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| In this entry, we prove the uniqueness of center of a circle in a slightly more general setting than the parent entry. In this more general setting, let $\mathfrak{G}$ be an ordered geometry satisfying the congruence axioms. For simplicity, let us write $oa$ for the \PMlinkname{line segment}{ClosedLineSegment2} $\overline{oa}$, and write $a:b:c$ to mean $b$ is between $a$ and $c$. |
In this entry, we prove the uniqueness of center of a circle in a slightly more general setting than the parent entry. In this more general setting, let $\mathfrak{G}$ be an ordered geometry satisfying the congruence axioms. For simplicity, let us write $oa$ for the \PMlinkname{line segment}{ClosedLineSegment2} $\overline{oa}$, and write $a:b:c$ to mean $b$ is between $a$ and $c$. |
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| Before proving the property that a circle in $\mathfrak{G}$ has a unique center, let us review some definitions. |
Before proving the property that a circle in $\mathfrak{G}$ has a unique center, let us review some definitions. |
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| Let $o$ and $a$ be points in $\mathfrak{G}$, a geometry in which the congruence axioms are defined. Let $\mathscr{C}(o,a)$ be the set of all points $p$ in $\mathfrak{G}$ such that the \PMlinkname{line segments}{ClosedLineSegment2} are congruent: $oa\cong op$. The set $\mathscr{C}(o,a)$ is called a \emph{circle}. When $a=o$, then $\mathscr{C}(o,a)$ is said to be \emph{degenerate}. Let $\mathscr{C}$ be a circle in $\mathfrak{G}$. A \emph{center} of $\mathscr{C}$ is a point $o$ such that for every pair of points $p,q$ in $\mathscr{C}$, $op \cong oq$. |
Let $o$ and $a$ be points in $\mathfrak{G}$, a geometry in which the congruence axioms are defined. Let $\mathscr{C}(o,a)$ be the set of all points $p$ in $\mathfrak{G}$ such that the \PMlinkname{line segments}{ClosedLineSegment2} are congruent: $oa\cong op$. The set $\mathscr{C}(o,a)$ is called a \emph{circle}. When $a=o$, then $\mathscr{C}(o,a)$ is said to be \emph{degenerate}. Let $\mathscr{C}$ be a circle in $\mathfrak{G}$. A \emph{center} of $\mathscr{C}$ is a point $o$ such that for every pair of points $p,q$ in $\mathscr{C}$, $op \cong oq$. |
| We say that $m$ is a \emph{midpoint} of two points $p$ and $q$ |
We say that $m$ is a \emph{midpoint} of two points $p$ and $q$ |
| if $pm \cong mq$ and $m,p,q$ are collinear. We say that $p$ is an \emph{interior point} of |
if $pm \cong mq$ and $m,p,q$ are collinear. |
| $\mathscr{C}(o,a)$ if $op < oa$. |
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| We collect some simple facts below. |
We collect some simple facts below. |
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| \begin{itemize} |
\begin{itemize} |
| \item In the circle $\mathscr{C}(o,a)$, $o$ is a center of $\mathscr{C}(o,a)$ (by definition). |
\item In the circle $\mathscr{C}(o,a)$, $o$ is a center of $\mathscr{C}(o,a)$ (by definition). |
| \item Let $\mathscr{C}$ be a circle. If $o$ is a center of $\mathscr{C}$ and $a$ is any point in $\mathscr{C}$, then $\mathscr{C}=\mathscr{C}(o,a)$, again by definition. |
\item Let $\mathscr{C}$ be a circle. If $o$ is a center of $\mathscr{C}$ and $a$ is any point in $\mathscr{C}$, then $\mathscr{C}=\mathscr{C}(o,a)$, again by definition. |
| \item A circle is degenerate if and only if it is a singleton. |
\item A circle is degenerate if and only if it is a singleton. |
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| If $p$ is in $\mathscr{C}(o,o)$, then $op\cong oo$, so that $p=o$, and $\mathscr{C}(o,o)=\lbrace o\rbrace$. Conversely, if $\mathscr{C}(o,a)=\lbrace b\rbrace$, then $b=a$. Let $L$ be any line passing through $o$. Choose a ray $\rho$ on $L$ emanating from $o$. Then there is a point $d$ on $\rho$ such that $od\cong oa$. So $d=a$ since $\mathscr{C}(o,a)$ is a singleton containing $a$. Similarly, there is a unique $e$ on $-\rho$, the opposite ray of $\rho$, with $oe\cong oa$. So $e=a$. Since $d:o:e$, we have that $a=d=o$. Therefore $\mathscr{C}(o,a)=\mathscr{C}(o,o)$. |
If $p$ is in $\mathscr{C}(o,o)$, then $op\cong oo$, so that $p=o$, and $\mathscr{C}(o,o)=\lbrace o\rbrace$. Conversely, if $\mathscr{C}(o,a)=\lbrace b\rbrace$, then $b=a$. Let $L$ be any line passing through $o$. Choose a ray $\rho$ on $L$ emanating from $o$. Then there is a point $d$ on $\rho$ such that $od\cong oa$. So $d=a$ since $\mathscr{C}(o,a)$ is a singleton containing $a$. Similarly, there is a unique $e$ on $-\rho$, the opposite ray of $\rho$, with $oe\cong oa$. So $e=a$. Since $d:o:e$, we have that $a=d=o$. Therefore $\mathscr{C}(o,a)=\mathscr{C}(o,o)$. |
| \item Suppose $\mathscr{C}$ is a non-degenerate circle. Then every line passing through a center $o$ of $\mathscr{C}$ is incident with at least two points $a,a'$ in $\mathscr{C}$. Furthermore, $o$ is the midpoint of $aa'$. |
\item Suppose $\mathscr{C}$ is a non-degenerate circle. Then every line passing through a center $o$ of $\mathscr{C}$ is incident with at least two points $a,a'$ in $\mathscr{C}$. Furthermore, $o$ is the midpoint of $aa'$. |
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| If on $L$ through $o$ lies only one point $a \in \mathscr{C}$, let $a'$ be the point on the opposite ray of $\ray{oa}$ such that $a'\in \mathscr{C}$. Then $a'=a$, which means that $o=a=a'$, implying that $\mathscr{C}$ is degenerate. Since $oa\cong oa'$, and $o,a,a'$ lie on the same line, $o$ is the midpoint of $aa'$. |
If on $L$ through $o$ lies only one point $a \in \mathscr{C}$, let $a'$ be the point on the opposite ray of $\ray{oa}$ such that $a'\in \mathscr{C}$. Then $a'=a$, which means that $o=a=a'$, implying that $\mathscr{C}$ is degenerate. Since $oa\cong oa'$, and $o,a,a'$ lie on the same line, $o$ is the midpoint of $aa'$. |
| \end{itemize} |
\end{itemize} |
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| Now, on to the main fact. |
Now, on to the main fact. |
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| \begin{thm} Every circle in $\mathfrak{G}$ has a unique center. \end{thm} |
\begin{thm} Every circle in $\mathfrak{G}$ has a unique center. \end{thm} |
| \begin{proof} Let $\mathscr{C}=\mathscr{C}(o,a)$ be a circle in $\mathfrak{G}$. Suppose $o'$ is another center of $\mathscr{C}$ and $o\ne o'$. Let $L$ be the line passing through $o$ and $o'$. Consider the (open) ray $\rho = \ray{oo'}$. By one of the congruence axioms, there is a unique point $b$ on $\rho$ such that $oa \cong ob$. So $b\in \mathscr{C}(o,a)$. |
\begin{proof} Let $\mathscr{C}=\mathscr{C}(o,a)$ be a circle in $\mathfrak{G}$. Suppose $o'$ is another center of $\mathscr{C}$ and $o\ne o'$. Let $L$ be the line passing through $o$ and $o'$. Consider the (open) ray $\rho = \ray{oo'}$. By one of the congruence axioms, there is a unique point $b$ on $\rho$ such that $oa \cong ob$. So $b\in \mathscr{C}(o,a)$. |
| \begin{itemize} |
\begin{itemize} |
| \item Case 1. Suppose $o'=b$. Consider the (open) opposite ray $-\rho$ of $\rho$. There is a unique point $d$ on $-\rho$ such that $od\cong oa$. So $d\in\mathscr{C}(o,a)$. Since $d,o,o'$ all lie on $L$, one must be between the other two. |
\item Case 1. Suppose $o'=b$. Consider the (open) opposite ray $-\rho$ of $\rho$. There is a unique point $d$ on $-\rho$ such that $od\cong oa$. So $d\in\mathscr{C}(o,a)$. Since $d,o,o'$ all lie on $L$, one must be between the other two. |
| \begin{itemize} |
\begin{itemize} |
| \item |
\item |
| Subcase 1. If $o:d:o'$, then $od < oo' = ob \cong oa$, contradiction. |
Subcase 1. If $o:d:o'$, then $od < oo' = ob \cong oa$, contradiction. |
| \item |
\item |
| Subcase 2. If $d:o':o$, then $oa \cong ob = oo' < od$, contradiction again. |
Subcase 2. If $d:o':o$, then $oa \cong ob = oo' < od$, contradiction again. |
| \item |
\item |
| Subcase 3. So suppose $o$ is between $d$ and $o'$. Now, since $o'$ is also center of $\mathscr{C}(o,a)$, we have that $bb = o'b \cong o'd$, which implies that $o'=d$ by another one of the congruence axioms. But $d:o:o'$, which forces $o'=o$, contradicting the assumption that $o'$ is not $o$ in the beginning. |
Subcase 3. So suppose $o$ is between $d$ and $o'$. Now, since $o'$ is also center of $\mathscr{C}(o,a)$, we have that $bb = o'b \cong o'd$, which implies that $o'=d$ by another one of the congruence axioms. But $d:o:o'$, which forces $o'=o$, contradicting the assumption that $o'$ is not $o$ in the beginning. |
| \end{itemize} |
\end{itemize} |
| \item |
\item |
| Case 2. If $o'$ is not $b$, then since $o,o',b$ lie on the same line $L$, one must be between the other two. Since $b$ also lies on the ray $\rho$ with $o$ as the source, $o$ cannot be between $o'$ and $b$. So we have only two subcases to deal with: either $o:o':b$, or $o:b:o'$. In either subcase, we need to again consider the opposite ray $-\rho$ of $\rho$ with $d$ on $-\rho$ such that $od\cong oa \cong ob$. From the properties of opposite rays, we also have the following two facts: |
Case 2. If $o'$ is not $b$, then since $o,o',b$ lie on the same line $L$, one must be between the other two. Since $b$ also lies on the ray $\rho$ with $o$ as the source, $o$ cannot be between $o'$ and $b$. So we have only two subcases to deal with: either $o:o':b$, or $o:b:o'$. In either subcase, we need to again consider the opposite ray $-\rho$ of $\rho$ with $d$ on $-\rho$ such that $od\cong oa \cong ob$. From the properties of opposite rays, we also have the following two facts: |
| \begin{enumerate} |
\begin{enumerate} |
| \item $d:o:o'$, implying $od < o'd$. |
\item $d:o:o'$, implying $od < o'd$. |
| \item $d:o:b$. |
\item $d:o:b$. |
| \end{enumerate} |
\end{enumerate} |
| \begin{itemize} |
\begin{itemize} |
| \item |
\item |
| Subcase 1. $o:o':b$. Then $o'b < ob \cong od < o'd$, contradiction. |
Subcase 1. $o:o':b$. Then $o'b < ob \cong od < o'd$, contradiction. |
| \item |
\item |
| Subcase 2. $o:b:o'$. Let us look at the betweenness relations among the points $b, d, o'$. |
Subcase 2. $o:b:o'$. Let us look at the betweenness relations among the points $b, d, o'$. |
| \begin{enumerate} |
\begin{enumerate} |
| \item If $b:o':d$, then $d:o':o$ by one of the conditions of the betweenness relations. But this forces $o'$ to be on $-\rho$. Since $o'$ is on $\rho$, this is a contradiction. |
\item If $b:o':d$, then $d:o':o$ by one of the conditions of the betweenness relations. But this forces $o'$ to be on $-\rho$. Since $o'$ is on $\rho$, this is a contradiction. |
| \item If $b:d:o'$, then $d$ would be on $\rho$. Since $d$ is on $-\rho$, we have another contradiction. |
\item If $b:d:o'$, then $d$ would be on $\rho$. Since $d$ is on $-\rho$, we have another contradiction. |
| \item If $d:b:o'$, then $o'b < o'd$. But $o'$ is a center of $\mathscr{C}(o,a)$, yet another contradiction. |
\item If $d:b:o'$, then $o'b < o'd$. But $o'$ is a center of $\mathscr{C}(o,a)$, yet another contradiction. |
| \end{enumerate} |
\end{enumerate} |
| Therefore, Subcase 2 is impossible also. |
Therefore, Subcase 2 is impossible also. |
| \end{itemize} |
\end{itemize} |
| This means that Case 2 is impossible. |
This means that Case 2 is impossible. |
| \end{itemize} |
\end{itemize} |
| Since both Case 1 and Case 2 are impossible, $o'=o$, and the proof is complete. \end{proof} |
Since both Case 1 and Case 2 are impossible, $o'=o$, and the proof is complete. \end{proof} |
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| \textbf{Remarks}. |
\textbf{Remarks}. |
| \begin{itemize} |
\begin{itemize} |
| \item |
\item |
| The assumption that $\mathfrak{G}$ is ordered cannot be dropped. Here is a simple counterexample. Consider an incidence geometry defined on a circle $C$ in the Euclidean plane. |
The assumption that $\mathfrak{G}$ is ordered cannot be dropped. Here is a simple counterexample. Consider an incidence geometry defined on a circle $C$ in the Euclidean plane. |
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| \begin{center} |
\begin{center} |
| \begin{pspicture}(-4,-2.5)(4,2.5) |
\begin{pspicture}(-4,-2.5)(4,2.5) |
| \rput[a](0,2){.} |
\rput[a](0,2){.} |
| \rput[a](0,-2){.} |
\rput[a](0,-2){.} |
| \pscircle(0,0){2} |
\pscircle(0,0){2} |
| \psdots(-2,0)(2,0)(-1.732,1)(-1.732,-1) |
\psdots(-2,0)(2,0)(-1.732,1)(-1.732,-1) |
| \rput[r](-2.2,0){$o$} |
\rput[r](-2.2,0){$o$} |
| \rput[l](2.2,0){$o'$} |
\rput[l](2.2,0){$o'$} |
| \rput[r](-1.9,1){$a$} |
\rput[r](-1.9,1){$a$} |
| \rput[r](-1.9,-1){$a'$} |
\rput[r](-1.9,-1){$a'$} |
| \end{pspicture} |
\end{pspicture} |
| \end{center} |
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| It is not possible to define a betweenness relation on $C$. However, it is still possible to define a congruence relation on $C$: $xy\cong zt$ if $xy$ and $zt$ have the same arc length. Given any points $o,a$ on $C$, the circle $\mathscr{C}(o,a)$ consists of exactly two points $a$ and $a'$ (see figure above). In addition, $\mathscr{C}(o,a)$ has two centers: $o$ and $o'$. |
It is not possible to define a betweenness relation on $C$. However, it is still possible to define a congruence relation on $C$: $xy\cong zt$ if $xy$ and $zt$ have the same arc length. Given any points $o,a$ on $C$, the circle $\mathscr{C}(o,a)$ consists of exactly two points $a$ and $a'$ (see figure above). In addition, $\mathscr{C}(o,a)$ has two centers: $o$ and $o'$. |
| \item |
\item |
| There is another definition of a circle, which is based on the concept of a metric. In this definition, examples can also be found where the uniqueness of center of a circle fails. The most commonly quoted example is found in the metric space of $p$-adic numbers. The metric defined is non-Archimedean, so every triangle is isosceles. From this it is not hard to see that every interior point of a circle is its center. |
There is another definition of a circle, which is based on the concept of a metric. In this definition, examples can also be found where the uniqueness of center of a circle fails. The most commonly quoted example is found in the metric space of $p$-adic numbers. The metric defined is non-Archimedean, so every triangle is isosceles. From this it is not hard to see that every interior point of a circle is its center. |
| \end{itemize} |
\end{itemize} |
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| \begin{thebibliography}{6} |
\begin{thebibliography}{6} |
| \bibitem{mg} M. J. Greenberg, {\it Euclidean and Non-Euclidean Geometries, Development and History}, W. H. Freeman and Company, San Francisco (1974) |
\bibitem{mg} M. J. Greenberg, {\it Euclidean and Non-Euclidean Geometries, Development and History}, W. H. Freeman and Company, San Francisco (1974) |
| \bibitem{nk} N. Koblitz, {\it p-adic Numbers, p-adic Analysis, and Zeta-Functions}, Springer-Verlag, New York (1977) |
\bibitem{nk} N. Koblitz, {\it p-adic Numbers, p-adic Analysis, and Zeta-Functions}, Springer-Verlag, New York (1977) |
| \end{thebibliography} |
\end{thebibliography} |
