Problems

Let \(ABC\) be a non-isosceles triangle. The point \(G\) is the point of intersection of the medians \(AE\), \(BF\), \(CD\), the point \(H\) is the point of intersection of all heights, the point \(I\) is the center of the circumscribed circle for \(ABC\), or the point of intersection of all perpendicular bisectors to the segments \(AB\), \(BC\), \(AC\).
Prove that points \(I,G,H\) lie on one line and the ratio \(IG:GH = 1:2\).

Paloma wrote digits from \(0\) to \(9\) in each of the \(9\) dots below, using each digit at most once. Since there are \(9\) dots and \(10\) digits, she must have missed one digit.

In the triangles, Paloma started writing either the three digits at the corners added together (the sum), or the three digits at the corners multiplied together (the product). She gave up before finishing the final two triangles.

What numbers could Paloma have written in the interior of the red triangle? Demonstrate that you’ve found all of the possibilities.

Let \(ABC\) be a triangle. Prove that the heights \(AD\), \(BE\), \(CF\) intersect in one point.

Let \(ABC\) be a triangle. Prove that the medians \(AD\), \(BE\), \(CF\) intersect in one point.

Let \(ABC\) be a triangle with medians \(AD\), \(BE\), \(CF\). Prove that the triangles \(ABC\) and \(DEF\) are similar. What is their similarity coefficient?

How many subsets are there of \(\{1,2,...,10\}\) (the integers from \(1\) to \(10\) inclusive) containing no consecutive digits? That is, we do count \(\{1,3,6,8\}\) but do not count \(\{1,3,6,7\}\).
For example, when \(n=3\), we have \(8\) subsets overall but only \(5\) contain no consecutive integers. The \(8\) subsets are \(\varnothing\) (the empty set), \(\{1\}\), \(\{2\}\), \(\{3\}\), \(\{1,3\}\), \(\{1,2\}\), \(\{2,3\}\) and \(\{1,2,3\}\), but we exclude the final three of these.

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 \(AH\), \(BH\), \(CH\).