You are given an irreflexive symmetric (but not necessarily transitive) “enemies” relation on a set of people.  In other words, if person A is an enemy of a person B, then B is also an enemy of A.  How can you divide up the people into two houses in such a way that every person has at least as many enemies in the other house as in their own house?

Hint:  Planar configuration of straight connecting lines puzzle may provide a hint to solving this puzzle.

Given an even number of points in general positions on the plane (that is, no three points co-linear), can you partition the points into pairs and connect the two points of each pair with a single straight line such that the straight lines do not overlap?

You are given one 44-meter piece of fence and 48 one-meter pieces of fence.  Assume each piece is a straight and unbendable.  What is the large area of (flat) land that you can enclose using these fence pieces?

Someone picks, at their will, two cards from a deck of cards. The cards have different numbers, so one is higher than the other. (In other words, the person picks two distinct numbers in the inclusive range 1 through 13.)  The cards are placed face down on a table in front of you.  You get to choose one of the cards and turn it face up.  Now, you will select one of the two cards (one of whose face you can see, the other one you can’t).  If you select the highest card, you win.  Design a card-selection strategy for which your chance of winning is strictly greater than 50%.

There is a polynomial and you have access to a function that evaluates that polynomial at a given number.  You don’t know the degree of the polynomial, nor do you know any of the coefficients of its terms.  However, you are told that all coefficients are non-negative integers.  How many times do you need to call the evaluation function in order to identify the polynomial (that is, to figure out the values of its coefficients)?

There is a table with a row of 2014 cards.  Each card has a red side and a blue side.  We’ll say that a card is red if the color on its visible face is red, and analogously for blue.  Two players take turns to do the following move:  select any 50 consecutive cards where the left-most card is red, then flip each of those 50 cards (thus, for those 50 cards, turning red cards into blue cards and blue cards into red cards).  Both players look at the cards from the same side of the table, so “left-most” means the same to both of the players (that is, you can think of one of the ends of the table as being designated as the left end).  When it is a player’s turn, if that player cannot make a move (that is, if there is no way to select 50 consecutive cards the left-most one of which is red), then that player loses and the other player wins.  If you are one of the players and all cards are initially red, can you be sure to win, and if so, do you want to be the player who goes first or second?

Two arbitrary rectangles are placed to form an “L”.  That is, the lower left-hand corner of the two rectangles share the same point. (What I’m trying to say is that there’s an “L” whose “I” and “_” parts have arbitrary widths and heights.) Using only a (pen and a) straightedge (that is, no measuring device and no compass), figure out a way to, with a single straight cut, divide the “L” into two pieces of equal area.