存在,
因为A1B1⊥OO1,C1O⊥A1B1,
所以A1B1⊥面OO1C1C,
所以A1B1⊥C1O,
因为B1M⊥C1O,
所以C1O⊥面A1B1M,
所以C1O⊥O1M,
如图,依题意可知O1C1=√2,OO1=2,
△OO1C1∽△O1C1M,
所以OO1/O1C1=O1C1/C1M,
解得C1M=1,.
存在,
因为A1B1⊥OO1,C1O⊥A1B1,
所以A1B1⊥面OO1C1C,
所以A1B1⊥C1O,
因为B1M⊥C1O,
所以C1O⊥面A1B1M,
所以C1O⊥O1M,
如图,依题意可知O1C1=√2,OO1=2,
△OO1C1∽△O1C1M,
所以OO1/O1C1=O1C1/C1M,
解得C1M=1,.