View source for 2000 Pan African MO Problems/Problem 5

Let <math>\gamma</math> be circle and let <math>P</math> be a point outside <math>\gamma</math>. Let <math>PA</math> and <math>PB</math> be the tangents from <math>P</math> to <math>\gamma</math> (where <math>A, B \in \gamma</math>). A line passing through <math>P</math> intersects <math>\gamma</math> at points <math>Q</math> and <math>R</math>. Let <math>S</math> be a point on <math>\gamma</math> such that <math>BS \parallel QR</math>. Prove that <math>SA</math> bisects <math>QR</math>.

== Solution ==

There is a projective transformation which maps <math>\gamma</math> to a circle and that maps the midpoint of <math>QR</math> to its center (EXPAND); therefore, we may assume without loss of generality that the midpoint of <math>QR</math> is the center of <math>\gamma</math>. But then <math>B</math> is the reflection of <math>A</math> across <math>QR</math>, so that <math>S</math> is the antipode of <math>A</math> on <math>\gamma</math>, and we are done.

== See also ==

{{Pan African MO box
|year=2000
|num-b=4
|num-a=6
}}

[[Category:Geometry Problems]] {{stub}}