Add problem sets, previous solutions

.gitignore

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+*.aux
+*.log
+*.synctex.gz
+

2019-Solutions.tex

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+\documentclass{amsart}
+
+\usepackage{graphicx}
+\usepackage{url}
+
+\setlength{\parindent}{0pt}
+\setlength{\parskip}{\baselineskip}
+
+\DeclareMathOperator{\tr}{tr}
+
+\begin{document}
+\begin{center}
+Science Atlantic Math Competition\\
+Solutions
+\end{center}
+
+\textbf{Problem 1}\\
+Show that the sum of five consecutive perfect squares is not itself a perfect square.
+
+\textit{Solution}\\
+Every set of five consecutive integers is of the form $\{n - 2, n - 1, n, n + 1, n + 2\}$.
+The sum of the squares of these integers is $5(n^2 + 2)$.
+If $5(n^2 + 2)$ is a perfect square, then $5 \mid (n^2 + 2)$.
+But $-2$ is not a quadratic residue mod 5, so $5(n^2 + 2)$ is not a perfect square.
+
+\textbf{Problem 2}\\
+Let $m$ and $n$ be positive integers.
+Suppose there are $m$ ants walking to the right and $n$ ants walking to the left along a line.
+All ants walk at a constant speed and initially the right-walking ants are all to the left of the left-walking ants.
+
+Whenever two ants collide, they immediately turn around and walk back in the opposite direction.
+(Of course, if an ant turns around, it may then collide with the ant that was following it and turn around again, and so on).
+Find (as a function of $m$ and $n$) the number of collisions that occur.
+
+\textit{Solution}\\
+The number of collisions is $mn$.
+
+A collision does not change the numbers of ants moving left and right.
+We may consider two ants reversing direction after a collision to instead have passed each other.
+Each of the $m$ ants moving right must pass each of the $n$ ants moving left, so the total number of passings (or collisions) is $mn$.
+
+\textbf{Problem 3}\\
+Squares are erected outward on all four sides of a parallelogram.
+Show that the centers of the squares themselves form a square.
+
+\textit{Solution}\\
+A parallelogram of width $w$ and height $h$ can be placed on the Cartesian plane with vertices at coordinates (counterclockwise)
+\[ (0,0), (w, 0), (a + w, h), (a, h) \]
+Now calculate that the centers of the erected squares (again counterclockwise) are
+\[ \left( \frac{w}{2}, -\frac{w}{2} \right), \left( \frac{a + h + 2w}{2}, \frac{-a + h}{2} \right), \left( \frac{2a + w}{2}, \frac{2h + w}{2} \right), \left( \frac{a - h}{2}, \frac{a + h}{2} \right) \]
+Finally using the Pythagorean theorem, the length of each side of the quadrilateral connecting these points is
+\[ \frac{\sqrt{(a + h + w)^2 + (-a + h - w)^2}}{2} \]
+and therefore the quadrilateral is a square.
+
+\pagebreak
+
+\textbf{Problem 4}\\
+Let $\displaystyle y = x + \frac{x}{x + \frac{x}{x + \frac{x}{x + \cdots}}}$.
+For what nonzero integer values of $x$ is $y$ also an integer?
+
+\textit{Solution}\\
+The only possible value of $x$ is $-4$.
+
+The equation can be re-written as $y = x + \frac{x}{y}$, or $y^2 = (y + 1)x$.
+Set $z = y + 1$; then $z^2 - 2z + 1 = zx$, implying that $z \mid 1$.
+If $z = 1$ then $y = 0$ and $x = 0$, which is not permitted.
+If $z = -1$ then $y = -2$ and $x = -4$, which is the only solution.
+
+\pagebreak
+
+\textbf{Problem 5}\\
+Let $C$ be a unit cube.
+Given any face $F$ of $C$, and any edge $e$ of $F$, let $V_{F, e}$ be the reflection of the center of $F$ in the edge $e$.
+Let $V$ be the set of all points that can be obtained in this manner.
+Sketch the convex polyhedron which has $V$ as its set of vertices, and describe it (with proof) in familiar terms.
+
+\textit{Solution}\\
+The polyhedron is a truncated octahedron with six square faces and eight regular hexagonal faces.
+
+\includegraphics[width=10cm]{Images/Truncated-Octahedron}
+
+\pagebreak
+
+\textbf{Problem 6}\\
+Determine all triples of rational numbers $(a, b, c)$ such that
+\[ a^3 + 2b^3 + 4c^3 = 6abc \]
+
+\textit{Solution}\\
+The only solution is $(0, 0, 0)$.
+
+Clearly $(0, 0, 0)$ is a solution.
+Suppose towards contradiction there exists another rational solution $(a, b, c)$.
+
+Let $q \in \mathbb{Q}$.
+Multiplying both sides of the equation by $q^3$ and grouping, we have that
+\[ (qa)^3 + 2(qb)^3 + 4(qc)^3 = 6(qa)(qb)(qc) \]
+Therefore any rational multiple of a solution is a solution.
+If $d_a, d_b, d_c$ are the least denominators of $a, b, c$, then $d_ad_bd_c(a, b, c)$ is an integer solution.
+Finally, dividing an integer solution $(a, b, c)$ by $\gcd(a, b, c)$, there is an integer solution for which the three terms are coprime.
+
+Without loss of generality let $(a, b, c)$ be an integer solution with $\gcd(a, b, c) = 1$.
+Note that by the uniqueness of prime factorization, if a prime $p$ divides $n^3$ then $p$ divides $n$ for any $n \in \mathbb{Z}$.
+Re-arrange the defining equation:
+\[ a^3 = 2(3abc - b^3 - 2c^3) \]
+Then $2 \mid a$ and we can write $a = 2a'$ for some integer $a'$.
+\[ 2b^3 = 4(3a'bc - 2a'^3 - c^3) \]
+As before, $2 \mid b$ and we can write $b = 2b'$ for some integer $b'$.
+But now we have
+\[ 4c^3 = 8(3a'b'c - a'^3 - 2b'^3) \]
+and $2 \mid c$.
+This is a contradiction because we assumed that $\gcd(a, b, c) = 1$.
+Therefore there is no other rational solution $(a, b, c)$ and $(0, 0, 0)$ is the only one.
+
+\pagebreak
+
+\textbf{Problem 7}\\
+An $n \times n$ matrix $A$ is said to have period $p$ if $p$ is the smallest positive integer such that $A^p = I_n$ (the $n \times n$ identity matrix).
+For what $p$ is there a $2 \times 2$ matrix with integer elements that has period $p$?
+
+\textit{Solution}\\
+The possible values of $p$ are 1, 2, 3, 4, and 6.
+
+Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ be a matrix with period $p$.
+
+Recall that $\det(B)^n = \det(B^n)$ for all square matrices $B$ and positive integers $n$.
+Therefore $\det(A)$ is a $p$-th root of unity, for the single eigenvalue of $I$ is 1.
+But since $A$ is an integer matrix, $\det(A)$ must be an integer, either -1 or 1.
+
+Recall that if $\lambda_1, \cdots, \lambda_m$ are the eigenvalues of $B$ for a square matrix $B$, then the eigenvalues of $B^n$ are $\lambda_1^n, \cdots, \lambda_m^n$.
+Therefore the eigenvalues of $A$ are $p$-th roots of unity.
+
+Calculate that the characteristic polynomial of $A$ is
+\[ \lambda^2 - \tr(A) \lambda + \det(A) \]
+Since the eigenvalues of $A$ are the roots of the characteristic polynomial of $A$, each eigenvalue $\lambda$ satisfies
+\begin{align*}
+\tr(A) \lambda &= \lambda^2 + \det(A) \\
+\lvert \tr(A) \lambda \rvert &= \lvert \lambda^2 + \det(A) \rvert \\
+\lvert \tr(A) \rvert &\leq \lvert \lambda^2 \rvert + \lvert \det(A) \rvert \\
+\lvert \tr(A) \rvert &\leq 2
+\end{align*}
+Therefore the linear term of the characteristic polynomial is one of $0, \pm 1, \pm 2$ and the constant term is one of $\pm 1$.
+There are ten polynomials which satisfy these conditions:
+\begin{align*}
+x^2 - 2x - 1 &= (x - (1 - \sqrt{2}))(x - (1 + \sqrt{2})) \\
+x^2 - 2x + 1 &= (x - 1)(x - 1) \\
+x^2 - x - 1 &= (x - \frac{1}{2}(1 - \sqrt{5}))(x - \frac{1}{2}(1 + \sqrt{5})) \\
+x^2 - x + 1 &= (x - e^\frac{\pi i}{3})(x - e^\frac{5 \pi i}{3}) \\
+x^2 - 1 &= (x - 1)(x + 1) \\
+x^2 + 1 &= (x - i)(x + i) \\
+x^2 + x - 1 &= (x - \frac{1}{2}(-1 - \sqrt{5}))(x - \frac{1}{2}(-1 + \sqrt{5})) \\
+x^2 + x + 1 &= (x - e^\frac{2 \pi i}{3})(x - e^\frac{4 \pi i}{3}) \\
+x^2 + 2x - 1 &= (x - (-1 - \sqrt{2}))(x - (-1 + \sqrt{2})) \\
+x^2 + 2x + 1 &= (x + 1)(x - 1)
+\end{align*}
+
+\pagebreak
+
+Picking from these roots only the roots of unity we find that the eigenvalues must be first, second, third, fourth, or sixth roots of unity.
+Therefore the only possibilities for $p$ are 1, 2, 3, 4, 6.
+It remains to give matrices which do indeed have this period:
+\vspace{-14cm}
+\begin{align*}
+p &= 1 & \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\
+p &= 2 & \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \\
+p &= 3 & \begin{pmatrix} -2 & -3 \\ 1 & 1 \end{pmatrix} \\
+p &= 4 & \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \\
+p &= 6 & \begin{pmatrix} 2 & 3 \\ -1 & -1 \end{pmatrix}
+\end{align*}
+
+\pagebreak
+
+\textbf{Problem 8}\\
+Determine the following quantity or prove that it does not exist:
+\[ \lim_{n \to \infty} \sum_{j = 1}^n \frac{2^\frac{j}{n} - 2^\frac{j - 1}{n}}{2^\frac{j}{n} + 1} \]
+
+\textit{Solution}\\
+The series converges to $\ln(3) - \ln(2)$.
+
+Note that given any $n$,
+\[ [1, 2] = \bigcup_{j = 1}^n \left[ 2^\frac{j - 1}{n}, 2^\frac{j}{n} \right] \]
+Also note that $2^\frac{j}{n} - 2^\frac{j - 1}{n}$ approaches 0 for all $1 \leq j \leq n$ as $n$ goes to infinity.
+Therefore the sum $\displaystyle \sum_{j = 1}^n \frac{2^\frac{j}{n} - 2^\frac{j - 1}{n}}{2^\frac{j}{n} + 1}$ is a Riemann sum for $\displaystyle \frac{1}{1 + x}$ on the interval $[1, 2]$.
+The limit as $n$ goes to infinity of the sum is then
+\[ \int_1^2 \frac{\text{dx}}{1 + x} = \ln(3) - \ln(2) \]
+\end{document}

2020-Solutions.tex

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+\documentclass{amsart}
+
+\setlength{\parindent}{0pt}
+\setlength{\parskip}{\baselineskip}
+
+\raggedbottom
+
+\begin{document}
+\begin{center}
+Science Atlantic Math Competition 2020 Solutions
+\end{center}
+
+\textbf{Problem 1}\\
+Write $\tan(x) + \cot(2x)$ in the form $bf(cx)$, where $b$ and $c$ are real numbers, and $f$ is a standard trigonometric function.
+
+\textit{Solution}\\
+\begin{align*}
+\tan(x) + \cot(2x) &= \frac{\sin(x)}{\cos(x)} + \frac{\cos(2x)}{\sin(2x)} \\
+&= \frac{\sin(x)}{\cos(x)} + \frac{\cos^2(x) - \sin^2(x)}{2\sin(x)\cos(x)} \\
+&= \frac{2 \sin^2(x) + \cos^2(x) - \sin^2(x)}{2\sin(x)\cos(x)} \\
+&= \frac{\sin^2(x) + \cos^2(x)}{2\sin(x)\cos(x)} \\
+&= \frac{1}{\sin(2x)} = \csc(2x)
+\end{align*}
+
+\pagebreak
+
+\textbf{Problem 2}\\
+A bug located at $(2, 0, 0)$ in $\mathbb{R}^3$ wants to get to $(-2, 0, 0)$ but is impeded by an impenetrable sphere of radius $1$ centred at the origin.
+Describe and give the length of the shortest path available to this bug.
+
+\textit{Solution}\\
+The projection of a path onto a plane is no longer than the path, so with no loss of generality, model the problem as the shortest path from $(2, 0)$ to $(-2, 0)$ which does not pass through the unit circle.
+If a string is stretched from $(2, 0)$ to $(-2, 0)$, its path consists of two line segments from $(-2, 0)$ and $(2, 0)$ meeting the unit circle at the points of tangency and the arc between those points.
+The points of tangency lie at $(-1/2, \sqrt{3}/2)$ and $(1/2, \sqrt{3}/2)$, or at angles of $2\pi/3$ and $\pi/3$ on the unit circle, respectively.
+The distances from $(-2, 0)$ to $(-1/2, \sqrt{3}/2)$ and from $(2, 0)$ to $(1/2, \sqrt{3}/2)$ are both $\sqrt{3}$, and the length of the arc between $\pi/3$ and $2\pi/3$ is $\pi/3$, so the length of the path is $2\sqrt{3} + \pi/3$.
+
+\pagebreak
+
+\textbf{Problem 3}\\
+A tetrahedron has vertices $ABCD$.
+Suppose that the plane $\gamma$ bisects the (internal) dihedral angle along edge $AB$ and meets edge $CD$ at $G$.
+Prove that
+\[ \frac{\lvert ABC \rvert}{\lvert ABD \rvert} = \frac{\lvert CG \rvert}{\lvert GD \rvert} \]
+
+\textit{Solution}\\
+
+\pagebreak
+
+\textbf{Problem 4}\\
+A sequence of natural numbers is \textit{eccentric} if no term can be written as a sum of terms that come before it (repetition is allowed).
+For instance, the finite sequence $12, 3, 13, 17, 2$ is eccentric, but $11, 3, 13, 17, 2$ is not because $11 + 3 + 3 = 17$.
+Does there exist an infinite eccentric sequence?
+
+\textit{Solution}\\
+Let $(a_n)_{n = 0}^\infty$ be a sequence of natural numbers.
+If any terms are repeated, the sequence is not eccentric (because a single term is considered a sum).
+
+Suppose no terms are repeated.
+By the pigeonhole principle, there exists $b$ such that $0 \leq b < a_0$ and infinitely many $a_n \equiv b \pmod{a_0}$.
+Then there exists a subsequence $(a_{n_i})_{i = 0}^\infty$ of $(a_n)$ such that each $a_{n_i} \equiv b \pmod{a_0}$.
+Then there exists $j$ such that $a_{n_0} = ja_0 + b$.
+Since $(a_n)$ has no repetitions, there exists $k \leq j + 1$ such that $a_{n_k} > a_{n_0}$, implying that there exists $\ell > j$ such that $a_{n_i} = \ell a_0 + b$.
+Then $a_{n_i} = a_{n_0} + (\ell - j) a_0$, so $(a_n)$ is not eccentric.
+
+Therefore no infinite eccentric sequences exist.
+
+\pagebreak
+
+\textbf{Problem 5}\\
+Define a \textit{factorial $M$-partition} of $N$ to be a set $\{a_1, a_2, \cdots, a_M\}$ of positive integers such that $\displaystyle N = \sum_{i = 1}^M a_i!$.
+Furthemore, define a factorial $M$-partition to be \textit{proper} if the $a_i$ are all unequal.
+Show that if for some $M, N$ there is a proper factorial $M$-partition of $N$, then every other proper factorial $M$-partition of $N$ is a permutation of it.
+
+\textit{Solution}\\
+First we show that $1! + \cdots + (n - 1)! < n!$ by induction.
+Indeed, $1! < 2!$ and if $n > 1$ and $1 + \cdots + (n - 1)! < n!$, then $1 + \cdots n! < 2n! < (n + 1)n! < (n + 1)!$.
+
+Now we show the main result by induction on $M$.
+For $M = 1$, $\{a!\} = \{b!\} \iff a = b$.
+Now suppose for a given $M$ that for every $N$ with a proper factorial $M$-partition, $N$ admits only one such partition up to ordering.
+Let $N$ be a number with a proper factorial $(M + 1)$-partition.
+Let $n$ be the unique natural number such that $n! \leq N < (n + 1)!$.
+Since $1! + \cdots + (n - 1)! < n! \leq N$, at least one element of the partition is $\geq n$.
+On the other hand, all terms of the partition are $<(n + 1)$ so the largest element is $n$.
+Now by supposition, $N - n$ has a unique ordered proper factorial $M$-partition, so appending $n$ to this gives a unique ordered proper factorial $(M + 1)$-partition of $N$.
+
+Therefore the result holds for all $M, N$ by induction.
+
+\pagebreak
+
+\textbf{Problem 6}\\
+Show for integers $\geq 1$ that
+\[ \frac{d^n}{dx^n} (1 + x)^{n - 1} e^\frac{x}{x + 1} = \frac{e^\frac{x}{x + 1}}{(1 + x)^{n + 1}} \]
+
+\textit{Solution}\\
+Let $y = 1 + x$ and note that $\displaystyle (1 + x)^{n - 1} e^\frac{x}{x + 1} = y^{n - 1} e^{1 - \frac{1}{y}}$.
+For each $m, n$ let $\displaystyle f_n^{(m)} = \frac{d^m}{dx^m} y^{n - 1} e^{1 - \frac{1}{y}}$.
+
+First we show by induction on $m$ that $f_{(n + 1)}^{(m)} = mf_n^{(m - 1)} + yf_n^{(m)}$.
+For the base case $m = 1$, $\displaystyle f_{(n + 1)}^{(1)} = \frac{d}{dx} yf_n = f_n + yf_n^{(1)}$.
+For the step case, if $f_{(n + 1)}^{(m)} = mf_n^{(m - 1)} + yf_n^{(m)}$, then $f_{(n + 1)}^{(m + 1)} = mf_n^{(m)} + f_n^{(m)} + yf_n^{(m + 1)} = (m + 1)f_n^{(m)} + yf_n^{(m + 1)}$.
+
+Now we show by induction on $n$ that $f_n^{(n)} = y^{-(n + 1)} e^{1 - \frac{1}{y}}$ as desired.
+The base case $n = 0$ is obvious.
+For the step case, $f_{(n + 1)}^{(n)} = nf_n^{(n - 1)} + yf_n^{(n)}$ and so $f_{(n + 1)}^{(n + 1)} = nf_n^{(n)} + f_n^{(n)} + yf_n^{(n + 1)} = (n + 1)f_n^{(n)} + yf_n^{(n + 1)}$.
+By supposition, $f_n^{(n)} = y^{-(n + 1)} e^{1 - \frac{1}{y}}$ and so $f_n^{(n + 1)} = (-(n + 1)y^{-(n + 2)} + y^{-(n + 3)}) e^{1 - \frac{1}{y}}$.
+Substituting, we have $f_{n + 1}^{(n + 1)} = y^{-(n + 2)} e^{1 - \frac{1}{y}}$ as desired.
+
+\pagebreak
+
+\textbf{Problem 7}\\
+Assess the convergence of
+\[ \sum_{n = 0}^\infty \arctan\left(\frac{2n - 1}{n^4 - 2n^3 + n^2 + 1}\right) \]
+If convergent, give the value of the series.
+
+\textit{Solution}\\
+By the rule
+\[ \displaystyle \tan(x - y) = \frac{\tan(x) - \tan(y)}{1 + \tan(x)\tan(y)} \]
+we have that
+\[ \displaystyle \arctan\left(\frac{2n - 1}{n^4 - 2n^3 + n^2 + 1}\right) = \arctan\left(\frac{n^2 - (n - 1)^2}{1 + n^2(n - 1)^2}\right) = \arctan(n^2) - \arctan((n - 1)^2) \]
+Then
+\begin{align*} \sum_{n = 0}^\infty \arctan\left(\frac{2n - 1}{n^4 - 2n^3 + n^2 + 1}\right) &= \sum_{n = 0}^\infty \arctan(n^2) - \arctan((n - 1)^2) \\
+&= -\arctan((-1)^2) + \lim_{n \to \infty} \arctan(n^2) \\
+&= -\frac{\pi}{4} + \frac{\pi}{2} = \frac{\pi}{4}
+\end{align*}
+
+\pagebreak
+
+\textbf{Problem 8}\\
+The symmetric $n \times n$ matrix $A_n$ has $ij$-th entry
+\[ \begin{cases} (3 - (-1)^i) / 2 & i = j \\ -1 & i = j \pm 1 \\ 0 & \text{otherwise} \end{cases} \]
+Determine $\det(A_n)$ for all positive integers $n$.
+
+\textit{Solution}\\
+Let $e_i = (3 - (-1)^i)/2$.
+An easy calculation shows that $A_1 = [2]$ and $A_2 = \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}$ have determinants 2 and 1 respectively.
+In the general case, $A_n$ has $A_{n - 2}$ as the top left $(n - 2) \times (n - 2)$ sub-block and $A_{n - 1}$ as the top left $(n - 1) \times (n - 1)$ sub-block.
+\[ \begin{bmatrix}
+& & & \cdots & \cdots \\
+& A_{n - 2} & & \cdots & \cdots \\
+& & & -1 & \cdots \\
+\cdots & \cdots & -1 & e_{n - 1} & -1 \\
+\cdots & \cdots & \cdots & -1 & e_n
+\end{bmatrix} \]
+Then $\det(A_n) = e_n \det(A_{n - 1} - \det(A_{n - 2})$ by evaluating first along the $n$-th row, then the $n$-th column of the resulting minors.
+This recurrence yields $\det(A_3) = 0$, $\det(A_4) = -1$, $\det(A_5) = -2$, $\det(A_6) = -1$, $\det(A_7) = 0$, $\det(A_8) = 1$, $\det(A_9) = 2$, and $\det(A_{10}) = 1$.
+Since $\det(A_9) = \det(A_1)$ and $\det(A_{10}) = \det(A_2)$, the sequence of determinants has period 8 and follows this pattern.
+
+\end{document}