Problem of the Week
Problem B
A-Maze-ing Mice!
Problem
Two mice, Charlie and Priscilla, run at the same speed in a small maze with two loops, as
shown below.
They can run counter-clockwise around loop 1 in
10 s, and clockwise around loop 2 in 4 s.
They start together from S, where the corners of
loop 1 and loop 2 meet, at time t = 0 s, with Charlie
running on loop 1 and Priscilla on loop 2.
a) If Charlie continues to run around loop 1, and Priscilla continues to run around loop 2,
when will they first meet again at S? When will the next meeting occur?
b) Suppose that, whenever they meet at S, they get confused and switch to the other loop,
continuing at the same speed. In which loop will Charlie be after 32 s? After 45 s?
c) If they continue as in part b), is the length of time between their meetings at S always
the same? Explain.
Solution
a) The table on the right gives the times
for both Charlie and Priscilla after each
of 5 laps. To meet at S, they must
have matching times at the end of a lap.
Clearly they will first meet at S after 20
seconds, and will continue to meet every
20 seconds thereafter, since Priscilla always runs 5 laps (in 20 seconds) around
Loop 2 while Charlie runs 2 laps (in 20
seconds) around Loop 1.
Laps C’s Time
P’s Time
1
10
4
2
20
8
3
30
12
4
40
16
5
50
20
b) The table on the right shows the times
at which Charlie and Priscilla complete
laps, assuming they switch loops whenever they meet at S. Their locations are
indicated by L1 for Loop 1, L2 for Loop
2, and S when they meet. Thus we see
that after 32 seconds, Charlie is in Loop
2, and after 45 seconds, he is in Loop 1.
c) The table entries indicate that Charlie
and Priscilla continue to meet every 20
seconds. This is due to the fact that
they both run at the same speed, so
it’s as if they did not switch loops upon
meeting at S, but just continued in the
same loop as in part a).
That they meet every 20 seconds is due
to the fact that Loop 1 takes either
mouse 10 seconds, while Loop 2 takes
four seconds. Thus whichever mouse is
running in Loop 2 can do five laps while
the other mouse does two laps of Loop
1. Since the smallest number that is a
whole number (of laps) times both 4 and
10 is 20, they meet every 20 seconds.
Laps C’s Time
P’s Time
1
10 (L1)
4 (L2)
2
20 (S)
8 (L2)
3
24 (L2)
12 (L2)
4
28 (L2)
16 (L2)
5
32 (L2)
20 (S)
6
36 (L2)
30 (L1)
7
40 (S)
40 (S)
8
50 (L1)
44 (L2)
9
60 (S)
48 (L2)
10
64 (L2)
52 (L2)
11
68 (L2)
56 (L2)
12
72 (L2)
60 (L2)