Taiwan IMO Selection Camp

Let $\omega_1, \omega_2$ be the inscribed circles of quadrilaterals $ABED$, $AECD$ respectively, and let $O_1, O_2$ be the centers of $\omega_1, \omega_2$ respectively. A sufficient condition for the existence of a point $F$ as stated in the problem is that $\omega_1, \omega_2$ are also the inscribed circles of quadrilaterals $ABCF$, $BCDF$ respectively.

From $B$ draw the tangent line to $\omega_2$ other than $BC$, and from $C$ draw the tangent line to $\omega_1$ other than $BC$; let these two tangent lines meet side $AD$ at points $F_1, F_2$ respectively. We need to prove that $F_1 = F_2$ if and only if $AB \parallel CD$.

Lemma: Let $\omega_1, \omega_2$ be two circles with centers $O_1, O_2$ respectively, both inscribed in the same angle, and let the vertex of this angle be $O$. Let points $P, S$ lie on one side of angle $O$, and points $Q, R$ lie on the other side of angle $O$, and suppose $\omega_1$ is the inscribed circle of triangle $PQO$, and $\omega_2$ is the excircle of triangle $RSO$ opposite to angle $O$. Let $p = OO_1 \cdot OO_2$. Then exactly one of the following relations holds:

$$OP \cdot OR < p < OQ \cdot OS, \quad OP \cdot OR > p > OQ \cdot OS, \quad OP \cdot OR = p = OQ \cdot OS.$$

Proof of the lemma: Let $\angle OPO_1 = \alpha, \angle OQO_1 = \beta, \angle OO_2R = \gamma, \angle OO_2S = \delta$, $\angle POQ = 2\varphi$. Since segments $PO_1, QO_1, RO_2, SO_2$ are respectively the internal or external angle bisectors of triangles $PQO, RSO$, we have
$$u + v = x + y(= 90^\circ - \varphi). \qquad (1)$$
By the law of sines,
$$\frac{OP}{OO_1} = \frac{\sin(u + \varphi)}{\sin u} \quad \text{and} \quad \frac{OO_2}{OR} = \frac{\sin(x + \varphi)}{\sin x}.$$
Since $x, u, \varphi$ are all acute angles,
$$\begin{aligned} OP \cdot OR \ge p & \Leftrightarrow \frac{OP}{OO_1} \ge \frac{OO_2}{OR} & \Leftrightarrow \sin x \sin(u + \varphi) \ge \sin u \sin(x + \varphi) \\ & \Leftrightarrow \sin(x - u) \ge 0 & \Leftrightarrow x \ge u. \end{aligned}$$
Thus $OP \cdot OR \ge p$ is equivalent to $x \ge u$, and $OP \cdot OR = p$ if and only if $x = u$.
Similarly, we can prove that $p \ge OQ \cdot OS$ is equivalent to $v \ge y$, and $p = OQ \cdot OS$ if and only if $v = y$. On the other hand, from (1) we know that $x \ge u$ and $v \ge y$ are equivalent, and $x = u$ is equivalent to $v = y$. This proves the lemma.

Returning to the problem itself, we apply the lemma to the following groups of four points: $\{B, E, D, F_1\}$, $\{A, B, C, D\}$, $\{A, E, C, F_2\}$. First suppose $OE \cdot OF_1 > p$; then we obtain
$$OE \cdot OF_1 > p \Rightarrow OB \cdot OD < p \Rightarrow OA \cdot OC > p \Rightarrow OE \cdot OF_2 < p.$$

In other words, $OE \cdot OF_1 > p$ implies
$OB \cdot OD < p < OA \cdot OC \quad \text{and} \quad OE \cdot OF_1 > p > OE \cdot OF_2$.
Similarly, if we suppose $OE \cdot OF_1 < p$, then we obtain
$OB \cdot OD > p > OA \cdot OC \quad \text{and} \quad OE \cdot OF_1 < p < OE \cdot OF_2$.
In these cases, $F_1 \neq F_2$, and moreover $OB \cdot OD \neq OA \cdot OC$, so $AB$ and $CD$ are not parallel.
Finally, there remains the case $OE \cdot OF_1 = p$. In this case, the lemma implies that $OB \cdot OD = p = OA \cdot OC$ and $OE \cdot OF_1 = p = OE \cdot OF_2$. Therefore $F_1 = F_2$ and $AB \parallel CD$.