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Tortoise and Hare are racing.

Tortoise moves one meter every minute.

Hare moves $2n − 1$ meters on the nth minute (which is equivalent to $n^2$ meters in the first minutes).

Prove that no matter how many meters ahead Tortoise starts, Hare will catch up given enough time.

This is my idea... Let $c$ be the head start distance tortoise gotten in meters. Let $z$ be the distance moved at nth minute.

For tortoise, @nth minute, $c$ + $z$. So e.g., after 10 minutes with 5$m$ head start, tortoise has moved 5$m$ + 10$m$ = 15$m$.

For hare, now I'm a bit confused.. how does $2n-1$ = $n^2$ meters in the first nth minute? This is how I picture it. At the 10th minute, hare moved 2(10)-1 = 19$m$ which means he COMPLETED 19metres at 10th minute but is COMPLETING the 20th meter at 10 minutes. But as for first nth minute, let's say $n$ = 10, he's already COMPLETED 20metres.

Given n minutes, $$z^2 > c + z$$ where z^2 is distance moved by hare after n minutes.

Then we can have $$z^2 - z > c = z(z-1) > c$$

The goal is so that I can choose ANY c and that after a certain time, the hare will catch up.

Say $c$ = 20 $n$ = 5, 5(5-1) > 20

Now, how could I prove it formally? Note, the class being taught up until now has been basic proofs (up to contradiction), now just starting on induction and sequences.

Don
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misheekoh
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  • Who was the intended audience? As the hare's speed is increasing and tortoises is constant, once you know the hare's speed will ever be greater than the tortoise then you know the hare will eventually win. There's really nothing to prove. – fleablood Nov 11 '15 at 03:53

2 Answers2

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The hare completes $2n - 1$ meters, counting from the beginning of the $n$th minute (in other words, the end of the $n-1$th minute) to the end of the $n$th minute. So, at the end of the first minute, it has gone 1 meter; between the beginning and the end of the second minute, it has gone 3 meters; and so on. So, at the end of two minutes (counting from the very start), it has gone $3+1=4=2^2$ meters. You can prove that this is $n^2$ inductively or in any other manner (see this Math.SE question, for example).

We want to show that for some head start $c$ and some number of minutes $k$, $k+c<k^2$. For $c=1$, $k$ can be anything greater than 1, and for $c=2$, anything greater than 2. For $c>2$, you can verify that $k=c$ satisfies the inequality.

Community
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shardulc
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Why make life difficult? In minute 3 the hare's speed is faster than the tortoise and it never becomes less than the tortoise again. Faster things always surpass slow things eventually.

fleablood
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  • "The hare's speed...": to make this rigorous you'd have to use calculus, which... well, doesn't make life difficult, but makes it more difficult than required :) – shardulc Nov 11 '15 at 03:59
  • gotta prove it formally.. – misheekoh Nov 11 '15 at 04:03
  • Had the puzzle been when does the hare pass the tortoise, I would have been more charitable but it asked to come up with a proof that the hare would catch up... – fleablood Nov 11 '15 at 04:06
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    Formally? let d be distance apart at 3 minutes. Let h_3 be hare's speed (> 1) at minute 3. Let t be minutes since minute 3. Let h(t) be distance traveled by rabble in minutes since minute 3. Then t + d equals position of tortoise aft t minutes past minute 3. h(t) > h_3*t > d + t whenever t > d/(h_3 - 1). – fleablood Nov 11 '15 at 04:24