（2008?天河区二模）如图1，已知长方体的长为BC=2cm，宽AC=1cm，高AA′=4cm．（1）一只蚂蚁如果沿长方体 已知长方体的长AC为2cm，宽BC为1cm，高AA'为4cm
\u5982\u56fe\uff0c\u5df2\u77e5\u957f\u65b9\u4f53\u7684\u957f\u4e3aAC=2cm\uff0c\u5bbdBC=1cm\uff0c\u9ad8AA\u2032=4\uff0e\u4e00\u53ea\u8682\u8681\u5982\u679c\u6cbf\u957f\u65b9\u4f53\u7684\u8868\u9762\u4eceA\u70b9\u722c\u5230B\u2032\u70b9\uff0c\u90a3\u4e48\u6cbf\u5982\u56fe\uff1a\u6839\u636e\u9898\u610f\uff0c\u5982\u4e0a\u56fe\u6240\u793a\uff0c\u6700\u77ed\u8def\u5f84\u6709\u4ee5\u4e0b\u4e09\u79cd\u60c5\u51b5\uff1a\uff081\uff09\u6cbfAA\u2032\uff0cA\u2032C\u2032\uff0cC\u2032B\u2032\uff0cB\u2032B\u526a\u5f00\uff0c\u5f97\u56fe\uff081\uff09AB\u20322=AB2+BB\u20322=\uff082+1\uff092+42=25\uff1b\uff083\u5206\uff09\uff082\uff09\u6cbfAC\uff0cCC\u2032\uff0cC\u2032B\u2032\uff0cB\u2032D\u2032\uff0cD\u2032A\u2032\uff0cA\u2032A\u526a\u5f00\uff0c\u5f97\u56fe\uff082\uff09AB\u20322=AC2+B\u2032C2=22+\uff084+1\uff092=4+25=29\uff1b\uff082\u5206\uff09\uff083\uff09\u6cbfAD\uff0cDD\u2032\uff0cB\u2032D\u2032\uff0cC\u2032B\u2032\uff0cC\u2032A\u2032\uff0cAA\u2032\u526a\u5f00\uff0c\u5f97\u56fe\uff083\uff09AB\u20322=AD2+B\u2032D2=12+\uff084+2\uff092=1+36=37\uff1b\uff082\u5206\uff09\u7efc\u4e0a\u6240\u8ff0\uff0c\u6700\u77ed\u8def\u5f84\u5e94\u4e3a\uff081\uff09\u6240\u793a\uff0c\u6240\u4ee5AB\u20322=25\uff0c\u5373AB\u2032=5cm\uff0e\uff081\u5206\uff09
\u4f60\u753b\u4e2a\u957f\u65b9\u4f53\u7684\u5c55\u5f00\u56fe\uff0c\u7136\u540e\u628a\u4e24\u70b9\u76f4\u7ebf\u76f8\u8fde\uff0c\u6839\u636e\u4e09\u89d2\u7b97\u51fa
解答：解：（1）根据题意，如下图所示，最短路径有以下三种情况：
①沿AA′，A′C′，C′B′，B′B剪开，得图（1）AB′2=AB2+BB′2=（2+1）2+42=25，
②沿AC，CC′，C′B′，B′D′，D'A'，A′A剪开，得图（2）AB′2=AC2+B′C2=12+（4+2）2=37，
③沿AD，DD′，B′D′，C′B′，C′A′，AA′剪开，得图（3）AB′2=AD2+B′D2=22+（4+1）2=29，
综上所述，最短路径应为（1）所示，
所以AB′2=25，
即AB′=5cm，
答：最短路径为（1）所示5cm；
（2）要保证底面圆最大，必须使得圆与长方形的两条长边相切，则此时圆的半径长为
| 1 |
| 2 |
当圆与A′C′相切时，∠A′0C′最大，此时为90°；
当圆与B′D′相切时，E为切点，∠A′0C′最小，此时，A′E=
| 1 |
| 2 |
| 3 |
| 2 |
| 1 |
| 3 |
所以∠A′0E≈18.4°，∠A′0C′≈36.8°，
∠A′OC′的度数范围为36.8°≤∠A′OC′≤90°．
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