Problem
GEO-B3-M03-P023 A Family of Tangents from Two Moving Points
B. New Original Problem. A circle \(\omega\), an external point \(P\), and a secant \(PAB\) are fixed. The tangents at \(A\) and \(B\) meet at \(K\). Through \(P\), draw an arbitrary line meeting the tangents \(KA\) and \(KB\) at \(M\) and \(N\). From \(M\) and \(N\), draw the second tangents to \(\omega\), different from \(KA\) and \(KB\); they meet at \(X\). Prove that all points \(X\) lie on one line passing through \(K\).
C. Hint 1. Send the line \(KP\) to infinity by a projective transformation preserving the circle as a circle.
D. Hint 2. In the new model, the statement becomes a symmetry with respect to a diameter.
E. Full Solution.
Use a projective transformation that sends \(\omega\) to a circle and the line \(KP\) to the line at infinity. Such transformations preserve tangency and incidence, so it is enough to prove the statement in the new model.
In the new model, the images of \(KA\) and \(KB\) are parallel. The points \(M\) and \(N\), lying on a line through the image of \(P\) at infinity, correspond to parallel cuts between two parallel tangents. The second tangents from \(M\) and \(N\) meet at a point \(X'\) moving along a line parallel to these tangents and passing through the image of \(K\). This follows from symmetry of the circle with respect to the perpendicular to the two parallel tangents.
Returning by the inverse projective transformation, we obtain that the original points \(X\) lie on one line, and this line passes through \(K\).
This is a strong task on projective simplification of a polar configuration; it is best given after the previous tasks are mastered.