Problem Analysis #49

[THOUGHT] Place P at the origin. Then vectors are simply coordinates: PA=A, PB=B, PQ=Q. Since P lies on the circle with diameter AB, ∠APB=90°, so A·B=0 with |A|=2 and |B|=14 (from right triangle with hypotenuse AB=10√2). Choose coordinates A=(2,0), B=(0,14). Let Q=(x,y). Use:

1. Circle with diameter AB: (Q−A)·(Q−B)=0.
2. Given dot condition: (A+B)·(Q+B)=2|Q|^2. Solve for (x,y), then compute |PA·QB| = |A·(B−Q)| = |−A·Q| = |2x|. [/THOUGHT] [PYTHON] import sympy as sp

x, y = sp.symbols('x y', real=True)

A = sp.Matrix([2, 0]) B = sp.Matrix([0, 14]) Q = sp.Matrix([x, y])

eq1 = sp.Eq((Q-A).dot(Q-B), 0) # circle condition eq2 = sp.Eq((A+B).dot(Q+B), 2*(Q.dot(Q))) # given vector condition

sol = sp.solve([eq1, eq2], [x, y], dict=True) print(sol)

vals = [] for s in sol: xv = sp.simplify(s[x]) yv = sp.simplify(s[y]) # compute |PA · QB| = |A · (B - Q)| = |2x| expr = sp.simplify(abs(2xv)) vals.append((xv, yv, sp.simplify(expr), sp.simplify((B-Q.subs(s)).norm()))) print(vals) [/PYTHON]

[THOUGHT] The solution with $|\overrightarrow{QB}|>0$ is $Q=\left(\frac{98}{25},\frac{336}{25}\right)$, giving $\left|\overrightarrow{PA}\cdot\overrightarrow{QB}\right|=\frac{196}{25}$. Thus $p=25,q=196$ and $p+q=221$. $\boxed{221}$