Slovenian Mathematical Olympiad, 2008

This problem has several interesting solutions. A nicely drawn figure suggests that quadrilaterals $ABH_1H_2$ and $H_2H_3DE$ 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 $L'$ and $K'$ be the feet of the altitudes from $L$ and $K$. Since $\angle ML'H = \angle MK'H$, points $M$, $L'$, $K'$, $H$ are concyclic and $\angle KML = \angle LHK$.
The following fact is also worth mentioning: since $K'KM$ is a right triangle the angle $K'KM$ is acute. Points $K'$, $K$, $L'$, $L$ are concyclic, so $\angle HLM = \pi - \angle K'KL$, 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 $\angle DH_1C = \angle CAD$ and $\angle DH_2C = \angle CBD$,
so $\angle DH_1C = \angle DH_2C$ and points $C$, $H_1$, $H_2$, $D$ are concyclic. But $H_1C$ is perpendicular to $AD$, $H_2D$ is perpendicular to $BC$ and lines $BC$ and $AD$ are parallel, so $H_1C$ and $H_2D$ are parallel as well. The cyclic quadrilateral $CH_1H_2D$ has a pair of parallel sides $H_1C$ and $H_2D$ and is therefore an isosceles trapezoid.
We have seen that $CH_1H_2D$ and $ABCD$ are isosceles trapezoids, so we have $|H_1H_2| = |CD| = |AB|$. The altitude $AH_1$ is perpendicular to the side $CD$ and so is the altitude $BH_2$, hence $AH_1$ and $BH_2$ are parallel. The quadrilateral $ABH_2H_1$ has a pair of parallel sides ($BH_2$ and $AH_1$) and a pair of sides of equal length ($AB$ and $H_1H_2$). Thus, it is either an isosceles trapezoid or a parallelogram. Let us demonstrate that it is, in fact, the latter.
Lines $AH_1$ and $BH_2$ are parallel, so the point $H_1$ lies on the side of the line $BH_2$ not containing $C$. As we have shown above the angle $\angle DCH_1$ is obtuse, so the angle $\angle CH_1H_2$, which is equal to $\angle DCH_1$, is also obtuse. But $\angle AH_1H_2 = \angle AH_1C + \angle CH_1H_2 > \frac{\pi}{2}$, so the angle $\angle AH_1H_2$ is obtuse. On the other hand, the angle $\angle CAH_1$ is acute, so $\angle BAH_1 = \angle CAH_1 - \angle CAB$ 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 $\angle BAH_1$ is acute and $\angle AH_1H_2$ is obtuse, so the quadrilateral $ABH_2H_1$ is a parallelogram. Hence, $H_1H_2$ is parallel to $AB$.
Similarly, we show that $H_2H_3$ is parallel to $ED$. If we wish to prove that $\angle H_1H_2H_3 = \frac{\pi}{2}$, 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.

Since $\angle CH_3B = \angle BEC = \angle BDC = \angle CH_2B$, points $C$, $H_3$, $H_2$, $B$ are concyclic. Similarly, we show that the quadrilateral $CH_1H_2D$ is cyclic.

We wish to see that $\angle H_3H_2H_1 = \frac{\pi}{2}$. But $\angle H_3H_2H_1 = \angle H_3H_2C + \angle CH_2H_1$ so $\angle H_3H_2C = \angle H_3BC$ and $\angle CH_2H_1 = CDH_1$. Let $F$ be the intersection of lines $BH_3$ and $EC$ and let $G$ be the intersection of $AC$ and $DH_2$. 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 $H_1H_2H_3$ is a right triangle.

Let the system of coordinates be centred at the circumcentre $O$. Since $H_1$ is the orthocentre of the triangle $ACD$ we have $\overrightarrow{OH_1} = \overrightarrow{OA} + \overrightarrow{OC} + \overrightarrow{OD}$. Similarly,
$$\overrightarrow{OH_2} = \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} \quad \text{and} \quad \overrightarrow{OH_3} = \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OE}.$$
So, $\overrightarrow{H_1H_2} = \overrightarrow{OH_2} - \overrightarrow{OH_1} = \overrightarrow{OB} - \overrightarrow{OA}$ and $\overrightarrow{H_2H_3} = \overrightarrow{OH_3} - \overrightarrow{OH_2} = \overrightarrow{OE} - \overrightarrow{OD}$.
The inner product of these two vectors is
$$\begin{aligned} \overrightarrow{H_1H_2} \cdot \overrightarrow{H_2H_3} &= (\overrightarrow{OB} - \overrightarrow{OA})(\overrightarrow{OE} - \overrightarrow{OD}) \\ &= \overrightarrow{OB} \cdot \overrightarrow{OE} - \overrightarrow{OB} \cdot \overrightarrow{OD} - \overrightarrow{OA} \cdot \overrightarrow{OE} + \overrightarrow{OA} \cdot \overrightarrow{OD} \\ &= |\overrightarrow{OB}| \cdot |\overrightarrow{OE}| \cos(\angle BOE) - |\overrightarrow{OB}| \cdot |\overrightarrow{OD}| \cos(\angle BOD) \end{aligned}$$
$$\begin{aligned} & - |OA| \cdot |OE| \cos(\angle AOE) + |OA| \cdot |OD| \cos(\angle AOD) \\ &= |OA|^2(\cos(\angle BOE) - \cos(\angle BOD) - \cos(\angle AOE) + \cos(\angle AOD)). \end{aligned}$$
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,
$$\begin{aligned} \overrightarrow{H_1H_2} \cdot \overrightarrow{H_2H_3} &= |OA|^2(\cos(\angle BOE) - \cos(60^\circ) - \cos(120^\circ) + \cos(\angle AOD)) \\ &= |OA|^2(\cos(\angle BOE) + \cos(\angle AOD)) \\ &= 2|OA|^2 \cos \left( \frac{\angle BOE + \angle AOD}{2} \right) \cos \left( \frac{\angle BOE - \angle AOD}{2} \right). \end{aligned}$$
Verify that $\angle BOE + \angle AOD = 2\angle BOD + \angle DOE + \angle AOB$, which implies
$$\begin{aligned} \frac{\angle BOE + \angle AOD}{2} &= \angle BOD + \frac{\angle DOE + \angle AOB}{2} = 60^\circ + \angle DCE + \angle ACB \\ &= 60^\circ + \angle DCB - \angle ECA = 60^\circ + (180^\circ - 30^\circ) - 120^\circ = 90^\circ = \frac{\pi}{2}, \end{aligned}$$
so $\cos(\frac{\angle BOE+\angle AOD}{2}) = 0$ and $\overrightarrow{H_1H_2} \cdot \overrightarrow{H_2H_3} = 0$. Hence $H_1H_2H_3$ is a right triangle.