Subfield: Geometry


Question: Let $A B C$ be a triangle with circumcircle $\Omega$ and incentre $I$. Let the line passing through $I$ and perpendicular to $C I$ intersect the segment $B C$ and the $\operatorname{arc} B C$ (not containing $A$ ) of $\Omega$ at points $U$ and $V$, respectively. Let the line passing through $U$ and parallel to $A I$ intersect $A V$ at $X$, and let the line passing through $V$ and parallel to $A I$ intersect $A B$ at $Y$. Let $W$ and $Z$ be the midpoints of $A X$ and $B C$, respectively. Prove that if the points $I, X$, and $Y$ are collinear, then the points $I, W$, and $Z$ are also collinear.

Solution:
We start with some general observations. Set $\alpha=\angle A / 2, \beta=\angle B / 2, \gamma=\angle C / 2$. Then obviously $\alpha+\beta+\gamma=90^{\circ}$. Since $\angle U I C=90^{\circ}$, we obtain $\angle I U C=\alpha+\beta$. Therefore $\angle B I V=\angle I U C-\angle I B C=\alpha=\angle B A I=\angle B Y V$, which implies that the points $B, Y, I$, and $V$ lie on a common circle (see Figure 1).

Assume now that the points $I, X$ and $Y$ are collinear. We prove that $\angle Y I A=90^{\circ}$.

Let the line $X U$ intersect $A B$ at $N$. Since the lines $A I, U X$, and $V Y$ are parallel, we get

$$
\frac{N X}{A I}=\frac{Y N}{Y A}=\frac{V U}{V I}=\frac{X U}{A I}
$$

implying $N X=X U$. Moreover, $\angle B I U=\alpha=\angle B N U$. This implies that the quadrilateral $B U I N$ is cyclic, and since $B I$ is the angle bisector of $\angle U B N$, we infer that $N I=U I$. Thus in the isosceles triangle $N I U$, the point $X$ is the midpoint of the base $N U$. This gives $\angle I X N=90^{\circ}$, i.e., $\angle Y I A=90^{\circ}$.

<img_3282>

Figure 1

Let $S$ be the midpoint of the segment $V C$. Let moreover $T$ be the intersection point of the lines $A X$ and $S I$, and set $x=\angle B A V=\angle B C V$. Since $\angle C I A=90^{\circ}+\beta$ and $S I=S C$, we obtain

$$
\angle T I A=180^{\circ}-\angle A I S=90^{\circ}-\beta-\angle C I S=90^{\circ}-\beta-\gamma-x=\alpha-x=\angle T A I,
$$

which implies that $T I=T A$. Therefore, since $\angle X I A=90^{\circ}$, the point $T$ is the midpoint of $A X$, i.e., $T=W$.

To complete our solution, it remains to show that the intersection point of the lines $I S$ and $B C$ coincide with the midpoint of the segment $B C$. But since $S$ is the midpoint of the segment $V C$, it suffices to show that the lines $B V$ and $I S$ are parallel.



Since the quadrilateral $B Y I V$ is cyclic, $\angle V B I=\angle V Y I=\angle Y I A=90^{\circ}$. This implies that $B V$ is the external angle bisector of the angle $A B C$, which yields $\angle V A C=\angle V C A$. Therefore $2 \alpha-x=2 \gamma+x$, which gives $\alpha=\gamma+x$. Hence $\angle S C I=\alpha$, so $\angle V S I=2 \alpha$.

On the other hand, $\angle B V C=180^{\circ}-\angle B A C=180^{\circ}-2 \alpha$, which implies that the lines $B V$ and $I S$ are parallel. This completes the solution.
As in
1, we first prove that the points $B, Y, I, V$ lie on a common circle and $\angle Y I A=90^{\circ}$. The remaining part of the solution is based on the following lemma, which holds true for any triangle $A B C$, not necessarily with the property that $I, X, Y$ are collinear.

Lemma. Let $A B C$ be the triangle inscribed in a circle $\Gamma$ and let $I$ be its incentre. Assume that the line passing through $I$ and perpendicular to the line $A I$ intersects the side $A B$ at the point $Y$. Let the circumcircle of the triangle $B Y I$ intersect the circle $\Gamma$ for the second time at $V$, and let the excircle of the triangle $A B C$ opposite to the vertex $A$ be tangent to the side $B C$ at $E$. Then

$$
\angle B A V=\angle C A E .
$$

Proof. Let $\rho$ be the composition of the inversion with centre $A$ and radius $\sqrt{A B \cdot A C}$, and the symmetry with respect to $A I$. Clearly, $\rho$ interchanges $B$ and $C$.

Let $J$ be the excentre of the triangle $A B C$ opposite to $A$ (see Figure 2). Then we have $\angle J A C=\angle B A I$ and $\angle J C A=90^{\circ}+\gamma=\angle B I A$, so the triangles $A C J$ and $A I B$ are similar, and therefore $A B \cdot A C=A I \cdot A J$. This means that $\rho$ interchanges $I$ and $J$. Moreover, since $Y$ lies on $A B$ and $\angle A I Y=90^{\circ}$, the point $Y^{\prime}=\rho(Y)$ lies on $A C$, and $\angle J Y^{\prime} A=90^{\circ}$. Thus $\rho$ maps the circumcircle $\gamma$ of the triangle $B Y I$ to a circle $\gamma^{\prime}$ with diameter $J C$.

Finally, since $V$ lies on both $\Gamma$ and $\gamma$, the point $V^{\prime}=\rho(V)$ lies on the line $\rho(\Gamma)=A B$ as well as on $\gamma^{\prime}$, which in turn means that $V^{\prime}=E$. This implies the desired result.

<img_3595>

Figure 2

<img_3954>

Figure 3

Now we turn to the solution of the problem.

Assume that the incircle $\omega_{1}$ of the triangle $A B C$ is tangent to $B C$ at $D$, and let the excircle $\omega_{2}$ of the triangle $A B C$ opposite to the vertex $A$ touch the side $B C$ at $E$ (see Figure 3 ). The homothety with centre $A$ that takes $\omega_{2}$ to $\omega_{1}$ takes the point $E$ to some point $F$, and the



tangent to $\omega_{1}$ at $F$ is parallel to $B C$. Therefore $D F$ is a diameter of $\omega_{1}$. Moreover, $Z$ is the midpoint of $D E$. This implies that the lines $I Z$ and $F E$ are parallel.

Let $K=Y I \cap A E$. Since $\angle Y I A=90^{\circ}$, the lemma yields that $I$ is the midpoint of $X K$. This implies that the segments $I W$ and $A K$ are parallel. Therefore, the points $W, I$ and $Z$ are collinear.