| The Running Dog |
| "Yes, when I take my dog for a walk," said a mathematical friend, "he frequently supplies me with some interesting puzzle to solve. One day, for example, he waited, as I left the door, to see which way I should go, and when I started he raced along to the end of the road, immediately returning to me; again racing to the end of the road and again returning. He did this four times in all, at a uniform speed, and then ran at my side the remaining distance, which according to my paces measured 27 yards. I afterwards measured the distance from my door to the end of the road and found it to be 625 feet. Now, if I walk 4 miles per hour, what is the speed of my dog when racing to and fro?" |
Problem Moderated by: Graeme |
| Problem Solution |
The first thing I notice when I start to solve this problem is that the distance traveled by the man and dog combined by the time of their first reunion is 1250, the distance from the house to the end of the road and back. Let's say the dog travels "f" times as fast as the human. Let d1 be the distance the dog travels in his first sortie, and h1 be the distance traveled by the human.
Solving these two equations for d1 and h1, we get
d1=(1250)(f)/(f+1)
h1=(1250)(1)/(f+1)
Now that the dog and the human are together again, the puzzle becomes a whole new game. The distance to the end of the road and back is now 1250-2h1, and that's the distance traveled by the dog plus the human during the dog's second sortie. Let d2 be the distance traveled by the dog in his second excursion, and h2 be the distance traveled by the human.
d2+h2 = 1250-2h1
d2+h2 = 1250-(1250)(2)/(f+1)
d2+h2 = (1250)(f-1)/(f+1)
d2/h2=f
Solving these, we get
d2=(1250) (f-1)/(f+1) (f)/(f+1)
h2=(1250) (f-1)/(f+1) (1)/(f+1)
Now the dog and human are together again after two dog-trips to the end of the road. Once again, it's a whole new game, where the distance to the end of the road and back is now 1250-2h1-2h2. Using the same logic once again, we get
d3+h3 = (1250) (f-1)2/(f+1)2
d3/h3 = f
By now, I'm sure you have spotted the pattern. Using a recursive proof, it's not hard to show that during the nth sortie of the dog, the distance covered by the dog plus the human, dn+hn is given by this formula:
dn+hn = (1250) (f-1)(n-1)/(f+1)(n-1)
The distance to the end of the road and back after the dog has made four sorties is given in the problem as twice 27 yards, or 162 feet. This is the same as the distance that would have been covered by dog and man combined during the fifth sortie,
d5+h5 = (1250) (f-1)4/(f+1)4 = 162
Solving this for (f-1)4/(f+1)4, we get the following:
162/1250 = (f-1)4/(f+1)4
81/625 = (f-1)4/(f+1)4
(81/625)(1/4) = (f-1)/(f+1)
3/5 = (f-1)/(f+1)
f = 4
Now we know the dog runs four times as fast as the man. The statement of the problem tells us the man goes 4 miles an hour. So the dog goes 16 miles an hour.
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