Subfield: Geometry


Question: Let $I$ be the incentre of a non-equilateral triangle $A B C, I_{A}$ be the $A$-excentre, $I_{A}^{\prime}$ be the reflection of $I_{A}$ in $B C$, and $l_{A}$ be the reflection of line $A I_{A}^{\prime}$ in $A I$. Define points $I_{B}, I_{B}^{\prime}$ and line $l_{B}$ analogously. Let $P$ be the intersection point of $l_{A}$ and $l_{B}$.
Prove that $P$ lies on line $O I$ where $O$ is the circumcentre of triangle $A B C$.

Solution:
Let $A^{\prime}$ be the reflection of $A$ in $B C$ and let $M$ be the second intersection of line $A I$ and the circumcircle $\Gamma$ of triangle $A B C$. As triangles $A B A^{\prime}$ and $A O C$ are isosceles with $\angle A B A^{\prime}=2 \angle A B C=\angle A O C$, they are similar to each other. Also, triangles $A B I_{A}$ and $A I C$ are similar. Therefore we have

$$
\frac{A A^{\prime}}{A I_{A}}=\frac{A A^{\prime}}{A B} \cdot \frac{A B}{A I_{A}}=\frac{A C}{A O} \cdot \frac{A I}{A C}=\frac{A I}{A O}
$$

Together with $\angle A^{\prime} A I_{A}=\angle I A O$, we find that triangles $A A^{\prime} I_{A}$ and $A I O$ are similar.

<img_4038>

Denote by $P^{\prime}$ the intersection of line $A P$ and line $O I$. Using directed angles, we have

$$
\begin{aligned}
\measuredangle M A P^{\prime} & =\measuredangle I_{A}^{\prime} A I_{A}=\measuredangle I_{A}^{\prime} A A^{\prime}-\measuredangle I_{A} A A^{\prime}=\measuredangle A A^{\prime} I_{A}-\measuredangle(A M, O M) \\
& =\measuredangle A I O-\measuredangle A M O=\measuredangle M O P^{\prime} .
\end{aligned}
$$

This shows $M, O, A, P^{\prime}$ are concyclic.



Denote by $R$ and $r$ the circumradius and inradius of triangle $A B C$. Then

$$
I P^{\prime}=\frac{I A \cdot I M}{I O}=\frac{I O^{2}-R^{2}}{I O}
$$

is independent of $A$. Hence, $B P$ also meets line $O I$ at the same point $P^{\prime}$ so that $P^{\prime}=P$, and $P$ lies on $O I$.
Note that triangles $A I_{B} C$ and $I_{A} B C$ are similar since their corresponding interior angles are equal. Therefore, the four triangles $A I_{B}^{\prime} C, A I_{B} C, I_{A} B C$ and $I_{A}^{\prime} B C$ are all similar. From $\triangle A I_{B}^{\prime} C \sim \triangle I_{A}^{\prime} B C$, we get $\triangle A I_{A}^{\prime} C \sim \triangle I_{B}^{\prime} B C$. From $\measuredangle A B P=\measuredangle I_{B}^{\prime} B C=\measuredangle A I_{A}^{\prime} C$ and $\measuredangle B A P=\measuredangle I_{A}^{\prime} A C$, the triangles $A B P$ and $A I_{A}^{\prime} C$ are directly similar.

<img_3530>

Consider the inversion with centre $A$ and radius $\sqrt{A B \cdot A C}$ followed by the reflection in $A I$. Then $B$ and $C$ are mapped to each other, and $I$ and $I_{A}$ are mapped to each other.



From the similar triangles obtained, we have $A P \cdot A I_{A}^{\prime}=A B \cdot A C$ so that $P$ is mapped to $I_{A}^{\prime}$ under the transformation. In addition, line $A O$ is mapped to the altitude from $A$, and hence $O$ is mapped to the reflection of $A$ in $B C$, which we call point $A^{\prime}$. Note that $A A^{\prime} I_{A} I_{A}^{\prime}$ is an isosceles trapezoid, which shows it is inscribed in a circle. The preimage of this circle is a straight line, meaning that $O, I, P$ are collinear.