Subfield: Geometry


Question: Let $\Omega$ be the circumcircle of the triangle $A B C$. The circle $\omega$ is tangent to the sides $A C$ and $B C$, and it is internally tangent to $\Omega$ at the point $P$. A line parallel to $A B$ and intersecting the interior of triangle $A B C$ is tangent to $\omega$ at $Q$.

Prove that $\angle A C P=\angle Q C B$.

Solution:
Assume that $\omega$ is tangent to $A C$ and $B C$ at $E$ and $F$, respectively and let $P E, P F, P Q$ meet $\Omega$ at $K, L, M$, respectively. Let $I$ and $O$ denote the respective centres of $\omega$ and $\Omega$, and consider the homethety $\mathscr{H}$ that maps $\omega$ onto $\Omega$. Now $K$ is the image of $E$ under $\mathscr{H}$, and $E I \perp A C$. Hence $O K \perp A C$, and thus $K$ is the midpoint of the arc $C A$. Similarly, $L$ is the midpoint of the $\operatorname{arc} B C$ and $M$ is the midpoint of the arc $B A$. It follows that arcs $L M$ and $C K$ are equal, because

$$
\begin{aligned}
\overparen{B M}=\overparen{M A} & \Longrightarrow \overparen{B L}+\overparen{L M}=\overparen{M K}+\overparen{K A} \Longrightarrow \overparen{L C}+\overparen{L M}=\overparen{M K}+\overparen{C K} \\
& \Longrightarrow 2 \overparen{L M}+\overparen{M C}=\overparen{M C}+2 \overparen{C K} \Longrightarrow \overparen{L M}=\overparen{C K}
\end{aligned}
$$

Thus arcs $F Q$ and $D E$ of $\omega$ are equal, too, where $D$ is the intersection of $C P$ with $\omega$. Since $C E$ and $C F$ are tangents to $\omega$, this implies that $\angle D E C=\angle C F Q$. Further, $C E=C F$, and thus triangles $C E D$ and $C F Q$ are congruent. In particular, $\angle E C D=\angle Q C F$, as required.

<img_4041>
Let $I$ and $O$ denote the respective centres of $\omega$ and $\Omega$. Observe that $C I$ is the angle bisector of angle $\angle C$, because $\omega$ is tangent to $A C$ and $B C$. Consider the homethety $\mathscr{H}$ that maps $\omega$ onto $\Omega$. Let $M$ be the image of $Q$ under $\mathscr{H}$. By construction, $I Q \perp A B$, so $O M \perp A B$. Thus the diameter $O M$ of $\Omega$ passes through the midpoint of the $\operatorname{arc} A B$ of $\Omega$, which also lies on the angle bisector $C I$. This implies that $\angle I C M=90^{\circ}$. We next show that $P, I, Q, C$ lie on a circle. Notice that

$$
\begin{aligned}
\angle P Q I & =90^{\circ}-\frac{1}{2} \angle Q I P=90^{\circ}-\frac{1}{2} \angle M O P=90^{\circ}-\left(180^{\circ}-\angle P C M\right) \\
& =(\angle P C I+\angle I C M)-90^{\circ}=\angle P C I .
\end{aligned}
$$

Hence $P, I, Q, C$ lie on a circle. But $P I=I Q$, so $C I$ is the angle bisector of $\angle P C Q$. Since $C I$ is also the angle bisector of angle $\angle C$, it follows that $\angle A C P=\angle Q C B$, as required.

<img_3450>
Let $I$ and $O$ denote the respective centres of $\omega$ and $\Omega$. Let $D$ be the second point of intersection of $C P$ with $\omega$, and let $\ell$ denote the tangent to $\omega$ at $D$, which meets $A C$ at $S$. Hence $I D \perp \ell$. By construction, $P, I, O$ lie one a line, and hence the isosceles triangles $P I D$ and $P O C$ are similar. In particular, it follows that $O C \perp \ell$, so $C$ is the midpoint of the arc of $\Omega$ defined by the points of intersection of $\ell$ with $\Omega$. It is easy to see that this implies that

$$
\angle D S C=\angle A B C .
$$

Under reflection in the angle bisector $C I$ of $\angle C, \ell$ is thus mapped to a tangent to $\omega$ parallel to $A B$ and intersecting the interior of $A B C$, since $\omega$ is mapped to itself under this reflection. In particular, $D$ is mapped to $Q$, and thus $\angle Q C B=\angle A C D$, as required.

<img_4014>
Let the tangent to $\omega$ at $Q$ meet $A C$ and $B C$ at $X$ and $Y$, respectively. Then $A C / X C=B C / Y C$, and thus there is a radius $r$ such that $r^{2}=A C \cdot Y C=B C \cdot X C$. Let $\Gamma$ denote the circle with centre $C$ and radius $r$, and consider the inversion $\mathscr{I}$ in the circle $\Gamma$. Under $\mathscr{I}$,

$A \longmapsto A^{\prime}$, the point on the ray $C A$ satisfying $C A^{\prime}=C Y$;

$B \longmapsto B^{\prime}$, the point on the ray $C B$ satisfying $C B^{\prime}=C X$;

$\Omega \longmapsto$ the line $A^{\prime} B^{\prime} ;$

$\omega \longmapsto \omega^{\prime}$, the excircle of $C A^{\prime} B^{\prime}$ opposite $C$;

$P \longmapsto P^{\prime}$, the point where $\omega^{\prime}$ touches $A^{\prime} B^{\prime}$;

In particular, $\omega^{\prime}$, the image of $\omega$, is a circle tangent to $A C, B C$ and $A^{\prime} B^{\prime}$, so it is either the excircle of $C A^{\prime} B^{\prime}$ opposite $C$, or the incircle of $C A^{\prime} B^{\prime}$. Let $\omega$ be tangent to $B C$ at $F$, and let $F^{\prime}$ be the image of $F$ under $\mathscr{I}$. Then $C F \cdot C F^{\prime}=B C \cdot X C$. Now $C F<B C$, so $C F^{\prime}>C X=C B^{\prime}$. Hence $\omega^{\prime}$ cannot be the incircle, so $\omega^{\prime}$ is indeed the excircle of $C A^{\prime} B^{\prime}$ opposite $C$.

Now note that $\omega$ is the excircle of $C X Y$ opposite $C$. The reflection about the angle bisector of $\angle C$ maps $X$ to $B^{\prime}, Y$ to $A^{\prime}$. It thus maps the triangle $C X Y$ to $C B^{\prime} A^{\prime}, \omega$ to $\omega^{\prime}$ and, finally, $Q$ to $P^{\prime}$. It follows that $\angle A C P=\angle A C P^{\prime}=\angle Q C B$, as required.

<img_3924>
Let $r$ be the radius such that $r^{2}=A C \cdot B C$. Let $\mathscr{J}$ denote the composition of the inversion $\mathscr{I}$ in the circle of centre $C$ and radius $r$, followed by the reflection in the angle bisector of $\angle C$. Under $\mathscr{J}$,

$A \longmapsto B, B \mapsto A ;$

$\Omega \longmapsto$ the line $A B$;

$\omega \longmapsto \omega^{\prime}$, the excircle of $A B C$ opposite the vertex $C$;

$P \longmapsto Q^{\prime}$, the point where $\omega^{\prime}$ touches $A B$;

In particular, note that the image $\omega^{\prime}$ of $\omega$ under $\mathscr{J}$ is a circle tangent to $A C, B C$ and $A B$, so it is either the incircle of $A B C$, or the excircle opposite vertex $C$. Observe that $r \geqslant \min \{A C, B C\}$, so the image of the points of tangency of $\omega$ must lie outside $A B C$, and thus $\omega^{\prime}$ cannot be the incircle. Thus $\omega^{\prime}$ is the excircle opposite vertex $C$ as claimed. Further, the point of tangency $P$ is mapped to $Q^{\prime}$.

Now, since the line $C P$ is mapped to itself under the inversion $\mathscr{I}$, and mapped onto $C Q^{\prime}$ under $\mathscr{J}, C P$ and $C Q^{\prime}$ are images of each other under reflection in the angle bisector of $\angle C$. But $C, Q, Q^{\prime}$ lie on a line for there is a homothety with centre $C$ that maps $\omega$ onto the excircle $\omega^{\prime}$. This completes the proof.

<img_3639>
$\quad$ Assume that $\omega$ is tangent to $A C$ and $B C$ at $E$ and $F$, respectively. Assume that $C P$ meets $\omega$ at $D$. Let $I$ and $O$ denote the respective centres of $\omega$ and $\Omega$. To set up a solution in the complex plane, we take the circle $\omega$ as the unit circle centered at the origin of the complex plane, and let $P O$ be the real axis with $o>0$, where we use the convention that lowercase letters denote complex coordinates of corresponding points in the plane denoted by uppercase letters.



Now, a point $Z$ on the circle $\Omega$ satisfies

$$
|z-o|^{2}=(o+1)^{2} \Longleftrightarrow \quad z z^{*}-o\left(z+z^{*}\right)-2 o-1=0
$$

The triangle $A B C$ is defined by the points $E$ and $F$ on $\omega$, the intersection $C$ of the corresponding tangents lying on $\Omega$. Thus $c=2 e f /(e+f)$, and further

$$
|c-o|^{2}=(o+1)^{2} \Longleftrightarrow \quad c c^{*}-o\left(c+c^{*}\right)-2 o-1=0
\tag{1}
$$

and this is the equality defining $o$. The points $A$ and $B$ are the second intersection points of $\Omega$ with the tangents to $\omega$ at $E$ and $F$ respectively. A point $Z$ on the tangent through $E$ is given by $z=2 e-e^{2} z^{*}$, and thus $A$ and $C$ satisfy

$$
\begin{aligned}
\left(2 e-e^{2} z^{*}\right) z^{*} & -o\left(2 e-e^{2} z^{*}+z^{*}\right)-2 o-1=0 \\
& \Longleftrightarrow \quad-e^{2} z^{* 2}+\left(2 e+o e^{2}-o\right) z^{*}-(2 e o+2 o+1)=0 \\
& \Longleftrightarrow \quad z^{* 2}-\left(2 e^{*}+o-o e^{* 2}\right) z^{*}+\left(2 e^{*} o+2 o e^{* 2}+e^{* 2}\right)=0
\end{aligned}
$$

since $|e|=1$. Thus

$$
a^{*}+c^{*}=2 e^{*}+o-o e^{* 2} \quad \Longrightarrow \quad a^{*}=\frac{2 e^{*} f}{e+f}+o\left(1-e^{* 2}\right)
$$

and similarly

$$
b^{*}=\frac{2 f^{*} e}{f+e}+o\left(1-f^{* 2}\right)
$$

Then

$$
\begin{aligned}
b^{*}-a^{*} & =\frac{2\left(e f^{*}-e^{*} f\right)}{e+f}+o\left(e^{* 2}-f^{* 2}\right) \\
& =\frac{2 e f\left(f^{* 2}-e^{* 2}\right)}{e+f}+o\left(e^{* 2}-f^{* 2}\right) \\
& =\left(f^{* 2}-e^{* 2}\right)\left(\frac{2 e f}{e+f}-o\right)=\left(f^{* 2}-e^{* 2}\right)(c-o) .
\end{aligned}
$$

Now let $Z$ be a point on the tangent to $\omega$ parallel to $A B$ passing through $Q$. Then

$$
z=2 q-q^{2} z^{*} \quad \Longleftrightarrow \quad z-q=q-q^{2} z^{*}=-q^{2}\left(z^{*}-q^{*}\right)
$$

for $|q|=1$, and thus

$$
\frac{b-a}{b^{*}-a^{*}}=\frac{z-q}{z^{*}-q^{*}}=\frac{-q^{2}\left(z^{*}-q^{*}\right)}{z^{*}-q^{*}}=-q^{2}
$$



It follows that

$$
\begin{aligned}
q^{2} & =-\frac{b-a}{b^{*}-a^{*}}=-\frac{\left(f^{2}-e^{2}\right)\left(c^{*}-o\right)}{\left(f^{* 2}-e^{* 2}\right)(c-o)}=e^{2} f^{2} \frac{c^{*}-o}{c-o} \\
& =e^{2} f^{2} \frac{\left(c^{*}-o\right)^{2}}{|c-o|^{2}}=e^{2} f^{2} \frac{\left(c^{*}-o\right)^{2}}{(1+o)^{2}}
\end{aligned}
$$

where we have used (1). In particular,

$$
q=e f \frac{c^{*}-o}{1+o}
$$

where the choice of sign is to be justified a posteriori. Further, the point $D$ satisfies

$$
-d p=\frac{d-p}{d^{*}-p^{*}}=\frac{c-p}{c^{*}-p^{*}} \quad \Longrightarrow \quad d=-\frac{c-p}{c^{*} p-1}=\frac{c+1}{c^{*}+1}
$$

using $p=-1$ to obtain the final equality.

Now, it suffices to show that (i) $D Q \| E F \perp C I$ and (ii) the midpoint of $[D Q]$ is on $C I$. The desired equality then follows by symmetry with respect to the angle bisector of the angle $\angle A C B$. Notice that (i) is equivalent with

$$
\frac{d-q}{d^{*}-q^{*}}=\frac{e-f}{e^{*}-f^{*}} \Longleftrightarrow d q=e f
$$

for $[D Q]$ and $[E F]$ are chords of $\omega$. But

$$
\begin{aligned}
d q=e f & \Longleftrightarrow \frac{c+1}{c^{*}+1} e f \frac{c^{*}-o}{1+o}=e f \quad \Longleftrightarrow \quad(c+1)\left(c^{*}-o\right)=\left(c^{*}+1\right)(1+o) \\
& \Longleftrightarrow \quad c c^{*}-o\left(c+c^{*}\right)-2 o-1=0 .
\end{aligned}
$$

The last equality is precisely the defining relation for $o$, (1). This proves (i). Further, the midpoint of $[D Q]$ is $\frac{1}{2}(d+q)$, so it remains to check that

$$
d q=\frac{d+q}{d^{*}+q^{*}}=\frac{c}{c^{*}}=e f
$$

where the first equality expresses that $[D Q]$ is a chord of $\omega$ (obviously) containing its midpoint, the second equality expresses the alignment of the midpoint of $[D Q], C$ and $I$, and the third equality follows from the expression for $c$. But we have just shown that $d q=e f$. This proves (ii), justifies the choice of sign for $q$ a posteriori, and thus completes the solution of the problem.