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已知正方形 $ABCD$ 的边长为 $2$ ，点 $Q$ 为边 $BC$ 的中点，点 $P$ 在正方形外部，且满足 $\angle APD=135^\circ$ ，则 $PQ$ 的最大值

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Multiply and divide integers

Full solution

Q. 已知正方形

ABCD

的边长为

2

，点

Q

为边

BC

的中点，点

P

在正方形外部，且满足

\angle APD=135^\circ

，则

PQ

的最大值

Identify Geometry and Given Values: Identify the geometry of the problem and the given values. $ABCD$ is a square with side length $2$ , so $BC = 2$ . Since $Q$ is the midpoint of $BC$ , $BQ = QC = 1$ .
Recognize Angle Configuration: Recognize that $\angle APD = 135^\circ$ suggests $P$ is located such that it forms an obtuse angle with points $A$ and $D$ . This configuration maximizes the distance $PQ$ when $P$ is on the circle centered at $D$ with radius $DA$ , passing through $A$ .
Calculate Circle Radius: Calculate the radius of the circle. Since $DA$ is a diagonal of the square, $DA = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$ .
Determine Position for Maximum PQ: Determine the position of P for maximum PQ. When P is on the extension of line AD, at a distance of $2\sqrt{2}$ from D, PQ will be maximized because it is the hypotenuse of triangle PQD.
Calculate PQ Using Pythagorean Theorem: Calculate $PQ$ using the Pythagorean theorem in triangle $PQD$ , where $PD = 2\sqrt{2}$ (radius) and $DQ = 1$ (half of $BC$ ). $PQ^2 = PD^2 + DQ^2 = (2\sqrt{2})^2 + 1^2 = 8 + 1 = 9$ .
Solve for PQ: Solve for PQ. $PQ = \sqrt{9} = 3$ .

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