Re: [中學] 圓相切的問題
※ 引述《SPEED72 (P.Pierce)》之銘言:
: 有三個圓兩兩外切,半徑分別為6,9,30,若有第四個圓與此三個圓均內切,求此圓的半
: 徑?
圓半徑r
36^2 + (r - 6)^2 - 2(r - 6)36 cos(a) = (r - 30)^2
15^2 + (r - 6)^2 - 2(r - 6)15 sin(a) = (r - 9)^2
=> cos(a) = [12 + 4r/3]/[2(r - 6)]
sin(a) = [12 + 2r/5]/[2(r - 6)]
(2r - 12)^2 - (12 + 2r/5)^2 = (12 + 4r/3)^2
整理化減 => r = 45, -45/29 (負不合)
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