半径为1的球面上有三点A,B,C,若A和B,A和C,B和C的球面距离都是 π 2 ,过A、B、C三点做截

\u5df2\u77e5A\u3001B\u3001C\u662f\u534a\u5f84\u4e3a1\u7684\u7403\u9762\u4e0a\u4e09\u70b9\uff0cO\u4e3a\u7403\u5fc3\uff0cA\u3001B\u548cA\u3001C\u7684\u7403\u9762\u8ddd\u79bb\u90fd\u662f\u03c0/2\uff0cB\u3001C\u7684\u7403\u9762\u8ddd\u79bb\u662f\u03c0/3

\u7403\u534a\u5f84R=1
\u89d2AOB=\u03c0/2,AB=2sin(\u03c0/4)=\u6839\u53f72,AC=AB=\u6839\u53f72
\u89d2BOC=\u03c0/3,BC=2sin(\u03c0/6)=1

sin(\u89d2BAC/2)=(1/2)/(\u6839\u53f72)=1/(2(\u6839\u53f72))=(\u6839\u53f72)/4
cos(\u89d2BAC/2)=(1-((\u6839\u53f72)/4)^2)^(1/2)=(1/4)(\u6839\u53f7(14)
sin(\u89d2BAC)=2sin(\u89d2BAC/2)cos(\u89d2BAC/2)=(1/4)(\u6839\u53f77)
\u8bbe\u4e09\u89d2\u5f62ABC\u7684\u5916\u63a5\u5706\u534a\u5f84=r
2r=BC/sin(\u89d2BAC)=4/(\u6839\u53f77)
r=2/(\u6839\u53f77)

\u7403\u5fc3O\u5230\u5e73\u9762ABC\u7684\u8ddd\u79bb=(R^2-r^2)^(1/2)=(1-(4/7))^(1/2)=(1/7)(\u6839\u53f721)

A\u548cC\u7684\u7403\u9762\u8ddd\u79bb\u90fd\u662f\u03c0\uff0f2\uff0c 2\u03c0r\u00f7\u03c0/2=4 \u6240\u4ee5A\u4e0eC\u5bf9\u5e94\u7684\u5706\u5fc3\u89d2\u662f360\u00b0\u00f74=90\u00b0\uff0c\u540c\u6837\u7684\u9053\u7406\uff0cA\u4e0eB\u5bf9\u5e94\u7684\u5706\u5fc3\u89d2\u4e5f\u662f360\u00b0\u00f74=90\u00b0\uff0c\u4ee5\u5706\u5fc3O\u4e3a\u539f\u70b9\u5efa\u7acb\u7a7a\u95f4\u76f4\u89d2\u5750\u6807\u7cfb\uff0c\u5e76\u4ee4A=(1,0,0), B=(0,1,0),\u56e0\u4e3aB\u548cC\u7684\u7403\u9762\u8ddd\u79bb\u662f\u03c0\uff0f3. B\u4e0eC\u5bf9\u5e94\u7684\u5706\u5fc3\u89d2\u662f360\u00b0\u00f76=60\u00b0 \u6240\u4ee5C=(0,1/2\uff0c2\u5206\u4e4b\u6839\u53f73)\uff0cA\u3001B\u3001C\u7684\u5750\u6807\u90fd\u51fa\u6765\u4e86\uff0c\u95ee\u9898\u81ea\u5df1\u89e3\u51b3\u597d\u4e0d\u597d\uff01\u795d\u4f60\u8003\u8bd5\u6109\u5feb\u54c8\uff01

球心O与A,B,C三点构成正三棱锥O-ABC,
已知OA=OB=OC=R=1,∠AOB=∠BOC=∠AOC=90°,
由此可得AO⊥面BOC.
S △BOC =
1
2
S △ABC =


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