10.6: Angles after inversion

Proposition $\PageIndex{1}$

In the inversive plane, the inverse of an arc is an arc.

Proof

Consider four distinct points $A, B, C$, and $D$; let $A', B', C'$, and $D'$ be their inverses. We need to show that $D$ lies on the arc $ABC$ if and only if $D'$ lies on the arc $A'B'C'$. According to Proposition 9.5.1, the latter is equivalent to the following:

$\measuredangle ADC = \measuredangle ABC \Leftrightarrow \measuredangle A'D'C' = \measuredangle A'B'C'.$

The latter follows from Theorem 10.2.1b.

The following theorem states that the angle between arcs changes only its sign after the inversion.

Theorem $\PageIndex{1}$

Let $AB_1C_1$, $AB_2C_2$ be two arcs in the inversive plane, and the arcs $A'B_1'C_1'$, $A'B_2'C_2'$ be their inverses. Let $[AX_1)$ and $[AX_2)$ be the half-lines tangent to $AB_1C_1$ and $AB_2C_2$ at $A$, and $[A'Y_1)$ and $[A'Y_2)$ be the half-lines tangent to $A'B_1'C_1'$ and $A'B_2'C_2'$ at $A'$. Then

$\measuredangle X_1AX_2 \equiv -\measuredangle Y_1A'Y_2$.

Proof

Applying to Proposition 9.6.1,

$\begin{array} {rcl} {\measuredangle X_1AX_2} & \equiv & {\measuredangle X_1AC_1 + \measuredangle C_1AC_2 + \measuredangle C_2AX_2 \equiv} \\ {} & \equiv & {(\pi - \measuredangle C_1B_1A) + \measuredangle C_1AC_2 + (\pi - \measuredangle AB_2C_2) \equiv} \\ {} & \equiv & {-(\measuredangle C_1B_1A + \measuredangle AB_2C_2 + \measuredangle C_2AC_1) \equiv} \\ {} & \equiv & {-(\measuredangle C_1B_1A + \measuredangle AB_2C_1) -(\measuredangle C_1B_2C_2 + \measuredangle C_2AC_1).} \end{array}$

The same way we get that

$\measuredangle Y_1A'Y_2 \equiv -(\measuredangle C_1'B_1'A' + \measuredangle A'B_2'C_1') - (\measuredangle C_1'B_2'C_2' + \measuredangle C_2'A'C_1').$

By Theorem 10.2.1b,

$\begin{array} {rcl} {\measuredangle C_1B_1A + \measuredangle AB_2C_1} & \equiv & {-(\measuredangle C_1'B_1'A' + \measuredangle A'B_2'C_1'),} \\ {\measuredangle C_1B_2C_2 + \measuredangle C_2AC_1} & \equiv & {-(\measuredangle C_1'B_2'C_2' + \measuredangle C_2'A'C_1')} \end{array}$.

and hence the result.

The angle between arcs can be defined as the angle between its tangent half-lines at the common endpoint. Therefore under inversion, the angles between arcs are preserved up to sign.

From Exercise 5.7.4, it follows that the angle between arcs with common endpoint A is the limit of $\measuredangle P_1AP_2$ where $P_1$ and $P_2$ are points approaching $A$ along the corresponding arcs. This observation can be used to define the angle between a pair of curves emerging from one point. It turns out that under inversion, angles between curves are also preserved up to sign.

Corollary $\PageIndex{1}$

Let $P$ be the inverse of point $Q$ in a circle $\Gamma$. Assume that $P'$, $Q'$, and $\Gamma'$ are the inverses of $P, Q$, and $\Gamma$ in another circle $\Omega$. Then $P'$ is the inverse of $Q'$ in $\Gamma'$.

Proof

If $P = Q$, then $P'=Q' \in \Gamma'$. Therefore, $P'$ is the inverse of $Q'$ in $\Gamma'$.

It remains to consider the case $P \ne Q$. Let $\Delta_2$ and $\Delta_2$ be two distinct circles that intersect at $P$ and $Q$. According to Corollary 10.5.2, $\Delta_1 \perp \Gamma$ and $\Delta_2 \perp \Gamma$.

Let $\Delta_1'$ and $\Delta_2'$ denote the inverses of $\Delta_1$ and $\Delta_2$ in $Omega$. Clearly, $\Delta_1'$ meets $\Delta_2'$ at $P'$ and $Q'$.

By Theorem $\PageIndex{1}$, $\Delta_1' \perp \Gamma'$ and $\Delta_2' \perp \Gamma'$. By Corollary 10.5.1, $P'$ is the inverse of $Q'$ in $\Gamma'$.