Is it possible to cover an equilateral triangle with two smaller equilateral triangles? Why or why not?
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\solution{
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. \par
\textcolor{gray}{\textit{Bonus question:}} Can you cover a square with three smaller squares?
Prove that at least two of them have a difference divisible by $n$.
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\solution{
$n~|~(a-b)\iff a \equiv b \pmod{n}$\par
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 $n$. \par
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 $n$. \par
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?
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\solution{
A domino covers two adjacent squares. Adjacent squares have different colors. \par
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. \par
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.
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\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, $W^c \not\subseteq D$. \par
\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)
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\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. \par
This is impossible, since we know that someone must know $n-1$. \par
Pick five points in $\mathbb{R}^2$ with integral coordinates. Show that two of these form a line segment that has an integral midpoint.
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\solution{
Let $e, o$ represent even and odd integers. \par
There are four possible classes of points: $(e, e)$, $(o, o)$, $(e, o)$, $(o, e)$. \par
$\text{midpoint}(a, b)=(\frac{a_x + b_x}{2}, \frac{a_y + b_y}{2})$. If $a_x + b_x$ and $a_y + b_y$ are both even, the midpoint of points $a$ and $b$ will have integer coordinates. \par
Since we pick five points from four classes, at least two must come from the same class. \par
$e + e = e$ and $o + o = e$, so the midpoint between two points of the same class must have integral coordinates. \par
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.
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\solution{
This is a proof by contradiction. We will try to construct a set of points where three points have such an arrangement. \par
We know that some two points on the line will have the same color:
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.
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\solution{Pick three points that form an equilateral triangle with side length 1.}
Choose $n +1$ integers between $1$ and $2n$. Show that you must select two numbers $a$ and $b$ such that $a$ divides $b$.
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\solution{
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. \par
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.
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\solution{
Rearrange the problem. Don't think about days, think about espressos. Consider the following picture:
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. \par
There are $499$ ``breaks'' between $500$ espressos. \par
In a year, there are $364$ clean breaks. This leaves $499-364=135$ ``dirty'' breaks. \par
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. \par
However, there are $401$ possible blocks, since we can start one at the $1^{\text{st}}, 2^{\text{nd}}, ..., 401^{\text{st}}$ espresso. \par
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.
