球心O在高线DE上,OA=OD=OB=OC=R,设正四面体的棱长为a,
则AE= √3a/3,DE= √6/3a,
在直角三角形AOE中,AO2=OE2+AE2,且AO+OE= √6/3a
解得a= 2/3√6AO= 2/3√6R
故答案为：2/3√6R．