In order to completely cover an equilateral triangle, the two smaller triangles must cover all three vertices. Since the longest length of an equilateral triangle is one of its sides, a smaller triangle cannot cover more than one vertex. Therefore, we cannot completely cover the triangle with two smaller copies. \\
\textcolor{gray}{\textit{Bonus question:}} Can you cover a square with three smaller squares?
Let $i_0 ... i_{n+1}$ be our set of integers. If we pick $i_0 ... i_{n+1}$ so that no two have a difference divisible by $n$, we must have $i_0\not\equiv i_k \pmod{n}$ for all $1\leq k \leq n+1$. There are $n$ such $i_k$, and there are $n$ equivalence classes mod $n$. \\
Therefore, either, $i_1 ... i_{n+1}$ must cover all equivalence classes mod $n$ (implying that $i_0\equiv i_k \pmod{n}$ for some k), or there exist two elements in $i_1 ... i_{n+1}$ that are equivalent mod $n$. \\
In either case, we can find $a, b$ so that $a \equiv b \pmod{n}$, which implies that $n$ divides $a-b$.
You have an $8\times8$ chess board with two opposing corner squares cut off. You also have a set of dominoes, each of which is the size of two squares. Is it possible to completely cover the the board with dominos, so that none overlap nor stick out?
A domino covers two adjacent squares. Adjacent squares have different colors. \\
If you remove two opposing corners of a chessboard, you remove two squares of the same color, and you're left with $32$ of one and $30$ of the other. \\
Since each domino must cover two colors, you cannot cover the modified board.
The ocean covers more than a half of the Earth's surface. Prove that the ocean has at least one pair of antipodal points.
\begin{solution}
Let $W$ be the set of wet points, and $W^c$ the set of points antipodal to those in $W$. $W$ and $W^c$ each contain more than half of the points on the earth. The set of dry points, $D$, contains less than half of the points on the earth. Therefore, $W^c \not\subseteq D$. \\
\textcolor{gray}{\textit{Note:}} This solution isn't very convincing. However, it is unlikely that the students know enough to provide a fully rigorous proof.
There are $n > 1$ people at a party. Prove that among them there are at least two people who have the same number of acquaintances at the gathering. (We assume that if A knows B, then B also knows A)
\begin{solution}
Assume that every attendee knows a different number of people. There is only one way this may happen: the most popular person knows $n-1$ people (that is, everyone but himself), the second-most popular knows $n-2$, etc. The least-popular person must then know $0$ people. \\
This is impossible, since we know that someone must know $n-1$. \\
There are four possible classes of points: $(e, e)$, $(o, o)$, $(e, o)$, $(o, e)$. \\
$\text{midpoint}(a, b)=(\frac{a_x + b_x}{2}, \frac{a_y + b_y}{2})$. If $a_x + b_x$ and $a_y + b_y$ are both even, the midpoint of points $a$ and $b$ will have integer coordinates. \\
Since we pick five points from four classes, at least two must come from the same class. \\
$e + e = e$ and $o + o = e$, so the midpoint between two points of the same class must have integral coordinates. \\
Every point on a line is painted black or white. Show that there exist three points of the same color where one is the midpoint of the line segment formed by the other two.
Every point on a plane is painted black or white. Show that there exist two points in the plane that have the same color and are located exactly one foot away from each other.
Split the the set $\{1, ..., 2n\}$ into classes defined by each integer's greatest odd divisor. There will be $n$ classes since there are $\frac{k}{2}$ odd numbers between $1$ and $n$. Because we pick $n +1$ numbers, at least two will come from the same class---they will be divisible. \\
Show that it is always possible to choose a subset of the set of integers $a_1, a_2, ... , a_n$ so that the sum of the numbers in the subset is divisible by $n$.
A stressed-out student consumes at least one espresso every day of a particular year, drinking $500$ overall. Show the student drinks exactly $100$ espressos on some consecutive sequence of days.
If there exists a sequence of days where the student drinks exactly 100 espressos, we must have at least one ``block'' (in orange, above) of 100 espressos that both begins and ends on a ``clean break'' between days. \\
There are $499$ ``breaks'' between $500$ espressos. \\
In a year, there are $364$ clean breaks. This leaves $499-364=135$ ``dirty'' breaks. \\
We therefore have $135$ places to start a block on a dirty break, and $135$ places to end a block on a dirty break. This gives us a maximum of $270$ dirty blocks. \\
However, there are $401$ possible blocks, since we can start one at the $1^{\text{st}}, 2^{\text{nd}}, ..., 401^{\text{st}}$ espresso. \\
Out of $401$ blocks, a maximum of $270$ can be dirty. We are therefore guaranteed at least $131$ clean blocks. This completes the problem---each clean block represents a set of consecutive, whole days during which exactly 100 espressos were consumed.
Given a table with a marked point, $O$, and with $2013$ properly working watches put down on the table, prove that there exists a moment in time when the sum of the distances from $O$ to the watches' centers is less than the sum of the distances from $O$ to the tips of the watches' minute hands.
