Puzzle 1:
Suppose that on Monday, we film the mountain from midnight to midnight, and that on Tuesday, we make a second film, again from midnight to midnight. Now superimpose the films, and start playing them. At the beginning of the film, the Monday climber is at the bottom, and the Tuesday descender is at the top. As the film progresses, the Monday climber ascends and the Tuesday descender...descends. When the film ends, they have switched places, and so obviously, at some point, their paths (or altitudes, really) had to cross.
Puzzle 2:
The synchronized elevation mountain climb cannot always be carried out. To show this, consider a mountain range that consists of three slopes, first, a 45 degree drop for 1000 horizontal feet, then a 45 degree rise for 2000 horizontal feet, then a 45 degree drop for 1000 horizontal feet. The rest of the world is flat. (and we do not allow the mountaineers to simply run the wrong way around the world!)
In order to begin the "climb", climber A must first go down. But if he does this, climber B will not be able to also go down. Therefore, climber A's first move cannot be to go down. But climber B, similarly, can only go up, to start the climb. Therefore, neither climber can make a move toward the mountain, and the problem cannot be solved, at least for this particular mountain.
If you find this mountain a little too contrived, you can make similar examples that have more wiggles in them, as long as, between climber A and climber B, there is one unmatched deep valley (deeper than all other elevations in the problem) followed by one unmatched high cliff. Such a pair constitutes an "insurmountainable" obstacle!
Puzzle 3:
The puzzle can be "repaired" by assuming that there is no valley lower than the position at which the climbers begin. That is, a line drawn from one climber to the other will go through solid rock (or just "skim" the ground at the beginning and end of the mountain range). In this case, the problem can always be solved.
I know of graph-theoretic arguments, but here is one that I think addresses the problem simply and correctly.
Begin by agreeing that the drawing of the mountain range always has n peaks and n-1 valleys, that we can completely ignore any horizontal regions (except for whatever we need at the beginning and end of the mountain range).
Agree further that the problem, when n=1, is trivial to solve.
Mathematical induction: We now show that if all problems with n peaks are soluble, then so are all problems with n+1 peaks.
Let us consider a mountain range M with n+1 peaks. Consider the highest valley, that is, the valley whose bottom is the highest of all valley bottoms. Consider the mountain range M' that is the same as M except that this highest valley has been omitted - that is, the drawing of range M is folded in such a way that exactly this valley disappears, leaving a "join" point P, and nothing else. Mountain range M' has n peaks (because the lower of the two limits of the valley has now disappeared, along with the bottom of the valley), so there is an itinerary that allows the two mountaineers to pass over the mountain range, maintaining equal elevation.
Take this itinerary for mountain range M', which has no information about the "missing valley", and construct an itinerary for M as follows. Whenever a climber reaches the point P in M' where the valley was removed, the climber must descend and ascend the valley. But it will always be possible for climber B to adjust his movements to match elevations with A. Why is this?
In one possiblity, climber A has just risen to point P, while climber B is standing at the point Q of equal elevation. In that case, climber A has walked upwards more than the depth of the missing valley, because that valley had the highest bottom, so the valley he is currently rising from is deeper. Let R be the point at which he most recently matched the depth of the hidden valley, and let S be the point occupied by B at that time. Then we modify the itinerary used on mountain range M' by adding the instruction that when climber A reaches point P, he is to go down the hidden valley while climber B retreats from point Q to point S. Then, as A climbs up the other side of the valley to the elevation of P again, climber B returns from point S to point Q. And now the old itinerary may be resumed.
In the second possibility, climber A is descending as he reaches P. But time reversal allows us to repeat the previous argument.
No new interesting things happen if climber B reaches point P. If both A and B reach point P at the same time in the itinerary for M', then in the new itinerary they are simply at opposing points of the hidden valley, and can easily descend and reascend, switching positions and proceeding.
Therefore, since we can solve the problem for n=1 peak, and since n+1 is solvable if n is solvable, the problem is always solvable.
This puzzle was found at "The Puzzle Toad", which refers back to Douglas West's book "Introduction to Graph Theory", which attributes the puzzle to DG Hoffman. The solution given by the Puzzle Toad involves forming a graph of those points on the profile of the mountain which are at the same height as any peak or valley.
This solution is one I really did think up all by myself, on the way to work.
Puzzle 4:
For the equator problem, pick some point on the equator, call it point A, and compare its temperature with that of its diametricallly opposite location, point B. If they are equal, we're done. Otherwise, let's suppose the temperature at point A is higher than at point B. (It doesn't really matter which it is, but it makes it easier to talk about the solution.). Now, imagine starting at point A, and traveling along the Equator to point B, while muttering to yourself "hotter, hotter, hotter" as long as your current location is hotter than its opposite point, or "colder, colder, colder" if it's colder. Well, you start your journey saying "hotter" and you'll end your journey saying "colder" (because you end at B which is colder than A!) so at least once you had to switch what you were saying, and that happened just at a place where the point and its opposite had the same temperature.
Puzzle 5:
I'm not claiming that hot water freezes faster than cold. Let's just suppose that this is the case. (This is reminding me of the people who flat out declare that bath water goes down the drain clockwise in one hemisphere and counterclockwise in the other...) Also, let's immediately discard some implausible explanations: that there is less of the hot water to freeze, because being hot it has expanded, or that a substantial amount of the hot water evaporates, saving us the trouble of freezing it. So if we have the same amount of water in two cups (and, let's say, two separate refrigerators, so the two cups of water don't influence each other) and if the hot water freezes first, then...at some point the two cups of water are at the same temperature. So how would Nature know which one to freeze first after that? In other words, there must be some property of the water aside from temperature which distinguishes the two cups. Think on it!
Back to Crossing Puzzle.