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\documentclass{article} | |
\usepackage{graphicx,amsmath,amsthm,amssymb} | |
\usepackage{fullpage} | |
\usepackage{eucal} | |
\usepackage{graphicx} | |
\usepackage{color} | |
\usepackage{tikz} | |
\usepackage{algorithm,algorithmic} | |
\newcommand{\question}[1]{\vspace{10pt}\noindent $\mathbf{#1}$} | |
\newcommand{\R}{\mathbb{R}} | |
\newcommand{\vect}[2]{\bigl[\begin{smallmatrix}#1\\#2\end{smallmatrix}\bigr]} | |
\newcommand{\vectt}[3]{\bigl[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\bigr]} | |
\newcommand{\gap}{{~~~~~~~~~~~~~~~~~~~~~}} | |
\newcommand{\vor}{\mathrm{Vor}} | |
\newcommand{\CC}{\mathrm{conv}} | |
\title{Computational Geometry Homework 3} | |
\author{Yiyun Tan} | |
\begin{document} | |
\maketitle | |
\section*{Projective Duality} | |
\question{1} | |
Let $a,b\in\R^2$. | |
Consider the lifted points $\bar a = \vect{a}{\|a\|^2}$ and $\bar b = \vect{b}{\|b\|^2}$ in $\R^3$. | |
Prove that the projection of $\bar{a}^*\cap \bar{b}^*$ is the perpendicular bisector between $a$ and $b$. | |
\vspace{1em} | |
If you are feeling ambitious, prove that for $a,b\in\R^d$, the same holds, where the perpendicular bisector of $2$ points in $\R^d$ is a hyperplane. | |
\vspace{1em} | |
Proof: Firstly we think $a,b\in\R^2$. | |
Suppose $a$=($x_{1}$,$y_{1}$), $b$=($x_{2}$,$y_{2}$) | |
And $\bar a = \vect{a}{\|a\|^2}$ and $\bar b = \vect{b}{\|b\|^2}$ in $\R^3$ | |
So $\bar a$=($x_{1}$,$y_{1}$,$x^{2}_{1}+y^{2}_{1}$), $\bar b$=($x_{2}$,$y_{2}$,$x^{2}_{2}+y^{2}_{2}$) | |
Then $\bar{a}^*$ is $z$=2$x_{1}$x+2$y_{1}$y-($x^{2}_{1}+y^{2}_{1}$), $\bar{b}^*$ is $z$=2$x_{2}$x+2$y_{2}$y-($x^{2}_{2}+y^{2}_{2}$) | |
So $\bar{a}^*\cap \bar{b}^*$ is 2$x_{1}$x+2$y_{1}$y-($x^{2}_{1}+y^{2}_{1}$) = 2$x_{2}$x+2$y_{2}$y-($x^{2}_{2}+y^{2}_{2}$) | |
Solve it we can get x=($x_{1}$+$y_{1}$)/2, y=($x_{2}$+$y_{2}$)/2 | |
It is the perpendicular bisector between $a$ and $b$ | |
\question{2} | |
Let $a,b,c\in \R^2$, not all collinear. | |
As in the previous question, we lift these points to $\bar a = \vect{a}{\|a\|^2}$, $\bar b = \vect{b}{\|b\|^2}$, and $\bar c = \vect{c}{\|c\|^2}$ in $\R^3$. | |
Let $h = \{\bar{x} = \vect{x}{x_z} \mid x_z = 2n^{\top}x - n_z\}$ for $n\in \R^2$ and $n_z\in \R$ be the plane passing thru $\bar{a}$, $\bar{b}$, and $\bar{c}$. | |
Prove that $n$ is the circumcenter of $a$, $b$, and $c$. | |
\vspace{1em} | |
\emph{Hint: dualize h, use incidence preservation, and recall that $2n^{\top} a - \|a\|^2 = \|n\|^2 - \|n-a\|^2$.} | |
\vspace{1em} | |
If you are feeling ambitious, consider the following high-dimensional version. | |
Let $a_0,\ldots a_d \in \R^d$ and the corresponding lifted points $a_i = \vect{a}{\|a\|^2}$ for $i=0\ldots d$. | |
Let $h = \{\bar{x} = \vect{x}{x_z} \mid x_z = 2n^{\top}x - n_z\}$ for $n\in \R^d$ and $n_z\in \R$ be the plane passing thru $\bar{a}$, $\bar{b}$, and $\bar{c}$. | |
Prove that $n$ is the circumcenter of $a_0,\ldots, a_d$. | |
\vspace{1em} | |
Proof: Firstly we think $a,b,c\in\R^2$. | |
Suppose $a$=($x_{1}$,$y_{1}$), $b$=($x_{2}$,$y_{2}$), $c$=($x_{3}$,$y_{3}$) | |
So $\bar a$=($x_{1}$,$y_{1}$,$x^{2}_{1}+y^{2}_{1}$), $\bar b$=($x_{2}$,$y_{2}$,$x^{2}_{2}+y^{2}_{2}$), $\bar c$=($x_{3}$,$y_{3}$,$x^{2}_{3}+y^{2}_{3}$) | |
And $h = \{\bar{x} = \vect{x}{x_z} \mid x_z = 2n^{\top}x - n_z\}$ is the plane passing thru $\bar{a}$, $\bar{b}$, and $\bar{c}$ | |
Also we know $2n^{\top} a - \|a\|^2 = \|n\|^2 - \|n-a\|^2$ | |
Let $a=x$ we have $2n^{\top} x = \|x\|^2 + \|n\|^2 - \|n-x\|^2$ | |
And by dualizing $h$, we have the point ${h}^*= \vect{n^{\top}}{n_{z}}$ | |
Use the conclusion in the previous question, we have the projection of $\bar{a}^*\cap \bar{b}^*$ is the perpendicular bisector between $a$ and $b$, the projection of $\bar{a}^*\cap \bar{c}^*$ is the perpendicular bisector between $a$ and $c$, and projection of $\bar{b}^*\cap \bar{c}^*$ is the perpendicular bisector between $b$ and $c$. | |
And $n$ also satisfies the above equations | |
So $n$ is the projection of $\bar{a}^*\cap \bar{b}^*\cap \bar{c}^*$ | |
Q.E.D. $n$ is the circumcenter of $a$, $b$, and $c$. | |
\end{document} |