Ships That Pass . . .
In the days when people crossed the Atlantic in passenger liners, a ship left London every day at 4.00 p.m. bound for New York, arriving exactly 7 days later.
Every day at the same instant (11.00 a.m. because of the time difference) a ship left New York bound for London, arriving exactly 7 days later.
All ships followed the same route, deviating slightly to avoid collisions when they met.
How many ships from London does each ship sailing from New York encounter during its transatlantic voyage, not counting any that arrive at the dock just as they leave, or leave the dock just as they arrive?
Eleanor Lovell
07 Dec 2009 11:24
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Tags: Mathschallenge
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Simon Whitehouse
13
At time, t=0 (t measured in days) one ship arrives in New York and another leaves. You don’t count that as an encounter.
Then at t = 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5, 5.5, 6, 6.5 the ship from New York encounters a ship coming from London. Note that it is on the half day that they meet.
This is demonstrated by the following:-
When the ship sets out from New York the closest one approaching is one day out. They travel towards each other at the same speed so they will meet when the first ship is a half day out of New York. At that point, the next closest ship approaching is a day behind the ship encountered. By the same logic as before it will encounter that ship in another half day’s time. And so on.
Am I right? More importantly, do I get a mince pie?
07 Dec 2009, 11:56
Eleanor Lovell
Simon you’re correct!
Here is Prof Ian Stewart’s explanation:
Suppose (the date doesn’t matter, but this choice makes the sums simpler) that the New York ship sets sail on 10 January. It arrives on 17 January, just as the 17 January ship from London departs.
Similarly, the ship that left London on 3 January arrives in New York on 10 January, just as the ship we’re talking about leaves.
So on the high seas, our ship encounters the ones that left London starting on 4 January and ending on 16 January. That’s 13 ships in all.
08 Dec 2009, 15:10
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