In the coordinate plane, let be a segment of length 10$$\sqrt{2}$$$, and consider the circle with diameter ABPQ(\overrightarrow{\text{PA}} + \overrightarrow{\text{PB}}) \cdot (\overrightarrow{\text{PQ}} + \overrightarrow{\text{PB}}) = 2\left|\overrightarrow{\text{PQ}}\right|^2\left|\overrightarrow{\text{PB}}\right| = 14\left|\overrightarrow{\text{QB}}\right| > 0\left|\overrightarrow{\text{PA}} \cdot \overrightarrow{\text{QB}}\right| = \dfrac{q}{p}pqp + q$. [4 points]
[THOUGHT] We set up coordinates with A = (-5√2, 0) and B = (5√2, 0), so the circle has radius 5√2 and center at the origin. Given |PB| = 14, we solved for P's coordinates and found |PA| = 2. The condition simplifies to P·(B-Q) = 0, meaning Q lies on the line through B perpendicular to OP. Solving this intersection with the circle gave Q's x-coordinate as x2 = (527√2)/125. Then PA·QB = 5√2(x2 - 5√2) = -196/25, so |PA·QB| = 196/25. Thus p = 25, q = 196, and p + q = 221.
[/THOUGHT] [PYTHON] import sympy as sp
Define symbols
sqrt2 = sp.sqrt(2) a = 5*sqrt2 # half of AB length
Coordinates of P
x1 = -24sqrt2/5 y1 = 7sqrt2/5 # positive case; negative will give same |PA·QB|
Coordinates of Q from condition P·(B-Q)=0
Q = B + t*(-y1, x1) with t = a*y1/25
t = ay1/25 x2 = a - ty1 y2 = t*x1
Verify Q is on circle
circle_eq = x22 + y22 assert sp.simplify(circle_eq - a**2) == 0, "Q not on circle"
Compute PA·QB
PA = (-a - x1, -y1) QB = (a - x2, -y2) dot_product = PA[0]*QB[0] + PA[1]*QB[1] abs_dot = sp.Abs(dot_product)
Simplify
abs_dot_simplified = sp.simplify(abs_dot) print(abs_dot_simplified) [/PYTHON]
[THOUGHT] The Python output confirms that |PA·QB| = 196/25. Since 196 and 25 are coprime, p = 25 and q = 196. Therefore, p + q = 221.
[/THOUGHT] \boxed{221}