就 广东中考 压轴 题23.(3) 个人解法 飨以诸君

有

ΔAEN∽ΔOCN

设

正方形边长a

OA与x轴正轴成角θ

有

相似比

a·tan(45°-θ)

/

a

=

1-tanθ

/

1+tanθ

即

n

/

√2a-n

=

1-tanθ

/

1+tanθ

即

√2a

/

√2a-n

=

2

/

1+tanθ

即

√2a-2n

=

√2a·tanθ

即

tanθ

=

1-√2n/a

且

A(cosθ·a，sinθ·a)

C(-sinθ·a，cosθ·a)

设

N点横坐标xN

有

cosθ·a-xN

/

xN+sinθ·a

=

1-tanθ

/

1+tanθ

即

cosθ·a+sinθ·a

/

xN+sinθ·a

=

2

/

1+tanθ

即

xN=(1+tanθ)(cosθ+sinθ)a/2-sinθ·a

即

S

=

1/2

a/cosθ

(1+tanθ)(cosθ+sinθ)a/2-sinθ·a-sinθ·a

=

1/2

a/cosθ

((1+2sinθcosθ）a

/

2cosθ

-

2sinθ·a)

=

1/2

a/cosθ

((sinθ-cosθ)²·a

/

2cosθ)

=

(sinθ-cosθ)²·a²

/

4cos²θ

=

(tanθ-1)²·a²

/

4

=

(1-√2n/a-1)²·a²

/

4

=

n²/2