Question

有关于数学球面体

Answer 1

设球心为O,三点为A,B,C.

则OA =OB = OC = R.

由A,B球面距离为πR/3,有∠AOB = π/3.

故△AOB为等边三角形,AB = OA = R.

同理BC = R,CA = R.

于是△ABC也为边长R的等边三角形.

其外接圆周长为4π,即外接圆半径r = 2.

由正弦定理,R = AB = 2rsin∠BCA = 2rsin(π/3) = 2√3.