There are 100 red dots and 100 blue dots on the plane (a lot of planar riddles lately). The dots are arranged such that no three are on the same line.

A pairing of the red dots and the blue dots is a one-to-one function that assigns one blue dot to each red dot.

A valid pairing is a pairing such that when paired dots are connected with a straight line segment, no line intersections occur. This is depicted here:

Prove that there exists a valid pairing.

I do not really like the solution I found for this riddle. If you find an elegant one, please post it.

Thanks to Nadav Sherman for giving me this one.

A very interesting riddle for those of you with some basic background in Set Theory.

Prove or Disprove the Following Claim:

There exists a subset B of P(N), such that |B| = A, and B is completely ordered by the subset relation.

Notes:

N denotes the set of natural numbers.

A denotes the cardinality of the continuum (i.e. A = aleph = 2^aleph 0).

“B is completely ordered by the subset relation” means that for every two elements a, b of B, either a is a subset of b or b is a subset of a.

A very easy riddle. Four frogs are sitting on the corners of the unit square (i.e. they have coordinates (0,0), (0,1), (1,1) and (1,0) ). Each turn, a frog can jump over any other frog, thereby transferring itself to the symmetrical point on the other side of the static frog. For example, if the frog at (0,0) jumps over the frog at (1,1) it will land on (2,2).

Prove that the frogs cannot transfer themselves to a square with a side of length 2.

I know of two different solutions to this one (both easy ).

Thanks to Nadav Sherman for asking me this riddle.

 This is a cute puzzle. Consider an infinite checkerboard divided in two with an infinite line lying along the x-axis, as depicted below:

You have several peons at your disposal. In each turn you can move the peons by making one of them jump over the other, thereby killing it (removing it from the board). The peon movement is demonstrated here:

When the game starts, all the peons are below the line. Your goal is to place a peon above the line. The example shows how to accomplish this starting with two peons. Indeed it is obvious that in order to make a peon cross the line, two peons are needed (clearly one is not enough as it cannot move at all).

How many peons are needed in order to not only cross the line, but place a peon two squares above it? As is demonstrated below, four peons are sufficient, and indeed this is the minimum (check that you see why 2 and 3 peons cannot do it).

Now for the riddle. How many peons are needed (again, starting below the line) in order to place a peon 5 squares above the line?