Slovenija 2008

This problem has several interesting solutions. A nicely drawn figure suggests that quadrilaterals $AB{H}_1{H}_2$ and $H_2{H}_{3}DE$ are parallelograms. We will prove this fact.



Triangles $ACD$, $BCD$ and $BDE$ are obtuse, so let us first make a short detour examining some properties of obtuse triangles. Let $KLM$ be a triangle with the obtuse angle $\angle KLM$ and let $H$ be the orthocentre. The point $H$ lies outside the triangle $KLM$ and on a different side of the line $KL$ than $M$. Let $L^{\prime }$ and $K^{\prime }$ be the feet of the altitudes from $L$ and $K$. Since $\angle M{L}^{\prime }H=\angle M{K}^{\prime }H$, points $M$, $L^{\prime }$, $K^{\prime }$, $H$ are concyclic and $\angle KML=\angle LHK$. The following fact is also worth mentioning: since $K^{\prime }KM$ is a right triangle the angle $K^{\prime }KM$ is acute. Points $K^{\prime }$, $K$, $L^{\prime }$, $L$ are concyclic, so $\angle HLM=\pi -\angle K^{\prime }KL$, and the angle $\angle HLM$ is obtuse.



Let us now return to the original problem. Here to help us are the facts stated above and a new and less cluttered figure. Since $|AC|=|BD|$, the inscribed angles over these two chords are equal, so $\angle ADC=\angle BAD$, which implies $\angle CBA=\pi -\angle ADC=\pi -\angle BAD=\angle DCB$, so $AD$ and $BC$ are parallel and $ABCD$ is an isosceles trapezoid. Points $A$, $B$, $C$ and $D$ are concyclic, so $\angle CAD=\angle CBD$. From the above consideration we see that $\angle D{H}_{1}C=\angle CAD$ and $\angle D{H}_{2}C=\angle CBD$, so $\angle D{H}_{1}C=\angle D{H}_{2}C$ and points $C$, $H_1$, $H_2$, $D$ are concyclic. But $H_{1}C$ is perpendicular to $AD$, $H_{2}D$ is perpendicular to $BC$ and lines $BC$ and $AD$ are parallel, so $H_{1}C$ and $H_{2}D$ are parallel as well. The cyclic quadrilateral $C{H}_1{H}_{2}D$ has a pair of parallel sides $H_{1}C$ and $H_{2}D$ and is therefore an isosceles trapezoid. We have seen that $C{H}_1{H}_{2}D$ and $ABCD$ are isosceles trapezoids, so we have $|H_1{H}_2|=|CD|=|AB|$. The altitude $A{H}_1$ is perpendicular to the side $CD$ and so is the altitude $B{H}_2$, hence $A{H}_1$ and $B{H}_2$ are parallel. The quadrilateral $AB{H}_2{H}_1$ has a pair of parallel sides ($B{H}_2$ and $A{H}_1$) and a pair of sides of equal length ($AB$ and $H_1{H}_2$). Thus, it is either an isosceles trapezoid or a parallelogram. Let us demonstrate that it is, in fact, the latter. Lines $A{H}_1$ and $B{H}_2$ are parallel, so the point $H_1$ lies on the side of the line $B{H}_2$ not containing $C$. As we have shown above the angle $\angle DC{H}_1$ is obtuse, so the angle $\angle C{H}_1{H}_2$, which is equal to $\angle DC{H}_1$, is also obtuse. But $\angle A{H}_1{H}_2=\angle A{H}_{1}C+\angle C{H}_1{H}_2>\frac{\pi }{2}$, so the angle $\angle A{H}_1{H}_2$ is obtuse. On the other hand, the angle $\angle CA{H}_1$ is acute, so $\angle BA{H}_1=\angle CA{H}_1-\angle CAB$ is acute as well (points $B$ and $H_1$ lie on the same side of the line $AC$, because neither of them lies on the same side as $D$). We have shown that $\angle BA{H}_1$ is acute and $\angle A{H}_1{H}_2$ is obtuse, so the quadrilateral $AB{H}_2{H}_1$ is a parallelogram. Hence, $H_1{H}_2$ is parallel to $AB$. Similarly, we show that $H_2{H}_3$ is parallel to $ED$. If we wish to prove that $\angle H_1{H}_2{H}_3=\frac{\pi }{2}$, it suffices to see that lines $AB$ and $ED$ intersect at a right angle. Denote the intersection of $AB$ and $ED$ by $T$. Since $\angle BET=\angle BED=\frac{1}{2}\angle BOD=30^{\circ }$ and $\angle TBE=\pi -\angle EBA=\pi -\angle ECA=\pi -120^{\circ }=60^{\circ }$, we have $\angle ETB=\pi -\angle TBE-\angle BET=90^{\circ }$. So $\angle ATE=90^{\circ }$ and this concludes our proof.



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Alternative solution.

Since $\angle C{H}_{3}B=\angle BEC=\angle BDC=\angle C{H}_{2}B$, points $C$, $H_3$, $H_2$, $B$ are concyclic. Similarly, we show that the quadrilateral $C{H}_1{H}_{2}D$ is cyclic.

We wish to see that $\angle H_3{H}_2{H}_1=\frac{\pi }{2}$. But $\angle H_3{H}_2{H}_1=\angle H_3{H}_{2}C+\angle C{H}_2{H}_1$ so $\angle H_3{H}_{2}C=\angle H_{3}BC$ and $\angle C{H}_2{H}_1=CD{H}_1$. Let $F$ be the intersection of lines $B{H}_3$ and $EC$ and let $G$ be the intersection of $AC$ and $D{H}_2$. We wish to show that $\angle FBC+\angle CDG=\frac{\pi }{2}$. We will not need the orthocentres this time, so let us draw another figure. Denote $\alpha =\angle FBC$ and $\beta =\angle CDG$. We have to show that $\alpha +\beta =\frac{\pi }{2}$. We have $\angle ECB=\frac{\pi }{2}+\alpha$ and $\angle DCA=\frac{\pi }{2}+\beta$. So, $\alpha +\beta =\angle ECB+\angle DCA-\pi =2\angle ECA+\angle DCE+\angle ACB-\pi =\frac{\pi }{3}+\angle DCE+\angle ACB$. In particular, $\angle ACB=\angle CAD$, because the quadrilateral $ABCD$ is an isosceles trapezoid, which implies $\angle DCE+\angle ACB=\angle DAE+\angle CAD=\angle CAE=\frac{1}{2}\angle COE=30^{\circ }$, so $\alpha +\beta =60^{\circ }+30^{\circ }=90^{\circ }$ and $H_1{H}_2{H}_3$ is a right triangle.



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Alternative solution.

Let the system of coordinates be centred at the circumcentre $O$. Since $H_1$ is the orthocentre of the triangle $ACD$ we have $\overrightarrow{O{H}_1}=\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OD}$. Similarly, $$\overrightarrow{O{H}_2}=\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}\text{and}\overrightarrow{O{H}_3}=\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OE}.$$ So, $\overrightarrow{H_1{H}_2}=\overrightarrow{O{H}_2}-\overrightarrow{O{H}_1}=\overrightarrow{OB}-\overrightarrow{OA}$ and $\overrightarrow{H_2{H}_3}=\overrightarrow{O{H}_3}-\overrightarrow{O{H}_2}=\overrightarrow{OE}-\overrightarrow{OD}$. The inner product of these two vectors is $$\begin{array}{rl}\overrightarrow{H_1{H}_2}\cdot \overrightarrow{H_2{H}_3}& =(\overrightarrow{OB}-\overrightarrow{OA})(\overrightarrow{OE}-\overrightarrow{OD})\\ & =\overrightarrow{OB}\cdot \overrightarrow{OE}-\overrightarrow{OB}\cdot \overrightarrow{OD}-\overrightarrow{OA}\cdot \overrightarrow{OE}+\overrightarrow{OA}\cdot \overrightarrow{OD}\\ & =|\overrightarrow{OB}|\cdot |\overrightarrow{OE}|\cos (\angle BOE)-|\overrightarrow{OB}|\cdot |\overrightarrow{OD}|\cos (\angle BOD)\end{array}$$ $$\begin{array}{rl}& -|OA|\cdot |OE|\cos (\angle AOE)+|OA|\cdot |OD|\cos (\angle AOD)\\ & =|OA{|}^2(\cos (\angle BOE)-\cos (\angle BOD)-\cos (\angle AOE)+\cos (\angle AOD)).\end{array}$$ Here, we used the fact that $O$ is the centre of the circle containing points $A$, $B$, $D$ and $E$. Triangles $AOC$ and $COE$ are equilateral, so $\angle AOE=120^{\circ }$. The triangle $BOD$ is also equilateral, so $\angle BOD=60^{\circ }$. Thus, $$\begin{array}{rl}\overrightarrow{H_1{H}_2}\cdot \overrightarrow{H_2{H}_3}& =|OA{|}^2(\cos (\angle BOE)-\cos (60^{\circ })-\cos (120^{\circ })+\cos (\angle AOD))\\ & =|OA{|}^2(\cos (\angle BOE)+\cos (\angle AOD))\\ & =2|OA{|}^2\cos \left(\frac{\angle BOE+\angle AOD}{2}\right)\cos \left(\frac{\angle BOE-\angle AOD}{2}\right).\end{array}$$ Verify that $\angle BOE+\angle AOD=2\angle BOD+\angle DOE+\angle AOB$, which implies $$\begin{array}{rl}\frac{\angle BOE+\angle AOD}{2}& =\angle BOD+\frac{\angle DOE+\angle AOB}{2}=60^{\circ }+\angle DCE+\angle ACB\\ & =60^{\circ }+\angle DCB-\angle ECA=60^{\circ }+(180^{\circ }-30^{\circ })-120^{\circ }=90^{\circ }=\frac{\pi }{2},\end{array}$$ so $\cos (\frac{\angle BOE+\angle AOD}{2})=0$ and $\overrightarrow{H_1{H}_2}\cdot \overrightarrow{H_2{H}_3}=0$. Hence $H_1{H}_2{H}_3$ is a right triangle.