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This is an exercise of foundations of geometry of Venema.

Prove: If $I_{O,r}$ is an inversion with center $O$ and radius $r$, and if $\alpha$ is a circle such that $O \in\alpha$, then $I_{O,r} (α−\{O\})$ is a line.

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Take two different points $A$ and $B$ on $\alpha$ and let their images be $A'$ and $B'$ and let $\ell$ be a line through $A'$ and $B'$. Let $s :=I_{O,r} (α−\{O\})$. We prove that $$s = \ell$$

First we show that $s\subseteq \ell$:

Take any $X'\in s$, then exist $X\in α−\{O\}$ such that $X'$ is it image. Since $$OA \cdot OA' = OX\cdot OX'= OB\cdot OB'$$ the quadrilateral $AXX'A'$ and $BXX'B'$ are cyclic. So by Miquel theorem $X',A,B$ are colinear. So $X'\in \ell$ and thus $s\subseteq \ell$.

Now we show that $\ell\subseteq s$:

Take any $Y\in \ell$ and let $Y'$ be it image. Then again we have: $$OA \cdot OA' = OY\cdot OY'= OB\cdot OB'$$ so the quadrilateral $AYY'A'$ and $BYY''B'$ are cyclic. So again by Miquel theorem circles $\alpha$, $(AYY'A')$ and $(BYY''B')$ must have common point and this is $Y'$. Thus $Y$ is image of the point of the circle $\alpha $ so $Y\in s$ and thus $\ell \subseteq s$ and we are done.

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