Please read rule #6: http://www.mathhelpforum.com/math-he...ly-151428.htmlHey, I'm pretty much as new as you can get to these forums. First off, I'm sorry if this isn't in the right subforum as this problem is kind of a mixture of calculus and algebra.. and my brain is just so fried at the moment o_O.

Secondlythis problem is my big semester-long-given-to-solve problem for a Problem Seminar class. I am in no way looking for the answers but only subtle insights or advice to give me the little boost I need to figure out the main ideas myself

So, the question in question:

Consider an oval race track.

(a)If the track is 21 miles in circumference and you have five stakes, prove that you can place the stakes around the track in such a way that for each n in N, one can run an n mile race by starting and ending at a pair of stakes.

(b) To what degree, if any, is the placement unique?

(c) Prove that 21 is a maximal number of miles for 5 stakes.

(d) Generalize for numbers of stakes. To go beyond 6 stakes is only for the brave.

(e) Given that a track of length l can be partitioned by k stakes and m < l, is it neccessarily possible to partition a track of length m with k stakes?

So for (a) I figured out that you can places the stakes at 2 miles, 7 miles, 8 miles and 11 miles to solve (a).

For (b) so far as I can figured the placement persay isn't unique but the spacing of 2,5,1,3 is and can also be used as 3,1,5,2. With fewer stakes and miles I have found that the spacing is unique and can be rotated around the the stakes with a space of 1 between them. (i.e. 3,6,1,2,7 yields the same results as 7,2,1,6,3). I'm not sure if that is too general or if I need to be more indepth with the idea of uniqueness in this case.

For (c) I merely need to show that 22 is not possible. I'm still somewhat stuck on how to go about that without running through hundreds of combinations of stakes.. (792 to be exact..).

With (d) I've found a pattern by finding the max number of miles per number of stakes for 1 stake, 2 stakes, 3 stakes and 4 stakes. The equation I've come up with so far is ((n-1)^2)+n = m where m is the maximal number of miles for n number of stakes. This implies that with 6 stakes, the maximal number of miles is 31. By all other patterns this seems to be correct as well. However I've tried dozens of stake positions and have yet to find one that yields up to 31 miles. I can't say that shows it's not possible as I spent probably 3-4 hours figuring out the stake placement for 5 stakes and 21 miles..

(e) seems like a cake walk for the fact that given the maximal miles for a number of stakes, using lesser miles is trivial as m would already be accounted for with a track of length l partitioned by k stakes.

Almost forget, thanks in advanced for anyone who takes the time to even look over this and muse.If this weren't for a grade, I'd still love to solve it so I hope it at least is interesting for other math inclined people to think about!

Thread closed.