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Geometric sequence, area of two triangles inscribing a circle

ÀÛ¼ºÀÚ Uploader : dutyfree ÀÛ¼ºÀÏ Upload Date: 2018-09-28º¯°æÀÏ Update Date: 2018-09-29Á¶È¸¼ö View : 397

Draw two triangles inscribing a circle having its radius R, positioning these two
triangles to make their intersection a regular hexagon of which area is S1. Then, draw a circle inscribing the regular
hexagon and, in a same manner as above, draw two triangles to make a regular hexagon having its area S2. Letting the area of intersection by repeating the above steps by n times as Sn, Sn is calculated as below.
a same manner above.

S1 is equal to the sum of areas of 12 triangles of height h = R/2. Thus,

S1 = 12*(1/2)*h*2*h/(3^(1/2))
    = 12/(3^(1/2))*h^2 = 3^(1/2)*R^2

S2 is inside the circle of radius R/2, then,

S2 = S1/4

Consequently, Sn becomes a geometric sequence of which the first term and common ratio are S1 and 1/4, respectively. The general term (nth term)
Sn is expressed :

Sn = S1*(1/4)^(n-1) = 3^(1/2)*R^2*(1/4)^(n-1)

*** Âü°í¹®Çå[References] ***

Sn = 3^(1/2)*R^2*(1/4)^(n-1)
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