[Math Puzzle] Day3

Is this an application of the intermediate value theorum?

If the circle is completely flat its trivial; with the random heights it has to go both up and down over the entire circumference, if we look from the perspective of an observer travelling around the track always looking at the point opposite him say that it starts lower than him, he must eventually go down to that height and the track must come up to meet his height and since it is a continuous track there must therefore be a point where they cross.

If you define the function f(x) (where x is a point on the circle) to be the height at x, then let g(x) = f(x) - f(-x), you can see that g(x) = -g(-x) and it is continuous, so g has to be 0 at some point (by the intermediate value theorem). This can be generalized to a sphere mapping onto two dimensions (for example the temperature and pressure at a point of the earth) and is a well known result in algebraic topology.

red dot is the proof.

actually this is just like that monk problem with the mountain 9am and 12noon going up and down stuff. so i suppose i have seen this before.
put the sol'n back in drawing form i guess if some haven't seen the monk riddle
let A start in green and B start in blue. those 2 paths are arbitrary paths they follow around half of the circle, always remaining opposite. as the red dot shows at some point the paths are same height.

dammit evan give me puzzles i haven't seen before !

Let h(t)=the height of the circle at angle t (pick an arbitary point as 0). h(t) is continuous. i(t)=h(t+pi) is continuous. j(t)=i(t)-h(t) is continous. j(t)=j(t+2*pi). There exists t1, t2 in [0,2*pi) such that j(t1)>=0 and j(t2)<=0. Therefore there exists t3 in [t1,t2] or [t2,t1] such that j(t3)=0.

Hamster1800 has the solution. For a number of the others, even if you show that the same height is reached at some point, you still haven't shown that your pair of points is at opposite ends of the projected circle.

Since "viewed from above it's a circle", can define a function t:[0, 2pi] -> R which represents the height of the track at each angle, which should be a continuous function. t(0) = t(2pi)

Define f:[0, pi] -> R by f(x) = t(pi + x) - t(x), which is a continuous function as well. Notice f(0) = -f(pi)
if f(0) = 0, done, otherwise can apply intermediate value theorem which says there is a y in (0, pi) such that f(y) = 0.
At this angle and the point opposite of it on the circle, the heights will be the same, since t(pi + y) - t(y) = 0
thus t(pi + y) = t(y).

uhm, if s=cos(s8pi)/cos(8pi), then t=(cos(s8pi)/cos(8pi) should be the same height. Then, it would be opposite of t=s+pi, though I might just be way off. Already had a math exam today, so I don't expect to be right *phuuw*

It nowhere says that the race track must be continuous, I construct my race track as
f(µ)=µ
0=<µ<2pi
The jump makes it a lot more interesting.

I believe a track by definition doesn't need to complete a circuit. However, if this is true, then the problem is unsolvable because you can just make a helix type track. Therefore, I believe it is to be assumed that the track forms a compelte circuit, and the solution is just using IVT.

No, all that is required to just have it be a defined function on the interval µ=[0,2pi), if you just employ the restriction that it needs to be continious you can still get into the problem where it have multiple values at the same angle.