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The group will be fastest when they ride

in one group of 11. With most of the time or all the time the 3 fastest in front.

The worst time will be at most

13 minutes 40 seconds

Proof

We know that

there is a clever energy preserving rotational strategy that works for 3 cyclists with time a<b<c such that the slowest time c is reduced to (a+b+c)/3

and

larger groups should be able to preserve the same or even more energy as a result of rotational strategy

Therefore

It must work at least as good for 11 cyclists.

We might go for the time by the top three cyclists

They could finish alone in 11 minutes.

But, we can not know

at which pace the slowest cyclists will be able to follow the pack.

Yet, we do know that they can finish in

at least in 13 minutes and 40 seconds, because the slowest cyclist is able to follow at that speed in a group of three, so certainly as well in a bigger group.

The group will be fastest when they ride

in one group of 11. With most of the time or all the time the 3 fastest in front.

The worst time will be at most

13 minutes 40 seconds

Proof

We know that

there is a clever energy preserving rotational strategy that works for 3 cyclists with time a<b<c such that the slowest time c is reduced to (a+b+c)/3

and common knowledge dictates that

larger groups should be able to preserve the same or even more energy as a result of energy preserving rotational strategies.

Therefore

It must work at least as good for 11 cyclists.

We might go for the time by the top three cyclists

They could finish alone in 11 minutes.

But, we can not know

at which pace the slowest cyclists will be able to follow the pack.

Yet, we do know that they can finish in

at least in 13 minutes and 40 seconds, because the slowest cyclist is able to follow at that speed in a group of three, so certainly as well in a bigger group.

The group will be fastest when they ride

in one group of 11. With most of the time or all the time the 3 fastest in front.

The best time will be at least

13 minutes 40 seconds

Proof

We know that

there is a clever energy preserving rotational strategy that works for 3 cyclists with time a<b<c such that the slowest time c is reduced to (a+b+c)/3

Therefore

It must work at least as good for 11 cyclists.

We might go for the time by the top three cyclists

They could finish alone in 11 minutes.

But, we can not know

at which pace the slowest cyclists will be able to follow the pack.

Yet, we do know that they can finish in

at least in 13 minutes and 40 seconds, because the slowest cyclist is able to follow at that speed in a group of three, so certainly as well in a bigger group.

The group will be fastest when they ride

in one group of 11. With most of the time or all the time the 3 fastest in front.

The worst time will be at most

13 minutes 40 seconds

Proof

We know that

there is a clever energy preserving rotational strategy that works for 3 cyclists with time a<b<c such that the slowest time c is reduced to (a+b+c)/3

and

larger groups should be able to preserve the same or even more energy as a result of rotational strategy

Therefore

It must work at least as good for 11 cyclists.

We might go for the time by the top three cyclists

They could finish alone in 11 minutes.

But, we can not know

at which pace the slowest cyclists will be able to follow the pack.

Yet, we do know that they can finish in

at least in 13 minutes and 40 seconds, because the slowest cyclist is able to follow at that speed in a group of three, so certainly as well in a bigger group.

The group will be fastest when they ride

in one paceline of 11.

The top three cyclists

could finish alone in 11 minutes. But, we can not know at which pace the slowest cyclists will be able to follow the pack.

Yet, we do know that they can finish at

least in 13 minutes and 40 seconds, because the slowest cyclist is able to do that in a group of three, so certainly as well in a bigger group.

The group will be fastest when they ride

in one group of 11. With most of the time or all the time the 3 fastest in front.

The best time will be at least

13 minutes 40 seconds

Proof

We know that

there is a clever energy preserving rotational strategy that works for 3 cyclists with time a<b<c such that the slowest time c is reduced to (a+b+c)/3

Therefore

It must work at least as good for 11 cyclists.

We might go for the time by the top three cyclists

They could finish alone in 11 minutes.

But, we can not know

at which pace the slowest cyclists will be able to follow the pack.

Yet, we do know that they can finish in

at least in 13 minutes and 40 seconds, because the slowest cyclist is able to follow at that speed in a group of three, so certainly as well in a bigger group.