10.1: Inversion

Let $\Omega$ be the circle with center $O$ and radius $r$. The inversion of a point $P$ in $\Omega$ is the point $P' \in [OP)$ such that

$OP \cdot OP' = r^2.$

In this case the circle $\Omega$ will be called the circle of inversion and its center $O$ is called the center of inversion.

The inverse of $O$ is undefined.

Note that if $P$ is inside $\Omega$, then $P'$ is outside and the other way around. Further, $P = P'$ if and only if $P \in \Omega$.

Note that the inversion maps $P'$ back to $P$.

Exercise $\PageIndex{1}$

Let $\Omega$ be a circle centered at $O$. Suppose that a line $(PT)$ is tangent to $\Omega$ at $T$. Let $P'$ be the foot point of $T$ on $(OP)$.

Show that $P'$ is the inverse of $P$ in $\Omega$.

Hint

By Lemma 5.6.2, $\angle OTP'$ is right. Therefore, $\triangle OPT \sim \triangle OTP'$ and in particular $OP \cdot OP' = OT^2$ and hence the result.

Lemma $\PageIndex{1}$

Let $\Gamma$ be a circle with the center $O$. Assume $A'$ and $B'$ are the inverses of $A$ and $B$ in $\Gamma$. Then

$\triangle OAB \sim \triangle OB'A'.$

Moreover

$$\begin{array} {rcl} {\measuredangle AOB} & \equiv & {-\measuredangle B'OA',} \\ {\measuredangle OBA} & \equiv & {-\measuredangle OA'B',} \\ {\measuredangle BAO} & \equiv & {-\measuredangle A'B'O.} \end{array}$$

Proof

Let $r$ be the radius of the circle of the inversion.

By the definition of inversion,

$OA \cdot OA' = OB \cdot OB' = r^2.$

Therefore,

$\dfrac{OA}{OB'} = \dfrac{OB}{OA'}.$

Clearly,

$\measuredangle AOB = \measuredangle A'OB' \equiv -\measuredangle B'OA'.$

From SAS, we get that

$\triangle OAB \sim \triangle OB'A'.$

Applying Theorem 3.3.1 and 10.1.2, we get 10.1.1.

Exercise $\PageIndex{2}$

Let $P'$ be the inverse of $P$ in the circle $\Gamma$. Assume that $P \ne P'$. Show that the value $\dfrac{PX}{P'X}$ is the same for all $X \in \Gamma$.

The converse to the exercise above also holds. Namely, given a positive real number $k \ne 1$ and two distinct points $P$ and $P'$ the locus of points $X$ such that $\dfrac{PX}{P'X} = k$ forms a circle which is called the Apollonian circle. In this case $P'$ is inverse of $P$ in the Apollonian circle.

Hint

Suppose that $O$ denotes the center of $\Gamma$. Assume that $X, Y \in \Gamma$; in particular, $OX = OY$.

Note that the inversion sends $X$ and $Y$ to themselves. By Lemma $\PageIndex{1}$,

$\triangle OPX \sim \triangle OXP'$ and $\triangle OPY \sim \triangle OYP'$.

Therefore, $\dfrac{PX}{P'X} = \dfrac{OP}{OX} = \dfrac{OP}{OY} = \dfrac{PY}{P'Y}$ and hence the result.

Exercise $\PageIndex{3}$

Let $A',B'$, and $C'$ be the images of $A,B$, and $C$ under the inversion in the cincircle of $\triangle ABC$. Show that the incenter of $\triangle ABC$ is the orthocenter of $\triangle A'B'C'$.

Hint

By Lemma $\PageIndex{1}$,

$\measuredangle IA'B' \equiv -\measuredangle IBA$, $\measuredangle IB'C' \equiv -\measuredangle ICB$, $\measuredangle IC'A' \equiv -\measuredangle IAC$,
$\measuredangle IB'A' \equiv -\measuredangle IAB$, $\measuredangle IC'B' \equiv -\measuredangle IBC$, $\measuredangle IA'C' \equiv -\measuredangle ICA$.

It remains to apply the theorem on the sum of angles of triangle (Theorem 7.4.1) to show that $(A'I) \perp (B'C')$, $(B'I) \perp (C'A')$ and $(C'I) \perp (B'A')$.

Exercise $\PageIndex{4}$

Make a ruler-and-compass construction of the inverse of a given point in a given circle.

Hint

Guess the construction from the diagram (the two nonintersecting lines on the diagram are parallel).