Subfield: Geometry


Question: Let $A B C D$ be a convex quadrilateral, and let $P, Q, R$, and $S$ be points on the sides $A B, B C, C D$, and $D A$, respectively. Let the line segments $P R$ and $Q S$ meet at $O$. Suppose that each of the quadrilaterals $A P O S, B Q O P, C R O Q$, and $D S O R$ has an incircle. Prove that the lines $A C, P Q$, and $R S$ are either concurrent or parallel to each other.

Solution:
Denote by $\gamma_{A}, \gamma_{B}, \gamma_{C}$, and $\gamma_{D}$ the incircles of the quadrilaterals $A P O S, B Q O P$, $C R O Q$, and $D S O R$, respectively.

We start with proving that the quadrilateral $A B C D$ also has an incircle which will be referred to as $\Omega$. Denote the points of tangency as in Figure 1. It is well-known that $Q Q_{1}=O O_{1}$ (if $B C \| P R$, this is obvious; otherwise, one may regard the two circles involved as the incircle and an excircle of the triangle formed by the lines $O Q, P R$, and $B C$ ). Similarly, $O O_{1}=P P_{1}$. Hence we have $Q Q_{1}=P P_{1}$. The other equalities of segment lengths marked in Figure 1 can be proved analogously. These equalities, together with $A P_{1}=A S_{1}$ and similar ones, yield $A B+C D=A D+B C$, as required.

<img_3437>

Figure 1

Next, let us draw the lines parallel to $Q S$ through $P$ and $R$, and also draw the lines parallel to $P R$ through $Q$ and $S$. These lines form a parallelogram; denote its vertices by $A^{\prime}, B^{\prime}, C^{\prime}$, and $D^{\prime}$ as shown in Figure 2.

Since the quadrilateral $A P O S$ has an incircle, we have $A P-A S=O P-O S=A^{\prime} S-A^{\prime} P$. It is well-known that in this case there also exists a circle $\omega_{A}$ tangent to the four rays $A P$, $A S, A^{\prime} P$, and $A^{\prime} S$. It is worth mentioning here that in case when, say, the lines $A B$ and $A^{\prime} B^{\prime}$ coincide, the circle $\omega_{A}$ is just tangent to $A B$ at $P$. We introduce the circles $\omega_{B}, \omega_{C}$, and $\omega_{D}$ in a similar manner.

Assume that the radii of the circles $\omega_{A}$ and $\omega_{C}$ are different. Let $X$ be the center of the homothety having a positive scale factor and mapping $\omega_{A}$ to $\omega_{C}$.

Now, Monge's theorem applied to the circles $\omega_{A}, \Omega$, and $\omega_{C}$ shows that the points $A, C$, and $X$ are collinear. Applying the same theorem to the circles $\omega_{A}, \omega_{B}$, and $\omega_{C}$, we see that the points $P, Q$, and $X$ are also collinear. Similarly, the points $R, S$, and $X$ are collinear, as required.

If the radii of $\omega_{A}$ and $\omega_{C}$ are equal but these circles do not coincide, then the degenerate version of the same theorem yields that the three lines $A C, P Q$, and $R S$ are parallel to the line of centers of $\omega_{A}$ and $\omega_{C}$.

Finally, we need to say a few words about the case when $\omega_{A}$ and $\omega_{C}$ coincide (and thus they also coincide with $\Omega, \omega_{B}$, and $\omega_{D}$ ). It may be regarded as the limit case in the following manner.



<img_3266>

Figure 2

Let us fix the positions of $A, P, O$, and $S$ (thus we also fix the circles $\omega_{A}, \gamma_{A}, \gamma_{B}$, and $\gamma_{D}$ ). Now we vary the circle $\gamma_{C}$ inscribed into $\angle Q O R$; for each of its positions, one may reconstruct the lines $B C$ and $C D$ as the external common tangents to $\gamma_{B}, \gamma_{C}$ and $\gamma_{C}, \gamma_{D}$ different from $P R$ and $Q S$, respectively. After such variation, the circle $\Omega$ changes, so the result obtained above may be applied.
Applying Menelaus' theorem to $\triangle A B C$ with the line $P Q$ and to $\triangle A C D$ with the line $R S$, we see that the line $A C$ meets $P Q$ and $R S$ at the same point (possibly at infinity) if and only if

$$
\frac{A P}{P B} \cdot \frac{B Q}{Q C} \cdot \frac{C R}{R D} \cdot \frac{D S}{S A}=1
\tag{1}
$$

So, it suffices to prove (1).

We start with the following result.

Lemma 1. Let $E F G H$ be a circumscribed quadrilateral, and let $M$ be its incenter. Then

$$
\frac{E F \cdot F G}{G H \cdot H E}=\frac{F M^{2}}{H M^{2}}
$$

Proof. Notice that $\angle E M H+\angle G M F=\angle F M E+\angle H M G=180^{\circ}, \angle F G M=\angle M G H$, and $\angle H E M=\angle M E F$ (see Figure 3). By the law of sines, we get

$$
\frac{E F}{F M} \cdot \frac{F G}{F M}=\frac{\sin \angle F M E \cdot \sin \angle G M F}{\sin \angle M E F \cdot \sin \angle F G M}=\frac{\sin \angle H M G \cdot \sin \angle E M H}{\sin \angle M G H \cdot \sin \angle H E M}=\frac{G H}{H M} \cdot \frac{H E}{H M} .
$$



<img_3451>

Figure 3

<img_4017>

Figure 4

We denote by $I, J, K$, and $L$ the incenters of the quadrilaterals $A P O S, B Q O P, C R O Q$, and $D S O R$, respectively. Applying Lemma 1 to these four quadrilaterals we get

$$
\frac{A P \cdot P O}{O S \cdot S A} \cdot \frac{B Q \cdot Q O}{O P \cdot P B} \cdot \frac{C R \cdot R O}{O Q \cdot Q C} \cdot \frac{D S \cdot S O}{O R \cdot R D}=\frac{P I^{2}}{S I^{2}} \cdot \frac{Q J^{2}}{P J^{2}} \cdot \frac{R K^{2}}{Q K^{2}} \cdot \frac{S L^{2}}{R L^{2}}
$$

which reduces to

$$
\frac{A P}{P B} \cdot \frac{B Q}{Q C} \cdot \frac{C R}{R D} \cdot \frac{D S}{S A}=\frac{P I^{2}}{P J^{2}} \cdot \frac{Q J^{2}}{Q K^{2}} \cdot \frac{R K^{2}}{R L^{2}} \cdot \frac{S L^{2}}{S I^{2}}
\tag{2}
$$

Next, we have $\angle I P J=\angle J O I=90^{\circ}$, and the line $O P$ separates $I$ and $J$ (see Figure 4). This means that the quadrilateral $I P J O$ is cyclic. Similarly, we get that the quadrilateral $J Q K O$ is cyclic with $\angle J Q K=90^{\circ}$. Thus, $\angle Q K J=\angle Q O J=\angle J O P=\angle J I P$. Hence, the right triangles $I P J$ and $K Q J$ are similar. Therefore, $\frac{P I}{P J}=\frac{Q K}{Q J}$. Likewise, we obtain $\frac{R K}{R L}=\frac{S I}{S L}$. These two equations together with (2) yield (1).
We present another approach for showing this equation:

$$
\frac{A P}{P B} \cdot \frac{B Q}{Q C} \cdot \frac{C R}{R D} \cdot \frac{D S}{S A}=1 \tag{1}
$$

Lemma 2. Let $E F G H$ and $E^{\prime} F^{\prime} G^{\prime} H^{\prime}$ be circumscribed quadrilaterals such that $\angle E+\angle E^{\prime}=$ $\angle F+\angle F^{\prime}=\angle G+\angle G^{\prime}=\angle H+\angle H^{\prime}=180^{\circ}$. Then

$$
\frac{E F \cdot G H}{F G \cdot H E}=\frac{E^{\prime} F^{\prime} \cdot G^{\prime} H^{\prime}}{F^{\prime} G^{\prime} \cdot H^{\prime} E^{\prime}}
$$

Proof. Let $M$ and $M^{\prime}$ be the incenters of $E F G H$ and $E^{\prime} F^{\prime} G^{\prime} H^{\prime}$, respectively. We use the notation $[X Y Z]$ for the area of a triangle $X Y Z$.

Taking into account the relation $\angle F M E+\angle F^{\prime} M^{\prime} E^{\prime}=180^{\circ}$ together with the analogous ones, we get

$$
\begin{aligned}
\frac{E F \cdot G H}{F G \cdot H E} & =\frac{[M E F] \cdot[M G H]}{[M F G] \cdot[M H E]}=\frac{M E \cdot M F \cdot \sin \angle F M E \cdot M G \cdot M H \cdot \sin \angle H M G}{M F \cdot M G \cdot \sin \angle G M F \cdot M H \cdot M E \cdot \sin \angle E M H} \\
& =\frac{M^{\prime} E^{\prime} \cdot M^{\prime} F^{\prime} \cdot \sin \angle F^{\prime} M^{\prime} E^{\prime} \cdot M^{\prime} G^{\prime} \cdot M^{\prime} H^{\prime} \cdot \sin \angle H^{\prime} M^{\prime} G^{\prime}}{M^{\prime} F^{\prime} \cdot M^{\prime} G^{\prime} \cdot \sin \angle G^{\prime} M^{\prime} F^{\prime} \cdot M^{\prime} H^{\prime} \cdot M^{\prime} E^{\prime} \cdot \sin \angle E^{\prime} M^{\prime} H^{\prime}}=\frac{E^{\prime} F^{\prime} \cdot G^{\prime} H^{\prime}}{F^{\prime} G^{\prime} \cdot H^{\prime} E^{\prime}}
\end{aligned}
$$

<img_3602>

Figure 6

Denote by $h$ the homothety centered at $O$ that maps the incircle of $C R O Q$ to the incircle of $A P O S$. Let $Q^{\prime}=h(Q), C^{\prime}=h(C), R^{\prime}=h(R), O^{\prime}=O, S^{\prime}=S, A^{\prime}=A$, and $P^{\prime}=P$. Furthermore, define $B^{\prime}=A^{\prime} P^{\prime} \cap C^{\prime} Q^{\prime}$ and $D^{\prime}=A^{\prime} S^{\prime} \cap C^{\prime} R^{\prime}$ as shown in Figure 6. Then

$$
\frac{A P \cdot O S}{P O \cdot S A}=\frac{A^{\prime} P^{\prime} \cdot O^{\prime} S^{\prime}}{P^{\prime} O^{\prime} \cdot S^{\prime} A^{\prime}}
$$

holds trivially. We also have

$$
\frac{C R \cdot O Q}{R O \cdot Q C}=\frac{C^{\prime} R^{\prime} \cdot O^{\prime} Q^{\prime}}{R^{\prime} O^{\prime} \cdot Q^{\prime} C^{\prime}}
$$

by the similarity of the quadrilaterals $C R O Q$ and $C^{\prime} R^{\prime} O^{\prime} Q^{\prime}$.



Next, consider the circumscribed quadrilaterals $B Q O P$ and $B^{\prime} Q^{\prime} O^{\prime} P^{\prime}$ whose incenters lie on different sides of the quadrilaterals' shared side line $O P=O^{\prime} P^{\prime}$. Observe that $B Q \| B^{\prime} Q^{\prime}$ and that $B^{\prime}$ and $Q^{\prime}$ lie on the lines $B P$ and $Q O$, respectively. It is now easy to see that the two quadrilaterals satisfy the hypotheses of Lemma 2 . Thus, we deduce

$$
\frac{B Q \cdot O P}{Q O \cdot P B}=\frac{B^{\prime} Q^{\prime} \cdot O^{\prime} P^{\prime}}{Q^{\prime} O^{\prime} \cdot P^{\prime} B^{\prime}}
$$

Similarly, we get

$$
\frac{D S \cdot O R}{S O \cdot R D}=\frac{D^{\prime} S^{\prime} \cdot O^{\prime} R^{\prime}}{S^{\prime} O^{\prime} \cdot R^{\prime} D^{\prime}}
$$

Multiplying these four equations, we obtain

$$
\frac{A P}{P B} \cdot \frac{B Q}{Q C} \cdot \frac{C R}{R D} \cdot \frac{D S}{S A}=\frac{A^{\prime} P^{\prime}}{P^{\prime} B^{\prime}} \cdot \frac{B^{\prime} Q^{\prime}}{Q^{\prime} C^{\prime}} \cdot \frac{C^{\prime} R^{\prime}}{R^{\prime} D^{\prime}} \cdot 
\frac{D^{\prime} S^{\prime}}{S^{\prime} A^{\prime}}
\tag{3}
$$

Finally, we apply BRianchon's theorem to the circumscribed hexagon $A^{\prime} P^{\prime} R^{\prime} C^{\prime} Q^{\prime} S^{\prime}$ and deduce that the lines $A^{\prime} C^{\prime}, P^{\prime} Q^{\prime}$, and $R^{\prime} S^{\prime}$ are either concurrent or parallel to each other. So, by Menelaus' theorem, we obtain

$$
\frac{A^{\prime} P^{\prime}}{P^{\prime} B^{\prime}} \cdot \frac{B^{\prime} Q^{\prime}}{Q^{\prime} C^{\prime}} \cdot \frac{C^{\prime} R^{\prime}}{R^{\prime} D^{\prime}} \cdot \frac{D^{\prime} S^{\prime}}{S^{\prime} A^{\prime}}=1
$$

This equation together with (3) yield (1).