6 people solved it during the test.

Here's How to Crack It From the converse of butterfly we have that $DE$ is perpendicular to $MH$. Let ray $MH$ intersect $\omega$ at $G$, it is well-known that $\angle{MGA}=90^{\circ}$, thus $DE$ is parallel to $AG$. Let $O$ and $T$ be the circumcenter of $\omega$ and the foot of the perpendicular from $A$ to $BC$. $$DE \mid\mid AG \implies \angle{GMB}=\angle{GMT}=\angle{GAT}=\angle{GAH}=\angle{AHE} \implies \triangle{BHM} \sim \triangle{AEH} \implies \frac{MH}{DE}=\frac{BC}{4AH}$$$$O_ADHE \sim OBMC \implies \frac{HO_A}{OM}=\frac{DE}{BC}\implies \frac{MH}{HO_A}=\frac{MH}{OM}\cdot \frac{BC}{DE}=\frac{MH}{DE}\cdot \frac{BC}{OM}=\frac{BC^2}{4OM\cdot AH}$$Thus $\frac{MH}{HO_A}$ is constant as $A$ varies, and because $H$ lies on a fixed circle while $A$ varies, $O_A$ must lie on a fixed circle as well.

Here's an different proof of proving that $\frac{MH}{HO_A}$ has a constant value.

My problem Official Wording wrote: Let $BC$ be a fixed chord of a circle $\omega$. Let $A$ be a variable point on the major arc $BC$ of $\omega$. Let $H$ be the orthocenter of $ABC$. Points $D$ and $E$ lying on sides $AB$ and $AC$ respectively are such that $H$ is the midpoint of segment $DE$. Let $O_A$ be the circumcenter of triangle $ADE$. Prove that, as $A$ varies, all points $O_A$ lies on a fixed circle. I wrote it forwards, so the official solution is quite involved. Solution. (mine, original) Let $M$ be the midpoint of $\overline{BC}$. Construct parallelogram $AHMK$; clearly $K$ is fixed. Let $Q$ be the reflection of $A$ in point $O_A$. By converse of Butterfly Theorem we know $\overline{MH} \perp \overline{DE}$. Lemma. Let $M$ be the midpoint of side $\overline{BC}$ in non-isosceles $\triangle ABC$. Points $D$ and $E$ lie on sides $\overline{AB}$ and $\overline{AC}$. Point $Q$ is such that $\overline{QD} \perp \overline{AB}$ and $\overline{QE} \perp \overline{AC}$. Then $\angle ABQ=\angle ACQ$ if and only if $MD=ME$. (Proof) Let $P$ be the isogonal conjugate of $Q$ in $\triangle ABC$. Suppose $\angle ABQ=\angle ACQ$. Let $J$ be the projection of $Q$ on side $\overline{BC}$. Now $\overline{PM} \perp \overline{BC}$. As pedal circles of $P$ and $Q$ coincide; $DEMJ$ is cyclic. Thus $\angle DJB=\angle EJC$ so $MD=ME$ follows. Assume $MD=ME$ and the claim is not true. Let $P'$ be the point on line $\overline{PA}$ with $\overline{MP'} \perp \overline{BC}$; clearly $P' \ne P$. Let $Q'$ be the isogonal conjugate of $P'$ in $\triangle ABC$; $D', E'$ be projections of $Q'$ on sides $\overline{AB}$ and $\overline{AC}$. Now $MD'=ME'$ as shown earlier. Since $\overline{D'E'} \parallel \overline{DE}$ we conclude $\overline{AM}$ bisects angle $DAE$; contradicting $AB \ne AC$. $\blacksquare$ Finally, $\overline{AK} \perp \overline{DE}$ so $\angle QAB=\angle CAK$. Accordingly, $Q$ and $K$ are isogonal conjugates in $\triangle ABC$ so $\angle BQC$ is fixed hence $\angle ABQ=\angle CBK$. As $A$ moves with constant angular velocity, $Q$ also moves with same velocity. By averaging principle $O_A$ moves on an arc. $\blacksquare$ EulerMacaroni noted that perpendicular bisectors of $\overline{AE}$ and $\overline{AD}$ pass through fixed points $T$ and $S$ on line $\overline{BC}$. This is quite direct to prove with trigonometry. Then $\angle TO_AS=180^{\circ}-\angle A$ solves the problem. I feel this claim is quite difficult to come up with without a proper diagram.

Let A' be the antipode of A. Butterfly theorem gives us JH|DE->Oa,H,A' collinear. A'BD=A'HD -> BA'HD cyclic similarly A'HEC concylic so DA'E=DBH+HCE=180-2A=DOaE -> A'DOaE cyclic. HA'=2HJ and HA'/HOa=(DOa/DA')^2=tanA^2 fixed since chord BC const -> The spiral send Oa to fixed circle

I wasn't able to complete the proof during the actual exam. Neat problem though. Let $M$ be the midpoint of $BC$. Claim: $MH \perp DE$. Let $\ell$ be the line through $A$ parallel to $DE$. Let $H_a, H_b, H_c$ be the foot of $A$, $B$ and $C$-altitude, respectively. \[(B, C; H_a, H_bH_c\cap BC) = - 1 = (D, E; H, \infty_{DE}) \stackrel{A}{=} (B, C; H_a, \ell\cap BC)\]Thus, $H_bH_c\cap BC \equiv \ell \cap BC$. By Brocard's Theorem on $\odot(H_bH_cBC)$, $MH\perp \ell$ and thus, $MH\perp DE$. [asy][asy] import geometry; unitsize(2.3cm); pair A = dir(108), B = dir(210), C = -1/B, A_ = -A, M = (B + C)/2; triangle t = triangle((point) A, (point) B, (point) C); pair H = orthocentercenter(t); line l = perpendicular(H, line(H, M)); pair D = intersectionpoint(l, t.AB), E = intersectionpoint(l, t.AC); pair O_A = circumcenter(A, D, E); line m = perpendicular(A, line(H, M)); pair L = intersectionpoint(t.BC, m); pair K = intersectionpoint(line(M, H), m); pair H_a = foot(t.VA), H_b = foot(t.VB), H_c = foot(t.VC); draw(t); draw(D--E); draw(L--B); draw(A--L); draw(M--K, dashed+linewidth(0.6)); draw(circle(M, abs(M-B))); draw(H_b--L, dashed+linewidth(0.4)); label("A", A, A); label("B", B, B*dir(10)); label("C", C, C); label("M", M, -dir(90)); label("E", E, dir(-100)); label("D", D, dir(-10)); label("H", H, dir(80)); label("H$_a$", H_a, -dir(90)); label("H$_b$", H_b, dir(45)); label("H$_c$", H_c, dir(135)); dot(A^^B^^C^^H^^D^^E^^M^^H_a^^H_b^^H_c); [/asy][/asy] Note that as $A$ moves on the circle $H$ moves on the reflection of $\omega$ in $\overline{BC}$, thus, $H$ moves on a circle. Now, $O_A, H, M$ are collinear. Let $A'$ be the antipode of $A$ in $\omega$, it's well-known that $M$ is the midpoint of $A'H$. Note that $BDHA'$ is cyclic since, $\angle DBA' = \angle ABA' = 90^{\circ} = \angle DHA$, similarly, for $EHA'C$. We claim that $\frac{O_AH}{HM}$ is fixed. \[\frac{O_AH}{HM} = \frac{O_AH/EH}{HM/EH} = \frac{2 \cot HO_AE}{A'H/EH} = \frac{2 \cot A}{\cot \angle HA'E} = \frac{2 \cot A}{\cot\angle HCA} = 2\cot^2 A.\]Thus, a homothety with constant scale at $M$ sends $H$ to $O_A$ and therefore, $O_A$ lies on a circle. [asy][asy] import geometry; unitsize(2.5cm); pair A = dir(100), B = dir(210), C = -1/B, A_ = -A, M = (B + C)/2; triangle t = triangle((point) A, (point) B, (point) C); pair H = orthocentercenter(t); line l = perpendicular(H, line(H, M)); pair D = intersectionpoint(l, t.AB), E = intersectionpoint(l, t.AC); pair O_A = circumcenter(A, D, E); draw(circumcircle(t)); draw(t); draw(D--E); draw(A_--O_A, dashed+linewidth(0.5)); draw(circumcircle(A, D, E), dashed+linewidth(0.4)); draw(circumcircle(B, D, A_)^^circumcircle(C, E, A_), dotted); label("A", A, A); label("B", B, B); label("C", C, C); label("A$'$", A_, A_); label("O$_A$", O_A, dir(0)); label("M", M, -dir(60)); label("E", E, dir(0)); label("D", D, -dir(20)); label("H", H, dir(70)); dot(A^^B^^C^^A_^^H^^D^^E^^M^^O_A); [/asy][/asy]