Two bugs sit at vertices $$A$$ and $$H$$ of a cube $$ABCDEFGH$$ with edge length $$4\sqrt{110}$$ units. The bugs start moving simultaneously along $$AC$$ and $$HF$$, with the speed of the first bug twice that of the other one. The shortest distance between the bugs is

Let the slower bug travel a fraction $$t$$ of $$HF$$, so before the faster bug reaches $$C$$ it travels a fraction $$2t$$ of $$AC$$, where $$0\leq t\leq\frac12$$. With suitable cube coordinates, the squared separation divided by the edge length squared is $$(3t-1)^2+t^2+1=10t^2-6t+2$$. This is minimised at $$t=\frac{3}{10}$$ with value $$\frac{11}{10}$$, so the minimum distance is $$4\sqrt{110}\sqrt{\frac{11}{10}}=44$$.