Runner's track distance error problem

[walking]

Here is the official solution:

I don't understand this solution. Firstly why can the author assume wlog that slower runner ran exactly 1 mile? Secondly, if 3.3ft is indeed the max error then worst case scenario is that first runner's track was 3.3 below actual length and 2nd was 3.3 over actual length. Then first runner ran ~22.166 ft/s and 2nd runner ran 22.18 ft/s which is faster. So it seems we cannot guarantee that 1st runner was indeed faster if error = 3.3?

Actually I'll be honest: I'm not sure I fully understand the problem statement.

 

[PeroK]

View attachment 276094

Here is the official solution:

View attachment 276095

I don't understand this solution. Firstly why can the author assume wlog that slower runner ran exactly 1 mile? Secondly, if 3.3ft is indeed the max error then worst case scenario is that first runner's track was 3.3 below actual length and 2nd was 3.3 over actual length. Then first runner ran ~22.166 ft/s and 2nd runner ran 22.18 ft/s which is faster. So it seems we cannot guarantee that 1st runner was indeed faster if error = 3.3?

Actually I'll be honest: I'm not sure I fully understand the problem statement.

The first runner must have run some distance close to a mile. You could try repeating the problem assuming he ran exactly one mile and one foot or one mile less one foot and see by how much the eventual answer varies.

I can check the detail if you want, but it looks like I agree with you. The tracks can be no more than 3.3 ft different in length, so the error on each track must be less than 1.65 ft.

 

[PeroK]

Okay, so here is an alternative solution. Suppose the first track is $x$ feet shorter than a mile and the second $x$ ft longer. We want to find the value of $x$ which makes the runners' speeds equal: $$ \frac{5280 - x}{t_1} = \frac{5280 + x}{t_2} $$
If that equation holds, then the runners are equal and only the different lengths of the track distinguishes them. Solving that equation gives: $$x = 5280 \frac{t_2 - t_1}{t_2 + t_1} = 1.66 ft$$

 

[walking]

Okay, so here is an alternative solution. Suppose the first track is $x$ feet shorter than a mile and the second $x$ ft longer. We want to find the value of $x$ which makes the runners' speeds equal: $$ \frac{5280 - x}{t_1} = \frac{5280 + x}{t_2} $$
If that equation holds, then the runners are equal and only the different lengths of the track distinguishes them. Solving that equation gives: $$x = 5280 \frac{t_2 - t_1}{t_2 + t_1} = 1.66 ft$$

Great, this is exactly what I had in mind. But then I am still wondering why this turns out to be exactly half of the author's solution. How do the two solutions relate? I think that is my main problem now.

 

[PeroK]

Great, this is exactly what I had in mind. But then I am still wondering why this turns out to be exactly half of the author's solution. How do the two solutions relate? I think that is my main problem now.

His solution is the maximum difference in the lengths of the tracks. He assumed the first track was exactly one mile and all the error was in the second track.

 

[walking]

His solution is the maximum difference in the lengths of the tracks. He assumed the first track was exactly one mile and all the error was in the second track.

But how did he know that method would work beforehand? The fact that 5280 feet is large and the error we are seeking is small doesn't seem to tell me intuitively that varying the 5280 slightly will still yield a similar answer. If anything, intuitively I would be wary of small changes since I would think it might affect the error we are seeking, which is already small.

So I guess my problem is with the author's intuition and how I can get my head around it!

 

[PeroK]

But how did he know that method would work beforehand? The fact that 5280 feet is large and the error we are seeking is small doesn't seem to tell me intuitively that varying the 5280 slightly will still yield a similar answer. If anything, intuitively I would be wary of small changes since I would think it might affect the error we are seeking, which is already small.

So I guess my problem is with the author's intuition and how I can get my head around it!

Try things out as I suggested in post #2.

 

[walking]

Ok so I have not been able to motivate the author's solution unfortunately. However I have found that it only works because the times are very close to each other. If the times were very different then letting one of the errors be 0 and only focusing on the error in the other track wouldn't lead to the same answer either way. The only reason the author got a very close answer to my one is because the times are extremely close.

I tried to use this as a way of motivating his solution. Because of how close the times were, we know that the error must be extremely bounded as well: obviously an error of 5 miles(!) in one of the tracks will not allow us to conclude that the first runner was actually faster. So we can say that if a,b are the errors in the tracks (where we take the "worst case scenario" as in PeroK's solution of subtracting from the faster one and adding for the slower one) then a+b must be bounded. So letting one of them be 0 leads to the same maximum as using both.

Then again, doesn't this argument work for times that are vastly different, e.g. one runner completed in 5s and the other in 1000s? We can still say that a+b is bounded. I think the crux here is that when the times are vastly different, we get something like a+100b rather than a+b. Then letting one of them be zero doesn't actually lead to the same maximum for the remaining value.

But anyway, this is too nuanced compared to the author's solution, so I am just wondering if there is a simpler motivation. I am convinced that it has something to do with the closeness of the times but I just can't seem to make the link (intuitively at least) to then realizing that we can wlog let one of the errors be 0. Symmetry perhaps?

Anyway, I'm done with this problem. :)