Problems

Let \(C_1\) and \(C_2\) be two concentric circles with \(C_1\) inside \(C_2\) and the center \(A\). Let \(B\) and \(D\) be two points on \(C_1\) that are not diametrically opposite. Extend the segment \(BD\) past \(D\) until it meets the circle \(C_2\) in \(C\). The tangent to \(C_2\) at \(C\) and the tangent to \(C_1\) at \(B\) meet in a point \(E\). Draw from \(E\) the second tangent to \(C_2\) which meets \(C_2\) at the point \(F\). Show that \(BE\) bisects angle \(\angle FBC\).

On a \(10\times 10\) board, a bacterium sits in one of the cells. In one move, the bacterium shifts to a cell adjacent to the side (i.e. not diagonal) and divides into two bacteria (both remain in the same new cell). Then, again, one of the bacteria sitting on the board shifts to a new adjacent cell, either horizontally or vertically, and divides into two, and so on. Is it possible for there to be an equal number of bacteria in all cells after several such moves?

Let \(p\) and \(q\) be two prime numbers such that \(q = p + 2\). Prove that \(p^q + q^p\) is divisible by \(p + q\).

For any real number \(x\), the absolute value of \(x\), written \(\left| x \right|\), is defined to be \(x\) if \(x>0\) and \(-x\) if \(x \leq 0\). What are \(\left| 3 \right|\), \(\left| -4.3 \right|\) and \(\left| 0 \right|\)?

Let \(x\) and \(y\) be real numbers. Prove that \(x \leq \left| x \right|\) and \(0 \leq \left| x \right|\). Then prove that the following inequality holds \(\left| x+y \right| \leq \left| x \right|+\left| y \right|\).

There are \(n\) balls labelled 1 to \(n\). If there are \(m\) boxes labelled 1 to \(m\) containing the \(n\) balls, a legal position is one in which the box containing the ball \(i\) has number at most the number on the box containing the ball \(i+1\), for every \(i\).

There are two types of legal moves: 1. Add a new empty box labelled \(m+1\) and pick a box from box 1 to \(m+1\), say the box \(j\). Move the balls in each box with (box) number at least \(j\) up by one box. 2. Pick a box \(j\), shift the balls in the boxes with (box) number strictly greater than \(j\) down by one box. Then remove the now empty box \(m\).

Prove it is possible to go from an initial position with \(n\) boxes with the ball \(i\) in the box \(i\) to any legal position with \(m\) boxes within \(n+m\) legal moves.

Given a natural number \(n\), find a formula for the number of \(k\) less than \(n\) such that \(k\) is coprime to \(n\). Prove that the formula works.

Today you saw two infinitely long buses with seats numbered as \(1,2,3,...\) carrying infinitely many guests each arriving at the full hotel. How do you accommodate everyone?

Now there are finitely many infinitely long buses with seats numbered as \(1,2,3,...\) carrying infinitely many guests each arriving at the full hotel. Now what do you do?

How about infinitely many very long buses with seats numbered \(1,2,3...\), each carrying infinitely many guests, all arriving at the hotel. Assume for now that the hotel is empty. But that seems like a lot of guests to accommodate. What should you do?

The whole idea of problems with Hilbert’s Hotel is about assigning numbers to elements of an infinite set. We say that a set of items is countable if and only if we can give all the items of the set as gifts to the guests at the Hilbert’s hotel, and each guest gets at most one gift. In other words, it means that we can assign a natural number to every item of the set. Evidently, the set of all the natural numbers is countable: we gift the number \(n\) to the guest in room \(n\).

The set of all integers, \(\mathbb{Z}\), is also countable. We gift the number \(n\) to the guest in room \(n\). Then we ask everyone to take their gift and move to the room double their original number. Rooms with odd numbers are now free (\(1, 3, 5, 7, \dots\)). We fill these rooms with guests from an infinite bus and gift the number \(-k\) to the guest in room \(2k+1\). Yes, that’s right: the person in the first room will be gifted the number \(0\).

Prove now that the set of all positive rational numbers, \(\mathbb{Q}^+\), is also countable. Recall that a rational number can be represented as a fraction \(\frac{p}{q}\) where the numbers \(p\) and \(q\) are coprime.