Achilles and the Gods Puzzle

[pdurrant]

The gods have heard rumours* that Achilles might not be as fast at running as they thought, so they decide to set him a task.

He must run a race on a special 100m racetrack. The special thing about this racetrack is that every time Achilles has run 10m along the race track, the whole track instantly stretches, as if made of rubber, by 100m.

Now Achilles can run indefinitely as a constant 10m/s. He really is a fast runner — good endurance too.

Have the gods set Achilles an impossible task? Can he ever finish the race?

Please give a reason for your answer, and don't forget to put your answer in

Spoiler:

spoiler tags.

*something about Achilles not being able to catch up to a tortoise.

This puzzle is adapted from the puzzle at http://www.smart-kit.com/s3222/jack-the-beanstalk/

[clarknova]

Originally Posted by pdurrant

The gods have heard rumours* that Achilles might not be as fast at running as they thought, so they decide to set him a task.

He must run a race on a special 100m racetrack. The special thing about this racetrack is that every time Achilles has run 10m along the race track, the whole track instantly stretches, as if made of rubber, by 100m.

Now Achilles can run indefinitely as a constant 10m/s. He really is a fast runner — good endurance too.

Have the gods set Achilles an impossible task? Can he ever finish the race?

Spoiler:

No, it's not impossible. Yes, he finishes the race 3 hours, 26 minutes, and 7 seconds after he starts.

At first I thought it was impossible, since for every 10m he ran, the track became 100m longer, thus he would be 90 meters further from the finish line every second. Then I re-read, and realized you said "the whole track stretches" which would mean it stretches with him on it. So after 1 second, he'd have run 10m, and then the track would stretch to 200m, and he would now be standing at the 20m mark (10% there). Another second, the track would stretch to 300m, and he'd now be at the 45m mark (15% there), after another second the track would be 400m, and he'd be at the 73 1/3m mark (18 1/3% there). Thus, he's slowly gaining.

/* achilles.c */
#include <stdio.h>

int main() {
  long track = 100;  // track length
  double distance = 0;  // position that Achilles is on the track.
  const int stretch = 100;  // stretch length
  const int speed = 10;  // Achilles speed in m/s
  int seconds = 0;

  while ((long)distance < track) {
    seconds++;
    distance += speed;
    distance += (distance / track) * stretch;
    track += stretch;
    printf("Running Time: %dsec / Achilles' Position: %fm / Track Length: %ldm\n",seconds,distance,track);
  }
  return(0);
}

[obs20]

Spoiler:

If the track is a straight line then he will gain 10 for every second and lose 50. If at the end of the first second he has run 10m the gods have stretched the 50m track ahead of him and 50m behind him. Thus after 1 second he has 140 meters to go and 55 meters behind him. After 2 seconds he will have 180 ahead of him.
I'm still working on the calculations.

[Sparrow]

Originally Posted by obs20

Spoiler:

... If at the end of the first second he has run 10m the gods have stretched the 50m track ahead of him and 50m behind him.

Spoiler:

Why would it be 50m each side of Achilles, wouldn't the track be stretched further in front of him?

[pdurrant]

Originally Posted by obs20

Spoiler:

If the track is a straight line then he will gain 10 for every second and lose 50. If at the end of the first second he has run 10m the gods have stretched the 50m track ahead of him and 50m behind him. Thus after 1 second he has 140 meters to go and 55 meters behind him. After 2 seconds he will have 180 ahead of him.
I'm still working on the calculations.

Spoiler:

I disagree with your assumption that the stretching of the track occurs equally in front and behind Achilles. Take a rubber band. Tie a piece of string around it, 1/10 of the way along. Stretch the band to twice the original length. How far along the band is the piece of string?

[clarknova]

To be more specific:

Spoiler:

Achilles spends 3 hours, 26 minutes, and 6.4681167 seconds (approx) to finish the race (12366.4681167 seconds).
He runs a total of 123.604681167 km (approx).
The track is 1,236.7 km long when he finishes.

[obs20]

Originally Posted by pdurrant

Spoiler:

I disagree with your assumption that the stretching of the track occurs equally in front and behind Achilles. Take a rubber band. Tie a piece of string around it, 1/10 of the way along. Stretch the band to twice the original length. How far along the band is the piece of string?

I see your point.

Spoiler:

He moves with the track

[HarryT]

Spoiler:

As long as the track stretches taking Achilles with it then it's clear that he will finish the race. Let us suppose that he is x% along the track when it stretches. He is still x% along it after the stretch, and every step he takes will increase "x", stretch or no stretch. The fact that the stretch is arithmetic rather than geometric ensures that he will complete the race in a finite time (which I'm afraid I can't be bothered to work out).

[LazyScot]

Spoiler:

Yes, he'll finish.

Observation 1

When the track stretches by 100m, he moves with it, so is position on the track as fraction of the total track stays the same. For example, after the first second he has travelled 10 meters, or one tenth of the track. When it expands, he is still at one tenth of the track.

Observation 2
Each second, the distance he runs, expressed as a fraction of the overall track length gets smaller. Mathematically for second n, he travels 1/ (10 *n). For example, for the first second, he travels 1/10 of the track, for the next second he travels 1/20 of the track, and so on.

Conclusion
At any time n, his position on the track is the sum of 1/10 + 1/20 + 1/30 ... + 1/(10*n).

This is equivalent to 1/10 * ( 1/1 + 1/2 + 1/3 + ... 1/n).

(1/1 + 1/2 + 1/3 + ... 1/n) is a sequence which is known to diverge, i.e. continually grow larger as n increases, and so any value will eventually be reached (google/wikipedia on Harmonic sequences).

[Shayne Parkinson]

Spoiler:

Yes, he'll get there eventually. Because the track stretches equally along its length, the percentage Achilles has covered is not changed by each stretch. After a stretch he then runs another 10 metres, increasing his percentage covered, before the track stretches again.

After one second, he's run 10m, which is 10% of the track. The track then stretches to 200m, so the son of Thetis is now at the 20m mark (10 * (200/100)), still at 10%.

After two seconds, he's at 30m, 15% of the track. The track stretches to 300m, taking Achilles to 45m (30 * (300/200)), still 15%.

After three seconds, 18.33%.

After four seconds, 20.83%.

And so on, gradually approaching 100%.

[pdurrant]

Originally Posted by clarknova

Spoiler:

No, it's not impossible. Yes, he finishes the race 3 hours, 26 minutes, and 7 seconds after he starts.

Spot on! Congratulations.

[pdurrant]

Originally Posted by HarryT

Spoiler:

As long as the track stretches taking Achilles with it then it's clear that he will finish the race. Let us suppose that he is x% along the track when it stretches. He is still x% along it after the stretch, and every step he takes will increase "x", stretch or no stretch. The fact that the stretch is arithmetic rather than geometric ensures that he will complete the race in a finite time (which I'm afraid I can't be bothered to work out).

The right answer, but I'm not convinced that you've properly shown thar it actually is the right answer.

Spoiler:

For instance, if, instead of extending it 100m, would Achilles finish if they extended it by 100m, 200m, 300m, etc,? After all, he'll still be increasing the percentage of the track he's covered all the time.

[pdurrant]

Originally Posted by LazyScot

Spoiler:

Yes, he'll finish.

Observation 1

When the track stretches by 100m, he moves with it, so is position on the track as fraction of the total track stays the same. For example, after the first second he has travelled 10 meters, or one tenth of the track. When it expands, he is still at one tenth of the track.

Observation 2
Each second, the distance he runs, expressed as a fraction of the overall track length gets smaller. Mathematically for second n, he travels 1/ (10 *n). For example, for the first second, he travels 1/10 of the track, for the next second he travels 1/20 of the track, and so on.

Conclusion
At any time n, his position on the track is the sum of 1/10 + 1/20 + 1/30 ... + 1/(10*n).

This is equivalent to 1/10 * ( 1/1 + 1/2 + 1/3 + ... 1/n).

(1/1 + 1/2 + 1/3 + ... 1/n) is a sequence which is known to diverge, i.e. continually grow larger as n increases, and so any value will eventually be reached (google/wikipedia on Harmonic sequences).

Very good indeed — the right answer, and a good proof that it's correct.

[pdurrant]

Originally Posted by Shayne Parkinson

Spoiler:

Yes, he'll get there eventually. Because the track stretches equally along its length, the percentage Achilles has covered is not changed by each stretch. After a stretch he then runs another 10 metres, increasing his percentage covered, before the track stretches again.

After one second, he's run 10m, which is 10% of the track. The track then stretches to 200m, so the son of Thetis is now at the 20m mark (10 * (200/100)), still at 10%.

After two seconds, he's at 30m, 15% of the track. The track stretches to 300m, taking Achilles to 45m (30 * (300/200)), still 15%.

After three seconds, 18.33%.

After four seconds, 20.83%.

And so on, gradually approaching 100%.

Yes, the right answer, but your reasoning is not sufficient.

Spoiler:

For instance, if, instead of extending it 100m each time, would Achilles finish if the gods extended it by 100m, 200m, 300m, etc,? After all, he'll still be increasing the percentage of the track he's covered all the time. — 10%, 15%, 17.5%, 18.9%, 19.8%,...

[Shayne Parkinson]

Originally Posted by pdurrant

Yes, the right answer, but your reasoning is not sufficient.

Spoiler:

For instance, if, instead of extending it 100m each time, would Achilles finish if the gods extended it by 100m, 200m, 300m, etc,? After all, he'll still be increasing the percentage of the track he's covered all the time. — 10%, 15%, 17.5%, 18.9%, 19.8%,...

A fair comment