Problems
Most magic tricks rely on some kind of sleight of hand. However, some tricks are powered by maths!
A fruitful way of analyzing card shuffles is by using the idea of “permutations". Permutations are important objects that occur in various parts of maths. 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 two-line notation. Say you have the cards from top to bottom Ace, two, three. Say Ace is 1. Suppose that after a shuffle \(p\), 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.
\[p = \left( \begin{array}{ccc} 1 & 2 & 3 \\ 3 & 1 & 2 \\ \end{array} \right).\]
A second way of writing permutations is function notation. In the same situation, we could write \(p(1)=3\), \(p(2)=1\) and \(p(3)=2\).
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 \(q\) be the permutation on the same three cards given by \(q(1)=2\), \(q(2)=3\) and \(q(3)=1\). Consider \(qp\) which is performing \(p\) first and then \(q\). To find out what the effect of this composite permutation is on \(1\), we can visualize it as follows: \[1\mapsto3=p(1)\mapsto q(p(1))=q(3)=1.\]
This shows that the function notation plays very nicely with composing permutations. By the way, if we work out the entire \(qp\) in this fashion, we find that \[qp = \left( \begin{array}{ccc} 1 & 2 & 3 \\ 1 & 2 & 3 \\ \end{array} \right).\]
In other words, \(q\) has “negated" the effect of \(p\)!
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.
What are the symmetries of an isosceles triangle (which is not equilateral)?
What are the symmetries of the reduce-reuse-recycle symbol?
What are the symmetries of an equilateral triangle?
