Selection Examinations for the IMO, 2012

First suppose that the lines $PQ$ and $RS$ intersect. Denote $T$ their point of intersection. Due to symmetry we may assume that $Q$ lies between $P$ and $T$ and $R$ lies between $S$ and $T$. We shall prove that the lines $EF$, $I_1I_2$ and $AC$ pass through the point $T$.

In the construction, $I_1$ is the intersection point of the bisectors of two angles in the triangle $PTS$, hence the center of the inscribed circle of the triangle $PTS$. The point $I_1$ lies on the bisector of the angle $\angle STP$. Similarly, $I_2$ is the intersection point of the bisectors of the exterior angles at $QR$ in the triangle $QTR$, hence the center of the inscribed circle to the side $QR$ of the triangle $QTR$. The point $I_2$ lies on the bisector of the angle $\angle RTQ$. Because $S$, $R$ and $T$ are collinear, and $P$, $Q$ and $T$ are collinear, $I_1$, $I_2$ and $T$ must also be collinear.

Denote $T'$ the intersection point of the lines $EF$ and $PQ$, and denote $T''$ the intersection point of the lines $EF$ and $RS$. According to the Menelaus' theorem, we have
$$\frac{BP}{PF} \cdot \frac{FT'}{T'E} \cdot \frac{EQ}{QB} = -1 \quad \text{and} \quad \frac{DR}{RF} \cdot \frac{FT''}{T''E} \cdot \frac{ES}{SD} = -1.$$
Consider the lengths of the tangent segments to get $|BP| = |BQ|$, $|DS| = |DR|$, $|FP| = |FR|$ and $|EQ| = |ES|$. Because of the above equalities and due to the obvious fact that $E$ and $F$ lie outside the quadrilateral $ABCD$, we have $\frac{FT'}{T'E} = \frac{|PF|}{|QE|} = \frac{|RF|}{|SE|} = \frac{FT''}{T''E}$. Hence $T' = T''$, and the lines $EF$, $PQ$ and $RS$ intersect in a common point $T \in EF$.

Let $T_1$ be the intersection point of the lines $AC$ and $PQ$, and let $T_2$ be the intersection point of the lines $AC$ and $RS$. We again apply the Menelaus' theorem to get
$$\frac{AP}{PB} \cdot \frac{BQ}{QC} \cdot \frac{CT_1}{T_1A} = -1 \quad \text{and} \quad \frac{AS}{SD} \cdot \frac{DR}{RC} \cdot \frac{CT_2}{T_2A} = -1.$$
Because $P$, $Q$, $R$ and $S$ lie each on a side of the quadrilateral $ABCD$, from the equality of the tangent segments $|AP| = |AS|$, $|BP| = |BQ|$, $|CQ| = |CR|$ and $|DR| = |DS|$ we derive $\frac{CT_1}{T_1A} = -\frac{|QC|}{|PA|} = -\frac{|RC|}{|SA|} = \frac{CT_2}{T_2A}$. Hence, $T_1 = T_2$ and the line $AC$ passes through the point $T$.

Now suppose $PQ \parallel RS$. Because of the equality of the tangent segments, we get $\angle BPQ = \frac{1}{2}(\pi - \angle QBP)$, $\angle SPA = \angle ASP = \frac{1}{2}(\pi - \angle PAS)$ and $\angle RSD = \frac{1}{2}(\pi - \angle SDR)$, hence $\angle QPS = \frac{1}{2}(\angle PAS + \angle QBP)$ and $\angle PSR = \frac{1}{2}(\angle PAS + \angle SDR)$. Since $PQ \parallel RS$, we have $\angle QPS + \angle PSR = \pi$, and hence $\angle BAD + \frac{1}{2}(\angle CBA + \angle ADC) = \pi$. We conclude $\angle BAD = \angle DCB$, which means that the quadrilateral $ABCD$ is a deltoid. The lines $EF$, $I_1I_2$ and $AC$ are thus parallel.