急 ! F6 circle計法1題 #718.

(a) Length of the altitude of the triangle with side length 1 = 1 x sin 60 = (sqrt 3)/2.
So radius of C1 = one - third of the altitude = (sqrt 3)/6.
Length of the altitude remaining for accomodating the other smaller circles = altitude of triangle - diameter of C1 = (sqrt 3)/2 - (sqrt 3)/3 = (sqrt 3)/6.
By similar figure :
altitude of 1st triangle/diameter of C1 = altitude of 2nd triangle/diameter of C2
[(sqrt 3)/2]/[(sqrt 3)/3] = [(sqrt 3)/6]/diameter of C2
diameter of C2 = [(sqrt 3)/6]/(2/3) = (sqrt 3)/9, so radius of C2 = (sqrt 3)/18.
(b)
Similarly, length of altitude remaining for the subsequent circles
= (sqrt 3)/6 - diameter of C2 = (sqrt 3)/6 - (sqrt 3)/9 = (sqrt 3)/18.
[(sqrt 3)/2]/[(sqrt 3)/3] = altitude of 3rd triangle/diameter of C3
diameter of C3 = (2/3)[(sqrt 3)/9] = (sqrt 3)/27, so radius of C3 = (sqrt 3)/54.
Based on the sequence : (sqrt 3)/6, (sqrt 3)/18, (sqrt 3)/54, ....
which is a G.S. with 1st term = (sqrt 3)/6 and common ratio = 1/3,
so radius of Cn = (sqrt 3)/6 (1/3)^(n - 1).
(c)
Sum of diameters = 2 x [(sqrt 3)/6]/(1 - 1/3) = (sqrt 3)/2.




2014-03-11 22:05:35 補充：
Correction : Line 8 should be [(sqrt 3)/6] x (2/3) = (sqrt 3)/9.

2014-03-11 22:09:32 補充：
Correction : Line 13 should be diameter of C3 = (2/3)[(sqrt 3)/18] = (sqrt 3)/27.