European Girls' Mathematical Olympiad

Solution:

As $DB = BC = CE$ we have $BI \perp CD$ and $CI \perp BE$. Hence $I$ is orthocenter of triangle $BFC$. Let $K$ be the point of intersection of the lines $BI$ and $CD$, and let $L$ be the point of intersection of the lines $CI$ and $BE$. Then we have the power relation $IB \cdot IK = IC \cdot IL$. Let $U$ and $V$ be the feet of the perpendiculars from $D$ to $EF$ and $E$ to $DF$, respectively. Now we have the power relation $DH \cdot HU = EH \cdot HV$.

Let $\omega_1$ and $\omega_2$ be the circles with diameters $BD$ and $CE$, respectively. From the power relations above we conclude that $IH$ is the radical axis of the circles $\omega_1$ and $\omega_2$.
Let $O_1$ and $O_2$ be centers of $\omega_1$ and $\omega_2$, respectively. Then $MB = MC$, $BO_1 = CO_2$ and $\angle MBO_1 = \angle MCO_2$, and the triangles $MBO_1$ and $MCO_2$ are congruent. Hence $MO_1 = MO_2$. Since radii of $\omega_1$ and $\omega_2$ are equal, this implies that $M$ lies on the radical axis of $\omega_1$ and $\omega_2$ and $M, I, H$ are collinear.

Let the points $K, L, U, V$ be as in Solution 1. Let $P$ be the point of intersection of $DU$ and $EI$, and let $Q$ be the point of intersection of $EV$ and $DI$.
Since $DB = BC = CE$, the points $CI$ and $BI$ are perpendicular to $BE$ and $CD$, respectively. Hence the lines $BI$ and $EV$ are parallel and $\angle IEB = \angle IBE = \angle UEH$. Similarly, the lines $CI$ and $DU$ are parallel and $\angle IDC = \angle ICD = \angle VDH$. Since $\angle UEH = \angle VDH$, the points $D, Q, F, P, E$ are concyclic. Hence $IP \cdot IE = IQ \cdot ID$.
Let $R$ be the second point intersection of the circumcircle of triangle $HEP$ and the line $HI$. As $IH \cdot IR = IP \cdot IE = IQ \cdot ID$, the points $D, Q, H, R$ are also concyclic. We have $\angle DQH = \angle EPH = \angle DFE = \angle BFC = 180^\circ - \angle BIC = 90^\circ - \angle BAC / 2$. Now using the concyclicity of $D, Q, H, R$, and $E, P, H, R$ we obtain $\angle DRH = \angle ERH = 180^\circ - (90^\circ - \angle BAC / 2) = 90^\circ + \angle BAC / 2$. Hence $R$ is inside the triangle $DEH$ and $\angle DRE = 360^\circ - \angle DRH - \angle ERH = 180^\circ - \angle BAC$ and it follows that the points $A, D, R, E$ are concyclic.

As $MB = MC$, $BD = CE$, $\angle MBD = \angle MCE$, the triangles $MBD$ and $MCE$ are congruent and $\angle MDA = \angle MEA$. Hence the points $M, D, E, A$ are concyclic. Therefore the points $M, D, R, E, A$ are concyclic. Now we have $\angle MRE = 180^\circ - \angle MAE = 180^\circ - (90^\circ + \angle BAC / 2) = 90^\circ - \angle BAC / 2$ and since $\angle ERH = 90^\circ + \angle BAC / 2$, we conclude that the points $I, H, R, M$ are collinear.

Suppose that we have a coordinate system and $(b_x, b_y), (c_x, c_y), (d_x, d_y), (e_x, e_y)$ are the coordinates of the points $B, C, D, E$, respectively. From $\overrightarrow{BI} \cdot \overrightarrow{CD} = 0$, $\overrightarrow{CI} \cdot \overrightarrow{BE} = 0$, $\overrightarrow{EH} \cdot \overrightarrow{CD} = 0$, $\overrightarrow{DH} \cdot \overrightarrow{BE} = 0$ we obtain $\overrightarrow{IH} \cdot (\vec{B} - \vec{C} - \vec{E} + \vec{D}) = 0$. Hence the slope of the line $IH$ is $(c_x + e_x - b_x - d_x)/(b_y + d_y - c_y - e_y)$.
Assume that the $x$-axis lies along the line $BC$, and let $\alpha = \angle BAC$, $\beta = \angle ABC$, $\theta = \angle ACB$. Since $DB = BC = CE$, we have $c_x - b_x = BC$, $e_x - d_x = BC - BC \cos \beta - BC \cos \theta$, $b_y = c_y = 0$, $d_y - e_y = BC \sin \beta - BC \sin \theta$. Therefore the slope of $IH$ is $(2 - \cos \beta - \cos \theta)/(\sin \beta - \sin \theta)$.
Now we will show that the slope of the line $MI$ is the same. Let $r$ and $R$ be the inradius and circumradius of the triangle $ABC$, respectively. As $\angle BMC = \angle BAC = \alpha$ and $BM = MC$, we have
$$m_y - i_y = \frac{BC}{2} \cot \left(\frac{\alpha}{2}\right) - r \text{ and } m_x - i_x = \frac{AC - AB}{2}$$
where $(m_x, m_y)$ and $(i_x, i_y)$ are the coordinates of $M$ and $I$, respectively. Therefore the slope of $MI$ is $(BC \cot (\alpha / 2) - 2r)/(AC - AB)$.
Now the equality of these slopes follows using
$$\frac{BC}{\sin \alpha} = \frac{AC}{\sin \beta} = \frac{AB}{\sin \theta} = 2R$$
hence
$$BC \cot \left(\frac{\alpha}{2}\right) = 4R \cos^2 \left(\frac{\alpha}{2}\right) = 2R(1 + \cos \alpha)$$
and
$$\frac{r}{R} = \cos \alpha + \cos \beta + \cos \theta - 1$$
as
$$\frac{BC \cot (\alpha / 2) - 2r}{AC - AB} = \frac{2R(1 + \cos \alpha) - 2r}{2R(\sin \beta - \sin \theta)} = \frac{2 - \cos \beta - \cos \theta}{\sin \beta - \sin \theta}$$
giving the collinearity of the points $I, H, M$.

Let the bisectors $BI$ and $CI$ meet the circumcircle of $ABC$ again at $P$ and $Q$, respectively. Let the altitude of $DEF$ belonging to $D$ meet $BI$ at $R$ and the one belonging to $E$ meet $CI$ at $S$.
Since $BI$ is angle bisector of the isosceles triangle $CBD$, $BI$ and $CD$ are perpendicular. Since $EH$ and $DF$ are also perpendicular, $HS$ and $RI$ are parallel. Similarly, $HR$ and $SI$ are parallel, and hence $HSIR$ is a parallelogram.
On the other hand, as $M$ is the midpoint of the $\operatorname{arc} BAC$, we have $\angle MPI = \angle MPB = \angle MQC = \angle MQI$, and $\angle PIQ = (\widehat{PA} + \widehat{CB} + \widehat{AQ}) / 2 = (\widehat{PC} + \widehat{CB} + \widehat{BQ}) / 2 = \angle PMQ$. Therefore $MPIQ$ is a parallelogram.
Since $CI$ is angle bisector of the isosceles triangle $BCE$, the triangle $BSE$ is also isosceles. Hence $\angle FBS = \angle EBS = \angle SEB = \angle HEF = \angle HDF = \angle RDF = \angle FCS$ and $B, S, F, C$ are concyclic. Similarly, $B, F, R, C$ are concyclic. Therefore $B, S, R, C$ are concyclic. As $B, Q, P, C$ are also concyclic, $SR$ and $QP$ are parallel.
Now it follows that $HSIR$ and $MQIP$ are homothetic parallelograms, and therefore $M, H, I$ are collinear.