Problems
Many magicians can perform what is known as the Faro shuffle. Actually there are two kinds of Faro shuffle: the in Faro shuffle and the out Faro shuffle.
Let us assume that the deck has an even number of cards. The first step of the Faro shuffle is to divide the deck into two smaller decks of equal size. One deck consist of the top half of the original deck in their original order. The other deck consists of the bottom half of the original deck in their original order.
The second step of the Faro shuffle is to interweave the two decks, so that each card is above and beneath a card of the opposite deck. This is where the in Faro and the out Faro differ: the in Faro changes the top and bottom cards of the original deck (from before step one) while the out Faro retains the original top and bottom cards.
Show that only 8 out Faro shuffles are needed to return a standard 52 card deck back into its original position.
How to move the top card to any position in an even size deck using only Faro shuffles?
You have a deck of \(n\) distinct cards. Deal out \(k\) cards from the top one by one and put the rest of the deck on top of the \(k\) cards. What is the minimum number of times you need to repeat the action to return every card back to its position?
Most magic tricks rely on some kind of sleight of hand. However, some tricks are powered by maths!
A very fruitful way of analyzing card shuffles is by using the idea of “permutations". Permutations are important objects that occur in various parts of maths. We will look at them in the context of card shuffling. Many interesting patterns emerge, and we will only touch the tip of the iceberg today.
Suppose you have a set of ordered objects. A permutation of this set is a reordering of the objects. For example, a permutation of a deck of cards ordered from top to bottom is simply a shuffle of the cards. Note that in general, a permutation can be defined as a relabelling of objects, so an order is not necessary.
Let’s discuss two ways of writing permutations.
The first way is the two-line notation. Say you have the cards from top to bottom Ace, two, three. Say Ace is 1. Suppose that after a shuffle \(\sigma\), we have from top to bottom two, three, Ace. The two-line notation keeps the original positions on the first line and the new positions in the second line.
\[\sigma = \left( \begin{array}{ccc} 1 & 2 & 3 \\ 3 & 1 & 2 \\ \end{array} \right).\]
A second way of writing permutations is the function notation. In the same situation, we could write \(\sigma(1)=3\), \(\sigma(2)=1\) and \(\sigma(3)=2\).
As a first indication of why permutations give a useful perspective, we note that permutations can be done after another and the result is still a permutation. Let \(\tau\) be the permutation on the same three cards given \(\tau(1)=2\), \(\tau(2)=3\) and \(\tau(3)=1\). Consider \(\tau\sigma\) which is performing \(\sigma\) first and then \(\tau\). To find out what the effect of this composite permutation is on \(1\), we can visualize it as follows: \[1\mapsto3=\sigma(1)\mapsto\tau(\sigma(1))=\tau(3)=1.\]
This shows that the function notation plays very nicely with composing permutations. By the way, if we work out the entire \(\tau\sigma\) in this fashion, we find that \[\tau\sigma = \left( \begin{array}{ccc} 1 & 2 & 3 \\ 1 & 2 & 3 \\ \end{array} \right).\]
In other words, \(\tau\) has “negated" the effect of \(\sigma\)!
We have a few more examples of how the notations are used. If you want to learn a magic trick or two - see problems 7 to 11.
Let \(ABC\) be a triangle with midpoints \(D\) on the side \(BC\), \(E\) on the side \(AC\), and \(F\) on the side \(AB\). Prove that the perpendicular bisectors to the sides \(AB\), \(BC\), \(AC\) intersect at one point.
Let \(ABC\) be a triangle with midpoints \(D\) on the side \(BC\), \(E\) on the side \(AC\), and \(F\) on the side \(AB\). Let \(M\) be the point of intersection of all medians of the triangle \(ABC\), let \(H\) be the point of intersection of the heights \(AJ\), \(BI\) and \(CK\). Prove that the points \(D,J,I,E,F,K\) lie on one circle. What can you say about the center of that circle?
Let \(ABC\) be a triangle with midpoints \(D\) on the side \(BC\), \(E\) on the side \(AC\), and \(F\) on the side \(AB\). Let \(M\) be the point of intersection of all medians of the triangle \(ABC\), let \(H\) be the point of intersection of the heights \(AJ\), \(BI\) and \(CK\). Consider the Euler circle of the triangle \(ABC\), the one that contains the points \(D,J,I,E,F,K\). This circle intersects the segments \(AH\), \(BH\), \(CH\) at points \(O\), \(P\), \(Q\) respectively. Prove that \(O\), \(P\), \(Q\) are the midpoints of the segments \(AH\), \(BH\), \(CH\).
Consider the point \(H\) of intersection of the heights of the triangle \(ABC\). Prove that Euler lines of the triangles \(ABC\), \(ABH\), \(BCH\), \(ACH\) intersect at one point. On the diagram below the points \(R,S,T\) are the points of intersection of medians in triangles \(ABH\), \(BCH\), and \(ACH\) correspondingly.
We often think of symmetry as a property of shapes. Another way of thinking about it is as something you do to an object which keeps the object looking the same. The example you’ve likely met is reflection. The other one that we’ll consider today is rotation. An important feature is that we consider ‘doing nothing’ as a symmetry - we call this the identity.
Note the two following features of symmetries
Applying a symmetry and then a second symmetry gives us another symmetry (which may be the same as one of the two we’ve used!)
Each symmetry has an inverse. Suppose we apply symmetry \(x\). Then there is some symmetry we can apply after \(x\), which means that overall, we’ve applied the identity.
