Problem
GEO-B1-M05-P033 Two Circles and the Tangent Point
Triangle \(ABC\) is inscribed in circle \(\Omega\) with center \(O\). The circle with diameter \(AO\) intersects the circumcircle of triangle \(OBC\) at point \(S\ne O\). The tangents to \(\Omega\) at \(B\) and \(C\) meet at \(P\). Prove that points \(A,S,P\) are collinear.
C. Hint 1. Prove that \(P\) lies on the circle \((OBC)\).
D. Hint 2. Compare two lines through \(S\) perpendicular to \(OS\).
E. Full solution.
Since \(PB\) and \(PC\) are tangents to \(\Omega\), radii \(OB\) and \(OC\) are perpendicular to them. Hence \(\angle OBP=\angle OCP=90^\circ\).
Thus points \(O,B,C,P\) lie on one circle, and \(OP\) is its diameter. Since \(S\) lies on this circle, \(\angle OSP=90^\circ\).
On the other hand, \(S\) lies on the circle with diameter \(AO\). Therefore \(\angle ASO=90^\circ\).
So both lines \(SP\) and \(SA\) pass through point \(S\) and are perpendicular to \(OS\). Through a point there is only one perpendicular to a given line; hence \(A,S,P\) are collinear.
A. Source analysis. Main objects: a circumcircle, a circle on a diameter, and tangents. The obvious approach is to look for angles at \(A\), but the hidden observation is that the tangent intersection point lies on circle \((OBC)\). Number of key ideas: 3.
F. Difficulty justification. This is Level 7: a regional-style problem with two diameter circles and a hidden cyclicity from tangents.
G. Check. This is not a one-step exercise: one first places \(P\) on circle \((OBC)\), then sees two right angles at \(S\).