Solution: The Tortoise and the Hare . . . Revisited Math Puzzle

The tortoise can run the 10-km course in 10 hours. The hare can do it in 2 hours. So to tie, he has to get going when the tortoise has 2 hours left. This is when the tortoise has run 8 km (5 mi) and has 2 km (1.2 mi) to go, so the hare can sleep for 8 hours, minus a few seconds to guarantee the win.

Let’s come up with a general solution. Let’s say

T = tortoise speed
H = hare speed
R = length of the race

Let’s assume that the hare is faster.

The time for the tortoise to run the race is

R / T

The time for the hare to run the race is

R / H

The hare has to get going when time elapsed in the race is

(R / T) – (R / H)

In our example we would have

(10 / 1) – (10 / 5) = 10 – 2 = 8

For a tie, the hare needs to start after the tortoise has been running for 8 hours.

Let’s say the race is 15 km (9.3 mi) and the two animals run at the same speeds as in the example above. We have

(15 / 1) – (15 / 5) = 15 – 3 = 12

The hare can nap for 12 hours.

If the tortoise runs at 2 km/h and the hare still runs at 5 km/h, a 10-km race looks like this:

(10 / 2) – (10 / 5) = 5 – 2 = 3

With this speedier tortoise, the hare gets only a 3-hour nap.

Our formula should work for any combination of tortoise speed, hare speed, and race length. The answers we calculated are for a tie between the two competitors. The hare would have to set his alarm clock slightly ahead of the specified time to win.