Friday, May 30, 2014

An Olympic Torch Puzzle

Two people are running around the Olympic stadium in the same direction. One of them is a much faster runner than another. They have only one torch and they pass this torch to each other every time one of them is crossing another. They start from the same spot and the faster runner holds the torch at the start. At the end f the race, the faster runner made 50 loops, the slower 35. How many loops made the torch?

Hint: assume that instead of one torch they have another item (say a newspaper) that is starting in the hands of a slower runner and is being exchanged for a torch every time the runners pass each other.

This puzzle came from Israeli newspaper Haaretz.

The answers are accepted any time until midnight Eastern Time on Sunday, on our Family Puzzle Marathon

Jerome said...

I have no idea if I'm doing this correctly, but I'll make a guess.

These runners will meet after a common denominator or times have passed. After the tenth time for the faster runner, the slower runner will have completed 7. The table goes as follows

Faster .... Slower
0 . . . . . . . 0
10 . . . . . . 7
20 . . . . . . 14
30 . . . . . . 21
40 . . . . . . 28
50 . . . . . . 35

If this is correct and the torch is not passed when they start then the distance traveled is 10 + 7 + 10 + 7 + 10 = 44 times around the track..

I have the runners meeting at the starting place. I'm not sure if they meet before the beginning.

Anonymous said...

I spent almost an hour trying to solve this one and I think the torch makes 45 loops.
Had the torch been held by the slower runner at the start it would make 40 loops.

Lulu

Anonymous said...

During the time period that it takes for the fast runner to do 50 laps around the track, & for the slow runner to do 35 laps around the track:
* the torch will switch from one runner to another 14 times (& maybe a 15th time just as time runs out and the fast runner completes their 50th lap).
* the torch will be carried a distance of between 42 and 43 laps

The torch first starts with the fast runner.
The fast runner passes the slow runner & the torch switches when the fast runner has gone an estimated 3 & 1/3 laps around the track & the slow runner has gone 2 & 1/3 laps.

The fast runner next passes the slow runner & the torch switches again when the fast runner has gone an estimated 6 & 2/3 laps around the track & the slow runner has gone 4 & 2/3 laps.
and so on.

-TracyZ

Maria said...

This was a hard one, I know. But the answer is actually simpler than it seems.
Why didn't anyone use my hint?? You still got it though:)

Take the hint and imagine another item (a newspaper) being passed in exchange for a torch. Both items start at the same position and also end at the same time in the same position. So, together they traveled 50+35=85 loops, and each traveled a whole number of loops. Too bad, because I first thought we could just divide 85 by 2 = 42.5

The torch and the newspaper were exchanged so many times, so they must have traveled very similar distances. Probably 42 and 43 as you all said. How to prove it?

First time the faster runner passes the slower, the torch had traveled 1 more loop in his hands than the newspaper in the hands of the slow runner. Next time the faster runner passes the slower, newspaper in his hands in catching up with the torch in travel distance. Third time it happens - the torch is again ahead by 1 loop. Forth time - newspaper catches up again. So, every time the runners meet the torch is either traveled 1 more loop than the newspaper or the same distance.

Same distance is not a good answer for us since 85/2=42.5 and they have to travel whole number of loops. So, the torch must be 1 loop ahead.

Beautiful, right?

Puzzle point for everyone who dared to answer this one.

Jerome said...

I don't see how you know that there is a 1 loop difference, but I know it is true. Taking Tracy's 3 1/3 2 1/3 starting distance which does have a difference of 1 loop and going through the loops until the fast runner gets to 50 loops gives 42.999998 loops for the torch which is just my sloppy computer language unable to round correctly.

What I would like to know is how was 3 1/3 and 2 1/3 found?

Maria said...

Jerome - forget about numbers. Imagine two runners starting at the same position and running in the same direction. When faster runner is passing the slower he has made one more loop than the slower one. Right? The torch in his hand made this extra loop as well.

Jerome said...

I can't help but work with the numbers. It bothers me greatly that a physical problem cannot be represented by an equation. And thank you Maria for posting your solution. The key to this is understanding that there has to be 1 loop difference between the fast and slow runner. If it is less or more, they simply will not meet.

The next thing you have to do is realize that the time taken to make that loop difference must be the same for both runners every time they do meet. So we can calculate their first meeting very simply and then add that amount for the 15 times that the meet to each meeting.

To calculate the first meeting

The number of loops for the slow runner is x
The time for the slow runner is t
The rate of travel is 35 loops per time

The number of loops for the fast runner is x + 1
The time traveled is the same as for the slow runner which is t
The rate of travel is 50 loops per time.

t = d/r

x/35 = (x + 1)/50 Cross multiply
50x = 35x + 35
15x = 35
x = 35/15
x = 2 and 1/3

So x + 1 is 3 and 1/3

I would sure like to know how Tracy found this result. It took me a very long time (since last Saturday) to realize it.

Anonymous said...

The answer is five loops (or rings) "made the torch", unless you're Russian, in which case four loops or rings and a snowflake made the torch.

Unknown said...

I thing, Jerome is correct. Both have completed 50 & 35 loops respectively. Hence, removing the common factor of 5; when the faster runner completes 10 loops, the slow runner completes 7 loops. At the beginning, the torch is with the faster runner. So, the torch travels for 10 loops and at the end of first phase, the torch will be with the slow runner. hence, in the second phase, the torch travels only 7 loops and so on... Finally, at the end of 5 phases, the torch might have travelled a total of 10+7+10+7+10=44. loops.

Anonymous said...

I tried this puzzle yesterday. My first quick guess was that (since this is a puzzle and likely to have a simple and clever solution) the torch would travel exactly half of the total laps run: 50 + 35 = 85 divided by 2 = 42.5. Then I figured that, alternatively, it could possibly be all 50 laps just not with the faster runner.

After my initial guesses I started figuring out some low and high figures, in order to get a range of possible answers. If we imagine that the slowest runner just sat on the start line the whole time, the faster runner would hand off the torch every other lap on the starting line and keep it on laps 1, 3, 5 etc up to 49, giving the torch 25 laps. On the other hand, if the faster runner sat out, the torch would travel 17 laps with the slower runner (2, 4, 6…34). So would the true answer lie somewhere between 17 and 25 inclusive? But then again, both runners are moving in this puzzle. So when the faster runner hands it off, the torch continues to advance with the slower runner, increasing the total. So perhaps it is additive or 25 + 17 = 42? I noticed that this result is close to my first guesstimate of 42.5.

Still unsure, I tried devising a whole bunch of rulers, charts, and spiral drawings to get a better idea of what exactly is going on. At first I found it challenging that we are not given actual lengths or times, leaving this scenario somewhat in the abstract. So I played with it some more. What if the track were actually 50 laps long? Then they would never cross paths. What if it were 5 laps long? 25 laps? Since actual times and distances are not given, I tried out several visual representations for showing the laps run. For starters I made a circle like a clock face and marked it with lap intervals (somewhat confusing). Then I drew double columns of ever-increasing spirals to represent and compare successive laps between the two runners (more helpful). I made a symmetrical chart of numbers with rows of 10 showing where the slower runner was at each completed lap of the faster runner, and circled the number of the interval on which he/she would be lapped. For the numbers given in this puzzle it totaled 15 times including when they meet again on the start/finish line at the very end of their run.

I also did a similar chart and/or spiral drawings for runners who ran at other ratios, for comparison. I found that if runner F (faster runner) runs 3 laps for every 1 by runner S (slower runner) then they would meet 2 times. With 6 laps by F and 2 laps by S, they meet 4 times. This seems to suggest a formula: (laps by F) - (laps by S) = number of encounters after the start. This would match with the figures of 50 - 35 = 15 hand-offs which was what I had come to using purely dogged calculations.

But where exactly do they meet each time? I drew a ruler and plotted points. Then some more points. For the numbers 50 and 35, you get intervals of 3.3333 and 2.3333 for runners F and S respectively. Reassuringly, when you divide 3.3333 into 50 you get 15, and 2.3333 into 35 is also 15. This makes sense! That is, with fifteen hand-offs you are dividing 50 laps into segments of 3.333 laps either with or without the torch for runner F, and 35 laps / 15 hand-offs = intervals of 2.3333 laps for runner S.

So I added up those interval-lengths using the chart from 0-50 laps for F and 0-35 laps for S. There are 8 intervals with the torch for F (since F starts and finishes with the torch) and 7 for S. That's 8 x 3.3333 and 7 x 2.3333 or 26.6666 + 16.3333 = 43.

The big epiphany I had with this puzzle came at the end, when I realized that the time interval between hand-offs would remain constant. That was not intuitively obvious to me in the beginning. In real life with real runners, you would have fatigue and other factors come into play--that's what makes a race interesting after all!

Margaret (of Margaret & Fiona)

Anonymous said...

I forgot to mention that of course 10 over 7 is the basic ration of their relative rates. So S goes 0.7 laps for every 1.0 lap by F. Those are the numbers I used for my charts.

You can use the same formulas I mentioned above to get your basic facts:

10 - 7 = 3 hand-offs
then 10/3 = intervals between hand-offs of 3.3333 laps for F
and 7/3 = intervals between hand-offs of 2.3333 laps for S

Multiplying laps and hand-offs by 5 gets you back to the numbers in the original puzzle, but of course the interval lengths between hand-offs remain the same since they are based on a ratio.

Margaret (of Fiona & Margaret)

Anonymous said...

Sorry Math Mom, I didn't proofread my last post and there were several typos. Could you post this version instead?
**

I forgot to mention that of course 10 over 7 is the basic ratio of their relative rates. So S goes 0.7 laps for every 1.0 lap by F. Those are the numbers I used for my charts.

You can use the same formulas I mentioned in my previous post to get your basic facts:

10 - 7 = 3 hand-offs
then 10/3 = interval between hand-offs for F of 3.3333 laps
and 7/3 = interval between hand-offs for S of 2.3333 laps

Multiplying laps and hand-offs by 5 gets you back to the numbers in the original puzzle, but of course the interval lengths between hand-offs remain the same since they are based on a ratio.

Margaret (of Fiona & Margaret)

Anonymous said...

Another way of understanding this puzzle is that, because the time interval between hand-offs remains exactly constant, each runner spends half their time with the torch, and the torch spends half its time with each runner.

There is a small irregularity in the result based on the circumstances that the number of intervals between hand-offs is an odd number (15) and the torch starts with runner F, which is what accounts for the half-lap increase from 42.5 to 43.

Margaret again :)

Anonymous said...

Maria, your hint about the newspaper was helpful to me in a tangential way. It got me thinking about the sum total of all the laps run.

Jerome, I don't know for sure how TracyZ got the figures 2 & 1/3 and 3 &1/3 but she said "an estimated 3 & 1/3 laps around the track" so my guess is that she eyeballed it using some rough calculations. This is what I did too except that I actually did some arithmetic going a few places out (using a self-drawn ruler and spiral loop drawings to help my intuition, and adding distances traveled by each runner in increments) and saw that it would end up as 3.3333 (a repeating number) for F. Then when the other parts of the math came together, it made sense because 50 laps / 15 hand-offs makes 3 & 1/3 laps, and 35 laps /15 hand-offs = 2 & 1/3 laps.

Finally, in retrospect it makes sense to me that you would take the difference between the number of laps each runner can go to get the number of hand-offs (50-35 = 15), because logically, if runner F is lapping runner S, he/she has to go around the track that many times more, and regardless of where it may occur, F is going to HAVE to pass runner S those 15 times, in order to rack up 50 laps.

Margaret again :)