance between two of a family of parallel lines in a plane.) The problem can be rewritten as follows: Given four distinct parallel planes $p_1$, $p_2$, $p_3$, $p_4$ in space, prove that there exists a regular tetrahedron XYZW such that $X\in p_1$, $Y\in p_2$, $Z\in p_3$, $W\in p_4$. In order to do this, it is enough to find a regular tetrahedron ABCD somewhere in space and four parallel planes $q_1$, $q_2$, $q_3$, $q_4$ such that $A\in q_1$, $B\in q_2$, $C\in q_3$, $D\in q_4$ and $d\left(q_1;\;q_2\right): d\left(q_2;\;q_3\right): d\left(q_3;\;q_4\right)=d\left(p_1;\;p_2\right): d\left(p_2;\;p_3\right): d\left(p_3;\;p_4\right)$. In fact, once we have found such a tetrahedron ABCD and such planes $q_1$, $q_2$, $q_3$, $q_4$, then, because of $p_1\parallel p_2\parallel p_3\parallel p_4$, $q_1\parallel q_2\parallel q_3\parallel q_4$ and $d\left(q_1;\;q_2\right): d\left(q_2;\;q_3\right): d\left(q_3;\;q_4\right)=d\left(p_1;\;p_2\right): d\left(p_2;\;p_3\right): d\left(p_3;\;p_4\right)$, there exists a similitude transformation which maps the planes $q_1$, $q_2$, $q_3$, $q_4$ to the planes $p_1$, $p_2$, $p_3$, $p_4$; this similitude transformation will then obviously map the regular tetrahedron ABCD with $A\in q_1$, $B\in q_2$, $C\in q_3$, $D\in q_4$ to a regular tetrahedron XYZW with $X\in p_1$, $Y\in p_2$, $Z\in p_3$, $W\in p_4$; hence, the existence of such a tetrahedron XYZW will be proven, and the problem will be solved. So consider a regular tetrahedron ABCD lying arbitrarily in space; we try to find four parallel planes $q_1$, $q_2$, $q_3$, $q_4$ such that $A\in q_1$, $B\in q_2$, $C\in q_3$, $D\in q_4$ and $d\left(q_1;\;q_2\right): d\left(q_2;\;q_3\right): d\left(q_3;\;q_4\right)=d\left(p_1;\;p_2\right): d\left(p_2;\;p_3\right): d\left(p_3;\;p_4\right)$. In fact, we start working in the plane ABC. Let T be the point on the line AC such that $AT: TC=d\left(p_1;\;p_2\right): d\left(p_2;\;p_3\right)$ (where the segments AT and TC are directed). Let $g_2$ be the line BT, and let $g_1$ and $g_3$ be the parallels to the line $g_2=BT$ through the points A and C, respectively. Then, the lines $g_1$, $g_2$, $g_3$ are parallel and, by Thales, $d\left(g_1;\;g_2\right): d\left(g_2;\;g_3\right)=AT: TC$. Thus, $d\left(g_1;\;g_2\right): d\left(g_2;\;g_3\right)=d\left(p_1;\;p_2\right): d\left(p_2;\;p_3\right)$. Now, denote by $g_4$ the line in the plane ABC which is parallel to the lines $g_1$, $g_2$, $g_3$ and satisfies $d\left(g_1;\;g_2\right): d\left(g_2;\;g_3\right): d\left(g_3;\;g_4\right)=d\left(p_1;\;p_2\right): d\left(p_2;\;p_3\right): d\left(p_3;\;p_4\right)$. Now, let $q_4$ be the plane passing through the line $g_4$ and the point D. Let $q_1$, $q_2$, $q_3$ be the planes parallel to $q_4$ and passing through the lines $g_1$, $g_2$, $g_3$, respectively (of course, we can construct such planes since the lines $g_1$, $g_2$, $g_3$ are parallel to $g_4$). Thus, we have found four parallel planes $q_1$, $q_2$, $q_3$, $q_4$ such that $A\in q_1$, $B\in q_2$, $C\in q_3$, $D\in q_4$, and these planes obviously satisfy $d\left(q_1;\;q_2\right): d\left(q_2;\;q_3\right): d\left(q_3;\;q_4\right)=d\left(g_1;\;g_2\right): d\left(g_2;\;g_3\right): d\left(g_3;\;g_4\right)$. Since $d\left(g_1;\;g_2\right): d\left(g_2;\;g_3\right): d\left(g_3;\;g_4\right)=d\left(p_1;\;p_2\right): d\left(p_2;\;p_3\right): d\left(p_3;\;p_4\right)$, we thus have $d\left(q_1;\;q_2\right): d\left(q_2;\;q_3\right): d\left(q_3;\;q_4\right)=d\left(p_1;\;p_2\right): d\left(p_2;\;p_3\right): d\left(p_3;\;p_4\right)$. Hence, according to the above, the problem is solved. Darij

How can I find new plane π in grobber's solution it isn't described?