Subfield: Geometry


Question: Let $A B C$ be a triangle with incentre $I$. The circle through $B$ tangent to $A I$ at $I$ meets side $A B$ again at $P$. The circle through $C$ tangent to $A I$ at $I$ meets side $A C$ again at $Q$. Prove that $P Q$ is tangent to the incircle of $A B C$.

Solution:
Let $Q X, P Y$ be tangent to the incircle of $A B C$, where $X, Y$ lie on the incircle and do not lie on $A C, A B$. Denote $\angle B A C=\alpha, \angle C B A=\beta, \angle A C B=\gamma$.

Since $A I$ is tangent to the circumcircle of $C Q I$ we get $\angle Q I A=\angle Q C I=\frac{\gamma}{2}$. Thus

$$
\angle I Q C=\angle I A Q+\angle Q I A=\frac{\alpha}{2}+\frac{\gamma}{2} .
$$

By the definition of $X$ we have $\angle I Q C=\angle X Q I$, therefore

$$
\angle A Q X=180^{\circ}-\angle X Q C=180^{\circ}-\alpha-\gamma=\beta .
$$

Similarly one can prove that $\angle A P Y=\gamma$. This means that $Q, P, X, Y$ are collinear which leads us to the conclusion that $X=Y$ and $Q P$ is tangent to the incircle at $X$.
<img_3357>
By the power of a point we have

$$
A D \cdot A C=A I^{2}=A P \cdot A B, \quad \text { which means that } \frac{A Q}{A P}=\frac{A B}{A C}
$$

and therefore triangles $A D P, A B C$ are similar. Let $J$ be the incenter of $A Q P$. We obtain

$$
\angle J P Q=\angle I C B=\angle Q C I=\angle Q I J,
$$

thus $J, P, I, Q$ are concyclic. Let $S$ be the intersection of $A I$ and $B C$. It follows that

$$
\angle I Q P=\angle I J P=\angle S I C=\angle I Q C .
$$

This means that $I Q$ is the angle bisector of $\angle C Q P$, so $Q P$ is indeed tangent to the incircle of $A B C$.
Like before, notice that $A Q \cdot A C=A P \cdot A B=A I^{2}$. Consider the positive inversion $\Psi$ with center $A$ and power $A I^{2}$. This maps $P$ to $B$ (and vice-versa), $Q$ to $C$ (and vice-versa), and keeps the incenter $I$ fixed. The problem statement will follow from the fact that the image of the incircle of triangle $A B C$ under $\Psi$ is the so-called mixtilinear incircle of $A B C$, which is defined to be the circle tangent to the lines $A B, A C$, and the circumcircle of $A B C$. Indeed, since the image of the line $Q P$ is the circumcircle of $A B C$, and inversion preserves tangencies, this implies that $Q P$ is tangent to the incircle of $A B C$.

We justify the claim as follows: let $\gamma$ be the incircle of $A B C$ and let $\Gamma_{A}$ be the $A$-mixtilinear incircle of $A B C$. Let $K$ and $L$ be the tangency points of $\gamma$ with the sides $A B$ and $A C$, and let $U$ and $V$ be the tangency points of $\Gamma_{A}$ with the sides $A B$ and $A C$, respectively. It is well-known that the incenter $I$ is the midpoint of segment $U V$. In particular, since also $A I \perp U V$, this implies that $A U=A V=\frac{A I}{\cos \frac{A}{2}}$. Note that $A K=A L=A I \cdot \cos \frac{A}{2}$. Therefore, $A U \cdot A K=A V \cdot A L=A I^{2}$, which means that $U$ and $V$ are the images of $K$ and $L$ under $\Psi$. Since $\Gamma_{A}$ is the unique circle simultaneously tangent to $A B$ at $U$ and to $A C$ at $V$, it follows that the image of $\gamma$ under $\Psi$ must be precisely $\Gamma_{A}$, as claimed.
From the power of a point theorem, we have

$$
A P \cdot A B=A I^{2}=A Q \cdot A C .
$$

Hence $P B C Q$ is cyclic, and so, $\angle A P Q=\angle B C A$. Let $K$ be the circumcenter of $\triangle B I P$ and let $L$ be the circumcenter of $\triangle Q I C$. Then $\overline{K L}$ is perpendicular to $\overline{A I}$ at $I$.

Let $N$ be the point of intersection of line $\overline{K L}$ with $\overline{A B}$. Then in the right triangle $\triangle N I A$, we have $\angle A N I=90^{\circ}-\frac{\angle B A C}{2}$ and from the external angle theorem for triangle $\triangle B N I$, we have $\angle A N I=\frac{\angle A B C}{2}+\angle N I B$. Hence

$$
\angle N I B=\angle A N I-\frac{\angle A B C}{2}=\left(90^{\circ}-\frac{\angle B A C}{2}\right)-\frac{\angle A B C}{2}=\frac{\angle B C A}{2} .
$$

Since $M I$ is tangent to the circumcircle of $\triangle B I P$ at $I$, we have

$$
\angle B P I=\angle B I M=\angle N I M-\angle N I B=90^{\circ}-\frac{\angle B C A}{2} .
$$

Also, since $\angle A P Q=\angle B C A$, we have

$\angle Q P I=180^{\circ}-\angle A P Q-\angle B P I=180^{\circ}-\angle B C A-\left(90^{\circ}-\frac{\angle B C A}{2}\right)=90^{\circ}-\frac{\angle B C A}{2}$,

as well. Hence $I$ lies on the angle bisector of $\angle B P Q$, and so it is equidistant from its sides $\overline{P Q}$ and $\overline{P B}$. Therefore, the distance of $I$ from $\overline{P Q}$ equals the inradius of $\triangle A B C$, as desired.



<img_3539>
Let $D$ be the point of intersection of $\overline{A I}$ and $\overline{B C}$ and let $R$ be the point of intersection of $\overline{A I}$ and $\overline{P Q}$. We have $\angle R I P=\angle P B I=\frac{\angle B}{2}$, $\angle R I Q=\angle I C Q=\frac{\angle C}{2}, \angle I Q C=\angle D I C=x$ and $\angle B P I=\angle B I D=\varphi$, since $\overline{A I}$ is tangent to both circles.

<img_3638>

From the angle bisector theorem, we have

$$
\frac{R Q}{R P}=\frac{A Q}{A P} \quad \text { and } \quad \frac{A C}{A B}=\frac{D C}{B D}
、
$$



Since $\overline{A I}$ is tangent to both circles at $I$, we have $A I^{2}=A Q \cdot A C$ and $A I^{2}=A P \cdot A B$. Therefore,

$$
\frac{R Q}{R P} \cdot \frac{D C}{B D}=\frac{A Q \cdot A C}{A B \cdot A P}=1
\tag{1}
$$

From the sine law in triangles $\triangle Q R I$ and $\triangle P R I$, it follows that $\frac{R Q}{\sin \frac{\angle C}{2}}=\frac{R I}{\sin y}$ and $\frac{R P}{\sin \frac{\angle B}{2}}=\frac{R I}{\sin \omega}$, respectively. Hence

$$
\frac{R Q}{R P} \cdot \frac{\sin \frac{\angle B}{2}}{\sin \frac{\angle C}{2}}=\frac{\sin \omega}{\sin y}
\tag{2}
$$

Similarly, from the sine law in triangles $\triangle I D C$ and $\triangle I D B$, it is $\frac{D C}{\sin x}=\frac{I D}{\sin \frac{\angle C}{2}}$ and $\frac{B D}{\sin \varphi}=\frac{I D}{\sin \frac{\angle B}{2}}$, and so

$$
\frac{D C}{B D} \cdot \frac{\sin \varphi}{\sin x}=\frac{\sin \frac{\angle B}{2}}{\sin \frac{\angle C}{2}}
\tag{3}
$$

By multiplying equations (2) with (3), we obtain $\frac{R Q}{R P} \cdot \frac{D C}{B D} \cdot \frac{\sin \varphi}{\sin x}=\frac{\sin \omega}{\sin y}$, which combined with (1) and cross-multiplying yields

$$
\sin \varphi \cdot \sin y=\sin \omega \cdot \sin x
\tag{4}
$$

Let $\theta=90^{\circ}+\frac{\angle A}{2}$. Since $I$ is the incenter of $\triangle A B C$, we have $x=90^{\circ}+\frac{\angle A}{2}-\varphi=\theta-\phi$. Also, in triangle $\triangle P I Q$, we see that $\omega+y+\frac{\angle B}{2}+\frac{\angle C}{2}=180^{\circ}$, and so $y=\theta-\omega$.

Therefore, equation (4) yields

$$
\sin \varphi \cdot \sin (\theta-\omega)=\sin \omega \cdot \sin (\theta-\varphi)
$$

or

$$
\frac{1}{2}(\cos (\varphi-\theta+\omega)-\cos (\varphi+\theta-\omega))=\frac{1}{2}(\cos (\omega-\theta+\varphi)-\cos (\omega+\theta-\varphi)),
$$

which is equivalent to

$$
\cos (\varphi+\theta-\omega)=\cos (\omega+\theta-\varphi) .
$$

So

$$
\varphi+\theta-\omega=2 k \cdot 180^{\circ} \pm(\omega+\theta-\varphi), \quad(k \in \mathbb{Z} .)
$$

If $\varphi+\theta-\omega=2 k \cdot 180^{\circ}+(\omega+\theta-\varphi)$, then $2(\varphi-\omega)=2 k \cdot 180^{\circ}$, with $|\varphi-\omega|<180^{\circ}$ forcing $k=0$ and $\varphi=\omega$. If $\varphi+\theta-\omega=2 k \cdot 180^{\circ}-(\omega+\theta-\varphi)$, then $2 \theta=2 k \cdot 180^{\circ}$, which contradicts the fact that $0^{\circ}<\theta<180^{\circ}$. Hence $\varphi=\omega$, and so $P I$ is the angle bisector of $\angle Q P B$.

Therefore the distance of $I$ from $\overline{P Q}$ is the same with the distance of $I$ from $A B$, which is equal to the inradius of $\triangle A B C$. Consequently, $\overline{P Q}$ is tangent to the incircle of $\triangle A B C$.