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Intermediate handouts

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31
src/Intermediate/An Introduction to Graph Theory/main.tex Executable file
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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\input{tikxset.tex}
\usepackage{adjustbox}
\uptitlel{Intermediate 2}
\uptitler{\smallurl{}}
\title{An Introduction to Graph Theory}
\subtitle{
Prepared by Mark on \today \\
Based on a handout by Oleg Gleizer
}
\begin{document}
\maketitle
\input{parts/0 intro.tex}
\input{parts/1 paths.tex}
\input{parts/2 planar.tex}
%\input{parts/3 counting.tex}
\end{document}

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src/Intermediate/An Introduction to Graph Theory/meta.toml Normal file
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[metadata]
title = "An Introduction to Graph Theory"
[publish]
handout = true
solutions = false

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src/Intermediate/An Introduction to Graph Theory/parts/0 intro.tex Normal file
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\section{Graphs}
\definition{}
A \textit{set} is an unordered collection of objects. \par
This means that the sets $\{1, 2, 3\}$ and $\{3, 2, 1\}$ are identical.
\definition{}
A \textit{graph} $G = (N, E)$ consists of two sets: a set of \textit{vertices} $V$, and a set of \textit{edges} $E$. \par
Vertices are simply named \say{points,} and edges are connections between pairs of vertices. \par
In the graph below, $V = \{a, b, c, d\}$ and $E = \{~ (a,b),~ (a,c),~ (a,d),~ (c,d) ~\}$.
\begin{center}
\begin{tikzpicture}
\begin{scope}[layer = nodes]
\node[main] (a) at (0, 0) {$a$};
\node[main] (b) at (0, -1) {$b$};
\node[main] (c) at (2, -1) {$c$};
\node[main] (d) at (4, 0) {$d$};
\end{scope}
\draw[-]
(a) edge (b)
(a) edge (c)
(a) edge (d)
(c) edge (d)
;
\end{tikzpicture}
\end{center}
Vertices are also sometimes called \textit{nodes}. You'll see both terms in this handout. \par
\problem{}
Draw the graph defined by the following vertex and edge sets: \par
$V = \{A,B,C,D,E\}$ \par
$E = \{~ (A,B),~ (A,C),~ (A,D),~ (A,E),~ (B,C),~ (C,D),~ (D,E) ~\}$\par
\vfill
We can use graphs to solve many different kinds of problems. \par
Most situations that involve some kind of \say{relation} between elements can be represented by a graph.
\pagebreak
Graphs are fully defined by their vertices and edges. The exact position of each vertex and edge doesn't matter---only which nodes are connected to each other. The same graph can be drawn in many different ways.
\problem{}
Show that the graphs below are equivalent by comparing the sets of their vertices and edges.
\begin{center}
\adjustbox{valign=c}{
\begin{tikzpicture}
\begin{scope}[layer = nodes]
\node[main] (a) at (0, 0) {$a$};
\node[main] (b) at (2, 0) {$b$};
\node[main] (c) at (2, -2) {$c$};
\node[main] (d) at (0, -2) {$d$};
\end{scope}
\draw[-]
(a) edge (b)
(b) edge (c)
(c) edge (d)
(d) edge (a)
(a) edge (c)
(b) edge (d)
;
\end{tikzpicture}
}
\hspace{20mm}
\adjustbox{valign=c}{
\begin{tikzpicture}
\begin{scope}[layer = nodes]
\node[main] (a) at (0, 0) {$a$};
\node[main] (b) at (-2, -2) {$b$};
\node[main] (c) at (0, -2) {$c$};
\node[main] (d) at (2, -2) {$d$};
\end{scope}
\draw[-]
(a) edge (b)
(b) edge (c)
(c) edge (d)
(d) edge (a)
(a) edge (c)
(b) edge[out=270, in=270, looseness=1] (d)
;
\end{tikzpicture}
}
\end{center}
\vfill
\pagebreak
\definition{}
The degree $D(v)$ of a vertex $v$ of a graph
is the number of the edges of the graph
connected to that vertex.
\theorem{Handshake Lemma}<handshake>
In any graph, the sum of the degrees of its vertices equals twice the number of the edges.
\problem{}
Prove \ref{handshake}.
\vfill
\problem{}
Show that all graphs have an even number number of vertices with odd degree.
\vfill
\problem{}
One girl tells another, \say{There are 25 kids
in my class. Isn't it funny that each of them
has 5 friends in the class?} \say{This cannot be true,} immediately replies the other girl.
How did she know?
\vfill
\problem{}
Say $G$ is a graph with nine vertices. Show that $G$ has at least five vertices of degree six or at least six vertices of degree 5.
\vfill
\pagebreak

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src/Intermediate/An Introduction to Graph Theory/parts/1 paths.tex Normal file
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\section{Paths and cycles}
A \textit{path} in a graph is, intuitively, a sequence of edges: $(x_1, x_2, x_4, ... )$. \par
I've highlighted one possible path in the graph below.
\begin{center}
\begin{tikzpicture}[
node distance={15mm},
thick,
main/.style = {draw, circle}
]
\node[main] (1) {$x_1$};
\node[main] (2) [above right of=1] {$x_2$};
\node[main] (3) [below right of=1] {$x_3$};
\node[main] (4) [above right of=3] {$x_4$};
\node[main] (5) [above right of=4] {$x_5$};
\node[main] (6) [below right of=4] {$x_6$};
\node[main] (7) [below right of=5] {$x_7$};
\draw[-] (1) -- (2);
\draw[-] (1) -- (3);
\draw[-] (2) -- (5);
\draw[-] (2) -- (4);
\draw[-] (3) -- (6);
\draw[-] (3) -- (4);
\draw[-] (4) -- (5);
\draw[-] (5) -- (7);
\draw[-] (6) -- (7);
\draw [
line width=2mm,
draw=black,
opacity=0.4
] (1) -- (2) -- (4) -- (3) -- (6);
\end{tikzpicture}
\end{center}
A \textit{cycle} is a path that starts and ends on the same vertex:
\begin{center}
\begin{tikzpicture}[
node distance={15mm},
thick,
main/.style = {draw, circle}
]
\node[main] (1) {$x_1$};
\node[main] (2) [above right of=1] {$x_2$};
\node[main] (3) [below right of=1] {$x_3$};
\node[main] (4) [above right of=3] {$x_4$};
\node[main] (5) [above right of=4] {$x_5$};
\node[main] (6) [below right of=4] {$x_6$};
\node[main] (7) [below right of=5] {$x_7$};
\draw[-] (1) -- (2);
\draw[-] (1) -- (3);
\draw[-] (2) -- (5);
\draw[-] (2) -- (4);
\draw[-] (3) -- (6);
\draw[-] (3) -- (4);
\draw[-] (4) -- (5);
\draw[-] (5) -- (7);
\draw[-] (6) -- (7);
\draw[
line width=2mm,
draw=black,
opacity=0.4
] (2) -- (4) -- (3) -- (6) -- (7) -- (5) -- (2);
\end{tikzpicture}
\end{center}
A \textit{Eulerian\footnotemark} path is a path that traverses each edge exactly once. \par
A Eulerian cycle is a cycle that does the same.
\footnotetext{Pronounced ``oiler-ian''. These terms are named after a Swiss mathematician, Leonhard Euler (1707-1783), who is usually considered the founder of graph theory.}
\vspace{2mm}
Similarly, a {\it Hamiltonian} path is a path in a graph that visits each vertex exactly once, \par
and a Hamiltonian cycle is a closed Hamiltonian path.
\medskip
An example of a Hamiltonian path is below.
\begin{center}
\begin{tikzpicture}[
node distance={15mm},
thick,
main/.style = {draw, circle}
]
\node[main] (1) {$x_1$};
\node[main] (2) [above right of=1] {$x_2$};
\node[main] (3) [below right of=1] {$x_3$};
\node[main] (4) [above right of=3] {$x_4$};
\node[main] (5) [above right of=4] {$x_5$};
\node[main] (6) [below right of=4] {$x_6$};
\node[main] (7) [below right of=5] {$x_7$};
\draw[-] (1) -- (2);
\draw[-] (1) -- (3);
\draw[-] (2) -- (5);
\draw[-] (2) -- (4);
\draw[-] (3) -- (6);
\draw[-] (3) -- (4);
\draw[-] (4) -- (5);
\draw[-] (5) -- (7);
\draw[-] (6) -- (7);
\draw [
line width=2mm,
draw=black,
opacity=0.4
] (1) -- (2) -- (4) -- (3) -- (6) -- (7) -- (5);
\end{tikzpicture}
\end{center}
\vfill
\pagebreak
\definition{}
We say a graph is \textit{connected} if there is a path between every pair of vertices. A graph is called \textit{disconnected} otherwise.
\problem{}
Draw a disconnected graph with four vertices. \par
Then, draw a graph with four vertices, all of degree one.
\vfill
\problem{}
Find a Hamiltonian cycle in the following graph.
\begin{center}
\begin{tikzpicture}[
node distance={20mm},
thick,
main/.style = {draw, circle}
]
\node[main] (1) {$x_1$};
\node[main] (2) [above right of=1] {$x_2$};
\node[main] (3) [below right of=1] {$x_3$};
\node[main] (4) [above right of=3] {$x_4$};
\node[main] (5) [above right of=4] {$x_5$};
\node[main] (6) [below right of=4] {$x_6$};
\node[main] (7) [below right of=5] {$x_7$};
\draw[-] (1) -- (2);
\draw[-] (1) -- (3);
\draw[-] (2) -- (5);
\draw[-] (2) -- (4);
\draw[-] (3) -- (6);
\draw[-] (3) -- (4);
\draw[-] (4) -- (5);
\draw[-] (5) -- (7);
\draw[-] (6) -- (7);
\end{tikzpicture}
\end{center}
\vfill
\pagebreak
\problem{}
Is there an Eulerian path in the following graph? \par
\begin{center}
\begin{tikzpicture}[
node distance={20mm},
thick,
main/.style = {draw, circle}
]
\node[main] (1) {$x_1$};
\node[main] (2) [above right of=1] {$x_2$};
\node[main] (3) [below right of=1] {$x_3$};
\node[main] (4) [above right of=3] {$x_4$};
\node[main] (5) [above right of=4] {$x_5$};
\node[main] (6) [below right of=4] {$x_6$};
\node[main] (7) [below right of=5] {$x_7$};
\draw[-] (1) -- (2);
\draw[-] (1) -- (3);
\draw[-] (2) -- (5);
\draw[-] (2) -- (4);
\draw[-] (3) -- (6);
\draw[-] (3) -- (4);
\draw[-] (4) -- (5);
\draw[-] (5) -- (7);
\draw[-] (6) -- (7);
\end{tikzpicture}
\end{center}
\vfill
\problem{}
Is there an Eulerian path in the following graph? \par
\begin{center}
\begin{tikzpicture}[
node distance={20mm},
thick,
main/.style = {draw, circle}
]
\node[main] (1) {$x_1$};
\node[main] (2) [above right of=1] {$x_2$};
\node[main] (3) [below right of=1] {$x_3$};
\node[main] (4) [above right of=3] {$x_4$};
\node[main] (5) [above right of=4] {$x_5$};
\node[main] (6) [below right of=4] {$x_6$};
\node[main] (7) [below right of=5] {$x_7$};
\draw[-] (1) -- (2);
\draw[-] (1) -- (3);
\draw[-] (2) -- (4);
\draw[-] (3) -- (6);
\draw[-] (3) -- (4);
\draw[-] (4) -- (5);
\draw[-] (5) -- (7);
\draw[-] (6) -- (7);
\end{tikzpicture}
\end{center}
\vfill
\problem{}
When does an Eulerian path exist? \par
\hint{Look at the degree of each node.}
\vfill
\pagebreak

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src/Intermediate/An Introduction to Graph Theory/parts/2 planar.tex Normal file
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\section{Planar Graphs}
\textbf{TODO.} Will feature planar graphs, euler's formula, utility problem, utility problem on a torus
\vfill
\pagebreak

157
src/Intermediate/An Introduction to Graph Theory/parts/3 counting.tex Normal file
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\section{Counting Graphs}
\definition{}
A graph is \textit{bipartite} if its nodes can be split into two groups, where no two nodes in the same group share an edge. One such graph is shown below.
\problem{}
Draw a bipartite graph with 5 vertices.
\vfill
\problem{}
Is the following graph bipartite? \par
\hint{Be careful.}
\begin{center}
\begin{tikzpicture}
% Nodes
\begin{scope}
\node[main] (A) at (0mm, 0mm) {$A$};
\node[main] (B) at (0mm, -10mm) {$B$};
\node[main] (C) at (0mm, -20mm) {$C$};
\node[main] (D) at (20mm, 0mm) {$D$};
\node[main] (E) at (20mm, -10mm) {$E$};
\node[main] (F) at (20mm, -20mm) {$F$};
\end{scope}
% Edges
\draw
(A) edge (D)
(A) edge (E)
(B) edge (F)
(C) edge (E)
(C) edge (D)
(E) edge (F)
;
\end{tikzpicture}
\end{center}
\vfill
\definition{}
A \textit{subgraph} is a graph inside another graph. \par
In the next problem, the left graph contains the left graph. \par
The triangle is a subgraph of the larger graph.
\problem{}
Find two subgraphs of the triangle in the larger graph.
\begin{center}
\adjustbox{valign=c}{
\begin{tikzpicture}
% Nodes
\begin{scope}
\node[main] (1) {1};
\node[main] (2) [right of=1] {2};
\node[main] (3) [below of=1] {3};
\end{scope}
% Edges
\draw
(1) edge (2)
(2) edge (3)
(3) edge (1)
;
\end{tikzpicture}
}
\hspace{20mm}
\adjustbox{valign=c}{
\begin{tikzpicture}
% Nodes
\begin{scope}
\node[main] (1) {1};
\node[main] (4) [below of=1] {4};
\node[main] (3) [left of=4] {3};
\node[main] (5) [right of=4] {5};
\node[main] (6) [right of=5] {6};
\node[main] (2) [above of=6] {2};
\node[main] (7) [below of=4] {7};
\end{scope}
% Edges
\draw
(1) edge (4)
(2) edge (5)
(2) edge (6)
(3) edge (4)
(4) edge (5)
(4) edge (7)
(5) edge (6)
(3) edge (7)
;
\end{tikzpicture}
}
\end{center}
\vfill
\pagebreak
A few special graphs have names. Here are a few you should know before we begin:
\definition{The path graph}
The \textit{path graph} on $n$ vertices (written $P_n$) is a straight line of vertices connected by edges. \par
$P_5$ is shown below.
\begin{center}
\begin{tikzpicture}
\node[main] (1) {1};
\node[main] (2) [right of=1] {2};
\node[main] (3) [right of=2] {3};
\node[main] (4) [right of=3] {4};
\node[main] (5) [right of=4] {5};
\draw[-] (1) -- (2);
\draw[-] (2) -- (3);
\draw[-] (3) -- (4);
\draw[-] (4) -- (5);
\end{tikzpicture}
\end{center}
\definition{The complete graph}
The \textit{complete graph} on $n$ vertices (written $K_n$) is the graph that has $n$ nodes, all of which share an edge.
$K_4$ is shown below.
\begin{center}
\begin{tikzpicture}
\node[main] (1) {A};
\node[main] (2) [above right of=1] {B};
\node[main] (3) [below right of=1] {C};
\node[main] (4) [above right of=3] {D};
\draw[-] (1) -- (2);
\draw[-] (1) -- (3);
\draw[-] (1) -- (4);
\draw[-] (2) -- (3);
\draw[-] (2) -- (4);
\draw[-] (3) -- (4);
\end{tikzpicture}
\end{center}
\problem{}
\begin{enumerate}
\item How many times does $P_4$ appear in $K_9$?
\item How many times does $C_4$ appear in $K_9$?
\item How many times does $K_{4,4}$ appear in $K_9$?
\item How many times does $C_5$ appear in $K_8$?
\item How many times does $K_{3,3}$ appear in $K_{12}$?
\item How many times does $K_{3,3}$ appear in $K_{6,6}$?
\end{enumerate}

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src/Intermediate/An Introduction to Graph Theory/tikxset.tex Normal file
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\usetikzlibrary{arrows.meta}
\usetikzlibrary{shapes.geometric}
\usetikzlibrary{patterns}
% We put nodes in a separate layer, so we can
% slightly overlap with paths for a perfect fit
\pgfdeclarelayer{nodes}
\pgfdeclarelayer{path}
\pgfsetlayers{main,nodes}
% Layer settings
\tikzset{
% Layer hack, lets us write
% later = * in scopes.
layer/.style = {
execute at begin scope={\pgfonlayer{#1}},
execute at end scope={\endpgfonlayer}
},
%
% Arrowhead tweak
>={Latex[ width=2mm, length=2mm ]},
%
% Labels inside edges
label/.style = {
rectangle,
% For automatic red background in solutions
fill = \ORMCbgcolor,
draw = none,
rounded corners = 0mm
},
%
% Nodes
main/.style = {
draw,
circle,
fill = white,
line width = 0.4mm
},
every path/.style = {
line width = 0.3mm
},
node distance={20mm},
thick,
main/.style = {draw, circle}
}

274
src/Intermediate/Combinatorics/main.tex Executable file
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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\newcommand{\nck}[2] {
\ensuremath{
{}_{#1}C_{#2}
}
}
\newcommand{\npk}[2] {
\ensuremath{
{}_{#1}P_{#2}
}
}
\uptitlel{Advanced 1}
\uptitler{\smallurl{}}
\title{Combinatorics}
\subtitle{Prepared by Mark on \today}
\begin{document}
\maketitle
\section{Getting started}
An \textbf{ordered} arrangement of objects is called a \textit{permutation}. \par
An \textbf{unordered} selection of objects is called a \textit{combination}\footnotemark{}\hspace{-1ex}. \par
All the following problems involve permutations.
\footnotetext{A \say{combination lock} cares about the order of its digits, so its name is inaccurate. Such an object is actually a \textit{permutation} lock!}
\problem{}<ABCDE>
How many different ways are there to rearrange the letters ABCDE?
\begin{solution}
$5 \times 4 \times 3 \times 2 \times 1 = 120$
\end{solution}
\vfill
\problem{}<AtoZ>
How many different ways are there to arrange the letters ABCDEFG...XYZ? \par
You don't need to fully evaluate your answer, it is a \textit{very} big number. \par
\hint{Look at \ref{ABCDE} again, and try to create a general strategy.}
\begin{instructornote}
A hint for students that are stuck: \par
In \ref{ABCDE}, start with five blank spaces. How many choices are there for A's position? \par
Once A is placed, how many are left for B?
\end{instructornote}
\vfill
\pagebreak
\definition{}
The \textit{factorial} of a positive integer $x$ is $x \times (x-1) \times ... \times 1$. We denote this $x!$. \par
For example, $8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320$.
\problem{}
Compute $\frac{10!}{8!}$
\vfill
\problem{}
Convince yourself that $(n+1)! = n! \times (n+1)$, and use this fact to show that $0! = 1$.
\vfill
\problem{}
How many ways are there to choose three student council officers from a class of 20 students? \par
How many ways are there to choose a president, a vice-president, and a treasurer from the same class? \par
\hint{You answers should be different. In which case does order matter?}
\begin{instructornote}
Have your students consider the non-unique arrangements and count how many are redundant.
\end{instructornote}
\vfill
\pagebreak
\problem{}
Say you have 4 red balls and 3 green balls. How many different ways can you arrange them on the table in font of you? \par
\begin{solution}
Consider the sequence RRRRGGG. There are $4!$ ways to rearrange the red balls, and $3!$ ways to rearrange the green balls. This is true for any sequence.
So, our solution is $\frac{7!}{3!4!}$.
\end{solution}
\vfill
\problem{}
How many \textit{unique} anagrams can we create from the word CRESCENDO?
\begin{solution}
CRESCENDO = CC EE RSNDO, our solution is $\frac{9!}{2!2!} = 90720$
\end{solution}
\vfill
\problem{}<3fromABCDE>
Given the letters ABCDE, how many different three-letter words can we make without repeating letters?
\vfill
\pagebreak
\section{Permutations}
It would be convenient to have a general tool for counting permutations. Let us try to create one. \par
(Remember, permutations are \textit{ordered} arrangements of objects.)
First, let's create a function $\npk{n}{k}$, which tells us how many $k$-object permutations we can choose from a group of $n$ objects.
\problem{}
What is $\npk{5}{3}$? \par
\hint{See \ref{3fromABCDE}}
\vfill
``Choosing $k$ items from $n$'' is a lot like splitting our $n$ objects into two groups: those we choose, and those we don't.
\begin{center}
\begin{tikzpicture}
% Points
\path [draw=black, fill=black] (0,0) circle (5pt);
\path [draw=black, fill=black] (1,0) circle (5pt);
\path [draw=black, fill=white] (2,0) circle (5pt);
% "Choose these" bracket
\draw[shift={(-0.5, -1)}, color=oblue, thick] (0pt,0pt) -- (0pt,3pt);
\draw[color=oblue, thick] (-0.5, -1) -- (2.5, -1) node[below, midway] {Choose $k$ objects};
\draw[shift={(2.5, -1)}, color=oblue, thick] (0pt,0pt) -- (0pt,3pt);
\draw[-] (3, -0.5) -- (3, 0.5);
\path [draw=black, fill=white] (4,0) circle (5pt);
\path [draw=black, fill=black] (5,0) circle (5pt);
% "Leave these" bracket
\draw[shift={(3.5, -1)}, color=oblue, thick] (0pt,0pt) -- (0pt,3pt);
\draw[color=oblue, thick] (3.5, -1) -- (5.5, -1) node[below, midway] {Leave the rest};
\draw[shift={(5.5, -1)}, color=oblue, thick] (0pt,0pt) -- (0pt,3pt);
\end{tikzpicture}
\end{center}
If we rearrange these, we get different permutations. How can we count them?
\problem{}<nPk>
Using the above diagram, create a formula for $\npk{n}{k}$. \par
\hint{We're counting \textit{permutations}, so the order of items in the first group matters.}
\begin{solution}
$\npk{n}{k}= \frac{n!}{(n-k)!}$
There are $n!$ possible arrangements of $n$ objects. However, since the order of the elements not chosen does not matter, we'll end up with $(n-k)!$ redundant orderings of each.
\end{solution}
\vfill
\pagebreak
\section{Combinations}
Now, let's count \textit{combinations}. \par
Here, we only care about \textit{which} items we choose---not the order in which we choose them.
We'll make a function $\nck{n}{k}$ (\say{n choose k}), which will tell us
how many different ways we can choose $k$ items from a set of $n$.
\problem{}
Find an expression for $\nck{n}{k}$ by modifying your definition of $\npk{n}{k}$.
\vfill
Usually, $\nck{n}{k}$ is written as $\binom{n}{k}$. This is also called the \textit{binomial coefficient}.
\problem{}<manyballs>
Say you have a few coins on the table in font of you:
\begin{itemize}
\item 8 identical 1-kopek\footnotemark{} coins
\item 3 identical 2-kopek coins
\item 6 identical 5-kopek coins
\item 4 identical 10-kopek coins
\end{itemize}
How many distinct ways are there to arrange these coins in a row?
\footnotetext{Russian currency. Comparable to a penny, since 100 kopeks make a ruble.}
\vfill
\problem{}
Now, derive the \textit{multinomial coefficient} $\binom{n}{k_1, k_2, ..., k_m}$. \par
\vspace{1mm}
The multinomial coefficient tells us how many distinct ways there to arrange $n$ objects
of $m$ classes, where each class $i$ contains $k_i$ identical objects. \par
\hint{
In \ref{manyballs}, $n = 21$ and $(k_1, k_2, k_3, k_4) = (8, 3, 6, 4)$. \\
So, the solution to \ref{manyballs} should be given by the multinomial coefficient $\binom{21}{8,3,6,4}$.
}
\vfill
\pagebreak
\section{Applications}
\problem{}
How many ways can a class of 27 people be seated in 30 seats?
\vfill
\problem{}
The following is the map of a city. Each line is a one-way road, you can only drive up or right. \par
How many different paths can you take from A to B? \par
How many of them go through the center point?
\begin{tikzpicture}
\draw [step=0.5,gray] (0,0) grid (7*0.5,4*0.5);
\path [draw=black, fill=black] (0 * 0.5, 0 * 0.5) circle (2pt) node[below] {A};
\path [draw=black, fill=black] (3 * 0.5, 2 * 0.5) circle (2pt);
\path [draw=black, fill=black] (7 * 0.5, 4 * 0.5) circle (2pt) node[above] {B};
\end{tikzpicture}
\vfill
\problem{}
How many ways can you put 19 identical balls into 6 bins, leaving no bin empty?
\vfill
\problem{}
Given an exam with 4 problems, how many ways are there to assign positive point values to each problem so that the exam contains a total of 100 points?
\vfill
\problem{}
How many ways can we split the number 2016 into a sum of positive integers? \par
\note{Consider $2016 + 1$ and $1 + 2016$ distinct sums. Order matters.}
\begin{solution}
Split 2016 into ones, and put a \say{bit} between each pair. \par
This gives us $2^{2015}$ positions to place a bar, and thus $2^{2016}$ possible sums.
\vspace{2mm}
You could also sum over the usual stars-and-bars technique to get the same result. \par
Showing that they're equal could be a good bonus problem!
\end{solution}
\vfill
\problem{}
A staircase must be built up a wall. It will start 4.5 meters away from the wall, which is 1.5 meters tall. The height of each step is exactly 30 centimeters. The width of each step must be an integer multiple of 50 centimeters. In how many ways can the staircase be constructed?
\vfill
\pagebreak
\end{document}

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title = "Combinatorics"
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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\uptitlel{Intermediate 2}
\uptitler{\smallurl{}}
\title{Newton's Laws of Motion}
\subtitle{
Prepared by Mark on \today \\
Based on a handout by Oleg Gleizer
}
\begin{document}
\maketitle
\section{Newton's First Law}
If the net force acting on an object is zero, the velocity of that object does not change. \\
Conversely, if the velocity of an object doesn't change, the net force acting on it is zero.
\medskip
In the context of vectors, the ``net force'' is the sum of all the force vectors acting on the object. ``Speed'' is the length (or \textit{magnitude}) of the velocity vector.
\problem{}
There are no forces acting on the object
$A$ below. The current velocity of the object,
in meters per second, is represented by the vector $\overrightarrow{v}$.
Draw the position of the object two seconds later.
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-11,-7) grid (1,1);
\draw [line width = 1.5pt, ->] (-9,0) -- (-5,-3);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-6.9,-1.5);
\filldraw (-9,0) circle (3pt);
\coordinate [label = left:{$A$}] (a) at (-9,0);
\end{tikzpicture}
\end{normalsize}
\end{center}
The sides of the grid squares
on the picture above are one metre long.
What is the speed of the object?
\vfill
What distance would the object cover
in two seconds? \\
\vfill
\pagebreak
\problem{}
There are no forces acting on the object
$A$ below.
The current velocity of this object,
in miles per hour,
is represented by the vector $\overrightarrow{v}$.
Draw the position of the object half an hour later.
\vspace{20pt}
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-11,-7) grid (1,1);
\draw [line width = 1.5pt, ->] (-9,0) -- (-1,-6);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-4.8,-3);
\filldraw (-9,0) circle (3pt);
\coordinate [label = left:{$A$}] (a) at (-9,0);
\end{tikzpicture}
\end{normalsize}
\end{center}
The sides of the grid squares
on the picture above are ten miles long.
What is the speed of this object?
\vfill
What distance would the object cover
in half an hour? \\
\vfill
What distance would the object cover
in three hours? \\
\vfill
\pagebreak
\definition{Acceleration}
\textit{Acceleration} is the rate at which velocity changes.
Let's represent acceleration by the vector $\overrightarrow{a}$.
If $\overrightarrow{a}$ does not change over time, then the speed
of an object at time $t$ is given by the following equation:
\begin{equation} \label{eq:ac}
\overrightarrow{v_t} = \overrightarrow{v_0} +
t \overrightarrow{a}.
\end{equation}
\problem{}
It takes a minivan seven seconds to accelerate from 0 to 60 miles per hour. Find its acceleration in meters per second squared. \\
\hint{1 mile $\approx$ 1600 meters}
\vfill
Note that in the previous problem, motion is one-dimensional (it happens on a straight line). In this case, both velocity and acceleration are one-dimensional vectors---in other words, (real) numbers! \\
In general, velocity and acceleration are \textit{not} numbers, but vectors. You'll see this in the next few problems.
\pagebreak
Now, let's try a few examples with vectors: \\
\medskip
Consider an object, currently at $P_0$, moving along the vector $\overrightarrow{v_0}$.
As before, let $\overrightarrow{a}$ represent the acceleration of the object. This could be caused by gravity, current, or any other constant force.
\medskip
One second later, the object will be at $P_1$, and has the velocity vector $\overrightarrow{v_1} = \overrightarrow{v_0} + \overrightarrow{a}$.
\medskip
Two seconds later, the object will be positioned at $P_2$ and will have the velocity vector $\overrightarrow{v_2} = \overrightarrow{v_0} + 2\overrightarrow{a}$.
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-19,-15) grid (1,1);
\draw [line width = 1.5pt, ->] (-18,-1) -- (-16,-4);
\coordinate [label = left:{$\overrightarrow{v_0}$}]
(v0) at (-17.1,-2.6);
\draw [line width = 1.5pt, ->] (-18,-1) -- (-15,-2);
\coordinate [label = above:{$\overrightarrow{a}$}]
(a) at (-16.4,-1.5);
\filldraw (-18,-1) circle (3pt);
\coordinate [label = above:{$P_0$}] (p0) at (-18,-1);
\filldraw (-14.5,-4.5) circle (3pt);
\coordinate [label = right:{$P_1$}] (p1) at (-14.45,-4.4);
\draw [line width = 1.5pt, ->] (-14.5,-4.5) -- (-9.5,-8.5);
\coordinate [label = right:{$\overrightarrow{v_1}$}]
(v1) at (-11.75,-6.5);
\draw [->] (-14.5,-4.5) -- (-12.5,-7.5);
\coordinate [label = left:{$\overrightarrow{v_0}$}]
(v01) at (-13.6,-6);
\draw (-12.5,-7.5) -- (-9.5,-8.5);
\coordinate [label = left:{$\overrightarrow{a}$}]
(a1) at (-11,-8.3);
\filldraw (-8,-9) circle (3pt);
\coordinate [label = above:{$P_2$}] (b) at (-8,-9);
\draw [->] (-8,-9) -- (-6,-12);
\coordinate [label = left:{$\overrightarrow{v_0}$}]
(v02) at (-7,-10.5);
\draw (-6,-12) -- (0,-14);
\coordinate [label = left:{$2\overrightarrow{a}$}]
(a1) at (-3.5,-13);
\draw [line width = 1.5pt, ->] (-8,-9) -- (0,-14);
\end{tikzpicture}
\end{normalsize}
\end{center}
$t$ seconds later, the object
will have the velocity
\[
\overrightarrow{v_t} = \overrightarrow{v_0} + t \overrightarrow{a}
\]
Note that the word {\it acceleration} has two different meanings.
One is a vector, as in the example above.
The other meaning is the length
of the vector that represents the magnitude
of the velocity change. In this case,
we do not put an arrow above the letter
$a$ representing acceleration.
In other words, $a = |\overrightarrow{a}|$.
Similarly, speed is the length
of the velocity vector,
$v = |\overrightarrow{v}|$.
\vfill
\pagebreak
\example{}<accel_example>
Moving with a constant acceleration,
an object moves from point $A$ to point $B$
in one second. The velocities of the motion,
in metres per second, are represented
by the vectors $\overrightarrow{v_0}$ and
$\overrightarrow{v_1}$.
Find the acceleration vector. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-18,-8) grid (1,1);
\draw [line width = 1.5pt, ->] (-17,-1) -- (-13,-4);
\coordinate [label = above:{$\overrightarrow{v_0}$}]
(v0) at (-14.9,-2.5);
\filldraw (-17,-1) circle (3pt);
\coordinate [label = left:{$A$}] (a) at (-17,-1);
\filldraw (-10,-3) circle (3pt);
\coordinate [label = above right:{$B$}] (b) at (-10,-3.1);
\draw [line width = 1.5pt, ->] (-10,-3) -- (0,-4);
\coordinate [label = above:{$\overrightarrow{v_1}$}]
(v1) at (-5,-3.5);
\end{tikzpicture}
\end{normalsize}
\end{center}
According to formula on the previous page,
$\overrightarrow{v_1} = \overrightarrow{v_0} + (1\text{ sec}) \times \overrightarrow{a}$. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-18,-8) grid (1,1);
\draw [line width = 1.5pt, ->] (-17,-1) -- (-13,-4);
\coordinate [label = above:{$\overrightarrow{v_0}$}]
(v0) at (-14.9,-2.5);
\filldraw (-17,-1) circle (3pt);
\coordinate [label = left:{$A$}] (a) at (-17,-1);
\filldraw (-10,-3) circle (3pt);
\coordinate [label = above right:{$B$}] (b) at (-10,-3.1);
\draw [line width = 1.5pt, ->] (-10,-3) -- (0,-4);
\coordinate [label = above:{$\overrightarrow{v_1}$}]
(v1) at (-5,-3.5);
\draw [line width = 1.5pt, ->] (-10,-3) -- (-6,-6);
\coordinate [label = below:{$\overrightarrow{v_0}$}]
(v01) at (-8.2,-4.5);
\end{tikzpicture}
\end{normalsize}
\end{center}
\problem{}
Complete \ref{accel_example}: \\
Draw the acceleration vector on the above diagram.
\vfill
\pagebreak
\problem{}
At an initial moment in time,
the object at point $A$
is moving with the velocity vector
$\overrightarrow{v_0}$,
measured in metres per second.
The constant acceleration
acting on the object
is represented by the vector
$\overrightarrow{a}$,
measured in meters per second squared.
Two seconds later, the object is located
at point $B$. Draw its velocity vector
at that moment.
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-19,-15) grid (1,1);
\draw [line width = 1.5pt, ->] (-18,-1) -- (-16,-4);
\coordinate [label = left:{$\overrightarrow{v_0}$}]
(v0) at (-17.1,-2.6);
\draw [line width = 1.5pt, ->] (-18,-1) -- (-15,-2);
\coordinate [label = above:{$\overrightarrow{a}$}]
(a) at (-16.4,-1.5);
\filldraw (-18,-1) circle (3pt);
\coordinate [label = left:{$A$}] (a) at (-18,-1);
\filldraw (-8,-9) circle (3pt);
\coordinate [label = left:{$B$}] (b) at (-8,-9);
%\draw [line width = 1.5pt, ->] (-8,-9) -- (0,-14);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vspace{12pt}
The side length of the grid squares
on the picture above is one metre.
Find the speed of the object when it is
at point $B$.
\vfill
\pagebreak
\section{Newton's Second Law}
The second law is simple:
\[
\overrightarrow{F} = m \times \overrightarrow{a}
\]
Here, $\overrightarrow{F}$ is the net force acting on an object with mass $m$, and $\overrightarrow{a}$ is the acceleration the object experiences as a result of this action. Mass is a measure of an object's \textit{inertia}: the heavier an object is, the more effort it takes to change its velocity. \\
In civilized countries, mass is measured in grams and force is measured in \textit{newtons}. One newton is the force it takes to accelerate 1 kg of mass to 1 meter per second. In other words,
\[
1\ N = (1\ kg) (1\ \frac{m}{s})
\]
\problem{}
The {\it Millennium Falcon}, at point $A$ at the moment,
is trying to escape from the Death Star, which is trying
to arrest the ship using its attracting beam.
The thrust of the Falcon's engines,
200,000 kN in total, is represented
by the vector $\overrightarrow{T}$.
The force of the beam is represented
by the vector $\overrightarrow{B}$.
In addition, a nearby star
exerts a gravitational force of
$\overrightarrow{G}$ on the ship.
Draw the vector of the net force
acting on the vessel. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-16,-9) grid (1,1);
\draw [line width = 1.5pt, ->] (-8,-5) -- (-12,-5);
\coordinate [label = below:{$\overrightarrow{T}$}] (v) at (-10,-5.1);
\filldraw (-8,-5) circle (3pt);
\coordinate [label = below:{$A$}] (a) at (-8,-5.1);
\draw [line width = 1.5pt, ->] (-8,-5) -- (-3,-6);
\coordinate [label = above:{$\overrightarrow{B}$}] (b) at (-5.4,-5.5);
\draw [line width = 1.5pt, ->] (-8,-5) -- (-9,-2);
\coordinate [label = left:{$\overrightarrow{G}$}] (g) at (-8.6,-3.5);
\end{tikzpicture}
\end{normalsize}
\end{center}
The mass of the ship
is 400,000 kg. What acceleration,
in metres per second squared,
does the ship experience?
\vfill
The purpose of this daring mission
was to find out the force
that the the attracting beam exerts.
However, Han Solo is not particularly
good with vectors. Please help him complete the mission.
\vfill
\pagebreak
\section{Newton's Third Law}
Newton's third law also concerns forces. It states that \textit{every action has an equal and opposite reaction}.
In other words, this means that when one object exerts force on another, the second simultaneously exerts a force equal in magnitude and opposite in direction to the force exerted on it by the first.
\medskip
In fact, we saw this in our first handout!
\begin{tcolorbox}[
colback=white,
colframe=gray!75!black,
title={ Handout 1, Page 7 }
]
Here is an important example of an inverse vector. When you stand still, the floor pushes you up with the force opposite to the force of the gravitational pull, a.k.a. \textit{weight}.
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw (0,0) -- (6,0);
\draw (2.5,0) -- (3,1);
\draw (3.5,0) -- (3,1);
\draw (3,1) -- (3,2);
\draw (3,2.3) circle (.3);
\draw (2.6,1.7) -- (3.4,1.7);
\draw [->, line width = 2pt] (2.5,0) -- (2.5,-1);
\coordinate [label = below: {weight}]
(g) at (2.5,-1.1);
\draw [->, line width = 2pt] (3.5,0) -- (3.5,1);
\coordinate [label = right: {floor reaction}]
(w) at (3.7,1);
\end{normalsize}
\end{tikzpicture}
\end{center}
The two opposing vectors add up to the zero vector, and therefore you don't move.
\end{tcolorbox}
\problem{}
In a second, the truck and car on the picture below will collide in a crash test.
The weight of the truck is 20 metric tons. The weight of the car is 2 metric tons.
Find the ratio of the accelerations,
$a_t$ (acceleration of the truck) and
$a_c$ (that of the car),
the vehicles will undergo. \\
\begin{center}
\begin{tikzpicture}
\draw [gray!50!black] (-7,0) -- (7,0);
\filldraw [gray!80!blue] (-3.6,1.5) -- (-4.4,1.5) --
(-4.4,2) -- (-6.8,2) -- (-6.8,.3) -- (-3.6,.3) --
(-3.6,1.5);
\filldraw (-6,.3) circle (.3);
\filldraw [gray] (-6,.3) circle (.15);
\filldraw (-4,.3) circle (.3);
\filldraw [gray] (-4,.3) circle (.15);
\filldraw [red!80!white] (4,.2) -- (6.6,.2) -- (6.55,.6) --
(6.2,.6) -- (6.1,.8) -- (4.9,.8) -- (4.7,.6)--
(4.2,.6) -- (4,.45) -- (4,.2);
\filldraw (6,.2) circle (.2);
\filldraw [gray] (6,.2) circle (.1);
\filldraw (4.6,.2) circle (.2);
\filldraw [gray] (4.6,.2) circle (.1);
\end{tikzpicture}
\end{center}
\medskip
\begin{huge}
$a_t \div a_c =$
\end{huge}
\vfill
\newpage
\section{Bonus Problems}
\problem{}
Use vector algebra to prove
that diagonals of a parallelogram
in the Euclidean plane
split each other in halves. \\
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw (0,0) -- (4,-2);
\coordinate [label = left: {$A$}] (a) at (0,0);
\coordinate [label = right: {$B$}] (b) at (4,-2);
\coordinate [label = above: {$\overrightarrow{v}$}] (v1) at (2,-.9);
\draw (-1,-3) -- (3,-5);
\coordinate [label = left: {$C$}] (c) at (-1,-3);
\coordinate [label = right: {$D$}] (d) at (3,-5);
\coordinate [label = below: {$\overrightarrow{v}$}] (v2) at (1,-4.1);
\draw (-1,-3) -- (0,0);
\draw (3,-5) -- (4,-2);
\coordinate [label = left: {$\overrightarrow{w}$}] (w1) at (-.55,-1.5);
\coordinate [label = right: {$\overrightarrow{w}$}] (w2) at (3.55,-3.5);
\draw (0,0) -- (3,-5);
\draw (-1,-3) -- (4,-2);
\coordinate [label = above: {$M$}] (m) at (1.7,-2.5);
\end{normalsize}
\end{tikzpicture}
\end{center}
\vfill
\newpage
\problem{}
Use vector algebra to prove that
all the three medians of a triangle
in the Euclidean plane intersect
at one point that splits each of the medians
in the ratio 2:1, counting from the vertices. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture}
\draw (0,0) -- (6,-1);
\draw (0,0) -- (3,-5);
\draw (3,-5) -- (6,-1);
\draw [color = white] (3,-5) -- (9,-6);
\draw [color = white] (6,-1) -- (9,-6);
\coordinate [label = left: {$A$}] (a) at (0,0);
\coordinate [label = right: {$B$}] (b) at (6,-1);
\coordinate [label = below: {$C$}] (c) at (3,-5);
\coordinate [label = above: {$\overrightarrow{v}$}]
(v) at (3,-.45);
\coordinate [label = left: {$\overrightarrow{w}$}]
(w) at (1.45,-2.5);
\filldraw (4.5,-3) circle (1.5pt);
\coordinate [label = right: {$M_A$}] (ma) at (4.65,-2.85);
\draw (0,0) -- (4.5,-3);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\newpage
\problem{}
Find the area of an equilateral triangle
with the side length $a$.
\vfill
\problem{}
Does there exist an equilateral triangle
in the Euclidean plane
such that all of its vertices
have integral coordinates?
Why or why not? \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-15,-9) grid (1,1);
\draw [line width = 1.5pt, ->] (-7,-9) -- (-7,1);
\coordinate [label = above:{$y$}] (y) at (-7,1);
\draw [line width = 1.5pt, ->] (-15,-4) -- (1,-4);
\coordinate [label = right:{$x$}] (x) at (1,-4);
\coordinate [label = below:{$1$}] (ox) at (-6,-4);
\coordinate [label = left:{$1$}] (oy) at (-7,-3);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\end{document}

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title = "Newton's Laws"
[publish]
handout = true
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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[solutions]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
% Quick P() macro.
% \P is already defined,
% but we won't need it in this doc.
\renewcommand{\P}[1] {P(#1)}
\newcommand{\Pt}[1] {
\ensuremath{\P{\text{#1}}}
}
\uptitlel{Intermediate 2}
\uptitler{\smallurl{}}
\title{Probability}
\subtitle{
Prepared by Mark on \today \\
Based on a handout by Oleg Gleizer
}
\begin{document}
\maketitle
\problem{}
A Zoo manager thinks of a way
to set up a new pavilion.
He has 7 different plants
and 12 different animals. \\
\begin{itemize}
\item How many ways are there to choose two animals and three plants?
\vfill
\item The manager finds that he only has 5 exhibits available. How many different sets of animals can he choose, if only one can be in each exhibit time?
\vfill
\end{itemize}
\problem{}
How many different 7-symbol license plates are possible if the first three symbols are letters and the remaining four are digits 0-9? \\
\hint{Symbols can repeat, but letters must be uppercase}
\vfill
\pagebreak
\problem{}
There are two plates on a table.
One plate has 10 identical candies,
the other has 8 different fruits. \\
\begin{itemize}
\item How many ways are there to choose one candy?
\vfill
\item How many ways are there to choose seven candies?
\vfill
\item How many ways are there to choose five fruits?
\vfill
\item How many ways are there to choose three candies and six fruits?
\vfill
\item Gregory chooses two fruits and two candies, and lines up the four objects on the table. In how many ways can he do it?
\vfill
\item Gregory chooses five fruits and seven candies, and lines up the twelve objects on the table. In how many ways can he do it?
\vfill
\end{itemize}
\pagebreak
\vspace{190pt}
\pagebreak
\begin{center}
\section{Probability}
\end{center}
\vspace{10pt}
A \textit{probability}, also known as a \textit{chance}, is a number showing how likely some event is to happen. Let us call the event $X$. Then the probability of $X$ taking place is
$$
\P{X} = \frac{
\text{The number of the outcomes such that $X$ happens.}
}{
\text{The number of all the possible outcomes.}
}
$$
Note that by definition, $0 \leq P(X) \leq 1$. \\
In some of the following problems, we will be flipping a coin. Let us use $H$ to represent the event of the coin landing heads, and $T$, the event of the coin landing tails.
\problem{}
Compute each of the following:
\begin{enumerate}
\item \Pt{Rolling a six-sided die and getting 2}
\item \Pt{Flipping a coin twice and getting the sequence HH}
\item \Pt{Flipping a coin twice and getting one head and one tail in any order}
\item \Pt{Rolling two six-sided dice and getting a sum of 5}
\end{enumerate}
\vfill
\pagebreak
Some parts of the previous problem involve repeated trials: two dice, or two coins. You may have solved these by listing out all the possible outcomes. Though this simple approach works for small problems, it isn't particularly useful for larger ones: ten coin flips create 1024 possible outcomes, and ten dice rolls, 60466174.
\medskip
A better way to think about repeated trials is as a ``tree,'' where each outcome represents a path. The following tree represents two coin flips, and the four paths down it (from left to right) correspond to the four possible outcomes: HH, HT, TH, TT.
% Ugly hack
\tikzstyle{level 1}=[level distance=3.5cm, sibling distance=3.5cm]
\tikzstyle{level 2}=[level distance=3.5cm, sibling distance=2cm]
\tikzstyle{split} = [text width=1em, text centered]
\tikzstyle{tsplit} = [text width=0, text centered]
\tikzstyle{end} = [minimum width=3pt, inner sep=0pt]
\begin{center}
\begin{tikzpicture}[grow=right]
\node[tsplit] {}
child {
node[split] {T}
child {
node[end, label=right:{T\ \ \ (TT)}] {}
edge from parent
}
child {
node[end, label=right:{H\ \ \ (TH)}] {}
edge from parent
}
edge from parent
}
child {
node[split] {H}
child {
node[end, label=right:{T\ \ \ (HT)}] {}
edge from parent
}
child {
node[end, label=right:{H\ \ \ (HH)}] {}
edge from parent
}
edge from parent
};
\end{tikzpicture}
\end{center}
If we label each edge with the probability of each event, we can calculate the probability of each outcome by multiplying the edges we passed:
\begin{center}
\begin{tikzpicture}[grow=right]
\node[tsplit] {}
child {
node[split] {T}
child {
node[end, label=right:{T\ \ \ ({TT, $\frac{1}{4}$})}] {}
edge from parent
node[below] {$\frac{1}{2}$}
}
child {
node[end, label=right:{H\ \ \ ({TH, $\frac{1}{4}$})}] {}
edge from parent
node[below] {$\frac{1}{2}$}
}
edge from parent
node[below] {$\frac{1}{2}$}
}
child {
node[split] {H}
child {
node[end, label=right:{T\ \ \ ({HT, $\frac{1}{4}$})}] {}
edge from parent
node[below] {$\frac{1}{2}$}
}
child {
node[end, label=right:{H\ \ \ ({HH, $\frac{1}{4}$})}] {}
edge from parent
node[below] {$\frac{1}{2}$}
}
edge from parent
node[below] {$\frac{1}{2}$}
};
\end{tikzpicture}
\end{center}
We can formalize this idea as follows:
\proposition{}
If we have two independent events $A$ and $B$, then $\Pt{A and B} = \P{A} \times \P{B}$. \\
Usually we write $\Pt{A and B}$ as $\P{A \cap B}$. \\
\vfill
Here's another important thought:
\proposition{}
If the probability of event $A$ happening is $\P{A}$, the probability of $A$ \textit{not} happening is $1 - \P{A}$
\pagebreak
\problem{}
There are three cans of white paint and two cans of black paint in a dark storage room. You take two cans out without looking. What is the probability that you'll choose two cans of the same color?
\vfill
\problem{}
Hospital records show that of patients
suffering from a certain disease,
75\% die of it. What is the probability
that of 5 randomly selected patients,
4 will recover? \\
\hint{What is the probability of a patient recovering?}
\vfill
\problem{}
When Oleg calls his daughter Anya,
the chance of the call getting through is 60\%.
How likely is it to have at least one connection
in four calls?
\vfill
\problem{}
The chance of a runner to improve
his own personal record in a race is $p$.
What is the probability that his record will improve after 3 races?
\vfill
\newpage
\problem{}
You toss a pair of fair dice five times.
What is the probability that you get a sum of ten exactly two times?
\vfill
\problem{}
You toss a pair of fair dice five times.
What is the probability that you get ten
at least twice?
\vfill
\problem{}
A fair coin is tossed 4 times. What is the chance of getting more heads than tails?
\vfill
\problem{}
A pharmaceutical study shows that a new drug causes negative side effects in 3 of every 100 patients.
To check the number, a researcher chooses 5 random people to survey.
Assuming the study is accurate, what is the probability of the following? \\
\begin{enumerate}
\item None of the five patients experience side effects.
\item At least two experience side effects.
\end{enumerate}
\vfill
\pagebreak
\problem{}
You pick up a natural number (positive integer)
at random. What is the probability
that the number is divisible by either two
or three?
\vfill
\problem{}
Three players are tossing a fair coin.
The first to have a HEAD wins.
What are the players' chances of winning?
\vfill
\section{Harder Probabilities}
\problem{}
Oleg wrote ten letters to Math Circle parents
and addressed the ten envelopes. However, he
left the final stages of mailing to a careless clerk who
didn't pay attention, inserting the letters
into the envelopes at random.
(However, he did manage to fit exactly
one letter in each envelope.)
What is the probability that exactly nine of
the ten letters is correctly addressed?
\vfill
\problem{}
On a sold-out flight, the first person to
board the plane forgot which seat was his
and chose a random seat. Subsequent passengers
took their assigned seat if available, or a
randomly chosen seat if not. When the last
person boards, there is only one seat left.
What is the probability that this was the
seat assigned to the last passenger?
\vfill
\pagebreak
\problem{}
Your new neighbor has two children.
Assuming that it is equally likely
for a child to be a boy or a girl,
what is the probability that both
of the neighbor's children are girls?
Does the probability change if you
discover that one of the children
is indeed a girl? If so, how?
\vfill
\problem{}
A bag contains a marble which is either
black or white --- but we don't know which.
We put a white marble into the bag and shake it.
We then draw out a white marble.
What is the probability that the marble
left inside the bag is also white?
\vfill
\problem{The Monty Hall Problem}
You are a contestant on a certain game show.
There are three doors.
Behind one door is a brand-new car.
Behind the other two doors are goats.
You are invited to choose one of the doors.
Before opening the selected door, the game show host
opens one of the other two doors, revealing a goat.
You can now either keep your (original)
choice, or switch to the other unopened door.
Which choice gives you a better chance of winning
the car? Does it matter? Explain your answer.
\vfill
\end{document}

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[metadata]
title = "Probability"
[publish]
handout = true
solutions = false

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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\usepackage{units}
\uptitlel{Intermediate}
\uptitler{\smallurl{}}
\title{Proof Techniques}
\subtitle{Prepared by Mark on \today{}}
% Default \implies is ugly
\let\implies\Rightarrow
\let\rimplies\Leftarrow
\let\iff\Leftrightarrow
\let\notimplies\nRightarrow
\begin{document}
\maketitle
\input{parts/0 intro}
\input{parts/1 contradiction}
\input{parts/2 induction}
\end{document}

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[metadata]
title = "Proof Techniques"
[publish]
handout = false
solutions = true

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\section{Introduction}
\definition{}
A \textit{proof} is a mathematical argument that irrefutably
demonstrates the truth of a given proposition.
\vspace{2mm}
Every proof involves some sort of \textit{implication}, denoted $\implies$. \par
The statement \say{$A$ implies $B$} (written $A \implies B$), means that $B$ is true whenever $A$ is true.
\problem{}<trueimplies>
Which of the following are true? \par
\note{You don't need to provide a proof.}
\begin{itemize}
\item $x$ is prime $\implies$ $x$ is odd.
\item $x$ is real $\implies$ $x$ is rational.
\item $x$ is odd $\implies$ $x$ is prime.
\end{itemize}
\vfill
\problem{}
As you saw above, $A \implies B$ does not guarantee that $B \implies A$. \par
Find two new statements $A$ and $B$ so that $A \implies B$ but $B \notimplies A$. \par
\hint{\say{new} as in \say{not from \ref{trueimplies}}}
\begin{solution}
A fairly trite example is below. \par
Note that \say{$X$ is a square} is a subset of the statement \say{$X$ is a rectangle.}
\vspace{2mm}
$X$ is a square $\implies$ $X$ is a rectangle, but $X$ is a rectangle $\notimplies$ $X$ is a square.
\end{solution}
\vfill
\pagebreak
\definition{}<iffdef>
As we just saw, implication is one-directional. \par
The statements $A \implies B$ and $B \implies A$ are independent of one another. \par
\vspace{1mm}
If both are true, we write $A \iff B$. This can be read as \say{$A$ if and only if $B$.} \par
In text, \say{if and only if} is often abbreviated as iff. \par
\vspace{1mm}
Bidirectional implication is the strongest relationship we can have between two statements: \par
If $A \iff B$, $A$ and $B$ are \textit{equivalent.} They are always either both true or both false.
\definition{}
The \textit{floor} of $x$ is the largest integer $a$ so that $a \leq x$. This is denoted $\lfloor x \rfloor$. \par
The \textit{ceiling} of $x$ is the largest integer $a$ so that $a \geq x$. This is denoted $\lceil x \rceil$.
\generic{Property:}
If $b_1 \leq a \leq b_2$ and $b_1 = b_2$, we must have that $b_1 = a = b_2$. \par
\vspace{1mm}
Also, if $a \leq b$ and $a \geq b$, we must have that $a = b$. \par
This is a trick we often use when showing that two quantities are equal.
\problem{}
Although $A \iff B$ looks like a single statement, we often need to prove each direction separately. \par
Show that $x \in \mathbb{Z}$ iff $\lfloor x \rfloor = \lceil x \rceil$
\begin{solution}
\textbf{Forwards:} $x \in \mathbb{Z} ~\implies~ \lfloor x \rfloor = \lceil x \rceil$ \par
If $x \in \mathbb{Z}$, by definition we have that $\lfloor x \rfloor = x$ and $\lceil x \rceil = x$ \par
So, $\lfloor x \rfloor = \lceil x \rceil$
\vspace{2mm}
\textbf{Backwards:} $x \in \mathbb{Z} ~\rimplies~ \lfloor x \rfloor = \lceil x \rceil$ \par
Assume that $\lfloor x \rfloor = \lceil x \rceil$, and show that $x$ is an integer. \par
Note that $\lfloor x \rfloor \leq x \leq \lceil x \rceil$ (by definition) \par
Since $\lfloor x \rfloor = \lceil x \rceil$, we must have that $\lfloor x \rfloor = x = \lceil x \rceil$. \par
$\lfloor x \rfloor$ is an integer, so $x$ must be an integer.
\end{solution}
\vfill
\pagebreak
\problem{}
We don't always need to prove each direction of an iff statement separately. \par
\begin{itemize}[itemsep = 1mm]
\item Convince yourself that we can \say{chain} iffs together: \par
If we show that $A \iff B \iff C \iff D$, do we know that $A \iff D$?
\item Does this still work if $A \iff B \implies C \iff D$?
\item Show that $x^2 - 6x - 6 = 3 \iff x = 3$ by building a chain of iffs. \par
\hint{You remember how to factor quadratics, right?}
\end{itemize}
\begin{solution}
Does this still work if $A \iff B \implies C \iff D$? \par
Of course not. $D \notimplies A$ since $C \notimplies B$.
We can only conclude that $A \implies D$.
\linehack{}
$x^2 - 6x - 6 = 3$ \par
$\iff x^2 - 6x - 9 = 0$ \par
$\iff (x-3)^2 = 0$ \par
$\iff x-3 = 0$ \par
$\iff x = 0$
Note that this is a well-defined argument. Every step is an iff statement we can rigorously justify.
We're not hand-wavily \say{rearranging} one equation into another,
we're building a chain of implications that eventually bring us to our result.
This is the logic behind most algebraic proofs.
\end{solution}
\vfill
\problem{}
Another trick you may find useful is the \say{implication cycle.} \par
Convince yourself that if $A \implies B \implies C \implies D \implies A$, \par
we can conclude that $A$, $B$, $C$, and $D$ are equivalent. \par
\note{$A \iff B$ means that $A$ and $B$ are equivalent. See \ref{iffdef}.}
\vfill
\problem{Bonus}
Use an implication cycle to show that the following definitions of a \textit{squarefree integer} are equivalent.
%\hint{Show that $A \implies B \implies C \implies D \implies A$}
\begin{enumerate}
\item $n^2$ does not divide $q$ for any $n \in \mathbb{Z}^+$, $n \neq 1$
\item $p^2$ does not divide $q$ for any prime $p$
\item $q$ is a product of distinct primes
\item $q ~|~ n^k \implies q ~|~ n$ for all $n, k \in \mathbb{Z}^+$
\end{enumerate}
\begin{solution}
Assume $q$ has a square factor, so that $q = an^2$ for some $a, n \in \mathbb{Z}^+$ \par
By D, we know that $q ~|~ (an)^2 \implies q ~|~ an$ \par
But $q ~|~ an \implies an^2 ~|~ an$ \par
$\implies n = 1$
\vspace{2mm}
So, $q$ cannot have a square factor that isn`t 1.
\end{solution}
\vfill
\pagebreak
Often enough, proving a statement is simply a matter of \say{definition chasing,}
where we expand the symbols used in the statement we're proving, and then do a bit of
rearranging to arrive at the result we want.
\definition{}
Let $n, x \in \mathbb{Z}$. \par
We say \say{$n$ divides $x$} if $x = kn$ for some $k \in \mathbb{Z}$
\definition{}
Let $a, b \in \mathbb{Z}$ and $n \in \mathbb{Z}^+$. \par
We say \say{$a$ is congruent to $b$ modulo $n$} (and write $a \equiv_{n} b$) if $n$ divides $a - b$. \par
\definition{}
Let $a, b \in \mathbb{Z}$. We define $a ~\%~ b$ as the remainder of $a \div b$.
\problem{}
Let $a, b, n$ be positive integers. \par
Show that $a + b ~\equiv_n (a ~\%~ n) + (b ~\%~ n)$
\begin{solution}
$a + b ~\equiv_n (a ~\%~ n) + (b ~\%~ n)$ \par
\vspace{2mm}
...can be rewritten as... \par
$n$ divides $a + b - (a ~\%~ n) - (b ~\%~ n)$ \par
\vspace{2mm}
...which can be rearranged to... \par
$n$ divides $(a - (a ~\%~ n)) + (b - (b ~\%~ n))$
\vspace{2mm}
...which is clearly true, if you think about the meaning of \say{$n$ divides $x$} and \say{$a ~\%~ b$}.
\end{solution}
\vfill
\pagebreak

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\section{Proofs by Contradiction}
\definition{}
A very common proof technique is \textit{proof by contradiction}.
It works as follows:
\vspace{2mm}
Say we want to prove a statement $P$. Assume that $P$ is false, and show that this implies a false statement.
In other words, we show that $P$ can't \textit{not} be true. \par
If it's false, we either get a known impossibility ($1 = 2$, pigs fly, et cetera), \par
or we find that (not $P$) $\implies$ (not (not $P$)), which is a contradiction in itself.
\problem{}
Show that the set of integers has no maximum using a proof by contradiction.
\begin{solution}
Assume there is a maximal integer $x$. \par
$x + 1$ is also an integer. \par
$x + 1$ is larger than $x$, which contradicts our original assumption!
\vspace{2mm}
This is a \textit{proof by infinite descent}, a special type of proof by contradiction.\par
Such proofs have the following structure:
\begin{itemize}
\item Assume there is a smallest (or largest) object with property $X$.
\item Show that we have an even smaller object that has the same property $X$.
\end{itemize}
\end{solution}
\vfill
\definition{}
We say a number $x \in \mathbb{R}$ is \textit{rational} if we can write $x$ as $\nicefrac{p}{q}$, \par
where $p, q$ are integers with no common factors.
\problem{}
Show that $\sqrt{2}$ is irrational. \par
\hint{Start by chasing definitions. \say{Not irrational} $=$ \say{rational.}}
\begin{solution}
Suppose $\sqrt{2}$ is rational. Then, there exist $p, q$ so that $\sqrt{2} = \frac{p}{q}$.
\vspace{2mm}
Multiply by $q$ and square to find that $2q^2 = p^2$, which implies that $p^2$ is even. \par
This then implies that $p$ is even, \par
which implies that $p^2$ is divisible by 4, \par
which implies that $q^2$ is divisible by 2, \par
and thus we see that $q$ is also even.
\vspace{2mm}
$p$ and $q$ are both even, so they cannot be coprime. \par
Therefore, we cannot write $\sqrt{2}$ as $\nicefrac{p}{q}$ for comprime $p, q$, \par
and $\sqrt{2}$ is therefore irrational.
\end{solution}
\vfill
\pagebreak

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\section{Proofs by Induction}
\definition{}
The last proof technique we'll discuss in this handout is \textit{induction.} \par
This is particularly useful when we have a \say{countable} variable, usually an integer. \par
Let's say we're proving a statement $A$ for all positive integers $n$. \par
We'll write the special case \say{$A$ holds for $n$} as $A_n$.
\vspace{2mm}
A proof by induction consists of two parts: a \textit{base case} and a \textit{inductive step}. \par
The base case is usually fairly simple: we show that our statement holds for $n = 0$. \par
In other words, the base case shows that $A_0$ is true.
The inductive step is a bit more confusing: we show that if our statement holds for
$n$, it must hold for $n = 1$. In other words, we show that $A_n \implies A_{n + 1}$.
\vspace{2mm}
In this way, we build an infinite implication chain: \par
The base case proves that $A_0$. By the inductive step,
$A_0 \implies A_1$, $A_1 \implies A_2$, and so on. \par
We can thus conclude that $A_n$ is true for all $n \in \{0, 1, 2, 3, ...\}$
\problem{}
Proof by induction will make a bit more sense with an example. \par
Read and understand the following proof.
\begin{examplesolution}
Show that $1 + 2 + ... + n = \frac{n(n+1)}{2}$
\linehack{}
\textbf{Base case:} $n = 1$ \par
Substitute $n = 1$ into the hypothesis:
$1 \qe \frac{1(1 + 1)}{2}$ \par
but we have that
$\frac{1(1 + 1)}{2} = \frac{2}{2} = 1$,
so this is of course true.
\vspace{2mm}
\textbf{Inductive step:} \par
Now, we assume our hypothesis is true for $n$, \par
and show it is true for $n + 1$.
\vspace{2mm}
Write the hypothesis for $n + 1$:
$$
1 + 2 + ... + n + (n + 1) \qe \frac{(n+1)(n+2)}{2}
$$
We know that $1 + 2 + ... + n = \frac{n(n+1)}{2}$, so:
$$
1 + 2 + ... + n + (n + 1) = \frac{n(n+1)}{2} + (n+1)
$$
now we can complete the proof with some algebra:
$$
\frac{n(n+1)}{2} + (n+1) = \frac{n(n+1) + 2(n+1)}{2} = \frac{(n + 2)(n+1)}{2}
$$
So, we've shown that the $n^\text{th}$ case implies the $(n+1)^\text{th}$ case.\par
We're therefore done, the hypothesis is true for $n \in \{1, 2, 3, ...\}$
\end{examplesolution}
\vfill
\pagebreak
\problem{}
Why do we need a base case when constructing a proof by induction? \par
\hint{Try to prove that $n = n + 1$ for all $n \in \mathbb{Z}^+$}
\begin{solution}
Consider the following example: \par
Say we want to prove that $n = n + 1$ for all $n \in \mathbb{Z}^+$.
\vspace{2mm}
\textbf{Inductive step:}
Assume our hypothesis is true for $n$ (that is $n = n + 1$). Is this true for $n + 1$? \par
Adding 1 to both sides, we get $n+1=n+1+1$, which means that $(n+1)=(n+1)+1$,
which is the statement we wanted to prove. We've thus completed the inductive step!
\vspace{2mm}
Our problem is as follows: we've shown that $A_1 \implies A_2 \implies ...$,
but we have no reason to believe that $A_1$ is true. If it was, our hypothesis
would be correct---but since it isn't, this is not a complete proof.
\end{solution}
\vfill
\problem{}
Show that $1^2 + 2^2 + 3^3 + ... + n^2 = \frac{1}{6}(n)(n+1)(2n+1)$.
\begin{solution}
\textbf{Base case:}\par
$1^2 = \frac{1}{6}(1)(1+1)(2 + 1) = 1$, which is true.
\vspace{2mm}
\textbf{Induction:}\par
Assume $1^2 + ... + n^2$ satisfies the equation above. \par
$$
1^2 + 2^2 + ... + n^2 + (n+1)^2 =
\frac{(n)(n+1)(2n+1)}{6} + (n + 1)^2
$$
which is equal to
$$
\frac{(n)(n+1)(2n+1) + 6(n+1)^2}{6}
$$
now expand and factor to get
$$
\frac{(n+1)(n+2)(2(n+1)+1)}{6}
$$
\end{solution}
\vfill
\problem{}
Show that $2^n - 1$ is divisible by 3 for all odd $n$ \par
\hint{If $n$ is odd, the next odd number is $n + 2$.}
\begin{solution}
\textbf{Base case:} \par
$2^2 - 1 = 3$, which is divisible by 3..
\vspace{2mm}
\textbf{Induction:}\par
Assume $2^n - 1$ is divisible by 3. \par
$2^{(n+2)} - 1 = 4(2^n) - 1 = 4(2^n - 1) + 3$ \par
By our induction hypothesis, $(2^n - 1)$ has a factor of 3. \par
Therefore, $4(2^n - 1) + 3$ must also have a factor of 3.
\end{solution}
\vfill
\pagebreak
\definition{}
As you may already know, \say{n choose k} is defined as follows: \par
$$
\binom{n}{k} = \frac{n!}{k!(n-k)!}
$$
This counts the number of ways to choose $k$ things from a set of $n$,
disregarding the order of the chosen items.
\theorem{Pascal's Identity}
The binomial coefficient defined above satisfies the following equality:
$$
\binom{n+1}{k} = \binom{n}{k-1} \times \binom{n}{k}
$$
\problem{}<binomsum>
Using induction, show that
$$
\binom{n}{0} + \binom{n}{1} + ... + \binom{n}{n} = 2^n.
$$
\vfill
Note that although induction is a powerful proof technique, it usually leads to uninteresting results. \par
If we prove a statement using induction, we conclude that it is true---but we get very little insight on
\textit{why} it is true.
\vspace{2mm}
Alternative proofs are take a bit more work than inductive proofs, but they are much more valuable. \par
For example, see the proof of the statement in \ref{binomsum} on the next page.
\pagebreak
\begin{ORMCbox}{Alternative Proof}{ogrape!10!white}{ogrape}
Consider the following problem: \par
How many ways are there to write a number $x$ as an ordered sum of positive integers? \par
\note{
An \say{ordered sum} means that the order of numbers in the sum matters. \\
For example, if $x = 5$, we will consider $4 + 1$ and $1 + 4$ as distinct sums.
}
\vspace{2mm}
First, we'll think of $x$ as an array of $1$s which we want to group
into positive integers. If $x = 3$, We have $x = 1~1~1$, which we can
group as $1+1+1$,~ $(1+1) + 1$,~ $1 + (1+1)$,~ and $(1+1+1)$.
\linehack{}
\textbf{Solution 1:}\par
One way to solve this is to use the usual \say{stars and bars} method, \par
where we count the number of ways we can place $n$ \say{bars} between
$x$ \say{stars}. Each bar corresponds to a \say{$+$} in the array of ones. \par
\note[Note]{Convince yourself that there are $\binom{x-1}{n}$ ways to place $n$ bars
between $x$ objects.}
If we add the number of ways to write $x$ as a sum of $n \in \{1, 2, ..., x\}$ integers, we get:
$$
\sum_{n = 1}^{x-1} \binom{x-1}{n} = \binom{x-1}{0} + \binom{n}{1} + ... + \binom{x-1}{x-1}
$$
\linehack{}
\textbf{Solution 2:}\par
We could also observe that there are $x - 1$ places to put a \say{bar} in
the array of ones. This corresponds to $x - 1$ binary positions, and thus
$2^{x-1}$ ways to separate our array of $1$s with bars.
\linehack{}
\textbf{Conclusion:}\par
We've found that the number of ways to split $x$ can be written as either
$\sum_{n = 1}^{x-1} \binom{x-1}{n}$ or $2^{x-1}$,
and therefore $\sum_{n = 1}^{x-1} \binom{x-1}{n} = 2^{x-1}$.
\end{ORMCbox}
\pagebreak

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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\usepackage{pdfpages}
\usepackage{sliderule}
\usepackage{changepage}
% Args:
% x, top scale y, label
\newcommand{\slideruleind}[3]{
\draw[
line width=1mm,
draw=black,
opacity=0.3,
text opacity=1
]
({#1}, {#2 + 1})
--
({#1}, {#2 - 1.1})
node [below] {#3};
}
\uptitlel{Intermediate 2}
\uptitler{\smallurl{}}
\title{Slide Rules}
\subtitle{Prepared by Mark on \today}
\begin{document}
\maketitle
\begin{center}
\begin{minipage}{6cm}
Dad says that anyone who can't use
a slide rule is a cultural illiterate
and should not be allowed to vote.
\vspace{1ex}
\textit{Have Space Suit --- Will Travel, 1958}
\end{minipage}
\end{center}
\hfill
\input{parts/0 logarithms.tex}
\input{parts/1 intro.tex}
\input{parts/2 multiplication.tex}
\input{parts/3 division.tex}
\input{parts/4 squares.tex}
\input{parts/5 inverses.tex}
\input{parts/6 log.tex}
% Make sure the slide rule is on an odd page,
% so that double-sided printing won't require
% students to tear off problems.
\checkoddpage
\ifoddpage\else
\vspace*{\fill}
\begin{center}
{
\Large
\textbf{This page unintentionally left blank.}
}
\end{center}
\vspace{\fill}
\pagebreak
\fi
\includepdf[
pages=1,
fitpaper=true
]{resources/rule.pdf}
\end{document}

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[metadata]
title = "Slide Rules"
[publish]
handout = true
solutions = true

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src/Intermediate/Slide Rules/parts/0 logarithms.tex Normal file
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\section{Logarithms}
\definition{}<logdef>
The \textit{logarithm} is the inverse of the exponent. That is, if $b^p = c$, then $\log_b{c} = p$. \\
In other words, $\log_b{c}$ asks the question ``what power do I need to raise $b$ to to get $c$?'' \\
\medskip
In both $b^p$ and $\log_b{c}$, the number $b$ is called the \textit{base}.
\problem{}
Evaluate the following by hand:
\begin{enumerate}
\item $\log_{10}{(1000)}$
\vfill
\item $\log_2{(64)}$
\vfill
\item $\log_2{(\frac{1}{4})}$
\vfill
\item $\log_x{(x)}$ for any $x$
\vfill
\item $log_x{(1)}$ for any $x$
\vfill
\end{enumerate}
\pagebreak
\definition{}
There are a few ways to write logarithms:
\begin{itemize}
\item[] $\log{x} = \log_{10}{x}$
\item[] $\lg{x} = \log_{10}{x}$
\item[] $\ln{x} = \log_e{x}$
\end{itemize}
\definition{}
The \textit{domain} of a function is the set of values it can take as inputs. \\
The \textit{range} of a function is the set of values it can produce.
\medskip
For example, the domain and range of $f(x) = x$ is $\mathbb{R}$, all real numbers. \\
The domain of $f(x) = |x|$ is $\mathbb{R}$, and its range is $\mathbb{R}^+ \cup \{0\}$, all positive real numbers and 0. \\
\medskip
Note that the domain and range of a function are not always equal.
\problem{}<expdomain>
What is the domain of $f(x) = 5^x$? \\
What is the range of $f(x) = 5^x$?
\vfill
\problem{}<logdomain>
What is the domain of $f(x) = \log{x}$? \\
What is the range of $f(x) = \log{x}$?
\vfill
\pagebreak
\problem{}<logids>
Prove the following identities: \\
\begin{enumerate}[itemsep=2mm]
\item $\log_b{(b^x)} = x$
\item $b^{\log_b{x}} = x$
\item $\log_b{(xy)} = \log_b{(x)} + \log_b{(y)}$
\item $\log_b{(\frac{x}{y})} = \log_b{(x)} - \log_b{(y)}$
\item $\log_b{(x^y)} = y \log_b{(x)}$
\end{enumerate}
\vfill
\begin{instructornote}
A good intro to the following sections is the linear slide rule:
\begin{center}
\begin{tikzpicture}[scale=1]
\linearscale{2}{1}{}
\linearscale{0}{0}{}
\slideruleind
{5}
{1}
{2 + 3 = 5}
\end{tikzpicture}
\end{center}
Take two linear rulers, offset one, and you add. \\
If you do the same with a log scale, you multiply! \\
\vspace{1ex}
Note that the slide rules above start at 0.
\linehack{}
After assembling the paper slide rule, you can make a visor with some transparent tape. Wrap a strip around the slide rule, sticky side out, and stick it to itself to form a ring. Cover the sticky side with another layer of tape, and trim the edges to make them straight. Use the edge of the visor to read your slide rule!
\end{instructornote}
\pagebreak

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\section{Introduction}
Mathematicians, physicists, and engineers needed to quickly solve complex equations even before computers were invented.
\medskip
The \textit{slide rule} is an instrument that uses the logarithm to solve this problem. Before you continue, cut out and assemble your slide rule.
\medskip
There are four scales on your slide rule, each labeled with a letter on the left side:
\def\sliderulewidth{13}
\begin{center}
\begin{tikzpicture}[scale=1]
\tscale{0}{9}{T}
\kscale{0}{8}{K}
\abscale{0}{7}{A}
\abscale{0}{5.5}{B}
\ciscale{0}{4.5}{CI}
\cdscale{0}{3.5}{C}
\cdscale{0}{2}{D}
\lscale{0}{1}{L}
\sscale{0}{0}{S}
\end{tikzpicture}
\end{center}
Each scale's ``generating function'' is on the right:
\begin{itemize}
\item T: $\tan$
\item K: $x^3$
\item A,B: $x^2$
\item CI: $\frac{1}{x}$
\item C, D: $x$
\item L: $\log_{10}(x)$
\item S: $\sin$
\end{itemize}
Once you understand the layout of your slide rule, move on to the next page.
\pagebreak

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\section{Multiplication}
We'll use the C and D scales of your slide rule to multiply. \\
Say we want to multiply $2 \times 3$. First, move the \textit{left-hand index} of the C scale over the smaller number, $2$:
\def\sliderulewidth{10}
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(2)}{1}{C}
\cdscale{0}{0}{D}
\end{tikzpicture}
\end{center}
Then we'll find the second number, $3$ on the C scale, and read the D scale under it:
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(2)}{1}{C}
\cdscale{0}{0}{D}
\slideruleind
{\cdscalefn(6)}
{1}
{6}
\end{tikzpicture}
\end{center}
Of course, our answer is 6.
\problem{}
What is $1.15 \times 2.1$? \\
Use your slide rule.
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(1.15)}{1}{C}
\cdscale{0}{0}{D}
\slideruleind
{\cdscalefn(1.15)}
{1}
{1.15}
\slideruleind
{\cdscalefn(1.15) + \cdscalefn(2.1)}
{1}
{2.415}
\end{tikzpicture}
\end{center}
\end{solution}
\vfill
Note that your answer isn't exact. $1.15 \times 2.1 = 2.415$, but an answer accurate within two decimal places is close enough for most practical applications. \\
\pagebreak
Look at your C and D scales again. They contain every number between 1 and 10, but no more than that.
What should we do if we want to calculate $32 \times 210$? \\
\problem{}
Using your slide rule, calculate $32 \times 210$. \\
%\hint{$32 = 3.2 \times 10^1$}
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(2.1)}{1}{C}
\cdscale{0}{0}{D}
\slideruleind
{\cdscalefn(2.1)}
{1}
{2.1}
\slideruleind
{\cdscalefn(2.1) + \cdscalefn(3.2)}
{1}
{6.72}
\end{tikzpicture}
\end{center}
Placing the decimal point correctly is your job. \\
$10^1 \times 10^2 = 10^3$, so our final answer is $6.72 \times 10^3 = 672$.
\end{solution}
\vfill
%This method of writing numbers is called \textit{scientific notation}. In the form $a \times 10^b$, $a$ is called the \textit{mantissa}, and $b$, the \textit{exponent}. \\
%You may also see expressions like $4.3\text{e}2$. This is equivalent to $4.3 \times 10^2$, but is more compact.
\problem{}
Compute the following:
\begin{enumerate}
\item $1.44 \times 52$
\item $0.38 \times 1.24$
\item $\pi \times 2.35$
\end{enumerate}
\begin{solution}
\begin{enumerate}
\item $1.44 \times 52 = 74.88$
\item $0.38 \times 1.24 = 0.4712$
\item $\pi \times 2.35 = 7.382$
\end{enumerate}
\end{solution}
\vfill
\pagebreak
\problem{}<provemult>
Note that the numbers on your C and D scales are logarithmically spaced.
\def\sliderulewidth{13}
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{0}{1}{C}
\cdscale{0}{0}{D}
\end{tikzpicture}
\end{center}
Why does our multiplication procedure work? \\
%\hint{See \ref{logids}}
\vfill
\pagebreak
Now we want to compute $7.2 \times 5.5$:
\def\sliderulewidth{10}
\begin{center}
\begin{tikzpicture}[scale=0.8]
\cdscale{\cdscalefn(5.5)}{1}{C}
\cdscale{0}{0}{D}
\slideruleind
{\cdscalefn(5.5)}
{1}
{5.5}
\slideruleind
{\cdscalefn(5.5) + \cdscalefn(7.2)}
{1}
{???}
\end{tikzpicture}
\end{center}
No matter what order we go in, the answer ends up off the scale. There must be another way. \\
\medskip
Look at the far right of your C scale. There's an arrow pointing to the $10$ tick, labeled \textit{right-hand index}. Move it over the \textit{larger} number, $7.2$:
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
\cdscale{0}{0}{D}
\slideruleind
{\cdscalefn(7.2)}
{1}
{7.2}
\end{tikzpicture}
\end{center}
Now find the smaller number, $5.5$, on the C scale, and read the D scale under it:
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
\cdscale{0}{0}{D}
\slideruleind
{\cdscalefn(7.2)}
{1}
{7.2}
\slideruleind
{\cdscalefn(3.96)}
{1}
{3.96}
\end{tikzpicture}
\end{center}
Our answer should be about $7 \times 5 = 35$, so let's move the decimal point: $5.5 \times 7.2 = 39.6$. We can do this by hand to verify our answer. \\
\medskip
\iftrue
\problem{}
Why does this work?
\else
Why does this work? \\
\medskip
Consider the following picture, where I've put two D scales next to each other:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
\cdscale{0}{0}{}
\cdscale{-10}{0}{}
\draw[
draw=black,
]
(0, 0)
--
(0, -0.3)
node [below] {D};
\draw[
draw=black,
]
(-10, 0)
--
(-10, -0.3)
node [below] {D};
\slideruleind
{-10 + \cdscalefn(7.2)}
{1}
{7.2}
\slideruleind
{\cdscalefn(7.2)}
{1}
{7.2}
\slideruleind
{\cdscalefn(3.96)}
{1}
{3.96}
\end{tikzpicture}
\end{center}
\medskip
The second D scale has been moved to the right by $(\log{10})$, so every value on it is $(\log{10})$ smaller than it should be.
\medskip
\medskip
In other words, the answer we get from reverse multiplication is the following: $\log{a} + \log{b} - \log{10}$. \\
This reduces to $\log{(\frac{a \times b}{10})}$, which explains the misplaced decimal point in $7.2 \times 5.5$.
\fi
\vfill
\pagebreak
\problem{}
Compute the following using your slide rule:
\begin{enumerate}
\item $9 \times 8$
\item $15 \times 35$
\item $42.1 \times 7.65$
\item $6.5^2$
\end{enumerate}
\begin{solution}
\begin{enumerate}
\item $9 \times 8 = 72$
\item $15 \times 35 = 525$
\item $42.1 \times 7.65 = 322.065$
\item $6.5^2 = 42.25$
\end{enumerate}
\end{solution}
\vfill
\pagebreak

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\section{Division}
Now that you can multiply, division should be easy. All you need to do is work backwards. \\
Let's look at our first example again: $3 \times 2 = 6$.
\medskip
We can easily see that $6 \div 3 = 2$
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(2)}{1}{C}
\cdscale{0}{0}{D}
\slideruleind
{\cdscalefn(6)}
{1}
{Align here}
\slideruleind
{\cdscalefn(2)}
{1}
{2}
\end{tikzpicture}
\end{center}
and that $6 \div 2 = 3$:
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(3)}{-3}{C}
\cdscale{0}{-4}{D}
\slideruleind
{\cdscalefn(6)}
{-3}
{Align here}
\slideruleind
{\cdscalefn(3)}
{-3}
{3}
\end{tikzpicture}
\end{center}
If your left-hand index is off the scale, read the right-hand one. \\
Consider $42.25 \div 6.5 = 6.5$:
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(6.5) - \cdscalefn(10)}{1}{C}
\cdscale{0}{0}{D}
\slideruleind
{\cdscalefn(4.225)}
{1}
{Align here}
\slideruleind
{\cdscalefn(6.5)}
{1}
{6.5}
\end{tikzpicture}
\end{center}
Place your decimal points carefully.
\vfill
\pagebreak
\problem{}
Compute the following using your slide rule. \\
\begin{enumerate}
\item $135 \div 15$
\item $68.2 \div 0.575$
\item $(118 \times 0.51) \div 6.6$
\end{enumerate}
\begin{solution}
\begin{enumerate}
\item $135 \div 15 = 9$
\item $68.2 \div 0.575 = 118.609$
\item $(118 \times 0.51) \div 6.6 = 9.118$
\end{enumerate}
\end{solution}
\vfill
\pagebreak

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\section{Squares, Cubes, and Roots}
Now, take a look at scales A and B, and note the label on the right: $x^2$. If C, D are $x$, A and B are $x^2$, and K is $x^3$.
\medskip
Finding squares of numbers up to ten is straightforward: just read the scale. \\
Square roots are also easy: find your number on B and read its pair on C. \\
\def\sliderulewidth{13}
\begin{center}
\begin{tikzpicture}[scale=1]
\abscale{0}{1}{B}
\cdscale{0}{0}{C}
\end{tikzpicture}
\end{center}
\problem{}
Compute the following.
\begin{enumerate}
\item $1.5^2$
\item $3.1^2$
\item $7^3$
\item $\sqrt{14}$
\item $\sqrt[3]{150}$
\end{enumerate}
\begin{solution}
\begin{enumerate}
\item $1.5^2 = 2.25$
\item $3.1^2 = 9.61$
\item $7^3 = 343$
\item $\sqrt{14} = 3.74$
\item $\sqrt[3]{150} = 5.313$
\end{enumerate}
\end{solution}
\vfill
\problem{}
Compute the following.
\begin{enumerate}
\item $42^2$
\item $\sqrt{200}$
\item $\sqrt{2000}$
\item $\sqrt{0.9}$
\item $\sqrt[3]{0.12}$
\end{enumerate}
\begin{solution}
\begin{enumerate}
\item $42^2 = 1,764$
\item $\sqrt{200} = 14.14$
\item $\sqrt{2000} = 44.72$
\item $\sqrt{0.9} = 0.948$
\item $\sqrt[3]{0.12} = 0.493$
\end{enumerate}
\end{solution}
\vfill
\pagebreak

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\section{Inverses}
Try finding $1 \div 32$ using your slide rule. \\
The procedure we learned before doesn't work!
\medskip
This is why we have the CI scale, or the ``C Inverse'' scale.
\problem{}
Figure out how the CI scale works and compute the following:
\begin{enumerate}[itemsep=1mm]
\item $\frac{1}{7}$
\item $\frac{1}{120}$
\item $\frac{1}{\pi}$
\end{enumerate}
\vfill
\pagebreak

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\section{Logarithms Base 10}
When we take a logarithm, the resulting number has two parts: the \textit{characteristic} and the \textit{mantissa}. \\
The characteristic is the integral (whole-numbered) part of the answer, and the mantissa is the fractional part (what comes after the decimal). \\
\medskip
For example, $\log_{10}{18} = 1.255$, so in this case the characteristic is $1$ and the mantissa is $0.255$.
\problem{}
Approximate the following logs without a slide rule. Find the exact characteristic, and approximate the mantissa.
\begin{enumerate}
\item $\log_{10}{20}$
\item $\log_{2}{18}$
\end{enumerate}
\begin{solution}
\begin{enumerate}
\item $\log_{10}{20} = 1.30$
\item $\log_{2}{18} = 4.17$
\end{enumerate}
\end{solution}
\vfill
Now, find the L scale on your slide rule. As you can see on the right, its generating function is $\log_{10}{x}$.
\problem{}
Compute the following logarithms using your slide rule. \\
You'll have to find the characteristic yourself, but your L scale will give you the mantissa. \\
Don't forget your log identities!
\begin{enumerate}
\item $\log_{10}{20}$
\item $\log_{10}{15}$
\item $\log_{10}{150}$
\item $\log_{10}{0.024}$
\end{enumerate}
\begin{solution}
Careful with number 4.
\begin{enumerate}
\item $\log_{10}{20} = 1.30$
\item $\log_{10}{15} = 1.176$
\item $\log_{10}{150} = 2.176$
\item $\log_{10}{0.024} = -1.6197$
\end{enumerate}
\end{solution}
\vfill
\pagebreak
%\problem{}
%Find the following.
%\begin{enumerate}[itemsep=2mm]
% \item $\frac{118 \times 0.51}{6.6}$
% \item $\sqrt{33.8} \times \sqrt[3]{226}$
% \item $\frac{\sqrt{152}}{\sqrt[3]{4.95}}$
% \item $\frac{\sqrt{96 \times 250}}{\sqrt{7 \times 0.88}}$
% \item The area of a circle with radius $1.47$
% \item The circumference of a circle with radius $31.4$
% \item The radius of a circle with area $6\pi$
% \item $\log_{10}{17.38}$
%\end{enumerate}
%\vfill
%\pagebreak
\section{Logarithms in Any Base}
Our slide rule easily computes logarithms in base 10, but we can also use it to find logarithms in \textit{any} base.
\proposition{}<logcob>
This is usually called the \textit{change-of-base} formula:
\[
\log_{b}{a} = \frac{\log_c{a}}{\log_c{b}}
\]
\problem{}
Using log identities, prove \ref{logcob}.
\vfill
\problem{}
Approximate the following:
\begin{enumerate}
\item $\log_{2}{56}$
\item $\log_{5.2}{26}$
\item $\log_{12}{500}$
\item $\log_{43}{134}$
\end{enumerate}
\begin{solution}
\begin{enumerate}
\item $\log_{2}{56} = 5.81$
\item $\log_{5.2}{26} = 1.97$
\item $\log_{12}{500} = 2.50$
\item $\log_{43}{134} = 1.30$
\end{enumerate}
\end{solution}
\vfill
\pagebreak

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\NeedsTeXFormat{LaTeX2e}
\ProvidesPackage{sliderule}[2022/08/22 Slide rule tools]
\RequirePackage{tikz}
\RequirePackage{ifthen}
% Scale functions:
% See https://sliderulemuseum.com/SR_Scales.htm
%
% l: length of the rule
% n: the number on the rule
%
% A/B: (l/2) * log(n)
% C/D: l / log(n)
% CI: abs(l * log(10 / n) - l)
% K: (l/3) * log(n)
%
% L: n * l
% T: l * log(10 * tan(n))
% S: l * log(10 * sin(n))
\def\sliderulewidth{10}
\def\abscalefn(#1){(\sliderulewidth/2) * log10(#1)}
\def\cdscalefn(#1){(\sliderulewidth * log10(#1))}
\def\ciscalefn(#1){(\sliderulewidth - \cdscalefn(#1))}
\def\kscalefn(#1){(\sliderulewidth/3) * log10(#1)}
\def\lscalefn(#1){(\sliderulewidth * #1)}
\def\tscalefn(#1){(\sliderulewidth * log10(10 * tan(#1)))}
\def\sscalefn(#1){(\sliderulewidth * log10(10 * sin(#1)))}
% Arguments:
% Label
% x of start
% y of start
\newcommand{\linearscale}[3]{
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
% Numbers and marks
\foreach \i in {0,..., 10}{
\draw[black]
({#1 + (\sliderulewidth / 10) * \i}, #2) --
({#1 + (\sliderulewidth / 10) * \i}, #2 + 0.3)
node[above] {\i};
}
% Submarks
\foreach \n in {0, ..., 9} {
\foreach \i in {1,..., 9} {
\ifthenelse{\i=5}{
\draw[black]
({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2) --
({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2 + 0.2);
} {
\draw[black]
({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2) --
({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2 + 0.1);
}
}
}
}
% Arguments:
% Label
% x of start
% y of start
\newcommand{\abscale}[3]{
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
% Numbers and marks 1 - 9
\foreach \i in {1,..., 9}{
\draw[black]
({#1 + \abscalefn(\i)}, #2) --
({#1 + \abscalefn(\i)}, #2 + 0.3)
node[above] {\i};
}
% Numbers and marks 10 - 100
\foreach \i in {1,..., 10}{
\draw[black]
({#1 + \abscalefn(10 * \i)}, #2) --
({#1 + \abscalefn(10 * \i)}, #2 + 0.3)
node[above] {\ifthenelse{\i=10}{1}{\i}};
}
% Submarks 1 - 9
\foreach \n in {1, ..., 9} {
\ifthenelse{\n<5}{
\foreach \i in {1,..., 9}
} {
\foreach \i in {2,4,6,8}
}
{
\ifthenelse{\i=5}{
\draw[black]
({#1 + \abscalefn(\n + \i / 10)}, #2) --
({#1 + \abscalefn(\n + \i / 10)}, #2 + 0.2);
} {
\draw[black]
({#1 + \abscalefn(\n + \i / 10)}, #2) --
({#1 + \abscalefn(\n + \i / 10)}, #2 + 0.1);
}
}
}
% Submarks 10 - 100
\foreach \n in {10,20,...,90} {
\ifthenelse{\n<50}{
\foreach \i in {1,..., 9}
} {
\foreach \i in {2,4,6,8}
}
{
\ifthenelse{\i=5}{
\draw[black]
({#1 + \abscalefn(\n + \i)}, #2) --
({#1 + \abscalefn(\n + \i)}, #2 + 0.2);
} {
\draw[black]
({#1 + \abscalefn(\n + \i)}, #2) --
({#1 + \abscalefn(\n + \i)}, #2 + 0.1);
}
}
}
}
\newcommand{\cdscale}[3]{
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
% Numbers and marks 1 - 10
\foreach \i in {1,..., 10}{
\draw[black]
({#1 + \cdscalefn(\i)}, #2) --
({#1 + \cdscalefn(\i)}, #2 + 0.3)
node[above] {\ifthenelse{\i=10}{1}{\i}};
}
% Submarks 1 - 9
\foreach \n in {1, ..., 9} {
\ifthenelse{\n<3}{
\foreach \i in {5,10,...,95}
} {
\foreach \i in {10,20,...,90}
}
{
\ifthenelse{\i=50}{
\draw[black]
({#1 + \cdscalefn(\n + \i / 100)}, #2) --
({#1 + \cdscalefn(\n + \i / 100)}, #2 + 0.2);
\ifthenelse{\n=1}{
\draw
({#1 + \cdscalefn(\n + \i / 100)}, #2 + 0.2)
node [above] {1.5};
}{}
} {
\ifthenelse{
\i=10 \OR \i=20 \OR \i=30 \OR \i=40 \OR
\i=60 \OR \i=70 \OR \i=80 \OR \i=90
}{
\draw[black]
({#1 + \cdscalefn(\n + \i / 100)}, #2) --
({#1 + \cdscalefn(\n + \i / 100)}, #2 + 0.15);
} {
\draw[black]
({#1 + \cdscalefn(\n + \i / 100)}, #2) --
({#1 + \cdscalefn(\n + \i / 100)}, #2 + 0.1);
}
}
}
}
}
\newcommand{\ciscale}[3]{
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
% Numbers and marks
\foreach \i in {1,...,10}{
\draw[black]
({#1 + \ciscalefn(\i)}, #2) --
({#1 + \ciscalefn(\i)}, #2 + 0.3)
node[above] {\ifthenelse{\i=10}{1}{\ifthenelse{\i=0}{0}{.\i}}};
}
% Submarks 1 - 9
\foreach \n in {1, ..., 9} {
\ifthenelse{\n<3}{
\foreach \i in {5,10,...,95}
} {
\foreach \i in {10,20,...,90}
}
{
\ifthenelse{\i=50}{
\draw[black]
({#1 + \ciscalefn(\n + \i / 100)}, #2) --
({#1 + \ciscalefn(\n + \i / 100)}, #2 + 0.2);
\ifthenelse{\n=1}{
\draw
({#1 + \ciscalefn(\n + \i / 100)}, #2 + 0.2)
node [above] {1.5};
}{}
} {
\ifthenelse{
\i=10 \OR \i=20 \OR \i=30 \OR \i=40 \OR
\i=60 \OR \i=70 \OR \i=80 \OR \i=90
}{
\draw[black]
({#1 + \ciscalefn(\n + \i / 100)}, #2) --
({#1 + \ciscalefn(\n + \i / 100)}, #2 + 0.15);
} {
\draw[black]
({#1 + \ciscalefn(\n + \i / 100)}, #2) --
({#1 + \ciscalefn(\n + \i / 100)}, #2 + 0.1);
}
}
}
}
}
\newcommand{\kscale}[3]{
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
% Numbers and marks 1 - 9
\foreach \i in {1,...,9}{
\draw[black]
({#1 + \kscalefn(\i)}, #2) --
({#1 + \kscalefn(\i)}, #2 + 0.3)
node[above] {\i};
}
% Numbers and marks 10 - 90
\foreach \i in {1,..., 9}{
\draw[black]
({#1 + \kscalefn(10 * \i)}, #2) --
({#1 + \kscalefn(10 * \i)}, #2 + 0.3)
node[above] {\ifthenelse{\i=10}{1}{\i}};
}
% Numbers and marks 100 - 1000
\foreach \i in {1,..., 10}{
\draw[black]
({#1 + \kscalefn(100 * \i)}, #2) --
({#1 + \kscalefn(100 * \i)}, #2 + 0.3)
node[above] {\ifthenelse{\i=10}{1}{\i}};
}
% Submarks 1 - 9
\foreach \n in {1, ..., 9} {
\ifthenelse{\n<4}{
\foreach \i in {1,..., 9}
} {
\foreach \i in {2,4,6,8}
}
{
\ifthenelse{\i=5}{
\draw[black]
({#1 + \kscalefn(\n + \i / 10)}, #2) --
({#1 + \kscalefn(\n + \i / 10)}, #2 + 0.2);
} {
\draw[black]
({#1 + \kscalefn(\n + \i / 10)}, #2) --
({#1 + \kscalefn(\n + \i / 10)}, #2 + 0.1);
}
}
}
% Submarks 10 - 90
\foreach \n in {10,20,...,90} {
\ifthenelse{\n<40}{
\foreach \i in {1,..., 9}
} {
\foreach \i in {2,4,6,8}
}
{
\ifthenelse{\i=5}{
\draw[black]
({#1 + \kscalefn(\n + \i)}, #2) --
({#1 + \kscalefn(\n + \i)}, #2 + 0.2);
} {
\draw[black]
({#1 + \kscalefn(\n + \i)}, #2) --
({#1 + \kscalefn(\n + \i)}, #2 + 0.1);
}
}
}
% Submarks 100 - 1000
\foreach \n in {100,200,...,900} {
\ifthenelse{\n<400}{
\foreach \i in {10,20,...,90}
} {
\foreach \i in {20,40,60,80}
}
{
\ifthenelse{\i=50}{
\draw[black]
({#1 + \kscalefn(\n + \i)}, #2) --
({#1 + \kscalefn(\n + \i)}, #2 + 0.2);
} {
\draw[black]
({#1 + \kscalefn(\n + \i)}, #2) --
({#1 + \kscalefn(\n + \i)}, #2 + 0.1);
}
}
}
}
\newcommand{\lscale}[3]{
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
% Numbers and marks
\foreach \i in {0,..., 10}{
\draw[black]
({#1 + \lscalefn(\i / 10)}, #2) --
({#1 + \lscalefn(\i / 10)}, #2 + 0.3)
node[above] {\ifthenelse{\i=10}{1}{\ifthenelse{\i=0}{0}{.\i}}};
}
% Submarks
\foreach \n in {0, ..., 9} {
\foreach \i in {1,...,19} {
\ifthenelse{\i=10}{
\draw[black]
({#1 + \lscalefn((\n + (\i / 20))/10)}, #2) --
({#1 + \lscalefn((\n + (\i / 20))/10)}, #2 + 0.2);
} {
\ifthenelse{
\i=1 \OR \i=3 \OR \i=5 \OR \i=7 \OR
\i=9 \OR \i=11 \OR \i=13 \OR \i=15 \OR
\i=17 \OR \i=19
}{
\draw[black]
({#1 + \lscalefn((\n + (\i / 20))/10)}, #2) --
({#1 + \lscalefn((\n + (\i / 20))/10)}, #2 + 0.1);
} {
\draw[black]
({#1 + \lscalefn((\n + (\i / 20))/10)}, #2) --
({#1 + \lscalefn((\n + (\i / 20))/10)}, #2 + 0.15);
}
}
}
}
}
\newcommand{\tscale}[3]{
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
% First line
\draw[black] ({#1}, #2) -- ({#1}, #2 + 0.2);
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
% Numbers and marks 6 - 10
\foreach \i in {6,...,9,10,15,...,45}{
\draw[black]
({#1 + \tscalefn(\i)}, #2) --
({#1 + \tscalefn(\i)}, #2 + 0.3)
node[above] {\i};
}
% Submarks 6 - 10
\foreach \n in {6, ..., 9} {
\foreach \i in {1,...,9}{
\ifthenelse{\i=5}{
\draw[black]
({#1 + \tscalefn(\n + \i / 10)}, #2) --
({#1 + \tscalefn(\n + \i / 10)}, #2 + 0.2);
} {
\draw[black]
({#1 + \tscalefn(\n + \i / 10)}, #2) --
({#1 + \tscalefn(\n + \i / 10)}, #2 + 0.1);
}
}
}
% Submarks 15 - 45
\foreach \n in {10, 15, ..., 40} {
\foreach \i in {1,...,24}{
\ifthenelse{
\i=5 \OR \i=10 \OR \i=15 \OR \i=20
} {
\draw[black]
({#1 + \tscalefn(\n + \i / 5)}, #2) --
({#1 + \tscalefn(\n + \i / 5)}, #2 + 0.2);
} {
\draw[black]
({#1 + \tscalefn(\n + \i / 5)}, #2) --
({#1 + \tscalefn(\n + \i / 5)}, #2 + 0.1);
}
}
}
}
\newcommand{\sscale}[3]{
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
% First line
\draw[black] ({#1}, #2) -- ({#1}, #2 + 0.2);
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
% Numbers and marks
\foreach \i in {6,...,9,10,15,...,30,40,50,...,60,90}{
\draw[black]
({#1 + \sscalefn(\i)}, #2) --
({#1 + \sscalefn(\i)}, #2 + 0.3)
node[above] {\i};
}
% Submarks 6 - 10
\foreach \n in {6, ..., 9} {
\foreach \i in {1,...,9}{
\ifthenelse{\i=5}{
\draw[black]
({#1 + \sscalefn(\n + \i / 10)}, #2) --
({#1 + \sscalefn(\n + \i / 10)}, #2 + 0.2);
} {
\draw[black]
({#1 + \sscalefn(\n + \i / 10)}, #2) --
({#1 + \sscalefn(\n + \i / 10)}, #2 + 0.1);
}
}
}
% Submarks 15 - 30
\foreach \n in {10, 15, ..., 25} {
\foreach \i in {1,...,24}{
\ifthenelse{
\i=5 \OR \i=10 \OR \i=15 \OR \i=20
} {
\draw[black]
({#1 + \sscalefn(\n + \i / 5)}, #2) --
({#1 + \sscalefn(\n + \i / 5)}, #2 + 0.2);
} {
\draw[black]
({#1 + \sscalefn(\n + \i / 5)}, #2) --
({#1 + \sscalefn(\n + \i / 5)}, #2 + 0.1);
}
}
}
% Submarks 30
\foreach \n in {30} {
\foreach \i in {1,...,19}{
\ifthenelse{
\i=2 \OR \i=4 \OR \i=6 \OR \i=8 \OR
\i=10 \OR \i=12 \OR \i=14 \OR \i=16 \OR
\i=18
} {
\draw[black]
({#1 + \sscalefn(\n + \i / 2)}, #2) --
({#1 + \sscalefn(\n + \i / 2)}, #2 + 0.2);
} {
\draw[black]
({#1 + \sscalefn(\n + \i / 2)}, #2) --
({#1 + \sscalefn(\n + \i / 2)}, #2 + 0.1);
}
}
}
% Submarks 40 - 50
\foreach \n in {40, 50} {
\foreach \i in {1,...,9}{
\ifthenelse{
\i=5 \OR \i=10 \OR \i=15 \OR \i=20
} {
\draw[black]
({#1 + \sscalefn(\n + \i)}, #2) --
({#1 + \sscalefn(\n + \i)}, #2 + 0.2);
} {
\draw[black]
({#1 + \sscalefn(\n + \i)}, #2) --
({#1 + \sscalefn(\n + \i)}, #2 + 0.1);
}
}
}
% Submarks 60
\foreach \i in {1,...,10}{
\ifthenelse{
\i=5 \OR \i=10
} {
\draw[black]
({#1 + \sscalefn(60 + \i * 2)}, #2) --
({#1 + \sscalefn(60 + \i * 2)}, #2 + 0.2);
} {
\draw[black]
({#1 + \sscalefn(60 + \i * 2)}, #2) --
({#1 + \sscalefn(60 + \i * 2)}, #2 + 0.1);
}
}
}

491
src/Intermediate/Vectors 1/main.tex Executable file
View File

@ -0,0 +1,491 @@
% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[solutions]{../../../lib/tex/ormc_handout}
\usepackage{adjustbox}
\usepackage{../../../lib/tex/macros}
\uptitlel{Intermediate 2}
\uptitler{\smallurl{}}
\title{Vectors 1}
\subtitle{
Prepared by Mark on \today \\
Based on a handout by Oleg Gleizer
}
\begin{document}
\maketitle
\section{Warm-Up}
\problem{}
Simplify the following fraction: \\
$\displaystyle{
\frac{
\displaystyle{\frac12}
}{
\phantom{..}
\displaystyle{\frac13} -
\displaystyle{\frac14}
\phantom{..}
}
} =$
\vfill
\problem{}
Simplify the following fraction: \\
$\displaystyle{
\frac{
\displaystyle{\frac{a}{b}} -
\displaystyle{\frac{c}{d}}
}{
\phantom{..}
\displaystyle{\frac{a}{d}} +
\displaystyle{\frac{c}{b}}
\phantom{..}
}
} =$
\vfill
\problem{}
The point $A$ is placed inside a circle.
\begin{center} \begin{normalsize}
\begin{tikzpicture}
\draw (0,0) circle (2cm);
\filldraw (40:1.2cm) circle (2pt);
\coordinate [label = left: {$A$}] (a) at (40:1.2cm);
\filldraw (0,0) circle (2pt);
\end{tikzpicture}
\end{normalsize} \end{center}
Cut the circle into two parts so that you can move one to make a circle centered at $A$.
\vfill
\pagebreak
\section{Vectors}
\definition{}
A vector in the Euclidean plane is a directed line segment.
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (4,-2);
\coordinate [label = left: {$A$}] (a) at (0,0);
\coordinate [label = right: {$B$}] (b) at (4,-2);
\coordinate [label = above: {$v$}] (v) at (2,-.9);
\end{normalsize}
\coordinate [label = left:
{$v = \overrightarrow{AB}$}] (c) at (10,-1);
\end{tikzpicture}
\end{center}
\vspace{30pt}
For the vector $v = \overrightarrow{AB}$, point $A$ is called {\it initial} and point $B$ is called {\it terminal}. \\
Two vectors, $v = \overrightarrow{AB}$ and $w = \overrightarrow{CD}$ are considered equivalent if the quadrilateral $ABDC$ is a parallelogram.
\vspace{20pt}
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (4,-2);
\coordinate [label = left: {$A$}] (a) at (0,0);
\coordinate [label = right: {$B$}] (b) at (4,-2);
\coordinate [label = above: {$v$}] (v) at (2,-.9);
\draw [->, line width = 2pt] (-1,-3) -- (3,-5);
\coordinate [label = left: {$C$}] (c) at (-1,-3);
\coordinate [label = right: {$D$}] (d) at (3,-5);
\coordinate [label = below: {$w$}] (w) at (1,-4.1);
\draw (-1,-3) -- (0,0);
\draw (3,-5) -- (4,-2);
\end{normalsize}
\end{tikzpicture}
\end{center}
\vspace{20pt}
In other words, two vectors are equivalent if they have the same length and direction. If this is the case, we write $v = w$.
\note[Note 1]{
Convince yourself that this is true. Why are these two definitions of vector equivalence interchangeable?
}
\note[Note 2]{
A vector is characterized by its direction and length. One cannot make a formal definition out of this observation, because a ``direction'' is formally defined in terms of a vector.
}
\vfill
\pagebreak
\theorem{}
If two distinct straight lines in the Euclidean plane form the angles of equal size with a third straight line in the plane, then they are parallel.
In other words, to check that the lines $a$ and $b$ on the picture below have no common point, you don't need to travel to infinity.
All you need to do is to measure the angles $\alpha$ and $\gamma$. If $\alpha = \gamma$, then $a$ is parallel to $b$. \\
\begin{center}
\begin{footnotesize}
\begin{tikzpicture}
\draw [line width=1pt] (-5,0) -- (5,0);
\draw [line width=1pt] (-4,-2) -- (2,4);
\draw[blue] (-1,0) arc (0:45:1);
\coordinate [label=right:{$\gamma$}] (g) at (-1.1,0.5);
\draw [line width=1pt] (-5,2) -- (5,2);
\draw[blue] (1,2) arc (0:45:1);
\coordinate [label=right:{$\alpha$}] (a) at (0.9,2.5);
\coordinate [label=right:{$b$}] (b) at (5,2);
\coordinate [label=right:{$a$}] (a) at (5,0);
\coordinate [label=right:{$c$}] (c) at (2,4);
\end{tikzpicture}
\end{footnotesize}
\end{center}
\vfill
\theorem{Euclid's 5th postulate:}
For any straight line in the (Euclidean) plane and for any point away from it, there exists a unique straight line that passes through the point and is parallel to the original line. \\
\begin{center}
\begin{footnotesize}
\begin{tikzpicture}
\draw [line width=1pt] (-6,0) -- (6,0);
\filldraw [black] (3,2) circle (2pt);
\draw [line width=1pt] (-6,2) -- (6,2);
\end{tikzpicture}
\end{footnotesize}
\end{center}
\vfill
\problem{}
Use a compass and a ruler to draw a straight line parallel to the one given below and passing through the given point not lying on the original straight line.
\begin{center}
\begin{footnotesize}
\begin{tikzpicture}
\draw [line width=1pt] (-6,0) -- (6,0);
\filldraw [black] (3,4) circle (2pt);
\end{tikzpicture}
\end{footnotesize}
\end{center}
\vfill
\newpage
\problem{}
Use a compass and a ruler to construct a vector $w$ with initial point $C$ equal to the vector $v$ below.
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (6,-2);
\coordinate [label = above: {$v$}] (v) at (3,-.9);
\filldraw (-4,-5) circle (1.5pt);
\coordinate [label = below left: {$C$}] (c) at (-4,-5.1);
\end{normalsize}
\end{tikzpicture}
\end{center}
\vfill
Physical forces, such as the force of gravity or the force that pulls together two magnets are vectors in three dimensions. The direction of a vector shows the direction in which the corresponding force is acting. The length of the vector shows the strength of the force. \\
\problem{}
On the picture below, draw the vectors of the gravitational pull the Earth exerts on you and on your Math Circle leader.
\begin{center}
\begin{normalsize}
\begin{tikzpicture}
\draw (0,0) circle (3);
\coordinate [label = above: {Earth}] (a) at (0,2);
\filldraw (0,0) circle (1pt);
\coordinate [label = below: {centre}] (c) at (0,-.1);
\draw [line width = 1pt] (30:3) -- (25:3.5);
\draw [line width = 1pt] (20:3) -- (25:3.5);
\draw [line width = 1pt] (25:4) -- (25:3.5);
\draw (25:4.15) circle (.15);
\draw [line width = 1pt] (3.3,1.9) -- (3.6,1.3);
\coordinate [label = above right: {Oleg}] (o) at (25:4.3);
\draw [line width = 1pt] (190:3) -- (188:3.2);
\draw [line width = 1pt] (186:3) -- (188:3.2);
\draw [line width = 1pt] (188:3.5) -- (188:3.2);
\draw (188:3.65) circle (.15);
\draw [line width = 1pt] (-3.35,-.23) -- (-3.28,-.68);
\coordinate [label = above left: {Me}] (m) at (188:3.85);
\end{tikzpicture}
\end{normalsize}
\end{center}
\begin{itemize}
\item Where do the gravitational force vectors point? Why?
\item If Oleg is twice as heavy as you are, how would you draw the gravitational pull vectors on the above picture?
\end{itemize}
\vfill
\pagebreak
\vspace{60pt}
Motion can be represented by a vector, too. The direction of the velocity vector shows the direction in which an object is moving at the moment. The length of the velocity vector represents the object's \textit{speed}. It shows how fast an object is moving at the moment. \\
\problem{}
The truck on the picture below
is going 30 mph.
The car on the same picture
is speeding at 90 mph the opposite way.
Draw the corresponding velocity vectors. \\
\begin{center}
\begin{tikzpicture}
\draw [gray!50!black] (-7,0) -- (7,0);
\filldraw [gray!80!blue] (-3.6,1.5) -- (-4.4,1.5) --
(-4.4,2) -- (-6.8,2) -- (-6.8,.3) -- (-3.6,.3) --
(-3.6,1.5);
\filldraw (-6,.3) circle (.3);
\filldraw [gray] (-6,.3) circle (.15);
\filldraw (-4,.3) circle (.3);
\filldraw [gray] (-4,.3) circle (.15);
\filldraw [red!80!white] (4,.2) -- (6.6,.2) -- (6.55,.6) --
(6.2,.6) -- (6.1,.8) -- (4.9,.8) -- (4.7,.6)--
(4.2,.6) -- (4,.45) -- (4,.2);
\filldraw (6,.2) circle (.2);
\filldraw [gray] (6,.2) circle (.1);
\filldraw (4.6,.2) circle (.2);
\filldraw [gray] (4.6,.2) circle (.1);
\end{tikzpicture}
\end{center}
\vspace{40pt}
Velocities and forces of the real world are vectors in three dimensions. To keep things simple, we'll begin our study of vectors in two dimensions. Everything we are going to learn about vectors in two-dimensional space is be valid in a Euclidean space of three---or more---dimensions.
\vfill
\section{Adding Vectors}
To find the sum of two vectors $v$ and $w$, one needs to take $w$ so that the initial point of $w$ coincides with the terminal point of $v$. The vector originating at the initial point of $v$ and terminating at the terminal point of $w$ is the sum $v + w$. \\
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (4,-2);
%\coordinate [label = left: {$A$}] (a) at (0,0);
%\coordinate [label = right: {$B$}] (b) at (4,-2);
\coordinate [label = above: {$v$}] (v) at (2,-.9);
\draw [->, line width = 2pt] (4,-2) -- (2,-4);
%\coordinate [label = left: {$D$}] (d) at (-2,-2);
%\coordinate [label = right: {$C$}] (c) at (2.15,-4);
\coordinate [label = right: {$w$}] (w) at (3.1,-3.1);
\draw [->, line width = 2pt] (0,0) -- (2,-4);
\coordinate [label = left:
{$v + w$}] (p) at (.9,-2.15);
\end{normalsize}
\end{tikzpicture}
\end{center}
\vfill
\pagebreak
\problem{}
Use a compass and a ruler to construct the sum $v + w$ of the vectors $v$ and $w$ given below. \\
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (5,-2);
\coordinate [label = above: {$v$}] (v) at (2.5,-.95);
\draw [->, line width = 2pt] (-3,-3) -- (-2,-6);
\coordinate [label = left: {$w$}] (w) at (-2.55,-4.5);
\end{normalsize}
\end{tikzpicture}
\end{center}
\definition{The Zero Vector}<zero_vector>
A vector that has coinciding initial and terminal points is called the {\it zero vector} and is denoted as $\overrightarrow{0}$. According to the above definition of the vector addition,
\begin{equation}
v + \overrightarrow{0} =
\overrightarrow{0} + v =
v
\end{equation}
for any vector $v$.
\vfill
\definition{Inverses of Vectors}<inverse>
A vector $w$ such that $w + v = \overrightarrow{0}$ is called the \textit{inverse} of $v$ and is denoted as $-v$. The vector $-v$ lies either on the same straight line as $v$ or on a parallel one, has the same length as $v$, but points in the opposite direction: \\
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (4,-2);
\coordinate [label = above: {$v$}] (v) at (2,-.9);
\draw [<-, line width = 2pt] (-1,-3) -- (3,-5);
\coordinate [label = below: {$-v$}] (n) at (.85,-4.1);
\end{normalsize}
\end{tikzpicture}
\end{center}
\vspace{20pt}
Note that $-v + v = \overrightarrow{0}$ by definition, but the validity of the equation $v + (-v) = \overrightarrow{0}$ follows from the definitions of vector addition and the zero vector. We can combine both into the following:
\begin{equation}
-v + v =
v + (-v) =
\overrightarrow{0}
\end{equation}
\vfill
\pagebreak
Here is an important example of an inverse vector. When you stand still, the floor pushes you up with the force opposite to the force of the gravitational pull, a.k.a. \textit{weight}. \\
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw (0,0) -- (6,0);
\draw (2.5,0) -- (3,1);
\draw (3.5,0) -- (3,1);
\draw (3,1) -- (3,2);
\draw (3,2.3) circle (.3);
\draw (2.6,1.7) -- (3.4,1.7);
\draw [->, line width = 2pt] (2.5,0) -- (2.5,-1);
\coordinate [label = below: {weight}]
(g) at (2.5,-1.1);
\draw [->, line width = 2pt] (3.5,0) -- (3.5,1);
\coordinate [label = right: {floor reaction}]
(w) at (3.7,1);
\end{normalsize}
\end{tikzpicture}
\end{center}
The two opposing vectors add up to the zero vector, and therefore you don't move.
\problem{}
Give an example of a different pair of opposite forces.
\vfill
The following is the last thing we'll mention about the opposite vectors. The formula $w - v$ is defined as $w + (-v)$ for any vectors $v$ and $w$:
\begin{equation}
w - v =
w + (-v)
\end{equation}
\newpage
\problem{Dividing a segment into parts}
Using your compass and ruler, divide a segment into three equal parts. How would you split it into four? five? Have an instructor check your answer before moving on.
\hint{Don't skip this problem, you'll need it later. Make sure you check your answer!}
\vfill
\begin{center}
\begin{tikzpicture} \begin{normalsize}
\filldraw (0,0) circle (1.5pt);
\coordinate [label = left: {$A$}] (a) at (0,0);
\filldraw (12,0) circle (1.5pt);
\coordinate [label = right: {$B$}] (b) at (12,0);
\draw [line width = 1.5 pt] (0,0) -- (12,0);
\end{normalsize} \end{tikzpicture}
\end{center}
\vfill
\begin{center}
\begin{tikzpicture} \begin{normalsize}
\filldraw (0,0) circle (1.5pt);
\coordinate [label = left: {$A$}] (a) at (0,0);
\filldraw (12,0) circle (1.5pt);
\coordinate [label = right: {$B$}] (b) at (12,0);
\draw [line width = 1.5 pt] (0,0) -- (12,0);
\end{normalsize} \end{tikzpicture}
\end{center}
\vfill
\newpage
\problem{}
Is it possible to check whether $v = w$ on the picture below using only a compass? Why or why not? \\
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (4,-2);
%\coordinate [label = left: {$A$}] (a) at (0,0);
%\coordinate [label = right: {$B$}] (b) at (4,-2);
\coordinate [label = above: {$v$}] (v) at (2,-.9);
\draw [->, line width = 2pt] (-1,-3) -- (3,-5);
%\coordinate [label = left: {$C$}] (c) at (-1,-3);
%\coordinate [label = right: {$D$}] (d) at (3,-5);
\coordinate [label = below: {$w$}] (w) at (1,-4.1);
\end{normalsize}
\end{tikzpicture}
\end{center}
\vfill
\problem{}
Use a compass and a ruler to construct the vector $w = -.75 v$ for the vector $v$ given below such that point $C$ is its terminal point.
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (6,-2);
\coordinate [label = above: {$v$}] (v) at (3,-.9);
\filldraw (-4,-5) circle (1.5pt);
\coordinate [label = below left: {$C$}] (c) at (-4,-5.1);
\end{normalsize}
\end{tikzpicture}
\end{center}
\vfill
\note[Note]{
With the tools we have thus far, we can multiply vectors by any rational number using only a compass and a ruler. Multiplying a vector by an irrational number is a bit more tricky, but it is doable...
}
\newpage
\problem{}
Use a compass and a ruler to construct the vector $w = \sqrt{3} v$ for the vector $v$ given below so that $C$ is its initial point. \\
\hint{Pythagoras.}
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (5,-2);
\coordinate [label = above: {$v$}] (v) at (2.5,-.9);
\filldraw (-5,-6) circle (1.5pt);
\coordinate [label = below left: {$C$}] (c) at (-5,-6.1);
\end{normalsize}
\end{tikzpicture}
\end{center}
\vfill
\newpage
\section{Bonus}
\problem{Oldaque de Freitas' Puzzle}
% spell:off
Two ladies are sitting in a street caf\'e, talking about their children. One lady says that she has three daughters. The product of the girls' ages equals 36 and the sum of their ages is the same as the number of the house across the street. The second lady replies that this information is not enough to figure out the age of each child. The first lady agrees and adds that her oldest daughter has beautiful blue eyes. The second lady then solves the puzzle. Please do the same.
% spell:on
\vfill
\problem{There must be a better way...}
Using pen and paper, sum up all the integers from $1$ to $1000$.
\vfill
\end{document}

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[metadata]
title = "Vectors 1"
[publish]
handout = true
solutions = false

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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[solutions]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\uptitlel{Intermediate 2}
\uptitler{\smallurl{}}
\title{Vectors 2}
\subtitle{
Prepared by Mark on \today \\
Based on a handout by Oleg Gleizer
}
\begin{document}
\maketitle
\section{Review}
\definition{}
A \textit{vector} is a directed line segment. In the vector below, $A$ is its initial point and $B$ is its terminal point.
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (4,-1);
\coordinate [label = left: {$A$}] (a) at (0,0);
\coordinate [label = right: {$B$}] (b) at (4,-1);
\coordinate [label = above: {$v$}] (v) at (2,-1);
\end{normalsize}
\end{tikzpicture}
\end{center}
\definition{Equivalence}<vec_eq>
We say two vectors are equal if the quadrilateral they form is a parallelogram. In other words, two vectors are equal if they have the same length and direction.
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (4, -1);
\coordinate [label = left: {$A$}] (a) at (0, 0);
\coordinate [label = right: {$B$}] (b) at (4, -1);
\coordinate [label = above: {$v$}] (v) at (2, -1);
\draw [->, line width = 2pt] (-1,-1) -- (3,-2);
\coordinate [label = left: {$C$}] (c) at (-1,-1);
\coordinate [label = right: {$D$}] (d) at (3,-2);
\coordinate [label = below: {$w$}] (w) at (1,-1.6);
\draw (-1,-1) -- (0,0);
\draw (3,-2) -- (4,-1);
\end{normalsize}
\end{tikzpicture}
\end{center}
\definition{Addition}
To add two vectors $v$ and $w$, we move $w$ so that the initial point of $w$ coincides with the terminal point of $v$. The vector originating at the initial point of $v$ and terminating at the terminal point of $w$ is the sum $v + w$. \\
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (4,-1);
\coordinate [label = above: {$v$}] (v) at (2,-.9);
\draw [->, line width = 2pt] (4,-1) -- (2,-2);
\coordinate [label = right: {$w$}] (w) at (3, -1.7);
\draw [->, line width = 2pt] (0,0) -- (2,-2);
\coordinate [label = left: {$v + w$}] (p) at (1,-1.2);
\end{normalsize}
\end{tikzpicture}
\end{center}
Note that $v+w = w+v$. If we create a parallelogram with sides $w$ and $v$ (\ref{vec_eq}), the sums create the same diagonal:
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 1pt] (0,0) -- (4,-1);
\coordinate [label = above: {$v$}] (v) at (2,-.9);
\draw [->, line width = 1pt] (4,-1) -- (2,-2);
\coordinate [label = right: {$w$}] (w) at (3, -1.7);
\draw [<-, line width = 1pt] (2,-2) -- (-2,-1);
\coordinate [label = above: {$v$}] (v) at (0.2,-2);
\draw [<-, line width = 1pt] (-2,-1) -- (0, 0);
\coordinate [label = right: {$w$}] (w) at (-1.5, -0.3);
\draw [->, line width = 2pt] (0,0) -- (2,-2);
\coordinate [label = left: {$v + w$}] (p) at (1,-1.2);
\end{normalsize}
\end{tikzpicture}
\end{center}
\vfill
\pagebreak
\definition{The Zero Vector}
A vector that has coinciding initial and terminal points is called the {\it zero vector} and is denoted as $\overrightarrow{0}$. According to the above definition of the vector addition,
\begin{equation*}
v + \overrightarrow{0} =
\overrightarrow{0} + v =
v
\end{equation*}
for any vector $v$.
\definition{Inverse Vectors}
A vector $w$ such that $w + v = \overrightarrow{0}$ is called the \textit{inverse} of $v$ and is denoted as $-v$. The vector $-v$ lies either on the same straight line as $v$ or on a parallel one, has the same length as $v$, but points in the opposite direction: \\
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (4, -1);
\coordinate [label = above: {$v$}] (v) at (2, -1);
\draw [<-, line width = 2pt] (-1,-1) -- (3,-2);
\coordinate [label = below: {$-v$}] (w) at (1,-1.6);
\end{normalsize}
\end{tikzpicture}
\end{center}
\vfill
\pagebreak
\section{Vectors}
\problem{}
For the given vector $\overrightarrow{v}$
and point $A$, construct the vector
$\overrightarrow{w} = \overrightarrow{v}$
having $A$ as its initial point
on the graph paper below.
Use the grid instead of a compass and ruler. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-8,-8) grid (1,1);
\draw [line width = 1.5pt, ->] (-5,-1) -- (-1,-3);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.8,-2.1);
\filldraw (-6,-4) circle (2pt);
\coordinate [label = left:{$A$}] (a) at (-6,-4);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\problem{}
For the given vector $\overrightarrow{v}$
and point $A$, construct the vector
$\overrightarrow{w} = -\overrightarrow{v}$
having $A$ as its initial point
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-12,-7) grid (1,1);
\draw [line width = 1.5pt, ->] (-5,-1) -- (-1,-3);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.8,-2.1);
\filldraw (-6,-4) circle (2pt);
\coordinate [label = below:{$A$}] (a) at (-6,-4.1);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\pagebreak
\problem{}
For the given vector $\overrightarrow{v}$
and point $A$, construct the vector
$\overrightarrow{w} = 1.5 \overrightarrow{v}$
having $A$ as its initial point
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-9,-9) grid (1,1);
\draw [line width = 1.5pt, ->] (-6,-1) -- (-2,-3);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-3.8,-2.1);
\filldraw (-7,-4) circle (2pt);
\coordinate [label = left:{$A$}] (a) at (-7,-4.1);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\problem{}
For the given vector $\overrightarrow{v}$
and point $A$, construct the vector
$\overrightarrow{w} = -2 \overrightarrow{v}$
having $A$ as its initial point
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-16,-8) grid (1,1);
\draw [line width = 1.5pt, ->] (-5,-2) -- (-1,-4);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.8,-3.1);
\filldraw (-6,-5) circle (2pt);
\coordinate [label = below:{$A$}] (a) at (-6,-5.1);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\pagebreak
\problem{}
For the given vector $\overrightarrow{v}$
and point $A$, construct the vector
$\overrightarrow{w} = -\frac13 \overrightarrow{v}$
having $A$ as its initial point
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-11,-6) grid (1,1);
\draw [line width = 1.5pt, ->] (-7,-1) -- (-1,-4);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-3.8,-2.4);
\filldraw (-6,-4) circle (2pt);
\coordinate [label = below:{$A$}] (a) at (-6,-4.1);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\problem{}
For the given vectors $\overrightarrow{v}$
and $\overrightarrow{w}$,
construct the vector
$\overrightarrow{w} + \overrightarrow{v}$
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-12,-10) grid (1,1);
\draw [line width = 1.5pt, ->] (-5,0) -- (0,-4);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.4,-2.1);
\draw [line width = 1.5pt, ->] (-10,-1) -- (-9,-5);
\coordinate [label = left:{$\overrightarrow{w}$}] (w) at (-9.5,-3.1);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\pagebreak
\problem{}
For the given vectors $\overrightarrow{v}$
and $\overrightarrow{w}$,
construct the vector
$\overrightarrow{w} - \overrightarrow{v}$
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-12,-6) grid (1,1);
\draw [line width = 1.5pt, ->] (-11,0) -- (-6,-4);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-8.4,-2.1);
\draw [line width = 1.5pt, ->] (-1,-1) -- (0,-5);
\coordinate [label = right:{$\overrightarrow{w}$}] (w) at (-.5,-3.1);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\problem{}
For the given vectors $\overrightarrow{v}$
and $\overrightarrow{w}$,
construct the vector
$2\overrightarrow{v} - 3\overrightarrow{w}$
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-13,-10) grid (1,1);
\draw [line width = 1.5pt, ->] (-12,-1) -- (-7,-5);
\coordinate [label = left:{$\overrightarrow{v}$}] (v) at (-9.6,-3.1);
\draw [line width = 1.5pt, ->] (-1,-2) -- (0,-5);
\coordinate [label = right:{$\overrightarrow{w}$}] (w) at (-.5,-3.6);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\pagebreak
\problem{}
For the given vectors $\overrightarrow{v}$
and $\overrightarrow{w}$,
construct the vector
$1.75\overrightarrow{v} - \frac23 \overrightarrow{w}$
on the graph paper below.
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-19,-9) grid (1,1);
\draw [line width = 1.5pt, ->] (-18,-1) -- (-10,-5);
\coordinate [label = left:{$\overrightarrow{v}$}] (v) at (-14.4,-3.1);
\draw [line width = 1.5pt, ->] (-3,0) -- (0,-6);
\coordinate [label = right:{$\overrightarrow{w}$}] (w) at (-1.5,-3);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\problem{}
For the given vectors $\overrightarrow{v}$
and $\overrightarrow{w}$,
construct the vector
$\overrightarrow{v} + \overrightarrow{w}$
originating at the same point
as the vector $\overrightarrow{v}$
and the vector
$\overrightarrow{w} + \overrightarrow{v}$
originating at the same point
as the vector $\overrightarrow{w}$.
Is $\overrightarrow{v} + \overrightarrow{w} =
\overrightarrow{w} + \overrightarrow{v}$?
Why or why not?
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-17,-7) grid (1,1);
\draw [line width = 1.5pt, ->] (-16,-1) -- (-11,-5);
\coordinate [label = left:{$\overrightarrow{v}$}] (v) at (-13.6,-3.1);
\draw [line width = 1.5pt, ->] (-8,-6) -- (-5,-1);
\coordinate [label = right:{$\overrightarrow{w}$}] (w) at (-6.5,-4.5);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\pagebreak
\section{Pythagoras' Theorem}
\problem{}
Formulate and prove the Pythagoras' theorem.
\vfill
\pagebreak
\problem{}
Use the Pythagorean theorem to find $x$
for the following right triangles. \\
\noindent a.~~
\begin{center}
\begin{normalsize}
\begin{tikzpicture}
\draw (0,0) -- (3,0) -- (0,3) -- (0,0);
\coordinate [label = left:{$1$}] (a) at (0,1.5);
\coordinate [label = below:{$1$}] (b) at (1.5,-.1);
\coordinate [label = right:{$x$}] (x) at (1.6,1.5);
\end{tikzpicture}
\end{normalsize}
\end{center}
\bigskip
\noindent b.~~
\begin{center}
\begin{normalsize}
\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (0,3) -- (0,0);
\coordinate [label = left:{$3$}] (a) at (0,1.5);
\coordinate [label = below:{$x$}] (x) at (2,-.1);
\coordinate [label = right:{$5$}] (b) at (2.1,1.6);
\end{tikzpicture}
\end{normalsize}
\end{center}
\bigskip
\noindent c.~~
\begin{center}
\begin{normalsize}
\begin{tikzpicture}
\draw (0,0) -- (4.5,0) -- (0,3) -- (0,0);
\coordinate [label = left:{$x$}] (x) at (0,1.5);
\coordinate [label = below:{$\sqrt{2}$}] (a) at (2,-.1);
\coordinate [label = right:{$\sqrt3$}] (b) at (2.2,1.7);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\pagebreak
\problem{}
Construct a segment of length $\sqrt{2} a$
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-7,-7) grid (1,1);
\draw [line width = 1.5pt] (-5,-1) -- (-1,-1);
\coordinate [label = above:{$a$}] (a) at (-3,-.9);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\problem{}
Construct a segment of length $\sqrt{5} a$
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-12,-6) grid (1,1);
\draw [line width = 1.5pt] (-5,-1) -- (-1,-1);
\coordinate [label = above:{$a$}] (a) at (-3,-.9);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\problem{}
The side length of a grid square below
is one unit. Find the length
the vector $\overrightarrow{v}$. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-5,-3) grid (1,1);
\draw [line width = 1.5pt, ->] (-4,0) -- (0,-2);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.7,-2.1);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\pagebreak
\problem{}
The side length of a grid square below
is three units. Find the length
the vector $\overrightarrow{w}$. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-6,-3) grid (1,1);
\draw [line width = 1.5pt, <-] (-5,0) -- (0,-2);
\coordinate [label = above:{$\overrightarrow{w}$}] (w) at (-2.5,-2);
\end{tikzpicture}
\end{normalsize}
\end{center}
\medskip
\section{Rationals}
A number is called {\it rational}
if it can be represented as a ratio $p/q$
of an integer $p$ and a positive integer $q$
such that $p$ and $q$ have no common
factors. Otherwise, that number is called {\it irrational} \\
\problem{}
Decide whether the following numbers
are rational or irrational.
In each case, give a reason. \\
\begin{enumerate}[itemsep=2mm]
\item $\displaystyle{\frac{375}{376}}$
\item $10$
\item $0.5$
\item $-5$
\item $1.2345$
\item $0.111111111...$
\item $\sqrt{2}$
\item $\sqrt[3]{10}$
\end{enumerate}
\vfill
\pagebreak
\problem{}
Find $\left\lfloor \sqrt[3]{10} \right\rfloor$
and $\left\lceil \sqrt[3]{10} \right\rceil$.
\vfill
\problem{}
Simplify $\sqrt{8}$.
\vfill
\pagebreak
\problem{}
A man is crossing a river in a boat.
The speed of the boat is three meters per second.
The speed of the water in the river
is one meter per second.
In what direction should the man steer the boat,
if he wants the vessel to move
perpendicular to the banks?
Construct the velocity vector. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture}
\draw (0,10) -- (10,10);
\draw (0,0) -- (10,0);
\draw (4.5,1) -- (5.5,1) --(5.5,3) -- (5,3.5) -- (4.5,3) --(4.5,1) ;
\filldraw (5,2.1) circle (2pt);
\draw [line width = 2pt, ->] (5,2.1) -- (7,2.1);
\coordinate [label = below: {$1\frac{m}{s}$}] (w) at (6.3,1.9);
\end{tikzpicture}
\end{normalsize}
\end{center}
\bigskip
The width of the river
is $10\sqrt{2}$ meters.
How long would it take the man
to cross the river?
\vfill
\pagebreak
\problem{}
You need to slide a heavy box over
the floor from point $A$ to point $B$.
The box is about twice as tall as you are.
Which way is easier, to push or to pull?
Why? \vspace{60pt}
\begin{center}
\begin{normalsize}
\begin{tikzpicture}
\filldraw [gray!90!blue] (1,0) -- (1,2) --
(4,2) -- (4,0) -- (1,0);
\draw (0,0) -- (10,0);
\filldraw (2.5,0) circle (1.5pt);
\coordinate [label = below: {$A$}] (a) at (2.5,-.1);
\filldraw (7.5,0) circle (1.5pt);
\coordinate [label = below: {$B$}] (b) at (7.5,-.1);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\problem{}
The dot on the picture below
represents a spaceship.
There are three forces acting on the ship.
$\overrightarrow{T}$ is the thrust
of the ship's engine.
$\overrightarrow{P}$ is the gravitational pull
of the neighboring planet.
$\overrightarrow{S}$ is the gravitational pull
of the planet's home star.
You are the captain.
Use a compass and a ruler to figure out
where the resulting force would steer the ship.
\vspace{100pt}
\begin{center}
\begin{normalsize}
\begin{tikzpicture}
\filldraw (0,0) circle (2pt);
\draw [line width = 1pt, ->] (0,0) -- (225:6);
\coordinate [label = below: {$\overrightarrow{T}$}]
(t) at (230:3);
\draw [line width = 1pt, ->] (0,0) -- (4,0);
\coordinate [label = above: {$\overrightarrow{P}$}]
(p) at (2,.1);
\draw [line width = 1pt, ->] (0,0) -- (150:3);
\coordinate [label = right: {$\overrightarrow{S}$}]
(s) at (140:1.5);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\end{document}

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src/Intermediate/Vectors 2/meta.toml Normal file
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[metadata]
title = "Vectors 2"
[publish]
handout = true
solutions = false