Subfield: Geometry


Question: Let $I$ and $I_{A}$ be the incenter and the $A$-excenter of an acute-angled triangle $A B C$ with $A B<A C$. Let the incircle meet $B C$ at $D$. The line $A D$ meets $B I_{A}$ and $C I_{A}$ at $E$ and $F$, respectively. Prove that the circumcircles of triangles $A I D$ and $I_{A} E F$ are tangent to each other.

Solution:
Let $\sphericalangle(p, q)$ denote the directed angle between lines $p$ and $q$.

The points $B, C, I$, and $I_{A}$ lie on the circle $\Gamma$ with diameter $I I_{A}$. Let $\omega$ and $\Omega$ denote the circles $\left(I_{A} E F\right)$ and $(A I D)$, respectively. Let $T$ be the second intersection point of $\omega$ and $\Gamma$. Then $T$ is the Miquel point of the complete quadrilateral formed by the lines $B C, B I_{A}, C I_{A}$, and $D E F$, so $T$ also lies on circle $(B D E)$ (as well as on circle $(C D F)$ ). We claim that $T$ is a desired tangency point of $\omega$ and $\Omega$.

In order to show that $T$ lies on $\Omega$, use cyclic quadrilaterals $B D E T$ and $B I I_{A} T$ to write

$$
\sphericalangle(D T, D A)=\sphericalangle(D T, D E)=\sphericalangle(B T, B E)=\sphericalangle\left(B T, B I_{A}\right)=\sphericalangle\left(I T, I I_{A}\right)=\sphericalangle(I T, I A) .
$$

<img_3849>

To show that $\omega$ and $\Omega$ are tangent at $T$, let $\ell$ be the tangent to $\omega$ at $T$, so that $\sphericalangle\left(T I_{A}, \ell\right)=$ $\sphericalangle\left(E I_{A}, E T\right)$. Using circles $(B D E T)$ and $\left(B I C I_{A}\right)$, we get

$$
\sphericalangle\left(E I_{A}, E T\right)=\sphericalangle(E B, E T)=\sphericalangle(D B, D T) .
$$

Therefore,

$$
\sphericalangle(T I, \ell)=90^{\circ}+\sphericalangle\left(T I_{A}, \ell\right)=90^{\circ}+\sphericalangle(D B, D T)=\sphericalangle(D I, D T)
$$

which shows that $\ell$ is tangent to $\Omega$ at $T$.
We use the notation of circles $\Gamma, \omega$, and $\Omega$ as in the previous solution.

Let $L$ be the point opposite to $I$ in circle $\Omega$. Then $\angle I A L=\angle I D L=90^{\circ}$, which means that $L$ is the foot of the external bisector of $\angle A$ in triangle $A B C$. Let $L I$ cross $\Gamma$ again at $M$.



Let $T$ be the foot of the perpendicular from $I$ onto $I_{A} L$. Then $T$ is the second intersection point of $\Gamma$ and $\Gamma$. We will show that $T$ is the desired tangency point.

First, we show that $T$ lies on circle $\omega$. Notice that

$$
\sphericalangle(L T, L M)=\sphericalangle(A T, A I) \text { and } \quad \sphericalangle(M T, M L)=\sphericalangle(M T, M I)=\sphericalangle\left(I_{A} T, I_{A} I\right) \text {, }
$$

which shows that triangles $T M L$ and $T I_{A} A$ are similar and equioriented. So there exists a rotational homothety $\tau$ mapping $T M L$ to $T I_{A} A$.

Since $\sphericalangle(M L, L D)=\sphericalangle(A I, A D)$, we get $\tau(B C)=A D$. Next, since

$$
\sphericalangle(M B, M L)=\sphericalangle(M B, M I)=\sphericalangle\left(I_{A} B, I_{A} I\right)=\sphericalangle\left(I_{A} E, I_{A} A\right)
$$

we get $\tau(B)=E$. Similarly, $\tau(C)=F$. Since the points $M, B, C$, and $T$ are concyclic, so are their $\tau$-images, which means that $T$ lies on $\omega=\tau(\Gamma)$.

<img_3304>

Finally, since $\tau(L)=A$ and $\tau(B)=E$, triangles $A T L$ and $E T B$ are similar so that

$$
\sphericalangle(A T, A L)=\sphericalangle(E T, E B)=\sphericalangle\left(E I_{A}, E T\right) .
$$

This means that the tangents to $\Omega$ and $\omega$ at $T$ make the same angle with the line $I_{A} T L$, so the circles are indeed tangent at $T$.
We also use the notation of circles $\omega$, and $\Omega$ from the previous solutions.

Perform an inversion centered at $D$. The images of the points will be denoted by primes, e.g., $A^{\prime}$ is the image of $A$.

For convenience, we use the notation $\angle B I D=\beta, \angle C I D=\gamma$, and $\alpha=180^{\circ}-\beta-\gamma=$ $90^{\circ}-\angle B A I$. We start with computing angles appearing after inversion. We get

$$
\begin{gathered}
\angle D B^{\prime} I^{\prime}=\beta, \quad \angle D C^{\prime} I^{\prime}=\gamma, \quad \text { and hence } \angle B^{\prime} I^{\prime} C^{\prime}=\alpha ; \\
\angle E^{\prime} I_{A}^{\prime} F^{\prime}=\angle E^{\prime} I_{A}^{\prime} D-\angle F^{\prime} I_{A}^{\prime} D=\angle I_{A} E D-\angle I_{A} F D=\angle E I_{A} F=180^{\circ}-\alpha .
\end{gathered}
$$

Next, we have

$$
\angle A^{\prime} E^{\prime} B^{\prime}=\angle D E^{\prime} B^{\prime}=\angle D B E=\beta=90^{\circ}-\frac{\angle D B A}{2}=90^{\circ}-\frac{\angle E^{\prime} A^{\prime} B^{\prime}}{2},
$$

which yields that triangle $A^{\prime} B^{\prime} E^{\prime}$ is isosceles with $A^{\prime} B^{\prime}=A^{\prime} E^{\prime}$. Similarly, $A^{\prime} F^{\prime}=A^{\prime} C^{\prime}$.

Finally, we get

$$
\begin{aligned}
\angle A^{\prime} B^{\prime} I^{\prime}=\angle I^{\prime} B^{\prime} D-\angle A^{\prime} B^{\prime} D=\beta- & \angle B A D=\beta-\left(90^{\circ}-\alpha\right)+\angle I A D \\
& =\angle I C D+\angle I A D=\angle C^{\prime} I^{\prime} D+\angle A^{\prime} I^{\prime} D=\angle C^{\prime} I^{\prime} A^{\prime}
\end{aligned}
$$

similarly, $\angle A^{\prime} C^{\prime} I^{\prime}=\angle A^{\prime} I^{\prime} B^{\prime}$, so that triangles $A^{\prime} B^{\prime} I^{\prime}$ and $A^{\prime} I^{\prime} C^{\prime}$ are similar. Therefore, $A^{\prime} I^{2}=A^{\prime} B^{\prime} \cdot A^{\prime} C^{\prime}$.

Recall that we need to prove the tangency of line $A^{\prime} I^{\prime}=\Omega^{\prime}$ with circle $\left(E^{\prime} F^{\prime} I_{A}^{\prime}\right)=\omega^{\prime}$. A desired tangency point $T^{\prime}$ must satisfy $A^{\prime} T^{\prime 2}=A^{\prime} E^{\prime} \cdot A^{\prime} F^{\prime}$; the relations obtained above yield

$$
A^{\prime} E^{\prime} \cdot A^{\prime} F^{\prime}=A^{\prime} B^{\prime} \cdot A^{\prime} C^{\prime}=A^{\prime} I^{\prime 2},
$$

so that $T^{\prime}$ should be symmetric to $I^{\prime}$ with respect to $A^{\prime}$.

Thus, let us define a point $T^{\prime}$ as the reflection of $I^{\prime}$ in $A^{\prime}$, and show that $T^{\prime}$ lies on circle $\Omega^{\prime}$; the equalities above will then imply that $A^{\prime} T^{\prime}$ is tangent to $\Omega^{\prime}$, as desired.
<img_4027>

The property that triangles $B^{\prime} A^{\prime} I^{\prime}$ and $I^{\prime} A^{\prime} C^{\prime}$ are similar means that quadrilateral $B^{\prime} I^{\prime} C^{\prime} T^{\prime}$ is harmonic. Indeed, let $C^{*}$ be the reflection of $C^{\prime}$ in the perpendicular bisector of $I^{\prime} T^{\prime}$; then $C^{*}$ lies on $B^{\prime} A^{\prime}$ by $\angle B^{\prime} A^{\prime} I^{\prime}=\angle A^{\prime} I^{\prime} C^{\prime}=\angle T^{\prime} I^{\prime} C^{*}$, and then $C^{*}$ lies on circle $\left(I^{\prime} B^{\prime} T^{\prime}\right)$ since $A^{\prime} B^{\prime} \cdot A^{\prime} C^{*}=A^{\prime} B^{\prime} \cdot A^{\prime} C^{\prime}=A^{\prime} I^{\prime 2}=A^{\prime} I^{\prime} \cdot A^{\prime} T^{\prime}$. Therefore, $C^{\prime}$ also lies on that circle (and the circle is $\left.\left(B^{\prime} I^{\prime} C^{\prime}\right)=\Gamma^{\prime}\right)$. Moreover, $B^{\prime} C^{*}$ is a median in triangle $B^{\prime} I^{\prime} T^{\prime}$, so $B^{\prime} C^{\prime}$ is its symmedian, which establishes harmonicity.

Now we have $\angle A^{\prime} B^{\prime} T^{\prime}=\angle I^{\prime} B^{\prime} C^{\prime}=\beta=\angle A^{\prime} B^{\prime} E^{\prime}$; which shows that $E^{\prime}$ lies on $B^{\prime} T^{\prime}$. Similarly, $F^{\prime}$ lies on $C^{\prime} T^{\prime}$. Hence, $\angle E^{\prime} T^{\prime} F^{\prime}=\angle B^{\prime} I^{\prime} C^{\prime}=180^{\circ}-\angle E^{\prime} I_{A}^{\prime} F^{\prime}$, which establishes $T^{\prime} \in \omega^{\prime}$.