There are four coins of equal size on an (infinite) table, forming
a square of sidelength 10cm. You are allowed to pick a coin A and
jump over another coin B - the coin A ends up on the other
side of coin B, at same distance. Formally, it ends up on the
unique spot such that B is the midpoint of the line segment
from A's old position to A's new position.

As you have seen, you can make several moves of this kind. Question: is it possible to move the coins such that they lie in a square of bigger size? If yes, how? If no, why not?

I got that problem from Dieter Mitsche. Suppose a group of more than two people make a tournament, i.e. each person plays some game against each other person. If every player wins at least one game, then there are three players A, B and C, such that A won agains B, B won against C, and C won against A. Why is that true?

On every birthday party, there are at least two people who know exactly the
*same number* of people (not necessarily knowing the
*same people*). Why is that true? Note that we assume that A knows B if
B knows A. Note also that this phenomenon is not unique to birthday parties.
Weddings are also a good example. Funerals not so much, since if the deceased
was famous, you might know him without him knowing you...

There is a stick of length 1 meter, on which several ants are walking.
The along the stick, i.e. not around it. An ant walks at 1 centimeter per
second. When it reaches the end of the stick, it drops off and will not enter
the stick again.^{[1]} When two ants meet, they
both turn around by 180 degree and walk in the direction they came from.
How long does it take until all ants have dropped off the stick? What is
the maximum time it can take?

In the beginning, in all holes except one there is a stone. A move consists of jumping with some stone

In infinite Solitaire, the board is the infinite integer grid

In this example we moved a stone to position

You have a rectangle of dimensions w*h, that is partitioned into several smaller rectangles. Claim: if every small rectangle has at least one side that has an integer length, then the same must hold for the whole rectangle. I know two proofs of this, one that is cumbersome and ugly (the one I found), and one that is really easy and beautiful (one that I am embarrassed not to have found)

An evil wizard has captured 100 elves, and threatens to kill them all. He
says "I will give you one chance. Tomorrow morning, I will line you up in
front of me house, one behind the other, face to the front. And I will put
hats onto your heads, they will either be white or black. Then everyone of
you, one at a time, will have to shout the color of his hat. But nothing
more, and no hidden signs. Who guesses his color correctly will be
free. Who not, will be killed immediately."

Now, the elves discuss what to do. Quite obviously, the elf that will be
last in line won't have any hint what the color of his hat will be. So he
has to guess. But when the wizard kills an elf, it will make a loud noise,
so the other elves will notice that one of them was killed.
Now, you have to invent a strategy that will save at least 99 elves.

You are a wizard and you want to make a potion that must be stirred exactly 15 minutes (well, a couple of seconds more or less do not matter). You don't have a clock. You only have two fuses, each of which burns for an hour. They do not burn at a constant speed, i.e. the first inch of a fuse might take 59 minutes to burn, while the rest burns down in a minute. So you simply don't know anything about how they will burn, just that they will burn one hour. How will you cook your potion?

I have this problem from my friend Stefan. There is an international conference, and every country sends two delegates. For dinner, there is a large round table, with a seat for every delegate and a name card on every plate. Now, the delegates talk and discuss and sit down in a random manner. The claim is that you always can rotate the table such that at least two delegates sit at their correct place, i.e. in front of their name cards.