When will she pass me for the first time? [solution]

Recently, my dad posed the following question here:

My wife and I walk on a circular track, starting at the same point.  She does m laps in the time that it takes me to do n laps.  She walks faster than I do, so m > n.  After how many laps will she catch up with me again?

If you haven’t solved it yet, give it a crack. It’s a fun problem that has surprising depth.

Here’s my solution (in it, I refer to “mom” rather than “my wife” for obvious reasons!):

Since mom’s lap rate is m laps per unit time, and dad’s lap rate is n laps per unit time, in time t, mom goes mt laps and dad goes nt laps.

They meet whenever their distance (measured in laps) is separated by an integer number of laps k. That is, mom and dad meet when

mt=nt+k, k\in\mathbb{Z}.

This happens at time

t=\frac{k}{m-n}.

Mom will have gone

mt=\frac{mk}{m-n}

laps and dad will have gone

nt=\frac{nk}{m-n}

laps when they meet for the kth time.

And that’s it! That’s the general solution. This means that:

  • At time t=0, dad and mom “meet” because they haven’t even started walking at all (they are k=0 laps apart).
  • At time t=\frac{1}{m-n}, dad and mom meet for their first time after having started walking (they are k=1 lap apart). This is the answer to the problem as it was originally stated. Mom will have gone mt=\frac{m}{m-n} laps and dad will have gone nt=\frac{n}{m-n} laps when they meet for the first time.
  • At time t=\frac{2}{m-n}, dad and mom meet for their second time (now k=2 laps apart).
  • At time t=\frac{k}{m-n}, dad and mom meet for their kth time.

Here are two examples:

  • If mom walks 15 laps in the time it takes dad to walk 10 laps, when they meet up for the first time, mom will have gone \frac{m}{m-n}=3 laps and dad will have gone \frac{n}{m-n}=2 laps.
  • If mom walks 12 laps in the time it takes dad to walk 5 laps, when they meet up for the first time, mom will have gone \frac{m}{m-n}=1\frac{5}{7} laps and dad will have gone \frac{n}{m-n}=\frac{5}{7} laps.

Boom! Problem solved! 🙂

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