设A、B、C是半径为1的球面上的三点，B、C两点间的球面距离为π3，点A与B、C两点间的球面距离均为π2，O为 半径为1的球面上的四点A，B，C，D是正四面体的顶点，则A与...
A\u3001B\u3001C\u662f\u534a\u5f84\u4e3a1\u7684\u7403\u9762\u4e0a\u4e09\u70b9\uff0cB\u3001C\u95f4\u7684\u7403\u9762\u8ddd\u79bb\u4e3a \u03c0 3 \uff0c\u70b9A\u4e0eB\u3001C\u4e24\u70b9\u95f4\u7684\u7403\u9762\u8ddd\u79bb\u5747\u4e3a\uff081\uff09\u2235\u7403\u9762\u8ddd\u79bb?=\u03b8?r\uff08\u03b8\u4e3a\u52a3\u5f27\u6240\u5bf9\u5706\u5fc3\u89d2\uff09\uff0c\u4e14B\u3001C\u95f4\u7684\u7403\u9762\u8ddd\u79bb\u4e3a \u03c0 3 \uff0c\u70b9A\u4e0eB\u3001C\u4e24\u70b9\u95f4\u7684\u7403\u9762\u8ddd\u79bb\u5747\u4e3a \u03c0 2 \uff0c\u6545\u5f97\u2220AOB= \u03c0 2 \uff0c\u2220BOC= \u03c0 3 \uff0c\u2220AOC= \u03c0 2 \uff1b\uff082\uff09\u2235OA=OB=OC=1\uff0c\u2234AB=AC= 2 \uff0cBC=1\uff0c\u2234S \u25b3OBC = 3 4 \uff0cS \u25b3ABC = 7 4 V 0 -ABC= 1 3 ? 3 4 ?1= 1 3 ? 7 4 ?d\uff0c\u2234d= 21 7 \uff0c\u7403\u5fc3\u5230\u622a\u9762ABC\u7684\u8ddd\u79bb\u4e3a 21 7 \uff0c\uff083\uff09\u8bbe\u7403\u7684\u5185\u63a5\u6b63\u65b9\u4f53\u68f1\u957f\u4e3aa\uff0c\u6839\u636e\u7403\u7684\u76f4\u5f84\u4e3a\u6b63\u65b9\u4f53\u7684\u5bf9\u89d2\u7ebf\uff0c\u5219 3 a=2\uff0c\u2234a= 2 3 3 \uff0c\u2234S \u6b63\u65b9\u4f53 \uff1aS \u7403\u9762 =6? ( 2 3 3 ) 2 \uff1a4\u041b=2\uff1a\u041b\uff0e
\u534a\u5f84\u4e3a1\u7684\u7403\u9762\u4e0a\u7684\u56db\u70b9A\uff0cB\uff0cC\uff0cD\u662f\u6b63\u56db\u9762\u4f53\u7684\u9876\u70b9\uff0c\u6240\u4ee5\u6b63\u56db\u9762\u4f53\u6269\u5c55\u4e3a\u6b63\u65b9\u4f53\u7684\u5916\u63a5\u7403\u4e0e\u5706\u67f1\u7403\u76f8\u540c\uff0c\u6b63\u65b9\u4f53\u7684\u5bf9\u89d2\u7ebf\u5c31\u662f\u5916\u63a5\u7403\u7684\u76f4\u5f84\uff0c\u6240\u4ee5\u6b63\u56db\u9762\u4f53\u7684\u68f1\u957f\u4e3a\uff1a263\uff1b(263)2\uff1d2?2cos\u2220AOBcos\u2220AOB\uff1d?13A\u4e0eB\u4e24\u70b9\u95f4\u7684\u7403\u9762\u8ddd\u79bb\u4e3a\uff1a1\u00d7arccos\uff08-13\uff09=arccos\uff08-13\uff09\u6545\u9009C\uff0e
解：如图，（1）因为球O的半径为1，B、C两点间的球面距离为| π |
| 3 |
点A与B、C两点间的球面距离均为
| π |
| 2 |
| π |
| 3 |
| π |
| 2 |
（2）因为BC=1，AC=AB=
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