Problem Analysis #geometry_30

Recorded Python call:

import sympy as sp

# Define symbols
R = 5*sp.sqrt(2)  # radius, since AB = 10√2
R_sq = R**2  # 50

# P coordinates: x_p = -24√2/5, y_p = ±7√2/5
x_p = -24*sp.sqrt(2)/5
y_p = 7*sp.sqrt(2)/5  # choose positive; the other sign gives same absolute value

# Check |PB|^2
PB_sq = (R - x_p)**2 + (0 - y_p)**2
print("PB^2 =", sp.simplify(PB_sq))

# Condition: P·Q = P·B
# P·B = R*x_p
P_dot_B = R * x_p
print("P·B =", sp.simplify(P_dot_B))

# Solve for Q = (x, y) on circle x^2+y^2 = R^2, with P·Q = P·B
x, y = sp.symbols('x y', real=True)
eq1 = sp.Eq(x**2 + y**2, R_sq)
eq2 = sp.Eq(x_p*x + y_p*y, P_dot_B)

# Solve for x
sol = sp.solve([eq1, eq2], (x, y), dict=True)
print("Solutions for Q:", sol)

# Compute PA·QB for each valid Q (exclude Q=B)
A = sp.Point(-R, 0)
B = sp.Point(R, 0)
P_pt = sp.Point(x_p, y_p)

for s in sol:
    Q_pt = sp.Point(s[x], s[y])
    PA_vec = sp.Matrix(A - P_pt)
    QB_vec = sp.Matrix(B - Q_pt)
    dot = PA_vec.dot(QB_vec)
    abs_dot = sp.simplify(abs(dot))
    QB_len = sp.sqrt((B.x - Q_pt.x)**2 + (B.y - Q_pt.y)**2)
    print(f"Q = ({sp.simplify(s[x])}, {sp.simplify(s[y])})")
    print(f"  PA·QB = {sp.simplify(dot)}")
    print(f"  |PA·QB| = {abs_dot}")
    print(f"  |QB| = {sp.simplify(QB_len)}")

The problem can be solved by placing the origin at the midpoint of the diameter $AB$. Let the circle have radius $R = 5\sqrt{2}$ (since $AB = 10\sqrt{2}$). Set $A = (-R, 0)$ and $B = (R, 0)$. Points $P$ and $Q$ lie on the circle $x^2 + y^2 = R^2 = 50$.

Given $|PB| = 14$, we find $P$. With $B = (5\sqrt{2}, 0)$, the condition $|PB|^2 = 196$ gives:

Using $x_P^2 + y_P^2 = 50$, this simplifies to $P \cdot B = -48$. Hence $x_P = -\frac{24\sqrt{2}}{5}$ and $y_P = \pm\frac{7\sqrt{2}}{5}$.

The vector equation $(\overrightarrow{PA}+\overrightarrow{PB})\cdot(\overrightarrow{PQ}+\overrightarrow{PB}) = 2|\overrightarrow{PQ}|^2$ is simplified using the origin at the midpoint. Since $A+B=0$, we have $\overrightarrow{PA}+\overrightarrow{PB} = -2P$ and $\overrightarrow{PQ}+\overrightarrow{PB} = Q+B-2P$. The equation reduces to:

Thus $Q$ satisfies $x_P x_Q + y_P y_Q = -48$ and $x_Q^2 + y_Q^2 = 50$.

Solving these yields two possible $Q$:

1. $Q = B$ (which gives $|QB| = 0$, excluded by the condition $|QB| > 0$),
2. $Q = \left(\frac{527\sqrt{2}}{125}, -\frac{336\sqrt{2}}{125}\right)$.

Now compute the required dot product $\overrightarrow{PA} \cdot \overrightarrow{QB} = (A-P)\cdot(B-Q)$. Expanding and using $P \cdot Q = P \cdot B$:

With $A = (-R, 0)$, $A\cdot B = -R^2 = -50$ and $A\cdot Q = -R\,x_Q$. Hence:

For the valid $Q$, $x_Q = \frac{527\sqrt{2}}{125}$, so $R\,x_Q = 5\sqrt{2} \cdot \frac{527\sqrt{2}}{125} = \frac{5270}{125} = \frac{1054}{25}$. Then:

Taking the absolute value gives $\frac{196}{25}$. Here $p=25$ and $q=196$ are coprime, so $p+q = 221$.