@article {MATHEDUC.06537468, author = {Izmestiev, Ivan}, title = {A porism for cyclic quadrilaterals, butterfly theorems, and hyperbolic geometry.}, year = {2015}, journal = {American Mathematical Monthly}, volume = {122}, number = {5}, issn = {0002-9890}, pages = {467-475}, publisher = {Mathematical Association of America (MAA), Washington, DC}, doi = {10.4169/amer.math.monthly.122.5.467}, abstract = {The author focuses on the following theorem obtained by J. Kocik: {Theorem 1 .} Let $p$, $q$, $r$, $s$ be 4-points lying in a line $\ell$. Let $C$ be a circle and choose a point $x\in C$ not collinear with the 4-points. If a chain of four chords starting at $x$ that draw consecutively through $p$, $q$, $r$ and $s$ closes (that is, goes back to $x$, thus the chain forms a cyclic quadrilateral), then a chain for any other choice of starting points passing through $p$, $q$, $r$, $s$ also closes. As a consequence, the configuration space of 4-verices on a round circle making cyclic quadrilaterals has a connected component. In this article, the author gives two proofs of the above theorem: one is based on the cross-ratios on a circle $S^1\subset {\mathbb R}^2$ (Theorem 2) and the other is due to the property of M\"obius transformations of a projective limit circle $tial \mathbb H^2$ (Theorem 3). First, the cross-ratio is usually defined by $\displaystyle {\mathrm{cr}}(a,b;c,d)=\frac {a-c}{b-c}/ \frac {a-d}{b-d}$ for any 4-distinct points. It is invariant under the projective transformation group ${\mathrm{PSL}}(2,\mathbb R)$. To prove Theorem 1. using the cross-ratio, the author shows the projective butterfly theorem: Let $p,q,r,s\in \ell$ be intersection points of a line $\ell$ with a cyclic quadrilateral. If $\ell$ intersects $C$ in $2$-points $a,b$, then $${\mathrm{cr}}(a,b;p,q)={\mathrm{cr}}(a,b;s,r). \tag1$$ If $\ell$ is tangent to $C$ at a point $a$, then $$\frac 1{a-p}-\frac1{a-q}=\frac 1{a-s}-\frac1{a-r}. \tag2$$ If $\ell$ and $C$ are disjoint, then $$\angle paq=\angle sar. \tag3$$ Here, $a$ is the tangential point prescribed precisely in the figure (of the paper). Conversely, if $C$ and four points $p,q,r,s\in \ell$ satisfy (1), (2), (3), then for every $x\in C$ there is a closed chain of four chords starting at $x$ and drawing consecutively through $p$, $q$, $r$ and $s$ constitutes a cyclic quadrilateral. Secondly, let a circle $C$ and a line $\ell$ be as above. Choose a point $p\notin C$. If $x\in C$, then define $I_p(x)$ to be the point of $C$ at which the chord of $x$ to $p$ intersects $C$. Then, the map $I_p:C\rightarrow C$ is observed to be a M\"obius transformation for each point $p$ of the plane $\mathbb R^2$. Here the M\"obius transformation group is a subgroup of the projective transformation group ${\mathrm{PGL}}(3,\mathbb R)$ preserving the circle $S^1$ in $\mathbb R\mathbb P^2$, which thus is isomorphic to the hyperbolic group ${\mathrm{PO}}(2,1)$. The author shows that Theorem 1 is equivalent with {Theorem 3.} Let $p,q,r,s\in \ell\smallsetminus C$ be any 4-points. If $x\in \ell\smallsetminus C$ satisfies $I_s\circ I_r\circ I_q\circ I_p(x)=x$, then $I_s\circ I_r\circ I_q\circ I_p$ is the identity element in ${\mathrm{PO}}(2,1)$. The author continues to study a generalization of the above theorem, an $n$-gone in $C$ (called Castillon's problem).}, reviewer = {Yoshinobu Kamishima (Saitama)}, msc2010 = {G45xx}, identifier = {2016c.00748}, }