Whiplash!

I have an alternative way of thinking about it.

I know that $\omega =2\pi f\quad ,\quad \omega =\frac { 2\pi }{ T } ,\quad T=\frac { 2\pi }{ \omega }$

I also know that $v=\omega r,\quad \omega =\frac { v }{ r }$

Rearranging I get the following:

For ${ T }_{ 1 }=\frac { 2\pi }{ \frac { v }{ r } } ,\quad { T }_{ 1 }=\frac { 2\pi r }{ v }$ .

As the motion progresses:

${ T }_{ 2 }=\frac { 2\pi \cdot \frac { 2 }{ 3 } r }{ v } ,{ T }_{ 3 }=\frac { 2\pi \cdot \frac { 1 }{ 3 } r }{ v } ,$

Here 'T' indicated the time period for a full circular path. However for T(1) only 1/12 of the period is completed and for T(2) and T(3) only 1/6 are completed. Hence

${ T }_{ 1 }=\frac { \pi r }{ 6v } ,{ T }_{ 2 }=\frac { 2\pi r }{ 9v } ,{ T }_{ 3 }=\frac { \pi r }{ 9v }$

$T=\frac { 1.5\pi r }{ 9v } +\frac { 2\pi r }{ 9v } +\frac { \pi r }{ 9v } =\frac { 9\pi r }{ 18v } =\frac { \pi r }{ 2v }$

Following rearrangement into the form of the answer, and because we know that 'v' is constant and clearly the length of 'r' decreases by a third for each edge 'completed':

$\frac { ({ T }_{ 2 })(T) }{ ({ T }_{ 1 }){ (T }_{ 3 }) } =\frac { \frac { 2\pi r }{ 9v } \cdot \frac { \pi r }{ 2v } }{ \frac { \pi r }{ 6v } \cdot \frac { \pi r }{ 9v } } =\frac { \frac { 2{ \pi }^{ 2 }{ r }^{ 2 } }{ 18{ v }^{ 2 } } }{ \frac { { \pi }^{ 2 }{ r }^{ 2 } }{ 54{ v }^{ 2 } } } =\frac { 108{ \pi }^{ 2 }{ r }^{ 2 }{ v }^{ 2 } }{ 18{ \pi }^{ 2 }{ r }^{ 2 }{ v }^{ 2 } } =6$

In this case the solution is once again 6. :)

Only one correction: the statement "The linear velocity '' of the free end of the rope will be the same throughout since no force is acting on it in that direction" doesn't seem right to me as this is not a dynamics problem. Linear velocity remains same because it is given so, not because of absense of some force. Nice solution otherwise. :)