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Two friends A and B set out for a morning walk around the Building on the defined Walking Path starting from the same point.

A will finish his walk when he completes 10 rounds.

A walks twice as fast as B.

If they walk in the opposite direction how many times they cross each other when A finishes his walk ?

If they walk in the same direction how many times A overtakes B before he finishes his walk.

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  • 2 months later...

There is a formula for this, it is the same one used to find synodic periods for planets etc. (for example, how long it takes between Earth and Mars passing each other as they orbit the Sun.)

The formula is T = 1/(1/p1-1/p2)

Where p1 id the time it takes for one plane to complete is orbit and p2 is the time for the other planet to complete is orbit.

In this case we are not given times, so we will simply give B a complete circuit walking time of 1  time unit and thus A has a complete circuit walking time of 1/2 time unit.

Thus for when they are walking in the same direction T = 1/(1/(1/2)-1/1) - 1/(2-1) =1

Meaning that A passes B every time B has completed a complete circuit. This works out to 5 times if you count the last time they meet up again at the end of the walk (4 if you don't.

For when they walk opposite of each other you make one of the times negative, giving T=1/(1/(1/2)-1/(-1)) = 1(2+1) =1/3

A and B meet when B has completed each 1/3 of his complete loop, and since B makes 5 complete loops, they meet 15 times if you count the last meeting ( 14 if not).

So Give Jon O'Starr a star.

 

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