有

AG=1

∠AGC=α+45°

即

AC=ED

=tan(α+45°)

=(1+tanα)/(1-tanα)

且

∠DFE

=4∠ACG

=180°-4α

即

EF

=ED/sin∠DFE

=ED/sin(180°-4α)

DF

=ED/tan∠DFE

=ED/tan(180°-4α)

且

C△DEF=a

即

ED+EF+DF

=ED(1+1/sin(180°-4α)+1/tan(180°-4α))

=ED(1+1/sin4α-1/tan4α)

=ED(1+(1-cos4α)/sin4α)

=ED(1+2sin22α/2sin2αcos2α)

=ED(1+tan2α)

=(1+tanα)/(1-tanα)

(1+2tanα-tan2α)/(1-tan2α)

=(1+2tanα-tan2α)/(1-2tanα+tan2α)

=a

即

（a+1)tan2α-(2a+2)tanα+a-1=0

且

0＜α＜45°

即

0＜tanα＜1

即

tanα

=(2a+2)-√(4(a+1)2-4(a2-1))

/(2a+2)

=1-√2/√(a+1)

即

AC

=(1+tanα)/(1-tanα)

=2-√2/√(a+1)

/(√2/√(a+1))

=√(2a+2)-1

ps.

未見答案

僅供參考