Note that $Z$ is the center of the spiral similarity taking $BY$ to $XC$. If $M_B, M_C$ are the midpoints of the altitudes from $B, C$ respectively then $M_B, M_C$ lie on $(HM)$, so it suffices to show that the spiral similarity taking $BY \rightarrow XC$ also takes $M_B \rightarrow M_C$, which is equivalent to $BM_B : M_BY = XM_C : M_CC$, or $\frac{BM_B}{BY} + \frac{CM_C}{CX} = 1$. If $E, F$ are the feet from $B, C$ then it suffices to show that $\frac{BE}{BY} + \frac{CF}{CX} = 2$, which by the Ratio Lemma simplifies to $\sin \frac{A}{2} \cdot \frac{\sin A}{\cos \frac{A}{2}} + \cos \frac{A}{2} \cdot \frac{\sin A}{\sin \frac{A}{2}} = 2 \sin^2 \frac{A}{2} + 2 \cos^2 \frac{A}{2} = 2$. $\square$

Let $AD,BE,CF$ be the altitudes of $\triangle ABC$, and denote $\omega_b = (BHX)$ and $\omega_c = (CHY)$. Showing $\angle HZM = 90^\circ$ is equivalent to showing $HDZM$ cyclic. Hence we'll show that $\frac{Pow(D,\omega_b)}{Pow(D,\omega_c)} = \frac{Pow(M,\omega_b)}{Pow(M,\omega_c)}$. Denote $\gamma= (ABC)$ and $T$ the area of $\triangle ABC$. By Internal and External Angle Bisector Theorem, \[\frac{FX}{CX} = \frac{AF}{AC} = \cos A \iff \frac{CF}{CX} = \cos A + 1 \iff CX = \frac{b \sin A}{1 + \cos A}\]\[\frac{YE}{YB} = \frac{AE}{AB} = \cos A \iff \frac{BE}{BY} = 1 - \cos A \iff BY = \frac{c \sin A}{1 - \cos A}\] By Linearity of Power of a Point ($Pow(X,\Omega_1, \Omega_2) := Pow(X,\Omega_1) - Pow(X,\Omega_2)$ is linear, for any point $X$ and circles $\Omega_1, \Omega_2$), \begin{align*} \frac{Pow(D,\omega_b)}{Pow(D,\omega_c)} &= \frac{Pow(M,\omega_b)}{Pow(M,\omega_c)}\\ \iff \frac{Pow(D,\omega_b,\gamma) - DB \cdot DC}{Pow(D,\omega_c,\gamma) - DB \cdot DC} &= \frac{Pow(M,\omega_b,\gamma) - MB \cdot MC}{Pow(M,\omega_c,\gamma) - MB \cdot MC}\\ \iff \frac{BD \cdot Pow(C,\omega_b, \gamma) + DC \cdot Pow(B,\omega_b,\gamma) - BC \cdot DB \cdot DC}{BD \cdot Pow(C,\omega_c,\gamma) + DC \cdot Pow(B,\omega_c,\gamma) - BC \cdot DB \cdot DC} &= \frac{Pow(B,\omega_b,\gamma) + Pow(C,\omega_b,\gamma) - 2 \cdot MB \cdot MC}{Pow(B,\omega_c,\gamma) + Pow(C,\omega_c,\gamma) - 2 \cdot MB \cdot MC}\\ \iff \frac{BD \cdot CX \cdot CH - BC \cdot DB \cdot DC}{DC \cdot BH \cdot BY - BC \cdot DB \cdot DC} &= \frac{CX \cdot CH - 2 \cdot MB \cdot MC}{BH \cdot BY - 2 \cdot MB \cdot MC}\\ \iff \frac{2cR \cos B \cos C \cdot CX - abc \cos B \cos C}{2bR \cos B \cos C \cdot BY - abc \cos B \cos C} &= \frac{2R \cos C \cdot CX - \frac{a^2}{2}}{2R \cos B \cdot BY - \frac{a^2}{2}}\\ \iff \frac{\frac{c}{2} \cdot CX - \frac{abc}{4R}}{\frac{b}{2} \cdot BY - \frac{abc}{4R}} &= \frac{\frac{bc\cos C}{a} \cdot CX - \frac{abc}{4R}}{\frac{bc \cos B}{a} \cdot BY - \frac{abc}{4R}}\\ \iff \frac{\frac{bc \sin A}{2(1+\cos A)} - T}{\frac{bc \sin A}{2(1-\cos A)} - T} &= \frac{\frac{bc\sin A}{2} \cdot \frac{2b \cos C}{a(1+\cos A)} - T}{\frac{bc\sin A}{2} \cdot \frac{2c \cos B}{a(1-\cos A)} - T}\\ \iff \frac{\frac{1}{1 + \cos A} - 1}{\frac{1}{1 - \cos A} - 1} &= \frac{\frac{2b\cos C}{a(1+ \cos A)} - 1}{\frac{2c\cos B}{a(1 - \cos A)} - 1}\\ \iff \frac{\cos A - 1}{\cos A + 1} &= \frac{(2b \cos C - a - a\cos A)(1-\cos A)}{(2c\cos B - a + a\cos A)(1+\cos A)}\\ \iff 2c \cos B - a + a\cos A &= a \cos A + a - 2b\cos C\\ \iff 2a &= 2b \cos C + 2c \cos B \end{align*}which is clearly true $(BC = BD + DC)$.

I like this problem very much because it is motivated by the most basic triangle constructions: altitudes, midpoints, angle bisectors, all came in nicely into this asymmetric configuration. Let $D$ be a point such that $ABDC$ is a parallelogram, and let $BE$ and $CF$ be altitudes. Let $BD$ intersect $(BHX)$ again at $P$ and $CD$ intersect $(CHY)$ again at $Q$. Note that $BD\perp BH$, so $\angle PZH=90^{\circ}$. Likewise $\angle QZH=90^{\circ}$ as well. So it suffice to prove that $P, M, Q$ are colinear. By Menelaus Theorem, it reduces to $$\frac{DP}{PB}\cdot\frac{BM}{MC}\cdot \frac{CQ}{QD}=1 \iff \frac{DP}{PB}=\frac{DQ}{QC}$$But note that $DC\parallel PX\parallel BF$, and since $AX$ is inner angle bisector of $\angle FAC$, then $$\frac{DP}{PB}=\frac{CX}{XF}=\frac{CA}{AF}$$Likewise, $DB\parallel QY\parallel CE$, and $AY$ is external angle bisector of $\angle EAB$, so $$\frac{DQ}{QC}=\frac{BY}{YE}=\frac{BA}{AE}$$Now the proposition is true since $\triangle CAF\sim \triangle BAE$, we have $\displaystyle\frac{CA}{AF}=\frac{BA}{AE}$. $\blacksquare$

Other solutions (from the official packet): Solution 2: In this solution, all angles used are directed angles. Let $BE$ and $CF$ be altitudes, and let $K, L$ be the midpoints of $BE$ and $CF$ respectively. Then $MK\parallel CE$ so $\angle HKM=90^{\circ}$. Likewise $L$ lies on the circle with diameter $HM$. It suffice to show that $Z$ lies on it as well, by showing $\angle ZKH=\angle ZLH$. But observe that $\triangle ZYB\sim \triangle ZCX$, so it suffice to prove that $E$ corresponds to $F$ in this similarity, so that $\triangle ZKB\sim \triangle ZLX$, that gives us $\angle ZKH=\angle ZLH$. Till this end it suffice to prove that $$\frac{LC}{XC}=\frac{KY}{YB}$$By the same argument as Solution $1$ we have $\displaystyle \frac{FX}{XC}=\frac{EY}{YB}$, so we have $$\frac{FX}{XC}=\frac{EY}{YB}\iff \frac{FC}{XC}=\frac{EY+YB}{YB}\iff \frac{LC}{XC}=\frac{BE+2EY}{2YB}\iff \frac{LC}{XC}=\frac{KE+EY}{YB}=\frac{KY}{YB}$$as we wanted to prove. $\blacksquare$ Solution 3: Let $HP$ and $HR$ be the diameters of the circles $(BHX)$ and $(CHY)$. Then $\angle HZP =\angle HZR = 90^{\circ}$. We want to show that $PR$ bisects $BC$. Since $\angle HBP = 90^{\circ}$, $BP \parallel AC$. Also $XP \parallel AB$, $CR \parallel AB$, $YR \parallel AC$. Let $W = PX \cup AC$, $T = RY \cup AB$, $S = PX \cup RY$. From the parallelograms, we find that $SP = BT$, $SR = CW$, $BP = AW = ST$ and $CR = WS = AT$. From the angle conditions, we easily get $WX = AW$ and $TY = AT$, moreover we also find that $\triangle WXC$ is similar to $\triangle TYB$. Then $$\frac{SP}{SR}=\frac{BT}{CW} = \frac{TY}{WX} = \frac{TA}{TS}$$so $PR$ is parallel to $AS$. Let $Q$ be the intersection of $AC$ and $PR$. Then triangles $\triangle CQR$ and $\triangle WAS$ are congruent. We get $CQ = BP$, so $CPBQ$ is a parallelogram. Then $PR$ bisects $BC$ as required. $\blacksquare$