Problem
GEO-B3-M04-P015 Tangency of a Family of Simson Lines
B. New Original Problem. On a circle, points \(P\) and \(C\) are fixed. Points \(A\) and \(B\) move on the circle so that \(\angle ACB\) is constant. Prove that the Simson lines of \(P\) with respect to triangles \(ABC\) are tangent to one fixed circle.
C. Hint 1. Consider the two projections of \(P\) onto \(CA\) and \(CB\).
D. Hint 2. Their distance on the circle with diameter \(PC\) remains constant.
E. Full Solution.
Let \(A_1\) and \(B_1\) be the projections of \(P\) onto \(CB\) and \(CA\). The points \(A_1\) and \(B_1\) lie on the circle with diameter \(PC\). The angle \(\angle A_1CB_1\) equals \(\angle ACB\), which is constant by assumption.
Therefore the chord \(A_1B_1\) of the circle with diameter \(PC\) has constant length. The Simson line is the line \(A_1B_1\). All chords of a fixed circle with the same length are tangent to a concentric circle of fixed radius. Hence all these Simson lines are tangent to one fixed circle.
This is olympiad-style: the Simson line becomes a family of chords of a fixed circle.