Subfield: Geometry


Question: Let $A B C D$ be a circumscribed quadrilateral. Let $g$ be a line through $A$ which meets the segment $B C$ in $M$ and the line $C D$ in $N$. Denote by $I_{1}, I_{2}$, and $I_{3}$ the incenters of $\triangle A B M$, $\triangle M N C$, and $\triangle N D A$, respectively. Show that the orthocenter of $\triangle I_{1} I_{2} I_{3}$ lies on $g$.

Solution:
Let $k_{1}, k_{2}$ and $k_{3}$ be the incircles of triangles $A B M, M N C$, and $N D A$, respectively (see Figure 1). We shall show that the tangent $h$ from $C$ to $k_{1}$ which is different from $C B$ is also tangent to $k_{3}$.

<img_3815>

Figure 1

To this end, let $X$ denote the point of intersection of $g$ and $h$. Then $A B C X$ and $A B C D$ are circumscribed quadrilaterals, whence

$$
C D-C X=(A B+C D)-(A B+C X)=(B C+A D)-(B C+A X)=A D-A X
$$

i.e.

$$
A X+C D=C X+A D
$$

which in turn reveals that the quadrilateral $A X C D$ is also circumscribed. Thus $h$ touches indeed the circle $k_{3}$.

Moreover, we find that $\angle I_{3} C I_{1}=\angle I_{3} C X+\angle X C I_{1}=\frac{1}{2}(\angle D C X+\angle X C B)=\frac{1}{2} \angle D C B=$ $\frac{1}{2}\left(180^{\circ}-\angle M C N\right)=180^{\circ}-\angle M I_{2} N=\angle I_{3} I_{2} I_{1}$, from which we conclude that $C, I_{1}, I_{2}, I_{3}$ are concyclic.

Let now $L_{1}$ and $L_{3}$ be the reflection points of $C$ with respect to the lines $I_{2} I_{3}$ and $I_{1} I_{2}$ respectively. Since $I_{1} I_{2}$ is the angle bisector of $\angle N M C$, it follows that $L_{3}$ lies on $g$. By analogous reasoning, $L_{1}$ lies on $g$.

Let $H$ be the orthocenter of $\triangle I_{1} I_{2} I_{3}$. We have $\angle I_{2} L_{3} I_{1}=\angle I_{1} C I_{2}=\angle I_{1} I_{3} I_{2}=180^{\circ}-\angle I_{1} H I_{2}$, which entails that the quadrilateral $I_{2} H I_{1} L_{3}$ is cyclic. Analogously, $I_{3} H L_{1} I_{2}$ is cyclic.



Then, working with oriented angles modulo $180^{\circ}$, we have

$$
\angle L_{3} H I_{2}=\angle L_{3} I_{1} I_{2}=\angle I_{2} I_{1} C=\angle I_{2} I_{3} C=\angle L_{1} I_{3} I_{2}=\angle L_{1} H I_{2},
$$

whence $L_{1}, L_{3}$, and $H$ are collinear. By $L_{1} \neq L_{3}$, the claim follows.
We start by proving that $C, I_{1}, I_{2}$, and $I_{3}$ are concyclic.

<img_3294>

Figure 2

To this end, notice first that $I_{2}, M, I_{1}$ are collinear, as are $N, I_{2}, I_{3}$ (see Figure 2). Denote by $\alpha, \beta, \gamma, \delta$ the internal angles of $A B C D$. By considerations in triangle $C M N$, it follows that $\angle I_{3} I_{2} I_{1}=\frac{\gamma}{2}$. We will show that $\angle I_{3} C I_{1}=\frac{\gamma}{2}$, too. Denote by $I$ the incenter of $A B C D$. Clearly, $I_{1} \in B I, I_{3} \in D I, \angle I_{1} A I_{3}=\frac{\alpha}{2}$.

Using the abbreviation $[X, Y Z]$ for the distance from point $X$ to the line $Y Z$, we have because of $\angle B A I_{1}=\angle I A I_{3}$ and $\angle I_{1} A I=\angle I_{3} A D$ that

$$
\frac{\left[I_{1}, A B\right]}{\left[I_{1}, A I\right]}=\frac{\left[I_{3}, A I\right]}{\left[I_{3}, A D\right]}
$$

Furthermore, consideration of the angle sums in $A I B, B I C, C I D$ and $D I A$ implies $\angle A I B+$ $\angle C I D=\angle B I C+\angle D I A=180^{\circ}$, from which we see

$$
\frac{\left[I_{1}, A I\right]}{\left[I_{3}, C I\right]}=\frac{I_{1} I}{I_{3} I}=\frac{\left[I_{1}, C I\right]}{\left[I_{3}, A I\right]}
$$

Because of $\left[I_{1}, A B\right]=\left[I_{1}, B C\right],\left[I_{3}, A D\right]=\left[I_{3}, C D\right]$, multiplication yields

$$
\frac{\left[I_{1}, B C\right]}{\left[I_{3}, C I\right]}=\frac{\left[I_{1}, C I\right]}{\left[I_{3}, C D\right]}
$$

By $\angle D C I=\angle I C B=\gamma / 2$ it follows that $\angle I_{1} C B=\angle I_{3} C I$ which concludes the proof of the



above statement.

Let the perpendicular from $I_{1}$ on $I_{2} I_{3}$ intersect $g$ at $Z$. Then $\angle M I_{1} Z=90^{\circ}-\angle I_{3} I_{2} I_{1}=$ $90^{\circ}-\gamma / 2=\angle M C I_{2}$. Since we have also $\angle Z M I_{1}=\angle I_{2} M C$, triangles $M Z I_{1}$ and $M I_{2} C$ are similar. From this one easily proves that also $M I_{2} Z$ and $M C I_{1}$ are similar. Because $C, I_{1}, I_{2}$, and $I_{3}$ are concyclic, $\angle M Z I_{2}=\angle M I_{1} C=\angle N I_{3} C$, thus $N I_{2} Z$ and $N C I_{3}$ are similar, hence $N C I_{2}$ and $N I_{3} Z$ are similar. We conclude $\angle Z I_{3} I_{2}=\angle I_{2} C N=90^{\circ}-\gamma / 2$, hence $I_{1} I_{2} \perp Z I_{3}$. This completes the proof.