Subfield: Geometry


Question: Let $\Gamma$ be the circumcircle of triangle $A B C$. A circle $\Omega$ is tangent to the line segment $A B$ and is tangent to $\Gamma$ at a point lying on the same side of the line $A B$ as $C$. The angle bisector of $\angle B C A$ intersects $\Omega$ at two different points $P$ and $Q$.

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

Solution:
Let $M$ be the midpoint of the arc $A B$ that does not contain $C$, let $V$ be the intersection of $\Omega$ and $\Gamma$, and let $U$ be the intersection of $\Omega$ and $A B$.

<img_3659>

The proof can be divided in two steps:

1. Proving that $M P \cdot M Q=M B^{2}$.

It is well-known that $V, U$ and $M$ are collinear (indeed the homothety with center in $V$ that sends $\Omega$ to $\Gamma$ sends $U$ to the point of $\Gamma$ where the tangent to $\Gamma$ is parallel to $A B$, and this point is $M$ ), and

$$
M V \cdot M U=M A^{2}=M B^{2} .
$$

This follows from the similitude between the triangles $\triangle M A V$ and $\triangle M U A$. Alternatively, it is a consequence of the following well-known lemma: Given a circle $\Gamma$ with a chord $A B$, let $M$ be the middle point of one of the two arcs $A B$. Take a line through $M$ which intersects $\Gamma$ again at $X$ and $A B$ at $Y$. Then $M X \cdot M Y$ is independent of the choice of the line.

Computing the power of $M$ with respect to $\Omega$ we obtain that

$$
M P \cdot M Q=M U \cdot M V=M B^{2}
$$

2. Conclude the proof given that $M P \cdot M Q=M B^{2}$.

The relation $M P \cdot M Q=M B^{2}$ in turn implies that triangle $\triangle M B P$ is similar to triangle $\triangle M Q B$, and in particular $\angle M B P=\angle M Q B$. Keeping into account that $\angle M C B=$ $\angle M B A$, we finally conclude that

$$
\angle Q B C=\angle M Q B-\angle M C B=\angle M B P-\angle M B A=\angle P B A,
$$

as required.
Let $M$ be the midpoint of the arc $A B$ that does not contain $C$, let $V$ be the intersection of $\Omega$ and $\Gamma$, and let $U$ be the intersection of $\Omega$ and $A B$.

<img_3659>

The proof can be divided in two steps:

1. Proving that $M P \cdot M Q=M B^{2}$.

Let us consider the inversion with respect to circle with center $M$ and radius $M A=M B$. This inversion switches $A B$ and $\Gamma$, and fixes the line passing through $M, U, V$. As a consequence, it keeps $\Omega$ fixed, and therefore it switches $P$ and $Q$. This is because they are the intersections between the fixed line $M C$ and $\Omega$, and the only fixed point on the segment $M C$ is its intersection with the inversion circle (thus $P$ and $Q$ are switched). This implies that $M P \cdot M Q=M B^{2}$.

2. Conclude the proof given that $M P \cdot M Q=M B^{2}$.

The relation $M P \cdot M Q=M B^{2}$ in turn implies that triangle $\triangle M B P$ is similar to triangle $\triangle M Q B$, and in particular $\angle M B P=\angle M Q B$. Keeping into account that $\angle M C B=$ $\angle M B A$, we finally conclude that

$$
\angle Q B C=\angle M Q B-\angle M C B=\angle M B P-\angle M B A=\angle P B A,
$$

as required.
Let $M$ be the midpoint of the arc $A B$ that does not contain $C$, let $V$ be the intersection of $\Omega$ and $\Gamma$, and let $U$ be the intersection of $\Omega$ and $A B$.

<img_3659>

The proof can be divided in two steps:

1. Proving that $M P \cdot M Q=M B^{2}$.

It is well-known that $V, U$ and $M$ are collinear (indeed the homothety with center in $V$ that sends $\Omega$ to $\Gamma$ sends $U$ to the point of $\Gamma$ where the tangent to $\Gamma$ is parallel to $A B$, and this point is $M$ ), and

$$
M V \cdot M U=M A^{2}=M B^{2} .
$$

This follows from the similitude between the triangles $\triangle M A V$ and $\triangle M U A$. Alternatively, it is a consequence of the following well-known lemma: Given a circle $\Gamma$ with a chord $A B$, let $M$ be the middle point of one of the two arcs $A B$. Take a line through $M$ which intersects $\Gamma$ again at $X$ and $A B$ at $Y$. Then $M X \cdot M Y$ is independent of the choice of the line.

Computing the power of $M$ with respect to $\Omega$ we obtain that

$$
M P \cdot M Q=M U \cdot M V=M B^{2}
$$

2. Conclude the proof given that $M P \cdot M Q=M B^{2}$.

Let $I$ and $J$ be the incenter and the $C$-excenter of $\triangle A B C$ respectively. It is well-known that $M A=M I=M J$, therefore the relation $M P \cdot M Q=M A^{2}$ implies that $(P, Q, I, J)=-1$.

Now observe that $\angle I B J=90^{\circ}$, thus $B I$ is the angle bisector of $\angle P B Q$ as it is well-known from the theory of harmonic pencils, and this leads easily to the conclusion.
Let $D$ denote the intersection of $A B$ and $C M$. Let us consider an inversion with respect to $B$, and let us use primes to denote corresponding points in the transformed diagram, with the gentlemen agreement that $B^{\prime}=B$.

<img_3283>

Since inversion preserves angles, it turns out that

$$
\angle A^{\prime} B^{\prime} M^{\prime}=\angle A^{\prime} M^{\prime} B^{\prime}=\angle A C B,
$$

and in particular triangle $A^{\prime} B^{\prime} M^{\prime}$ is isosceles with basis $B^{\prime} M^{\prime}$.

The image of $C M$ is the circumcircle of $B^{\prime} C^{\prime} M^{\prime}$, which we denote by $\omega^{\prime}$. It follows that the centers of both $\omega^{\prime}$ and the image $\Omega^{\prime}$ of $\Omega$ lie on the perpendicular bisector of $B^{\prime} M^{\prime}$. Therefore, the whole transformed diagram is symmetric with respect to the perpendicular bisector of $B^{\prime} M^{\prime}$, and in particular the $\operatorname{arcs} D^{\prime} P^{\prime}$ and $Q^{\prime} C^{\prime}$ of $\omega^{\prime}$ are equal.

This is enough to conclude that $\angle D^{\prime} B^{\prime} P^{\prime}=\angle Q^{\prime} B^{\prime} C^{\prime}$, which implies the conclusion.