题目
外切于半径为R的球的圆锥,侧面积与球面积之比为3:2,求圆锥底面半径r
提问时间：2020-10-22
答案
设t为圆锥侧面与底面夹角,
则母线长l = r/cos(t)
R = r*tan(t/2)
圆锥侧面积s1 = pi*l*r = pi*r/cos(t)*r
球的表面积s2 = 4*pi*R^2 = 4*pi*r^2*tan(t/2)^2
s1/s2 = 1/cos(t)/4tan(t/2)^2 = 3/2
=>1/cos(t) * (1+cos(t)/(1-cos(t)) = 6 (tan(t/2)^2 = (1-cos(t))/(1+cos(t))
=>cos(t) = 1/2或者1/3
=>tan(t/2) = sqrt((1-cos(t))/(1+cos(t))) = √3/3或者√2/2
=>r = R/tan(t/2) =√2R 或者 √3R
则母线长l = r/cos(t)
R = r*tan(t/2)
圆锥侧面积s1 = pi*l*r = pi*r/cos(t)*r
球的表面积s2 = 4*pi*R^2 = 4*pi*r^2*tan(t/2)^2
s1/s2 = 1/cos(t)/4tan(t/2)^2 = 3/2
=>1/cos(t) * (1+cos(t)/(1-cos(t)) = 6 (tan(t/2)^2 = (1-cos(t))/(1+cos(t))
=>cos(t) = 1/2或者1/3
=>tan(t/2) = sqrt((1-cos(t))/(1+cos(t))) = √3/3或者√2/2
=>r = R/tan(t/2) =√2R 或者 √3R
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