On the edges $ AB,BC,CD,DA$ of a parallelepiped $ ABCDA_1 B_1 C_1 D_1$ points $ K,L,M,N$ are selected, respectively. Prove that the circumcenters of the tetrahedra $ A_1 AKN, B_1 BKL, C_1 CLM, D_1 DMN$ are vertices of a parallelogram.
Problem
Source: Ukraine 1997 grade 11
Tags: geometry, circumcircle, parallelogram, modular arithmetic, 3D geometry, sphere, geometry proposed
Luis González
12.03.2010 01:14
Let $ \mathcal{E}_1,\mathcal{E}_2,\mathcal{E}_3,\mathcal{E}_4$ be the circumspheres of the tetrahedra $ A_1AKN,B_1BKL,C_1CLM$ and $ D_1DMN$ with centers $ O_1,O_2,O_3,O_4.$ $ \mathcal{E}_1,\mathcal{E}_2,\mathcal{E}_3,\mathcal{E}_4$ intersect the plane $ ABCD$ through the circles $ \omega_1, \omega_2, \omega_3,\omega_4$ with centers $ P_1,P_2,P_3,P_4$ and let $ E \equiv \omega_3 \cap \omega_4,$ $ F\equiv \omega_1 \cap \omega_2,$ different from $ M,K,$ respectively. Since $ AD \parallel BC,$ by Reim's theorem for the pairs of secant circles $ \omega_3,\omega_4$ and $ \omega_1,\omega_2,$ the intersections $ E,F,$ different from $ M,K,$ lie on the straight line $ NL.$ Consequently, we have $ \angle CME = \angle DNE = \angle FKA$ $ \Longrightarrow$ $ EM \parallel FK.$ Radical axes of the pairs of circles $ \omega_3,\omega_4$ and $ \omega_1,\omega_2$ are parallel, which implies that $ P_1P_2 \parallel P_3P_4.$ Analogously, we'll have $ P_2P_3 \parallel P_4P_1$ $ \Longrightarrow$ $ P_1P_2P_3P_4$ is a parallelogram, thus $ O_1,O_2,O_3,O_4$ lie on four parallel lines $ \ell_1,\ell_2,\ell_3,\ell_4$ respectively such that the distance between $ \ell_1,\ell_4$ equals the distance between $ \ell_2,\ell_3$ $ (\star).$ On the other hand, let $ \mathcal{E}_1,\mathcal{E}_4$ intersect the plane $ ADD_1A_1$ through the circles $ \delta_1,\delta_4.$ Since $ AA_1 \parallel DD_1,$ by Reim's theorem the intersection $ U\equiv \delta_1 \cap \delta_4,$ different from $ N,$ lies on the edge $ A_1D_1.$ Similarly if $ \mathcal{E}_2,\mathcal{E}_3$ intersect the plane $ BCC_1B_1$ through the circles $ \delta_2,\delta_3,$ the intersection $ V \equiv \delta_2 \cap \delta_3,$ different from $ L,$ lies on the edge $ B_1C_1.$ Hence $ \angle D_1UN = \angle NAA_1 = \angle LBB_1 = \angle C_1VL$ $ \pmod\pi$ $ \Longrightarrow$ $ UN \parallel VL.$ Radical axes of $ \delta_1 , \delta_4$ and $ \delta_2 , \delta_3$ are parallel, but the radical axes of $ \omega_1,\omega_4$ and $ \omega_2,\omega_3$ are parallel as well. Hence the radical planes of the spheres $ \mathcal{E}_1,\mathcal{E}_4$ and $ \mathcal{E}_2,\mathcal{E}_3$ are parallel $ \Longrightarrow$ $ O_1O_4 \parallel O_2O_3.$ Together with $ (\star),$ it follows that $ O_1O_2O_3O_4$ is a parallelogram.