Added Sunny's solutions

src/Advanced/Geometry of Masses/parts/0 balance 1d.tex

@@ -5,6 +5,7 @@ Consider a mass $m_1$ on top of a pin. \par
 Due to gravity, the mass exerts a force on the pin at the point of contact. \par
 For simplicity, we'll say that the magnitude of this force is equal the mass of the object---
 that is, $m_1$.
+
 \begin{center}
 	\begin{tikzpicture}[scale=2]
 		\fill[color = black] (0, 0.1) circle[radius=0.1];
@@ -12,9 +13,10 @@ that is, $m_1$.
 
 		\draw[line width = 0.25mm, pattern=north west lines] (0, 0) -- (-0.15, -0.3) -- (0.15, -0.3) -- cycle;
 
-
 		\draw[color = black, opacity = 0.5] (1, 0.1) circle[radius=0.1];
-		\draw[line width = 0.25mm, pattern=north west lines, opacity = 0.5] (1, 0) -- (0.85, -0.3) -- (1.15, -0.3) -- cycle;
+
+		\draw[line width = 0.25mm, pattern=north west lines, opacity = 0.5]
+			(1, 0) -- (0.85, -0.3) -- (1.15, -0.3) -- cycle;
 
 		\draw[->, line width = 0.5mm] (1, 0) -- (1, -0.5) node[below] {$m_1$};
 		%\draw[->, line width = 0.5mm, dashed] (1, 0) -- (1, 0.5) node[above] {$-m_1$};
@@ -44,8 +46,8 @@ As you might expect, it is not stable: the rod pivots and falls down.
 		\draw[color = black, opacity = 0.5] (1.2, 0.0) circle[radius=0.1];
 		\draw[-, line width = 0.5mm, opacity = 0.5] (0.7, 0) -- (1.9, 0);
 
-		\draw[line width = 0.25mm, pattern=north west lines, opacity = 0.5] (1.5, 0) -- (1.35, -0.3) -- (1.65, -0.3) -- cycle;
-
+		\draw[line width = 0.25mm, pattern=north west lines, opacity = 0.5]
+			(1.5, 0) -- (1.35, -0.3) -- (1.65, -0.3) -- cycle;
 
 		\draw[->, line width = 0.5mm] (1.2, 0) -- (1.2, -0.5) node[below] {$m_1$};
 		%\draw[->, line width = 0.5mm, dashed] (1.5, 0) -- (1.5, 0.5) node[above] {$f_p$};
@@ -57,6 +59,7 @@ It therefore exerts a \textit{torque} on the mass-rod system, causing it to rota
 \pagebreak
 
 
+
 \definition{Torque}
 Consider a rod on a single pivot point.
 If a force with magnitude $m_1$ is applied at an offset $d$ from the pivot point,
@@ -110,7 +113,6 @@ Find the position of this system's center of mass. \par
 
 		\draw[-, line width = 0.5mm] (-0.5, 0) -- (1.5, 0);
 
-
 		\fill[color = black] (-0.5, 0) circle[radius=0.1];
 		\node[above] at (-0.5, 0.2) {$3$};
 
@@ -139,9 +141,24 @@ Do the same for the following system, where $m_1$ and $m_2$ are arbitrary masses
 	\end{tikzpicture}
 \end{center}
 
+\begin{solution}
+	The CoM will be such that the rod is split into $d_1+d_2=1$
+	according to the relation $m_1 d_1 = m_2 d_2$.
+	This is sufficient, but if you want to solve for one of the $d$,
+	you get $d_1 = \frac{m_2}{m_1+m_2}$.
+
+	\vspace{2mm}
+
+	This should be intuitive; the distance of each mass from the CoM
+	is proportional to the other mass's share of the total mass.
+\end{solution}
+
 \vfill
 \pagebreak
 
+
+
+
 \definition{}
 Consider a massless, horizontal rod of infinite length. \par
 Affix a finite number of point masses to this rod. \par
@@ -172,5 +189,12 @@ Consider a one-dimensional system of masses consisting of $n$ masses $m_1, m_2,
 with each $m_i$ positioned at $x_i$. Show that the resulting system always has a unique center of mass. \par
 \hint{Prove this by construction: find the point!}
 
+\begin{solution}
+	\begin{equation*}
+		x_0 = \frac{1}{M}\sum_{i=1}^n m_i x_i
+	\end{equation*}
+	where $M = \sum_{i=1}^n m_i$
+\end{solution}
+
 \vfill
 \pagebreak

src/Advanced/Geometry of Masses/parts/1 balance 2d.tex

@@ -140,6 +140,13 @@ Show that any two-dimensional system of masses has a unique center of mass. \par
 	\end{tikzpicture}
 \end{center}
 
+\begin{solution}
+	\begin{equation*}
+		x_0 = \frac{\sum_{i=1}^n m_i x_i}{\sum_{i=1}^n m_i}
+		\qquad
+		y_0 = \frac{\sum_{i=1}^n m_i y_i}{\sum_{i=1}^n m_i}
+	\end{equation*}
+\end{solution}
 
 \vfill
 \pagebreak

src/Advanced/Geometry of Masses/parts/1 continuous.tex

@@ -2,19 +2,69 @@
 
 Now let's extend this idea to a \textit{continuous distribution} of masses rather than discrete point masses. This isn't so different; a continuous distribution of mass is really just a lot of point-masses, only that there are so many of them so close together that you can't even count them\footnote{For example, your pencil might seem like a continuous distribution of mass, but it's really just a whole lot of atoms.}. In general, finding the CoM requires integral calculus, but not always...\footnote{Many of the following problems can be solved with integration even though you're meant to solve them without it. But remember, in math, whenever you accomplish the same task two different ways, that really means that they're somehow the same thing.}
 
+\begin{figure}[htp]
+	\centering
+	\includegraphics[width=0.3\linewidth]{img/seahorse.jpg}
+	\label{seahorse}
+\end{figure}
 
 \problem{}
-You are given a cardboard cutout of Figure \ref{seahorse} and some office supplies. How might you determine the CoM? Does your strategy also work in 3D?
+You are given a cardboard cutout of a seahorse and some office supplies. \par
+How might you determine its center of mass? Does your strategy also work in 3D?
+
+\begin{solution}
+	Many correct answers. One example:
+	\begin{enumerate}[label={(\arabic*)}]
+		\item Stick a thumb tack into the horse and let it come to equilibrium
+		\item Use a ruler or string to draw a straight line through that point along the direction of gravity
+		\item Repeat (1) and (2) at a different point
+		\item The intersection of the two lines marks the CoM
+	\end{enumerate}
+\end{solution}
+
+\vfill
+
+\definition{Centroid}
+Centroids are closely related to, and often synonymous with, centers of mass.
+A centroid is the geometric center of an object, regardless of the mass distribution.
+Thus, the centroid and center of mass are the same when the mass is uniformly distributed.
+
+\problem{}<rightiso>
+Where is the center of a right isosceles triangle? What about any isosceles triangle?
+
+\begin{solution}
+	There are probably some other clever ways of doing this without calculus but here's one way:
+	\begin{center}
+		\includegraphics[width=0.3\linewidth]{img/right_isos.png}
+	\end{center}
+	Clearly, the centroid $O$ must be somewhere along $SC$.
+	Now we just need to find $s$ in terms of $x$, that is, the balancing point along either of the shorter sides.
+	To do this, we split the triangle into three regions: $\triangle{AVT}$ on the left,
+	the rectangle $TUCV$ on the right, and $\triangle{TUB}$ also on the right.
+
+	Each region exerts a torque proportional to its area times the horizontal distance from $VT$ of that region's
+	centroid. Note that, even though we don't know what $x$ is yet, we can use it to find itself.
+	By similar triangles, the centroids of $\triangle{AVT}$ and $\triangle{TUB}$ are both located $x(1-x)$ away from $VT$.
+	The centroid of $TUVC$ is trivially $\frac{1-x}{2}$. So we get the following equation:
+	\begin{align*}
+		\frac{1}{2}x^2 \cdot x(1-x) &= x(1-x) \cdot \frac{1-x}{2} + \frac{1}{2}(1-x)^2 \cdot x(1-x) \quad .
+	\end{align*}
+	We easily find that $x=\frac{1}{3}$. Remarkably, the ratio $\frac{SO}{SC}$ is also $\frac{1}{3}$.
+
+	Any right triangle is just an isosceles right triangle that's been scaled along some axis, so the centroid scales with it and this one-third rule still applies.
+
+	Any isosceles triangle is just two right triangles, so \textit{its} centroid will be in between its two "sub-centroids" from each right triangle, that is, one-third the altitude.
+\end{solution}
 
 \vfill
 
 \problem{}
-Where is the CoM of a right isosceles triangle? What about any isosceles triangle?
+How can you easily find the center of mass of any triangle? Why does this work?
 
-\vfill
 
-\problem{}
-How can you easily find the CoM of any triangle? Why does this work?
+\begin{solution}
+    It turns out that all three medians of a triangle always intersect at a single point. That point is the centroid. You could feasibly guess this by taking what you learned from Problem 13 and applying Cavalieri's Theorem. Otherwise, I'm interested to see what students come up with.
+\end{solution}
 
 \vfill
 
@@ -30,16 +80,16 @@ you can get it to balance on the beveled edge, as seen in Figure
 \begin{figure}[htp]
 \centering
 \begin{minipage}{0.5\textwidth}
-    \centering
-    \includegraphics[width=0.6\linewidth]{img/soda.png}
-    \caption{}
-    \label{soda}
+	\centering
+	\includegraphics[width=0.6\linewidth]{img/soda.png}
+	\caption{}
+	\label{soda}
 \end{minipage}\hfill
 \begin{minipage}{0.5\textwidth}
-    \centering
-    \includegraphics[width=0.6 \linewidth]{img/soda_filled.png}
-    \caption{}
-    \label{soda filled}
+	\centering
+	\includegraphics[width=0.6 \linewidth]{img/soda_filled.png}
+	\caption{}
+	\label{soda filled}
 \end{minipage}
 \end{figure}
 
@@ -50,17 +100,35 @@ Let's also assume that we live in two dimensions.
 We start slowly filling it up with soda to a vertical height $h$.
 What is $h$ just before the can tips over?
 
+\begin{solution}
+	Similar to our solution to \ref{rightiso}, we draw a vertical line from the
+	desired balancing point and split the regions into triangles and rectangles.
+	Using symmetry and simple trigonometry, we find that:
+	$h = \sqrt{\sqrt{2}+\frac{8}{9}} + 3\sqrt{2}$
+\end{solution}
+
 \vfill
 
 \problem{}<3D soda>
 Think about how you might approach this problem in 3D. Does $h$ become larger or smaller?
 
+\begin{solution}
+	This is a pretty open-ended question and is meant simply to make students think about
+	how the problem would change in 3D. I believe that $h$ gets smaller.
+\end{solution}
+
+
 \vfill
 
 \pagebreak
 
 So far we've made the assumption our shapes have mass that is \textit{uniformly distributed}. But that doesn't have to be the case.
 
+\begin{solution}
+	This problem is really just \ref{rightiso} again in disguise.
+	So, the balancing point is at $1/3$ the length of the staff measured from the dense-end.
+\end{solution}
+
 \problem{}
 A mathemagical wizard will give you his staff if you can balance it horizontally on your finger. The strange magical staff has unit length and it's mass is distributed in a very special way. It's density decreases linearly from $\lambda_0$ at one end to $0$ at the other. Where is the staff's balancing point?
 

src/Advanced/Geometry of Masses/parts/2 pappus.tex

@@ -17,6 +17,11 @@ Figure \ref{pappus1} depicts three different surfaces constructed by revolving a
 \textit{Pappus's First Centroid Theorem} allows you to determine the area of a surface of revolution using information about the line segment and the axis of rotation.
 Can you intuitively come up with Pappus's First Centroid Theorem for yourself? Figure \ref{pappus1} is very helpful. It may also help to draw from surface area formulae you already know. What limitations are there on the theorem?
 
+\begin{solution}
+	\url{https://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem}
+	\url{https://mathworld.wolfram.com/PappussCentroidTheorem.html}
+\end{solution}
+
 \vfill
 \pagebreak
 
@@ -40,27 +45,51 @@ The centroid of a semi-circular line segment is already given in Figure \ref{pap
 	\label{arc}
 \end{figure}
 
-\label{arc centroid} Given arc $AB$ with radius $r$ and subtended by $2\alpha$, determine $OG$, the distance from the centre of the circle to the centroid of the arc.
+\problem{}
+Given arc $AB$ with radius $r$ and subtended by $2\alpha$, determine $OG$, the distance from the centre of the circle to the centroid of the arc.
+
+\begin{solution}
+	\url{https://mathspanda.com/A2FM/Lessons/Centres_of_mass_of_standard_shapes_LESSON.pdf}
+\end{solution}
 
 \vfill
 
 \pagebreak
 
-\problem{}
-Using your answers from Problem \ref{isosceles centroid} and Problem \ref{arc centroid}. Where is the centroid of the \textit{sector} of the circle in Figure \ref{arc}. (Hint: Cut it up.)
+\problem{}<sector>
+Where is the centroid of the \textit{sector} of the circle in Figure \ref{arc}?
+\hint{cut it up.}
+
 
 \vfill
 
 \problem{}
-Seeing your success with his linear staff, the wizard challenges you with another magical staff to balance. It looks identical to the first one, but you're told that the density decreases from $\lambda_0$ to $0$ according to the function $\lambda(x) = \lambda_0\sqrt{1-x^2}$.
+Seeing your success with his linear staff, the wizard challenges you with another magical staff to balance.
+It looks identical to the first one, but you're told that the density decreases from $\lambda_0$ to $0$
+according to the function $\lambda(x) = \lambda_0\sqrt{1-x^2}$.
+
+\begin{solution}
+	This is equivalent to finding the x-coordinate of the centroid of a quarter-circle. See \ref{sector}.
+\end{solution}
 
 \vfill
 \problem{}
 Infinitely many masses $m_i$ are placed at $x_i$ along the positive $x$-axis, starting with $m_0 = 1$ placed at $x_0 = 1$. Each successive mass is placed twice as far from the origin compared to the previous one. But also, each successive mass has a quarter the weight of the previous one. Find the CoM if it exists.
 
+\begin{solution}
+	We have $m_i = 1/4^i$ and $x_i = 2^i$ so
+	\begin{align*}
+		\sum_{i=0}^\infty m_i x_i &= \sum \frac{2^i}{4^i} \\
+		&= \sum \frac{1}{2^i} \\
+		&= \frac{1}{1-1/2} = 2 \qquad \text{as this is just a geometric series} \\
+	\end{align*}
+	\[ M = \sum m_i = \frac{1}{1-1/4} = 4/3 \]
+	Then
+	\[ x_{CM} = \frac{\sum m_i x_i}{M} = 3/2 \]
+\end{solution}
+
 \vfill
 
-\problem{}
-(Bonus) Try to actually find $h$ from Problem \ref{3D soda}. Good luck.
+\problem{Bonus}
+Try to actually find $h$ from Problem \ref{3D soda}. Good luck.
 
-%\item (Bonus, not related to the packet) Spongebob, Patrick, and Squidward are all hiding from the Sea Bear. Initially, Spongebob and Patrick are keeping watch 100 yards apart with Squidward halfway in between them. When Spongebob gets scared, he runs to hide halfway between Squidward and Patrick. Then Patrick, not wanting to be the farthest from the centre, runs to be halfway between Squidward and Patrick. Squidward, seeing this, quickly finds the new halfway point between Spongebob and Patrick. This pattern keeps repeating until all three of them are pointlessly clambering on top of each other. Where do they end up relative to their initial positions?