AS / GS 2

(a).
tan30=CB/d
CB=d/√3
Since ∠CDB=∠DCB=45,
So, CB=DB
a=d-DB
a=d-d/√3
a=(√3-1)d/√3

DB=d/√3
AD=(√3-1)d/√3
tan30=(DA1)/(AD)
DA1=(√3-1)d/3
tan(∠A1DB-45)=(A1B1)/(DA1)
A1B1=[(√3-1)d/3]tan 45
A1B1=(√3-1)d/3
tan∠A 2A 1B1=(B 1A 2)/(A1B1)
B 1A 2=[(√3-1)d/3](tan30)
B 1A 2=(√3-1)d/(3√3) (b).AD=(√3-1)d/√3=(√3-1)d/(√3^1)DA1=(√3-1)d/3=(√3-1)d/(√3^2)A1B1=(√3-1)d/3=(√3-1)d/(√3^2)B 1A 2=(√3-1)d/(3√3)=(√3-1)d/(√3^3)A2B2=(√3-1)d/(3√3)=(√3-1)d/(√3^3)…B_[n-1]A_[n]=(√3-1)d/[√3^(n+1)]A_[n]B_[n]=(√3-1)d/[√3^(n+1)] So, thesum of AD+DA1+A1B1+…+B_[n-1]A_[n]+A_[n]B_[n] is, http://s1099.photobucket.com/albums/g395/jasoncube/?action=view¤t=c1ce45fc.jpg

圖片參考：http://imgcld.yimg.com/8/n/HA00016323/o/701109110065313873456590.jpg

So, the answer is (√3+1)d/√3.
(c).The sum of area of ∆ADA1+∆A1B 1A 2+A2B 2A 3+…+A_[n-1]B_[n-1]A_n is,
http://s1099.photobucket.com/albums/g395/jasoncube/?action=view¤t=2e5f54b9.jpg


圖片參考：http://imgcld.yimg.com/8/n/HA00016323/o/701109110065313873456601.jpg

So, the answer is (3-√3)d/4.