Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 5.1 AIME, harder Prove it Brazil

Let $P_1P_2...P_n$ be a polygon inscribed in a circle and $Q$ a point outside the plane of the circle. Let $\beta_i$ be the plane perpendicular to $QP_i$ and passing through $P_i$ , $i = 1, 2, ..., n$ . Prove that the intersection of all planes $\beta_i$ is a point.

Solution

Let $S$ be the sphere that contains the circumcircle of the polygon and $Q$ and let $Q'$ be the point diametrically opposite to $Q$ in $S$ . Then $\angle Q'P_iQ = 90^\circ$ , that is, $Q'P_i$ is perpendicular to $QP_i$ and since $\beta_i$ contains all lines perpendicular to $QP_i$ passing through $P_i$ , $Q'$ belongs to $\beta_i$ .

It remains to prove that the intersection of the planes is not a line. Let $t$ be the intersection of $\beta_1$ and $\beta_2$ . It is a line, since the intersection is non-empty. This line is perpendicular to both $QP_1$ and $QP_2$ , so it is perpendicular to plane $QP_1P_2$ . Analogously, the intersection $u$ of $\beta_2$ and $\beta_3$ is perpendicular to the plane $QP_2P_3$ . Since the planes $QP_1P_2$ and $QP_2P_3$ are secant in $QP_2$ , $t \neq u$ and the intersection of all planes is the intersection of $t$ and $u$ , which is a single point.

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