Problems

Imagine you have now a general finite number of new guests arriving to the full hotel. What do you do?

A circle is inscribed into the triangle \(ABC\) with sides \(BC=6, AC=10\) and \(AB= 12\). A line tangent to the circle intersects two longer sides of the triangle \(AB\) and \(AC\) at the points \(F\) and \(G\) respectively. Find the perimeter of the triangle \(AFG\).

Liam saw an unusual clock in the museum: the clock had no digits, and it’s not clear how the clock should be rotated. That is, we know that \(1\) is the next digit clockwise from \(12\), \(2\) is the next digit clockwise from \(1\), and so on. Moreover all the arrows (hour, minute, and second) have the same length, so it’s not clear which is which. What time does the clock show?

Two circles are tangent to each other and the smaller circle with the center \(A\) is located inside the larger circle with the center \(C\). The radii \(CD\) and \(CE\) are tangent to the smaller circle and the angle \(\angle DCE = 60^{\circ}\). Find the ratio of the radii of the circles.

For positive real numbers \(a,b,c\) prove the inequality: \[(a^2b + b^2c + c^2a)(ab^2 + bc^2 + ca^2)\geq 9a^2b^2c^2.\]

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 \(p + q\).

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 \(\angle FBC\).

Let’s play some games today! We will play a classic game known as nim, which is thought to be one of the oldest games.

Typically people play nim using matchsticks, though stones and coins are popular too. There are a few heaps of matchsticks in nim. Players take turns to remove matchsticks from a heap of their choosing. The player can remove any number of matchsticks they wish from that heap. Whoever has no matchsticks left to take loses.

This following position will be written as \(\text{Nim}(3,3,3)\):

As another example, this is \(\text{Nim}(1,2,3,4)\):

We will omit heaps of size zero, so \(\text{Nim}(3,0,3,0,3)\) is the same as \(\text{Nim}(3,3,3)\).

Nim is important because a large class of games are equivalent to it despite its simple appearance. The interested reader should look up “Sprague-Grundy Theorem".

Let us introduce a few terms that will be helpful for analyzing games. A game \(G\) consists of some positions and a set of rules. A position \(g\) in the game \(G\) is called a winning position if the player starting this turn has a winning strategy. This means as long as the player starting this turn continues to play optimally, the second player has to lose. Conversely, a position \(g\) is a losing position if the player starting this turn has no winning strategy.

Let us define XOR (or addition mod 2). XOR is defined for 0 and 1 only. Here is a table recording the values of XOR:

Now we define the important concept of nim-sum. Given two natural numbers \(x\) and \(y\), we first convert them into binary representations and then compute XOR on individual digits. The resulting number, denoted \(x \oplus y\), is the nim-sum of \(x\) and \(y\). Here is an example.

This is simply saying \(22 \oplus 5 = 19\). Note that \(22=(10110)_2\) and \(5=(00101)_2\).

Verify \((x \oplus y) \oplus z = x \oplus (y \oplus z)\), so we can speak of \(x \oplus y \oplus z\) with no ambiguity.