Subfield: Geometry


Question: Let $A B C D$ be a cyclic quadrilateral with circumcentre $O$. Let the internal angle bisectors at $A$ and $B$ meet at $X$, the internal angle bisectors at $B$ and $C$ meet at $Y$, the internal angle bisectors at $C$ and $D$ meet at $Z$, and the internal angle bisectors at $D$ and $A$ meet at $W$. Further, let $A C$ and $B D$ meet at $P$. Suppose that the points $X, Y, Z, W, O$ and $P$ are distinct.

Prove that $O, X, Y, Z$ and $W$ lie on the same circle if and only if $P, X, Y, Z$ and $W$ lie on the same circle.

Solution:
Let $\Omega$ be the circumcircle of the quadrilateral $A B C D$ and let $r$ be its radius.

First, notice that the points $X, Y, Z, W$ are concyclic. Indeed, using oriented (modulo $180^{\circ}$ ) angles,

$$
\angle(X W, X Y)+\angle(Z Y, Z W)=\angle(X A, X B)+\angle(Z C, Z D)=-\frac{\angle A+\angle B}{2}-\frac{\angle C+\angle D}{2}=0 .
$$

Let $\omega$ be the circle passing through these four points. Our goal is to prove that $O \in \omega$ occurs if and only if $P \in \omega$.

Next, we rule out the case when $A B C D$ is a trapezium. Suppose that if $A B C D$ is an isosceles trapezium; without loss of generality say $A B \| C D$. By symmetry, points $X, Z, O$ and $P$ lie on the symmetry axis (that is, the common perpendicular bisector of $A B$ and $C D$ ), therefore they are collinear. By the conditions of the problem, these four points are distinct and $X$ and $Z$ are on $\omega$, so neither $O$, nor $P$ can lie on $\omega$; therefore the problem statement is obvious. From now on we assume that $A B \nparallel C D$ and $B C \nparallel A D$.

<img_3441>



Let $A B$ and $C D$ meet at $Q$, and let $B C$ and $A D$ meet at $R$. Without loss of generality, suppose that $B$ lies between $A$ and $Q$, and $D$ lies between $A$ and $R$. Now we show that line $Q R$ is the radical axis between circles $\Omega$ and $\omega$.

Point $W$ is the intersection point of the bisectors of $\angle A$ and $\angle D$, so in triangle $A D Q$, point $W$ is the incentre. Similary, $B Y$ and $C Y$ are the external bisectors of $\angle Q B C$ and $\angle B C Q$, so in triangle $B C Q$, point $Y$ is the excentre opposite to $Q$. Hence, both $Y$ and $W$ lie on the bisector of $\angle D Q A$.

By $\angle D Q A=180^{\circ}-\angle D-\angle A=\angle B-\angle A$ we get

$$
\angle B Y Q=\angle Y B A-\angle Y Q B=\frac{\angle B}{2}-\frac{\angle D Q A}{2}=\frac{\angle B}{2}-\frac{\angle B-\angle A}{2}=\frac{\angle A}{2}=\angle B A W
$$

so the points $A, B, Y, W$ are concyclic, and therefore $Q A \cdot Q B=Q Y \cdot Q W$; hence, $Q$ has equal power with respect to $\Omega$ and $\omega$. It can be seen similarly that $R$ has equal power with respect to the circles $\Omega$ and $\omega$. Hence, the radical axis of $\Omega$ and $\omega$ is line $Q R$.

Let lines $O P$ and $Q R$ meet at point $S$. It is well-known that with respect to circle $A B C D$, the diagonal triangle $P Q R$ is autopolar. As consequences we have $O P \perp Q R$, and the points $P$ and $S$ are symmetric to circle $A B C D$, so $O S \cdot O P=r^{2}$.

Notice that $P$ and $O$ lie inside $\Omega$, so the polar line $Q R$ lies entirely outside, so $S$ is different from $P$ and $O$. Moreover,

$$
S O \cdot S P=O S \cdot(O S-O P)=O S^{2}-O S \cdot O P=S O^{2}-r^{2}
$$

so $S O \cdot S P$ is equal to the power of $S$ with respect to $\Omega$. Since $S$ lies on the radical axis $Q R$, it has equal power with respect to the two circles; therefore, $S O \cdot S P$ is equal to the power of $S$ with respect to $\omega$. From this, it follows that $O \in \omega$ if and only if $P \in \omega$.
We will prove that the points $X, Y, Z, W, O, P$ lie on a conic section $\Gamma$. Five distinct points uniquely determine a conic section, therefore

$X, Y, Z, W, O$ are concyclic $\Leftrightarrow \Gamma$ is a circle $\Leftrightarrow X, Y, Z, W, P$ are concyclic.

<img_3375>



Let $\alpha=\angle A C B=\angle A D B, \beta=\angle B D C=\angle B A C, \gamma=\angle C A D=\angle C B D$ and $\delta=\angle D B A=$ $\angle D C A$, where of course $\alpha+\beta+\gamma+\delta=180^{\circ}$. For every point $\xi$ in the plane, let $a(\xi), b(\xi), c(\xi)$ and $d(\xi)$ be the signed distances between $\xi$ and the lines $A B, B C, C D$ and $D A$, respectively, such that the quadrilateral lies on the positive sides of these lines.

The equations of the bisectors of $\angle A X W, \angle B X Y, \angle C Y Z$ and $\angle D Z W$ are $a(\xi)-d(\xi)=0$, $b(\xi)-a(\xi)=0, c(\xi)-b(\xi)=0$ and $d(\xi)-c(\xi)=0$, respectively. Notice that the points $X, Y, Z, W$ satisfy the quadratic equations $(a(\xi)-d(\xi))(b(\xi)-c(\xi))=0$ and $(a(\xi)-b(\xi))(c(\xi)-d(\xi))=0$, so

$$
a(\xi) b(\xi)+c(\xi) d(\xi)=a(\xi) c(\xi)+b(\xi) d(\xi)=a(\xi) d(\xi)+b(\xi) c(\xi) \quad \text { for } \xi=X, Y, Z, W \tag{1}
$$

Without loss of generality, suppose that $A B C D$ is inscribed in a unit circle. Then

$$
a(O)=\cos \alpha, \quad b(O)=\cos \beta, \quad c(O)=\cos \gamma, \quad d(O)=\cos \delta \tag{2a}
$$

By $\frac{a(P)}{b(P)}=\frac{B P \cdot \sin \delta}{B P \cdot \sin \gamma}=\frac{1 / \sin \gamma}{1 / \sin \delta}$ and the analogous relations we can see that

$$
a(P)=\frac{k}{\sin \gamma}, \quad b(P)=\frac{k}{\sin \delta}, \quad c(P)=\frac{k}{\sin \alpha}, \quad d(P)=\frac{k}{\sin \beta} \tag{2b}
$$

with some positive number $k$.

Now let $s=\frac{\sin \alpha \sin \beta \sin \gamma \sin \delta}{k^{2}}$; then, by $(2 a)$ and $(2 b)$,

$$
\begin{aligned}
a(O) b(O)-s \cdot a(P) b(P) & =\cos \alpha \cos \beta-\sin \alpha \sin \beta=\cos (\alpha+\beta) \\
c(O) d(O)-s \cdot c(P) d(P) & =\cos (\gamma+\delta)=-\cos (\alpha+\beta) \\
a(O) b(O)+c(O) d(O) & =s \cdot(a(P) b(P)+c(P) d(P))
\end{aligned} \tag{3a}
$$

analogous calculation provides

$$
\begin{aligned}
& a(O) c(O)+b(O) d(O)=s \cdot(a(P) c(P)+b(P) d(P))
\end{aligned} \tag{3b}
$$
$$
\begin{aligned}
& a(O) d(O)+b(O) c(O)=s \cdot(a(P) d(P)+b(P) c(P))
\end{aligned} \tag{3c}
$$

In order to find the equation of the curve $\Gamma$, choose real numbers $u, v, w$, not all zero, such that

$$
\begin{gathered}
u+v+w=0
\end{gathered}
\tag{4}
$$
$$
\begin{gathered}
u \cdot(a(P) b(P)+c(P) d(P))+v \cdot(a(P) c(P)+b(P) d(P))+w \cdot(a(P) d(P)+b(P) c(P))=0
\end{gathered} \tag{5}
$$

This is always possible, because this a system of two homogeneous linear equations with three variables. Then the equation of $\Gamma$ will be

$$
f(\xi)=u \cdot(a(\xi) b(\xi)+c(\xi) d(\xi))+v \cdot(a(\xi) c(\xi)+b(\xi) d(\xi))+w \cdot(a(\xi) d(\xi)+b(\xi) c(\xi))=0 \tag{*}
$$

As can be seen, $f(X)=f(Y)=f(Z)=f(W)=0$ follows from (1) and (4), $f(P)=0$ follows from (5), then $f(O)=s \cdot f(P)=0$ follows from $(3 a-3 c)$. So, the points $X, Y, Z, W, O, P$ all satisfy the equation $f(\xi)=0$.

Notice that $f(B)=u \cdot c(B) d(B)$ and $f(A)=w \cdot b(A) c(A)$; since at least one of $u$ and $w$ is nonzero, either $A$ or $B$ does not satisfy $(*)$. Therefore, the equation cannot degenerate to an identity.

Hence, the equation $f(\xi)=0$ is at most quadratic, it is not an identity, but satisfied by $X, Y, Z, W, O, P$, so these six points lie on a conic section.
Let $A^{\prime}, B^{\prime}, C^{\prime}$ and $D^{\prime}$ be the second intersection of the angle bisectors of $\angle D A C$, $\angle A B C, \angle B C D$ and $\angle C D A$ with the circumcircle of $A B C D$, respectively. Notice that $A^{\prime}$ and $C^{\prime}$ are the midpoints of the two arcs of $B D$ and $B^{\prime}$ and $D^{\prime}$ are the midpoints of arcs corresponding to $A C$ so the lines $A^{\prime} C^{\prime}$ and $B^{\prime} D^{\prime}$ intersect in $O$.

<img_3552>

We will prove that the points $X, Y, Z, W, P$ and $O$ lie on a conic section. This is a more general statement because five points uniquely determine a conic sections so if $X, Y, Z, W$ and $P$ lie on a circle then this circle is the conic section so $O$ also lies on this circle. In the same way if $X, Y, Z$, $W$ and $O$ are on a circle then it also contains $P$.

Let $E$ be the intersection of the lines $A^{\prime} B$ and $C D^{\prime}$. Let $F$ be the intersection of the lines $A^{\prime} D$ and $B^{\prime} C$. Finally, let $G$ be the intesection of $X Y$ and $Z W$.

Now we use Pascal's theorem five times:

1. From $B D D^{\prime} C A A^{\prime}$ we see that the points $P, W$ and $E$ are collinear.
2. From $B B^{\prime} D^{\prime} C C^{\prime} A^{\prime}$ we see that the points $Y, O$ and $E$ are collinear.
3. From $B^{\prime} B D A^{\prime} A C$ we see that the points $X, P$ and $F$ are collinear.
4. From $B^{\prime} D^{\prime} D A^{\prime} C^{\prime} C$ we see that the points $O, Z$ and $F$ are collinear.
5. From $D^{\prime} C B^{\prime} B A^{\prime} D$ we see that the points $E, F$ and $G$ are collinear.

So the opposite sides of the hexagon $P X Y O Z W$ intersect at points $F, G$ and $E$, which are collinear, so by the converse of Pascal's theorem the points $X, Y, Z, W, P$ and $O$ indeed lie on a conic section.