Subfield: Geometry


Question: The incircle $\omega$ of acute-angled scalene triangle $A B C$ has centre $I$ and meets sides $B C$, $C A$, and $A B$ at $D, E$, and $F$, respectively. The line through $D$ perpendicular to $E F$ meets $\omega$ again at $R$. Line $A R$ meets $\omega$ again at $P$. The circumcircles of triangles $P C E$ and $P B F$ meet again at $Q \neq P$. Prove that lines $D I$ and $P Q$ meet on the external bisector of angle $B A C$.

$\angle(a, b)$ denotes the directed angle between lines $a$ and $b$, measured modulo $\pi$.

Solution:
Step 1. The external bisector of $\angle B A C$ is the line through $A$ perpendicular to $I A$. Let $D I$ meet this line at $L$ and let $D I$ meet $\omega$ at $K$. Let $N$ be the midpoint of $E F$, which lies on $I A$ and is the pole of line $A L$ with respect to $\omega$. Since $A N \cdot A I=A E^{2}=A R \cdot A P$, the points $R$, $N, I$, and $P$ are concyclic. As $I R=I P$, the line $N I$ is the external bisector of $\angle P N R$, so $P N$ meets $\omega$ again at the point symmetric to $R$ with respect to $A N$ - i.e. at $K$.

Let $D N$ cross $\omega$ again at $S$. Opposite sides of any quadrilateral inscribed in the circle $\omega$ meet on the polar line of the intersection of the diagonals with respect to $\omega$. Since $L$ lies on the polar line $A L$ of $N$ with respect to $\omega$, the line $P S$ must pass through $L$. Thus it suffices to prove that the points $S, Q$, and $P$ are collinear.
<img_3407>

Step 2. Let $\Gamma$ be the circumcircle of $\triangle B I C$. Notice that

$$
\begin{aligned}
\angle(B Q, Q C)=\angle & (B Q, Q P)+\angle(P Q, Q C)=\angle(B F, F P)+\angle(P E, E C) \\
& =\angle(E F, E P)+\angle(F P, F E)=\angle(F P, E P)=\angle(D F, D E)=\angle(B I, I C),
\end{aligned}
$$

so $Q$ lies on $\Gamma$. Let $Q P$ meet $\Gamma$ again at $T$. It will now suffice to prove that $S, P$, and $T$ are collinear. Notice that $\angle(B I, I T)=\angle(B Q, Q T)=\angle(B F, F P)=\angle(F K, K P)$. Note $F D \perp F K$ and $F D \perp B I$ so $F K \| B I$ and hence $I T$ is parallel to the line $K N P$. Since $D I=I K$, the line $I T$ crosses $D N$ at its midpoint $M$.

Step 3. Let $F^{\prime}$ and $E^{\prime}$ be the midpoints of $D E$ and $D F$, respectively. Since $D E^{\prime} \cdot E^{\prime} F=D E^{\prime 2}=$ $B E^{\prime} \cdot E^{\prime} I$, the point $E^{\prime}$ lies on the radical axis of $\omega$ and $\Gamma$; the same holds for $F^{\prime}$. Therefore, this radical axis is $E^{\prime} F^{\prime}$, and it passes through $M$. Thus $I M \cdot M T=D M \cdot M S$, so $S, I, D$, and $T$ are concyclic. This shows $\angle(D S, S T)=\angle(D I, I T)=\angle(D K, K P)=\angle(D S, S P)$, whence the points $S, P$, and $T$ are collinear, as desired.



<img_3779>
Namely, we introduce the points $K, L, N$, and $S$, and show that the triples $(P, N, K)$ and $(P, S, L)$ are collinear. We conclude that $K$ and $R$ are symmetric in $A I$, and reduce the problem statement to showing that $P, Q$, and $S$ are collinear.

Step 1. Let $A R$ meet the circumcircle $\Omega$ of $A B C$ again at $X$. The lines $A R$ and $A K$ are isogonal in the angle $B A C$; it is well known that in this case $X$ is the tangency point of $\Omega$ with the $A$-mixtilinear circle. It is also well known that for this point $X$, the line $X I$ crosses $\Omega$ again at the midpoint $M^{\prime}$ of arc $B A C$.

Step 2. Denote the circles $B F P$ and $C E P$ by $\Omega_{B}$ and $\Omega_{C}$, respectively. Let $\Omega_{B}$ cross $A R$ and $E F$ again at $U$ and $Y$, respectively. We have

$$
\angle(U B, B F)=\angle(U P, P F)=\angle(R P, P F)=\angle(R F, F A),
$$

so $U B \| R F$.

<img_3555>

Next, we show that the points $B, I, U$, and $X$ are concyclic. Since

$$
\angle(U B, U X)=\angle(R F, R X)=\angle(A F, A R)+\angle(F R, F A)=\angle\left(M^{\prime} B, M^{\prime} X\right)+\angle(D R, D F),
$$

it suffices to prove $\angle(I B, I X)=\angle\left(M^{\prime} B, M^{\prime} X\right)+\angle(D R, D F)$, or $\angle\left(I B, M^{\prime} B\right)=\angle(D R, D F)$. But both angles equal $\angle(C I, C B)$, as desired. (This is where we used the fact that $M^{\prime}$ is the midpoint of $\operatorname{arc} B A C$ of $\Omega$.)

It follows now from circles $B U I X$ and $B P U F Y$ that

$$
\begin{aligned}
\angle(I U, U B)=\angle(I X, B X)=\angle\left(M^{\prime} X, B X\right)= & \frac{\pi-\angle A}{2} \\
& =\angle(E F, A F)=\angle(Y F, B F)=\angle(Y U, B U),
\end{aligned}
$$

so the points $Y, U$, and $I$ are collinear.

Let $E F$ meet $B C$ at $W$. We have

$$
\angle(I Y, Y W)=\angle(U Y, F Y)=\angle(U B, F B)=\angle(R F, A F)=\angle(C I, C W)
$$

so the points $W, Y, I$, and $C$ are concyclic.



Similarly, if $V$ and $Z$ are the second meeting points of $\Omega_{C}$ with $A R$ and $E F$, we get that the 4-tuples $(C, V, I, X)$ and $(B, I, Z, W)$ are both concyclic.

Step 3. Let $Q^{\prime}=C Y \cap B Z$. We will show that $Q^{\prime}=Q$.

First of all, we have

$$
\begin{aligned}
& \angle\left(Q^{\prime} Y, Q^{\prime} B\right)=\angle(C Y, Z B)=\angle(C Y, Z Y)+\angle(Z Y, B Z) \\
& \quad=\angle(C I, I W)+\angle(I W, I B)=\angle(C I, I B)=\frac{\pi-\angle A}{2}=\angle(F Y, F B)
\end{aligned}
$$

so $Q^{\prime} \in \Omega_{B}$. Similarly, $Q^{\prime} \in \Omega_{C}$. Thus $Q^{\prime} \in \Omega_{B} \cap \Omega_{C}=\{P, Q\}$ and it remains to prove that $Q^{\prime} \neq P$. If we had $Q^{\prime}=P$, we would have $\angle(P Y, P Z)=\angle\left(Q^{\prime} Y, Q^{\prime} Z\right)=\angle(I C, I B)$. This would imply

$$
\angle(P Y, Y F)+\angle(E Z, Z P)=\angle(P Y, P Z)=\angle(I C, I B)=\angle(P E, P F)
$$

so circles $\Omega_{B}$ and $\Omega_{C}$ would be tangent at $P$. That is excluded in the problem conditions, so $Q^{\prime}=Q$.

<img_3410>

Step 4. Now we are ready to show that $P, Q$, and $S$ are collinear.

Notice that $A$ and $D$ are the poles of $E W$ and $D W$ with respect to $\omega$, so $W$ is the pole of $A D$. Hence, $W I \perp A D$. Since $C I \perp D E$, this yields $\angle(I C, W I)=\angle(D E, D A)$. On the other hand, $D A$ is a symmedian in $\triangle D E F$, so $\angle(D E, D A)=\angle(D N, D F)=\angle(D S, D F)$. Therefore,

$$
\begin{aligned}
\angle(P S, P F)=\angle(D S, D F)=\angle(D E, D A)= & \angle(I C, I W) \\
& =\angle(Y C, Y W)=\angle(Y Q, Y F)=\angle(P Q, P F)
\end{aligned}
$$

which yields the desired collinearity.