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@ -2,9 +2,9 @@
\index{geometry} \index{geometry}
In geometric problems, a usual challenge is In geometric problems, it is often challenging
to realize how to approach the problem so that to find a way to approach the problem so that
the solution can be conveniently implemented the solution to the problem can be conveniently implemented
and the number of special cases is small. and the number of special cases is small.
As an example, consider a problem where As an example, consider a problem where
@ -51,10 +51,10 @@ $s=(a+b+c)/2$.
This is a possible way to solve the problem, This is a possible way to solve the problem,
but there is one pitfall: but there is one pitfall:
how to divide the quadrilateral into triangles? how to divide the quadrilateral into triangles?
It turns out that sometimes we can't pick It turns out that sometimes we cannot just pick
arbitrary opposite vertices. two arbitrary vertices.
For example, in the following quadrilateral, For example, in the following situation,
the division line is outside the quadrilateral: the division line lies outside the quadrilateral:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.45] \begin{tikzpicture}[scale=0.45]
@ -80,13 +80,13 @@ However, another way to draw the line works:
\draw[dashed,thick] (3,2) -- (1,1); \draw[dashed,thick] (3,2) -- (1,1);
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
For a human, it is clear which of the lines is the correct It is clear for a human which of the lines is the correct
choice, but for a computer the situation is difficult. choice, but the situation is difficult for a computer.
However, it turns out that we can solve the problem using However, it turns out that we can solve the problem using
another method which is much easier to implement another method that is much easier to implement
and doesn't contain any special cases. and does not involve any special cases.
There is a general formula Namely, there is a general formula
\[x_1y_2-x_2y_1+x_2y_3-x_3y_2+x_3y_4-x_4y_3+x_4y_1-x_1y_4,\] \[x_1y_2-x_2y_1+x_2y_3-x_3y_2+x_3y_4-x_4y_3+x_4y_1-x_1y_4,\]
that calculates the area of a quadrilateral that calculates the area of a quadrilateral
whose vertices are whose vertices are
@ -94,9 +94,9 @@ $(x_1,y_1)$,
$(x_2,y_2)$, $(x_2,y_2)$,
$(x_3,y_3)$ and $(x_3,y_3)$ and
$(x_4,y_4)$. $(x_4,y_4)$.
This formula is easy to calculate, there are no special This formula is easy to implement, there are no special
cases, and it turns out that we can generalize the formula cases, and it turns out that we can even generalize the formula
for \emph{all} polygons. to \emph{all} polygons.
\section{Complex numbers} \section{Complex numbers}
@ -106,9 +106,9 @@ for \emph{all} polygons.
A \key{complex number} is a number of the form $x+y i$, A \key{complex number} is a number of the form $x+y i$,
where $i = \sqrt{-1}$ is the \key{imaginary unit}. where $i = \sqrt{-1}$ is the \key{imaginary unit}.
A geometric interpretation for a complex number is A geometric interpretation of a complex number is
that it represents a two-dimensional point $(x,y)$ that it represents a two-dimensional point $(x,y)$
or a vector from the origin to point $(x,y)$. or a vector from the origin to a point $(x,y)$.
For example, $4+2i$ corresponds to the For example, $4+2i$ corresponds to the
following point and vector: following point and vector:
@ -129,14 +129,14 @@ following point and vector:
\index{complex@\texttt{complex}} \index{complex@\texttt{complex}}
The complex number class \texttt{complex} in C++ is The complex number class \texttt{complex} in C++ is
useful when solving geometry problems. useful when solving geometric problems.
Using the class, we can store points and vectors Using the class we can represent points and vectors
as complex numbers, and the class also contains tools as complex numbers, and the class also contains tools
that are useful in geometry. that are useful in geometry.
In the following code, \texttt{C} is the type of In the following code, \texttt{C} is the type of
a coordinate, and \texttt{P} is the type of a point or vector. a coordinate, and \texttt{P} is the type of a point or vector.
In addition, the code defines shortcuts \texttt{X} and \texttt{Y} In addition, the code defines the macros \texttt{X} and \texttt{Y}
that can be used to refer to x and y coordinates. that can be used to refer to x and y coordinates.
\begin{lstlisting} \begin{lstlisting}
@ -164,17 +164,16 @@ P s = v+u;
cout << s.X << " " << s.Y << "\n"; // 5 3 cout << s.X << " " << s.Y << "\n"; // 5 3
\end{lstlisting} \end{lstlisting}
The appropriate type for \texttt{C} is An appropriate coordinate type is
\texttt{long long} (integer) or \texttt{long double} \texttt{long long} (integer) or \texttt{long double}
(real number), depending on the situation. (real number), depending on the situation.
It is a good idea to use integers whenever possible, Integers should be used whenever possible,
because the integer calculations are exact. because calculations with integers are exact.
If real numbers are needed,
If coordinates are real numbers, precision errors should be taken into account
precision error should be taken into account
when comparing them. when comparing them.
A safe way to check if numbers $a$ and $b$ are equal A safe way to check if numbers $a$ and $b$ are equal
is the comparison $|a-b|<\epsilon$, is to compare them using $|a-b|<\epsilon$,
where $\epsilon$ is a small number (for example, $\epsilon=10^{-9}$). where $\epsilon$ is a small number (for example, $\epsilon=10^{-9}$).
\subsubsection*{Functions} \subsubsection*{Functions}
@ -192,7 +191,7 @@ because that distance equals the length
of the vector $(x_2-x_1,y_2-y_1)$. of the vector $(x_2-x_1,y_2-y_1)$.
The following code calculates the distance The following code calculates the distance
of points $(4,2)$ and $(3,-1)$: between points $(4,2)$ and $(3,-1)$:
\begin{lstlisting} \begin{lstlisting}
P a = {4,2}; P a = {4,2};
P b = {3,-1}; P b = {3,-1};
@ -200,16 +199,16 @@ cout << abs(b-a) << "\n"; // 3.60555
\end{lstlisting} \end{lstlisting}
The function \texttt{arg(v)} calculates the The function \texttt{arg(v)} calculates the
angle of a vector $v=(x,y)$ with respect to the x axel. angle of a vector $v=(x,y)$ with respect to the x axis.
The function gives the angle in radians, The function gives the angle in radians,
where $r$ radians equals $180 r/\pi$ degrees. where $r$ radians equals $180 r/\pi$ degrees.
The angle of a vector that points to the right is 0, The angle of a vector that points to the right is 0,
and the angle decreases clockwise and increases and angles decrease clockwise and increase
counterclockwise. counterclockwise.
The function \texttt{polar(s,a)} constructs a vector The function \texttt{polar(s,a)} constructs a vector
whose length is $s$ and that points to angle $a$. whose length is $s$ and that points to an angle $a$.
In addition, a vector can be rotated by angle $a$ In addition, a vector can be rotated by an angle $a$
by multiplying it by a vector with length 1 and angle $a$. by multiplying it by a vector with length 1 and angle $a$.
The following code calculates the angle of The following code calculates the angle of
@ -229,11 +228,12 @@ cout << arg(v) << "\n"; // 0.963648
The \key{cross product} $a \times b$ of vectors The \key{cross product} $a \times b$ of vectors
$a=(x_1,y_1)$ and $b=(x_2,y_2)$ equals $x_1 y_2 - x_2 y_1$. $a=(x_1,y_1)$ and $b=(x_2,y_2)$ equals $x_1 y_2 - x_2 y_1$.
The cross product indicates whether the vector $b$ The cross product tells us whether $b$
turns to the left (positive value) or to the right (negative value) turns left (positive value), does not turn (zero)
when it is placed directly after the vector $a$. or turns to right (negative value)
when it is placed directly after $a$.
The following picture shows three examples: The following picture illustrates the above cases:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.45] \begin{tikzpicture}[scale=0.45]
@ -264,7 +264,7 @@ The following picture shows three examples:
\end{center} \end{center}
\noindent \noindent
For example, in the left picture For example, in the first picture
$a=(4,2)$ and $b=(1,2)$. $a=(4,2)$ and $b=(1,2)$.
The following code calculates the cross product The following code calculates the cross product
using the class \texttt{complex}: using the class \texttt{complex}:
@ -275,7 +275,8 @@ P b = {1,2};
C r = (conj(a)*b).Y; // 6 C r = (conj(a)*b).Y; // 6
\end{lstlisting} \end{lstlisting}
The function \texttt{conj} negates the y coordinate The above code works, because
the function \texttt{conj} negates the y coordinate
of a vector, of a vector,
and when the vectors $(x_1,-y_1)$ and $(x_2,y_2)$ and when the vectors $(x_1,-y_1)$ and $(x_2,y_2)$
are multiplied together, the y coordinate are multiplied together, the y coordinate
@ -304,8 +305,9 @@ $p$ is on the left side of the line:
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
Now the cross product $(p-s_1) \times (p-s_2)$ In this situation,
indicates the location of the point $p$. the cross product $(p-s_1) \times (p-s_2)$
tells us the location of the point $p$.
If the cross product is positive, If the cross product is positive,
$p$ is located on the left side, $p$ is located on the left side,
and if the cross product is negative, and if the cross product is negative,
@ -369,9 +371,10 @@ intersection point is $b=c$:
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
This case is easy to test because the This case is easy to check, because
possibilities for the common intersection point there are only four possibilities
are $a=c$, $a=d$, $b=c$ and $b=d$. for the intersection point:
$a=c$, $a=d$, $b=c$ and $b=d$.
\textit{Case 3:} \textit{Case 3:}
There is exactly one intersection point There is exactly one intersection point
@ -410,10 +413,10 @@ using the formula
where $a$, $b$ and $c$ are the vertices of the triangle. where $a$, $b$ and $c$ are the vertices of the triangle.
Using this formula, it is possible to calculate the Using this formula, it is possible to calculate the
shortest distance of a point from a line. shortest distance between a point and a line.
For example, in the following picture $d$ is the For example, in the following picture $d$ is the
shortest distance between point $p$ and the line shortest distance between the point $p$ and the line
that is defined by points $s_1$ and $s_2$: that is defined by the points $s_1$ and $s_2$:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.75] \begin{tikzpicture}[scale=0.75]
\draw (-2,-1)--(6,3); \draw (-2,-1)--(6,3);
@ -461,7 +464,7 @@ the polygon.
A convenient way to solve the problem is to A convenient way to solve the problem is to
send a ray from the point to an arbitrary direction send a ray from the point to an arbitrary direction
and calculate the number of times it touches and calculate the number of times it touches
the border of the polygon. the boundary of the polygon.
If the number is odd, If the number is odd,
the point is inside the polygon, the point is inside the polygon,
and if the number is even, and if the number is even,
@ -488,10 +491,10 @@ For example, we could send the following rays:
\end{samepage} \end{samepage}
The rays from $a$ touch 1 and 3 times The rays from $a$ touch 1 and 3 times
the border of the polygon, the boundary of the polygon,
so $a$ is inside the polygon. so $a$ is inside the polygon.
Correspondingly, the rays from $b$ Correspondingly, the rays from $b$
touch 0 and 2 times the border of the polygon, touch 0 and 2 times the boundary of the polygon,
so $b$ is outside the polygon. so $b$ is outside the polygon.
\section{Polygon area} \section{Polygon area}
@ -500,10 +503,10 @@ A general formula for calculating the area
of a polygon is of a polygon is
\[\frac{1}{2} |\sum_{i=1}^{n-1} (p_i \times p_{i+1})| = \[\frac{1}{2} |\sum_{i=1}^{n-1} (p_i \times p_{i+1})| =
\frac{1}{2} |\sum_{i=1}^{n-1} (x_i y_{i+1} - x_{i+1} y_i)|, \] \frac{1}{2} |\sum_{i=1}^{n-1} (x_i y_{i+1} - x_{i+1} y_i)|, \]
when the vertices are where the vertices are
$p_1=(x_1,y_1)$, $p_2=(x_2,y_2)$, $\ldots$, $p_n=(x_n,y_n)$ $p_1=(x_1,y_1)$, $p_2=(x_2,y_2)$, $\ldots$, $p_n=(x_n,y_n)$
sorted so that in such an order that
$p_i$ and $p_{i+1}$ are adjacent vertices on the border $p_i$ and $p_{i+1}$ are adjacent vertices on the boundary
of the polygon, of the polygon,
and the first and last vertex is the same, i.e., $p_1=p_n$. and the first and last vertex is the same, i.e., $p_1=p_n$.
@ -527,8 +530,8 @@ is
\[\frac{|(2\cdot5-4\cdot5)+(5\cdot3-5\cdot7)+(7\cdot1-3\cdot4)+(4\cdot3-1\cdot4)+(4\cdot4-3\cdot2)|}{2} = 17/2.\] \[\frac{|(2\cdot5-4\cdot5)+(5\cdot3-5\cdot7)+(7\cdot1-3\cdot4)+(4\cdot3-1\cdot4)+(4\cdot4-3\cdot2)|}{2} = 17/2.\]
The idea in the formula is to go through trapezoids The idea in the formula is to go through trapezoids
where one side is a side of the polygon, whose one side is a side of the polygon,
and another side is on the horizontal line $y=0$. and another side lies on the horizontal line $y=0$.
For example: For example:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.7] \begin{tikzpicture}[scale=0.7]
@ -547,9 +550,9 @@ For example:
\draw (0,0) -- (10,0); \draw (0,0) -- (10,0);
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
The are of such a trapezoid is The area of such a trapezoid is
\[(x_{i+1}-x_{i}) \frac{y_i+y_{i+1}}{2},\] \[(x_{i+1}-x_{i}) \frac{y_i+y_{i+1}}{2},\]
when the vertices of the polygon are $p_i$ and $p_{i+1}$. where the vertices of the polygon are $p_i$ and $p_{i+1}$.
If $x_{i+1}>x_{i}$, the area is positive, If $x_{i+1}>x_{i}$, the area is positive,
and if $x_{i+1}<x_{i}$, the area is negative. and if $x_{i+1}<x_{i}$, the area is negative.
@ -559,21 +562,22 @@ all such trapezoids, which yields the formula
\[|\sum_{i=1}^{n-1} (x_{i+1}-x_{i}) \frac{y_i+y_{i+1}}{2}| = \[|\sum_{i=1}^{n-1} (x_{i+1}-x_{i}) \frac{y_i+y_{i+1}}{2}| =
\frac{1}{2} |\sum_{i=1}^{n-1} (x_i y_{i+1} - x_{i+1} y_i)|.\] \frac{1}{2} |\sum_{i=1}^{n-1} (x_i y_{i+1} - x_{i+1} y_i)|.\]
Note that the absolute value of the sum is calculated, Note that the absolute value of the sum is taken,
because the value of the sum may be positive or negative because the value of the sum may be positive or negative
depending on whether we walk clockwise or counterclockwise depending on whether we walk clockwise or counterclockwise
along the sides of the the polygon. along the perimeter of the polygon.
\subsubsection{Pick's theorem} \subsubsection{Pick's theorem}
\index{Pick's theorem} \index{Pick's theorem}
\key{Pick's theorem} provides another way to calculate \key{Pick's theorem} provides another way to calculate
the area of a polygon where are vertices have integer coordinates. the area of a polygon provided that all vertices
of the polygon have integer coordinates.
According to Pick's theorem, the area of the polygon is According to Pick's theorem, the area of the polygon is
\[ a + b/2 -1,\] \[ a + b/2 -1,\]
where $a$ is the number of integer points inside the polygon where $a$ is the number of integer points inside the polygon
and $b$ is the number of integer points on the border of the polygon. and $b$ is the number of integer points on the boundary of the polygon.
For example, the area of the polygon For example, the area of the polygon
\begin{center} \begin{center}
@ -656,7 +660,7 @@ The Euclidean distance between the points is
and the Manhattan distance is and the Manhattan distance is
\[|5-2|+|2-1|=4.\] \[|5-2|+|2-1|=4.\]
The following picture shows regions that are within a distance of 1 The following picture shows regions that are within a distance of 1
from the center point, using Euclidean and Manhattan distances: from the center point, using the Euclidean and Manhattan distances:
\begin{center} \begin{center}
\begin{tikzpicture} \begin{tikzpicture}
@ -673,18 +677,18 @@ from the center point, using Euclidean and Manhattan distances:
\subsubsection{Furthest points} \subsubsection{Furthest points}
Some problems are easier to solve if we use the Some problems are easier to solve if the
Manhattan distance instead of the Euclidean distance. Manhattan distance is used instead of the Euclidean distance.
For example, consider a problem where we are given For example, consider a problem where we are given
$n$ points $(x_1,y_1),(x_2,y_2),\ldots,(x_n,y_n)$ $n$ points $(x_1,y_1),(x_2,y_2),\ldots,(x_n,y_n)$
and our task is to calculate the maximum distance and our task is to calculate the maximum distance
between any two points. between any two points.
This is a difficult problem if we should maximize This is a difficult problem if the Euclidean distance
the Euclidean distance, should be maximized,
but calculating but it is easy to maximize the
the maximum Manhattan distance is easy Manhattan distance,
because it either because it is either
\[\max A - \min A \hspace{20px} \textrm{or} \hspace{20px} \max B - \min B,\] \[\max A - \min A \hspace{20px} \textrm{or} \hspace{20px} \max B - \min B,\]
where where
\[A = \{x_i+y_i : i = 1,2,\ldots,n\}\] \[A = \{x_i+y_i : i = 1,2,\ldots,n\}\]
@ -710,8 +714,8 @@ A useful technique related to the Manhattan distance
is to rotate all coordinates 45 degrees so that is to rotate all coordinates 45 degrees so that
a point $(x,y)$ becomes $(a(x+y),a(y-x))$, a point $(x,y)$ becomes $(a(x+y),a(y-x))$,
where $a=1/\sqrt{2}$. where $a=1/\sqrt{2}$.
The multiplier $a$ is so chosen that The coefficient $a$ is chosen so that
the distances of the points remain the same. the distances between the points remain the same.
After the rotation, the region within a distance of $d$ After the rotation, the region within a distance of $d$
from a point is a square with horizontal and vertical sides: from a point is a square with horizontal and vertical sides: