Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.1 National olympiad, round 2 Prove it Russia

In the space, three segments $A_1A_2$ , $B_1B_2$ , and $C_1C_2$ are chosen.

They are not coplanar, and they meet at point $P$ . Denote by $O_{ijk}$ the center of the sphere passing through $A_i, B_j, C_k$ , and $P$ . Prove that the four lines $O_{111}O_{222}, O_{112}O_{221}, O_{121}O_{212}$ , and $O_{211}O_{122}$ have a common point.

В пространстве даны три отрезка $A_1A_2$ , $B_1B_2$ и $C_1C_2$ , не лежащие в одной плоскости и пересекающиеся в одной точке $P$ . Обозначим через $O_{ijk}$ центр сферы, проходящей через точки $A_i, B_j, C_k$ и $P$ . Докажите, что прямые $O_{111}O_{222}$ , $O_{112}O_{221}$ , $O_{121}O_{212}$ и $O_{211}O_{122}$ пересекаются в одной точке.

Solution

For a fixed $i$ , all the points $O_{ijk}$ belong to the perpendicular bisector $\alpha_i$ of $PA_i$ ; notice that $\alpha_1 \parallel \alpha_2$ . By similar reasons, the points $O_{ijk}$ appear to be eight vertices of some parallelepiped.

For any segment $XY$ , the perpendicular bisector to this segment is the plane perpendicular to it and passing through its midpoint, i.e., the locus of points equidistant from $X$ and $Y$ .

All points of the form $O_{1jk}$ lie in the perpendicular bisector $\alpha_1$ to the segment $PA_1$ . Similarly, all points $O_{2jk}$ lie in the perpendicular bisector $\alpha_2$ to the segment $PA_2$ ; note that $\alpha_1 \parallel \alpha_2$ .

Similarly, introduce the planes $\beta_j$ — perpendicular bisectors to the segments $PB_j$ , and the planes $\gamma_k$ — perpendicular bisectors to the segments $PC_k$ . Then the points $O_{ijk}$ are the vertices of a parallelepiped formed by the planes $\alpha_i, \beta_j$ , and $\gamma_k$ . Now the statement of the problem follows from the fact that the diagonals of this parallelepiped intersect at one point — its center of symmetry.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.