IMO 2006 Shortlisted Problems

Point $E$ is the centre of a homothety $h$ which takes circle ${\omega }_1$ to circle $\omega$. The radii $O_{1}D$ and $OB$ of these circles are parallel as both are perpendicular to line $t$. Also, $O_{1}D$ and $OB$ are on the same side of line $EO$, hence $h$ takes $O_{1}D$ to $OB$. Consequently, points $E$, $D$ and $B$ are collinear. Likewise, points $F$, $D$ and $A$ are collinear as well. Let lines $AE$ and $BF$ intersect at $C$. Since $AF$ and $BE$ are altitudes in triangle $ABC$, their common point $D$ is the orthocentre of this triangle. So $CD$ is perpendicular to $AB$, implying that $C$ lies on line $t$. Note that triangle $ABC$ is acute-angled. We mention the well-known fact that triangles $FEC$ and $ABC$ are similar in ratio $\cos \gamma$, where $\gamma =\angle ACB$. In addition, points $C$, $E$, $D$ and $F$ lie on the circle with diameter $CD$. Let $P$ be the common point of lines $EF$ and $t$. We are going to prove that $P$ lies on line $A{O}_1$. Denote by $N$ the second common point of circle ${\omega }_1$ and $AC$; this is the point of ${\omega }_1$ diametrically opposite to $D$. By Menelaus' theorem for triangle $DCN$, points $A$, $O_1$ and $P$ are collinear if and only if $$\frac{CA}{AN}\cdot \frac{N{O}_1}{O_{1}D}\cdot \frac{DP}{PC}=1.$$ Because $N{O}_1=O_{1}D$, this reduces to $CA/AN=CP/PD$. Let line $t$ meet $AB$ at $K$. Then $CA/AN=CK/KD$, so it suffices to show that $$\begin{array}{c}\frac{CP}{PD}=\frac{CK}{KD}.\end{array}\text{\hspace{1em}(1)}$$ To verify (1), consider the circumcircle $\Omega$ of triangle $ABC$. Draw its diameter $CU$ through $C$, and let $CU$ meet $AB$ at $V$. Extend $CK$ to meet $\Omega$ at $L$. Since $AB$ is parallel to $UL$, we have $\angle ACU=\angle BCL$. On the other hand $\angle EFC=\angle BAC$, $\angle FEC=\angle ABC$ and $EF/AB=\cos \gamma$, as stated above. So reflection in the bisector of $\angle ACB$ followed by a homothety with centre $C$ and ratio $1/\cos \gamma$ takes triangle $FEC$ to triangle $ABC$. Consequently, this transformation takes $CD$ to $CU$, which implies $CP/PD=CV/VU$. Next, we have $KL=KD$, because $D$ is the orthocentre of triangle $ABC$. Hence $CK/KD=CK/KL$. Finally, $CV/VU=CK/KL$ because $AB$ is parallel to $UL$. Relation (1) follows, proving that $P$ lies on line $A{O}_1$. By symmetry, $P$ also lies on line $A{O}_2$ which completes the solution.

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Alternative solution.

We proceed as in the first solution to define a triangle $ABC$ with orthocentre $D$, in which $AF$ and $BE$ are altitudes. Denote by $M$ the midpoint of $CD$. The quadrilateral $CEDF$ is inscribed in a circle with centre $M$, hence $MC=ME=MD=MF$. Consider triangles $ABC$ and $O_1{O}_{2}M$. Lines $O_1{O}_2$ and $AB$ are parallel, both of them being perpendicular to line $t$. Next, $M{O}_1$ is the line of centres of circles ( $CEF$ ) and ${\omega }_1$ whose common chord is $DE$. Hence $M{O}_1$ bisects $\angle DME$ which is the external angle at $M$ in the isosceles triangle $CEM$. It follows that $\angle DM{O}_1=\angle DCA$, so that $M{O}_1$ is parallel to $AC$. Likewise, $M{O}_2$ is parallel to $BC$. Thus the respective sides of triangles $ABC$ and $O_1{O}_{2}M$ are parallel; in addition, these triangles are not congruent. Hence there is a homothety taking $ABC$ to $O_1{O}_{2}M$. The lines $A{O}_1$, $B{O}_2$ and $CM=t$ are concurrent at the centre $Q$ of this homothety. Finally, apply Pappus' theorem to the triples of collinear points $A,O,B$ and $O_2,D,O_1$. The theorem implies that the points $AD\cap O{O}_2=F$, $A{O}_1\cap B{O}_2=Q$ and $O{O}_1\cap BD=E$ are collinear. In other words, line $EF$ passes through the common point $Q$ of $A{O}_1$, $B{O}_2$ and $t$.