Solution:

Denote by $B'$ and $C'$ the reflections of $B$ and $C$ in $M$ and $N$, respectively. Points $C'$, $A$, $B'$ are clearly collinear and $DEB'A$ is a parallelogram. Since $EH \perp AD$, we have $EH \perp EB'$. Also $HA \perp AB'$, so points $H, E, B', A$ are concyclic. This gives

$$\angle C'QH = \angle NQH = \angle NEH = \angle AEH = \angle AB'H = \angle C'B'H,$$
and so points $C'$, $B'$, $Q$, $H$ are concyclic. Analogously points $C'$, $B'$, $P$, $H$ are concyclic, and so all points $B'$, $C'$, $P$, $Q$, $H$ are. Now we have

$$\angle NMB' = \angle AB'M = \angle C'B'P = \angle C'QP = \angle NQP,$$
which proves that $P, Q, N, M$ are also concyclic.

Introduce points $B'$, $C'$ as above. Also define $A' = B'E \cap C'D$, so that $EA DA'$ is a parallelogram and $ADE$ is the medial triangle of $A'B'C'$. It that follows that the orthocentre $H$ of $ADE$ is the circumcentre of $A'B'C'$, and in particular

$$\angle C'B'H = 90^{\circ} - \angle B'A'C' = 90^{\circ} - \angle DAE = \angle AEH = \angle NEH = \angle NQH = \angle C'QH.$$
So again we have that $C'$, $B'$, $Q$, $H$ are concyclic and conclude as in Solution 1.

Solution:

Let $X$ be the second intersection of $(DHM)$ and $(EHN)$ and let $O'$ be the circumcentre of $(AMN)$. Note that $\angle MXN = \angle MDH + \angle NEH = 180^{\circ} - 2\angle DAE$ and since $\angle MO'N = 2\angle DAE$ we have that $X, M, O', N$ is cyclic and since $\angle MXH = \angle NXH$ it means that $HX$ is the angle bisector of $\angle MXN$ but since $O'M = O'N$ it means that $H, X, O'$ are collinear. Let $BM$ and $CN$ intersect at $T$ and let $K$ and $L$ be the midpoints of $MN$ and $BC$. Note that $L$ is also the midpoint of $DE$. Since $MN$ is parallel to $BC$ it means that $T$, $K$, and $L$ are collinear, but since $A$, $K$, and $L$ are collinear we get that $A$, $T$, $K$, and $L$ are collinear. Now, $\frac{TL}{TK} = \frac{BC}{MN} = 6$. Since $KL = KA$ it means that $\frac{AT}{TK} = \frac{TL - 2TK}{TK} = 4$ so by the lemma below, $T$ lies on $HO'$. Since $HO'$ is the radical axis of $(DHM)$ and $(EHN)$ we finish the problem using the Radical Axes Theorem $(TM \cdot TP = TN \cdot TQ)$.

Lemma. Let $A'$ be the reflection of $A$ around the orthocentre $H$ of $\triangle ABC$ and $O$ and $M$ be the circumcentre of $\triangle ABC$ and the midpoint of $BC$, respectively. Let $T$ be the intersection of $A'O$ and $AM$. Then $\frac{AT}{TM} = 4$.

Proof. Since $OM \parallel AA'$ we have $\frac{AT}{TM} = \frac{AA'}{OM} = \frac{2AH}{OM} = \frac{4OM}{OM} = 4$. We used here that $AH = 2OM$. $\square$

As in solution 2, we will prove that $O'$ is both on line $HT$ and the radical axis of the circles, hence $T$ is on the radical axis, from which we conclude the required concyclicity. We present alternative proofs of both facts, discovered by contestants.

Let $M_{1}$, $N_{1}$ be the midpoints of $AM$, $AN$, respectively, so that $AM_{1}:M_{1}D = AN_{1}:N_{1}E = 1:3$. It is easy to verify (e.g. by applying the theorems of Ceva or Menelaus, or by computing in barycentric coordinates as in Solution 3) that $T$ lies on $EM_{1}$ and $DN_{1}$. Note that $M_{1}N_{1} \parallel DE$, and also $M_{1}O' \parallel HE$ as both are perpendicular to $AMD$, and similarly $N_{1}O' \parallel HD$. It follows that $DEH$ and $N_{1}M_{1}O'$ are homothetically similar triangles, and the center of their (negative) homothety is $T = DN_{1} \cap EM_{1}$. Therefore $T$ also lies on $HO'$, as claimed. (We also have that the homothety is by factor $\frac{M_{1}N_{1}}{ED} = -\frac{1}{4}$.)

Now, let $M_{2}$, $N_{2}$ be the second intersection points of $O'M$, $O'N$ with the circumcircles of $HMD$, $HNE$, respectively. To prove that $O'$ is on the radical axis, it suffices to show that $O'M \cdot O'M_{2} = O'N \cdot O'N_{2}$. But by definition of $O'$ we have $O'M = O'N$, so we must show $O'M_{2} = O'N_{2}$, which is equivalent to $M_{2}N_{2} \parallel MN$. Angle chasing in circle $MM_{2}DH$ gives

$$\angle OM_{2}H = \angle MM_{2}H = \angle MDH = \angle ADH = 90^{\circ} - \angle EAD = 90^{\circ} - \angle NAM = \angle O'MN,$$
from which it follows that $M_{2}H \parallel MN$. Similarly, we have $N_{2}H \parallel MN$, and the two facts together imply that $M_{2}, H, N_{2}$ are collinear and the line through them is parallel to $MN$, as claimed. $\square$

Solution:

We compute using linear combinations with respect to $ADE$. We have $B = 2D - E$, $C = 2E - D$, $M = \frac{A + D}{2}$, and $N = \frac{A + E}{2}$, from which we immediately obtain that the intersection $T = BM \cap CN$ is $T = \frac{3A + D + E}{5} = \frac{6M - B}{5} = \frac{6N - C}{5}$. As in solution 2, we reduce to showing that $T$ is on the radical axis of $HDM$ and $HEN$, whence $TM \cdot TP = TN \cdot TQ$ proves the concyclicity of $P, Q, N, M$.

Synthetic finish, similar to Solution 2. Let $O$ be the circumcentre of $ADE$ and $O' = \frac{A + O}{2}$ be the circumcentre of $AMN$. As in Solution 2, we have that $O'$ is on the desired radical axis, so it is enough to show $T \in O'H$. Let $G = \frac{A + D + E}{3}$ be the barycentre of $ADE$. By properties of the Euler line, we also have $G = \frac{H + 2O}{3}$. Now using our known identities we find

$$T = \frac{3A + D + E}{5} = \frac{2A + 3G}{5} = \frac{2A + H + 2O}{5} = \frac{H + 4O'}{5}$$
and in particular $T \in HO'$, as we wanted to show.

Computational finish. Let $f(T)$ be the power difference at $T$ w.r.t. $DHM$ and $HEN$. We compute $f(A)$ and $f(L)$ where $L = \frac{D + E}{2}$. Since $T = \frac{3A + 2L}{5}$, it is enough to show that $3P(A) + 2P(L) = 0$. In the following all lengths are directed. We compute trigonometrically: Let $\alpha, \beta, \gamma$ be the angles of $ADE$ and assume the diameter of its circumcircle is $1$. We have

$$f(A) = \frac{AD^{2} - AE^{2}}{2} = \frac{\sin^{2}(\gamma) - \sin^{2}(\beta)}{2}.$$
To compute $P(L)$, let $D', E'$ be the second intersection points of $HMD, HNE$ with $DE$, and let $M', N', F$ be the feet of the perpendiculars from $H$ to $AD, AE, DE$, respectively. Note that $DM' = \sin \alpha \cos \beta$, thus

$$M'M = DM - DM' = \frac{\sin(\alpha + \beta)}{2} - \sin \alpha \cos \beta = \frac{\sin(\beta - \alpha)}{2}.$$
We also have $HM' = \cos \alpha \cos \beta$, $HF = \cos \beta \cos \gamma$, and $HM'M \sim HFD'$, therefore

$$FD' = \frac{HF}{HM'} M'M = \frac{\cos \gamma \sin(\beta - \alpha)}{\cos \alpha}$$
and similarly
$$E'F = \frac{\cos \beta \sin(\gamma - \alpha)}{\cos \alpha}.$$
We also have the standard $FD = \sin \gamma \cos \beta$ and $EF = \sin \beta \cos \gamma$. We can finally compute

$$f(L) = LD \cdot LD' - LE \cdot LE' = \frac{\sin \alpha}{2} (LD' + LE') = \frac{\sin \alpha}{2} (FD' + FE' - FD - FE)$$
$$\qquad = \frac{\sin \alpha}{4 \cos \alpha} (\cos \gamma \sin (\beta - \alpha) - \cos \beta \sin (\gamma - \alpha) - 2 \cos \alpha (\sin \gamma \cos \beta - \sin \beta \cos \gamma))$$
$$\qquad = \frac{3 \sin \alpha \sin(\beta - \gamma)}{4} = \frac{3}{4} (\sin (\beta + \gamma) \sin (\beta - \gamma)) = \frac{3}{8} (\cos (2\gamma) - \cos (2\beta))$$
$$\qquad = \frac{3}{4} (\sin^{2}(\beta) - \sin^{2}(\gamma)) = -\frac{3}{2} f(A).\quad \square$$

Solution:

Let $T = BM \cap CN$, let $F$ be the foot of the altitude from $A$ to $BC$, let $O$ be the circumcentre of $(ADE)$, let $D' \neq D$ and $E' \neq E$ be the second intersections of $(DHMP)$ and $(EHNQ)$ with line $BC$ and let $U$ and $V$ be the antipodes of $D$ and $E$ on $(DHMP)$ and $(EHNQ)$, respectively.

We begin with a bit of barycentric coordinates. Set barycentric coordinates in $\triangle ABC$, set so that $A = (1,0,0)$, $B = (0,1,0)$, and $C = (0,0,1)$. The definitions of $D$ and $E$ give $D = (0,2/3,1/3)$ and $E = (0,1/3,2/3)$, whence $M = (1/2,1/3,1/6)$ and $N = (1/2,1/6,1/3)$. This means that line $BM$ is given by $(1/2:y:1/6)$ while line $CN$ is given by $(1/2:1/6:z)$. So their intersection $T$ is $(1/2:1/6:1/6) = (3:1:1)$, giving $T = \frac{3A + B + C}{5} = \frac{3A + D + E}{5}$.

Our next tool is the linearity of the power of a point. Let $f:\mathbb{R}^{2} \to \mathbb{R}$ be defined by

$$f(Z) = \mathrm{Pow}_{(DHMP)}(Z) - \mathrm{Pow}_{(EHNQ)}(Z).$$
It suffices to show that $f(T) = 0$; from there, the required concyclicity will follow from $TM \cdot TP = TN \cdot TQ$. The key observation is that $f$ is a linear function. In particular, $f(T) = \frac{3f(A) + f(D) + f(E)}{5}$. So, we need only show that $3f(A) + f(D) + f(E) = 0$. Pick an arbitrary one-dimensional coordinate system on the line $BC$ and let $\tau$ be the map from a point on $BC$ to its coordinate. We compute

$$f(A) = AM \cdot AD - AN \cdot AE = \frac{AD^2 - AE^2}{2} = \frac{FD^2 - FE^2}{2}$$
$$\qquad = (\tau(E) - \tau(D))\left(\tau(F) - \tau\left(\frac{D + E}{2}\right)\right),$$
$$f(D) = -DE \cdot DE' = (\tau(E) - \tau(D))(\tau(D) - \tau(E')),\quad f(E) = ED \cdot ED' = (\tau(E) - \tau(D))(\tau(E) - \tau(D')).$$
Rearranging, it suffices to show that $3\tau(F) = \tau(D') + \tau(E') + \tau((D + E)/2)$. This can be rewritten as $3F = D' + E' + (D + E)/2$. By projecting down to line $BC$, it suffices to show that the displacement vector $H + 2A - (O + U + V)$ is perpendicular to line $BC$.

We do this using complex numbers. Let $(ADE)$ be the unit circle with $A = a$, $D = d$, and $E = e$, so that $O = 0$ and $H = h = a + d + e$. Note that $U$ satisfies $UM \perp AD$ and $UH \perp DH \perp AE$. Translating these conditions into equations, we have $u = a d\overline{u}$ and $u + a e\overline{u} = h + a e\overline{h}$. Rearranging gives

$$v := H + 2A - (O + U + V) = h + 2a - \frac{(d + e)h + 2a d e\overline{h}}{d + e} = 2\left(a - \frac{a d e\overline{h}}{d + e}\right) = -\frac{2de}{d + e}.$$
This displacement vector $v$ satisfies $v = de\overline{v}$ and so it is orthogonal to line $DE$, as desired.

Solution:

This solution uses almost exclusively complex numbers. As in other solutions, we reduce to showing that $H$, $S := (DHM) \cap (EHN)$, and $T := BM \cap CN$ are collinear; this is all of the synthetic information we shall use. (If one computes $T = \frac{3A + D + E}{5}$ using means other than complex numbers, the solution becomes much shorter.)

We use complex numbers with $A = a$, $D = d$, and $E = e$ on the unit circle. Note that $H = a + d + e$, $M = \frac{a + d}{2}$, and $B = 2d - e$. We will make great use of the "arbitrary line intersection formula," which says that the intersection between lines $WX$ and $YZ$ can be written explicitly as

$$\frac{(\overline{u} x - w\overline{x})(y - z) - (w - x)(\overline{y} z - y\overline{z})}{(\overline{w} - \overline{x})(y - z) - (w - x)(\overline{y} - \overline{z})}.$$

We first use this to find $T = t$. We compute

$$b - m = (2d - e) - \frac{a + d}{2} = \frac{3d - a - 2e}{2}$$
$$\overline{b} - m = \frac{3a e - d e - 2a d}{2a d e}$$
$$\overline{b} m - b\overline{m} = \frac{2e - d}{de} \cdot \frac{a + d}{2} - (2d - e) \cdot \frac{a + d}{2ad} = \frac{(a + d)(2a e + e^{2} - a d - 2d e)}{2ad e}$$

If $\mathcal{E}$ is some expression, we use the notation $\mathcal{E} - \{\sim\}$ to denote $\mathcal{E}$ minus the expression formed by swapping $d$ and $e$ in $\mathcal{E}$. We now have

$$t = \frac{(\overline{b}m - b\overline{m})(c - n) - \{\sim\}}{(\overline{b} - \overline{m})(c - n) - \{\sim\}}$$
$$\quad = \frac{(a + d)(2a e + e^{2} - a d - 2d e)(3e - a - 2d) - \{\sim\}}{(3a e - d e - 2a d)(3e - a - 2d) - \{\sim\}}$$
$$\quad = \frac{[a^{3}(d - 2e) + a^{2}(3d^{2} - 7d e + 5e^{2}) + a(2d^{3} - d^{2}e - 3d e^{2} + 3e^{3}) + d e(4d^{2} - 8d e + 3e^{2})] - \{\sim\}}{[a^{2}(2d - 3e) + a(4d^{2} - 11d e + 9e^{2}) + d e(2d - 3e)] - \{\sim\}}$$
$$\quad = \frac{a^{3}(3(d - e)) - a^{2}(2(d^{2} - e^{2})) + a(-(d^{3} - e^{3}) + 2d e(d - e)) + d e(d^{2} - e^{2})}{a^{2}(5(d - e)) - a(5(d^{2} - e^{2})) + d e(5(d - e))}$$
$$\quad = \frac{3a^{3} - 2a^{2}(d + e) - a(d^{2} - d e + e^{2}) + d e(d + e)}{5(a^{2} - a(d + e) + d e)}$$
$$\quad = \frac{(a - d)(a - e)(3a + d + e)}{5(a - d)(a - e)} = \frac{3a + d + e}{5}.$$
(The factorization in the last line can be motivated by noting that the expression, while cubic in $a$, is only quadratic in $d$. When written out as a polynomial in $d$, each coefficient is divisible by $a - e$; by symmetry, the numerator is divisible by $a - d$ as well, and the factorization follows.)

Computing $S = s$ is slightly harder, as it is the intersection of two circles rather than of two lines. We get around this by noting that $\{h, d, m, s\}$ are concyclic if and only if $\{0, h - d, h - m, h - s\}$ are concyclic, which happens if and only if $\{\frac{1}{h - d}, \frac{1}{h - m}, \frac{1}{h - s}\}$ are collinear. (One can see this by inversion, or just by writing out the cross-ratio in the special case when one of the points is zero.) Thus $\frac{1}{h - s}$ is the intersection of the line through $w := \frac{1}{h - d}$ and $x := \frac{1}{h - m}$ and the line through $y := \frac{1}{h - e}$ and $z := \frac{1}{h - n}$. We compute

$$w - x = \frac{1}{a + e} - \frac{1}{\frac{a + d}{2} + e} = -\frac{a - d}{(a + e)(a + d + 2e)}$$
$$\overline{w - x} = \frac{a e^{2}(a - d)}{(a + e)(2ad + a e + d e)}$$
$$\overline{w} x - w\overline{x} = \frac{a e}{a + e} \cdot \frac{2}{a + d + 2e} - \frac{1}{a + e} \cdot \frac{2ad e}{2ad + a e + d e}$$
$$\qquad = 2a e \frac{(2ad + a e + d e) - d(a + d + 2e)}{(a + e)(a + d + 2e)(2ad + a e + d e)}$$
$$\qquad = \frac{2a e(a - d)(d + e)}{(a + e)(a + d + 2e)(2ad + a e + d e)}.$$

Using the line intersection formula, we have

$$\frac{1}{h - s} = \frac{(\overline{w} x - w\overline{x})(y - z) - \{\sim\}}{(\overline{w} - \overline{x})(y - z) - \{\sim\}}$$
$$= 2(d + e) \frac{[e(ad + 2ae + de)] - \{\sim\}}{[e^{2}(a + d + 2e)(ad + 2ae + de)] - \{\sim\}}$$
$$= 2(d + e) \frac{[a(de + 2e^{2}) + de^{2}] - \{\sim\}}{[a^{2}(de^{2} + 2e^{3}) + a(d^{2}e^{2} + 5de^{3} + 4e^{4}) + de(de^{2} + 2e^{3})] - \{\sim\}}$$
$$= \frac{2(d + e)(a(2(e^{2} - d^{2})) + de(e - d))}{(a^{2} + de)(2(e^{3} - d^{3}) + de(e - d)) + a(4(e^{4} - d^{4}) + 5de(e^{2} - e^{2}))}$$
$$= \frac{2(d + e)(2a(d + e) + de)}{(a^{2} + de)(2d^{2} + 3de + 2e^{2}) + a(d + e)(4d^{2} + 5de + 4e^{2})}.$$

Since $h - t = \frac{2a + 4d + 4e}{5}$, this gives us

$$\frac{h - s}{h - t} = \frac{5}{4} \cdot \frac{(a^{2} + de)(2d^{2} + 3de + 2e^{2}) + a(d + e)(4d^{2} + 5de + 4e^{2})}{(d + e)(2ad + 2ae + de)(a + 2d + 2e)}.$$

It is easy to see that this is real by using the symmetry of the expressions (both the numerator and denominator satisfy $\mathcal{E} = a^{2}d^{3}e^{3}\overline{\mathcal{E}}$). We conclude that $H$, $S$, and $T$ are collinear, as desired.

Solution:

As usual, we reduce to proving $T = (3A + D + E)/5$ is on the radical axis and compute; this time in Cartesian coordinates.

Let $A(0,h)$, $D(b,0)$, $E(c,0)$ be coordinates for $ADE$. Then $H(0, -\frac{bc}{h})$, $M(\frac{b}{2}, \frac{h}{2})$, $N(\frac{c}{2}, \frac{h}{2})$ and $T(\frac{b + c}{5}, \frac{3h}{5})$. We compute $O_{D}(x_{D}, y_{D})$ the circumcentre of $DMH$ and obtain $O_{E}$ by symmetry. We then have to verify that $O_{D}O_{E} \perp HT$, which can be done by comparing slopes.

The centre $O_{D}$ can be given by the intersection of perpendicular bisectors of $DM$ and $DH$. This gives the following system of equations on $x_{D}, y_{D}$:

$$h \cdot x_{D} + c \cdot y_{D} = \frac{b h}{2} - \frac{b c^{2}}{2 h}$$
$$-b \cdot x_{D} + h \cdot y_{D} = -\frac{3 b^{2}}{4} + \frac{h^{2}}{4}$$
Solving the system gives

$$(h^{2} + b c)(x_{D} - x_{E}) = \frac{b - c}{4} (5b c + 3h^{2})$$
$$(h^{2} + b c)(y_{D} - y_{E}) = -\frac{b - c}{4} h(b + c)$$
So $\frac{y_{D} - y_{E}}{x_{D} - x_{E}} = -\frac{h(b + c)}{5b c + 3h^{2}}$.

The other slope is more immediate:

$$\frac{y_{T} - y_{H}}{x_{T} - x_{H}} = \frac{3h / 5 + b c / h}{(b + c) / 5} = \frac{5b c + 3h^{2}}{h(b + c)} = -\frac{x_{D} - x_{E}}{y_{D} - y_{E}}$$
so indeed the two slopes correspond to perpendicular lines.