Solution:
Let \(X\) be the second intersection of \((D H M)\) and \((E H N)\) and let \(O^{\prime}\) be the circumcentre of \((A M N)\) Note that \(\angle M X N = \angle M D H + \angle N E H = 180^{\circ} - 2\angle D A E\) and since \(\angle M O^{\prime}N = 2\angle D A E\) we have that \(X,M,O^{\prime},N\) is cyclic and since \(\angle M X H = \angle N X H\) it means that \(H X\) is the angle bisector of \(\angle M X N\) but since \(O^{\prime}M = O^{\prime}N\) it means that \(H,X,O^{\prime}\) are collinear. Let \(B M\) and \(C N\) intersect at \(T\) and let \(K\) and \(L\) be the midpoints of \(M N\) and \(B C\) . Note that \(L\) is also the midpoint of \(D E\) . Since \(M N\) is parallel to \(B C\) 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, \(\frac{T L}{T K} = \frac{B C}{M N} = 6\) . Since \(K L = K A\) it means that \(\frac{A T}{T K} = \frac{T L - 2T K}{T K} = 4\) so by the lemma below, \(T\) lies on \(H O^{\prime}\) . Since \(H O^{\prime}\) is the radical axis of \((D H M)\) and \((E H N)\) we finish the problem using the Radical Axes Theorem \((T M\cdot T P = T N\cdot T Q)\) .

Lemma. Let \(A^{\prime}\) 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 \(A^{\prime}O\) 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 \(AH = 2OM\) . \(\square\)

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Let \(A B C\) be an acute triangle. Points \(B\) , \(D\) , \(E\) , and \(C\) lie on a line in this order and satisfy \(B D =\) \(D E = E C\) . Let \(M\) and \(N\) be the midpoints of \(A D\) and \(A E\) , respectively. Let \(H\) be the orthocentre of triangle \(A D E\) . Let \(P\) and \(Q\) be points on lines \(B M\) and \(C N\) , respectively, such that \(D\) , \(H\) , \(M\) , and \(P\) are concyclic and \(E\) , \(H\) , \(N\) , and \(Q\) are concyclic. Prove that \(P\) , \(Q\) , \(N\) , and \(M\) are concyclic.

The orthocentre of a triangle is the point of intersection of its altitudes.

Solution:
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

\[\angle OM_{2}H = \angle MM_{2}H = \angle MDH = \angle ADH = 90^{\circ} - \angle EAD = 90^{\circ} - \angle NAM = \angle O'M N,\]

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. \(\square\)

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Solution:
We compute using linear combinations with respect to \(A D E\) . 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 of \(T = B M\cap C N\) 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 \(H D M\) and \(H E N\) , whence \(T M\cdot T P = T N\cdot T Q\) proves the concyclicity of \(P,Q,N,M\)

Synthetic finish, similar to Solution 2. Let \(O\) be the circumcentre of \(A D E\) and \(O^{\prime} = \frac{A + O}{2}\) be the circumcentre of \(A M N\) . As in Solution 2, we have that \(O^{\prime}\) is on the desired radical axis, so it is enough to show \(T\in O^{\prime}H\) . Let \(G = \frac{A + D + E}{3}\) be the barycentre of \(A D E\) . By properties of the Euler line, we also have \(G = \frac{H + 2O}{3}\) . Now using our known identities we find

\[T = \frac{3A + D + E}{5} = \frac{2A + 3G}{5} = \frac{2A + H + 2O}{5} = \frac{H + 4O^{\prime}}{5}\]

and in particular \(T\in H O^{\prime}\) , as we wanted to show.

Computational finish. Let \(f(T)\) be the power difference at \(T\) w.r.t. \(D H M\) and \(E H N\) . 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 \(A D E\) and assume the diameter of its circumcircle is 1. We have

\[f(A) = \frac{A D^{2} - A E^{2}}{2} = \frac{\sin^{2}(\gamma) - \sin^{2}(\beta)}{2}.\]

To compute \(P(L)\) , let \(D^{\prime},E^{\prime}\) be the second intersection points of \(H M D,H N E\) with \(D E\) , and let \(M^{\prime},N^{\prime},F\) be the feet of the perpendiculars from \(H\) to \(A D,A E,D E\) , respectively. Note that \(D M^{\prime} =\) \(\sin \alpha \cos \beta\) , thus

\[M^{\prime}M = D M - D M^{\prime} = \frac{\sin(\alpha + \beta)}{2} - \sin \alpha \cos \beta = \frac{\sin(\beta - \alpha)}{2}.\]

We also have \(H M^{\prime} = \cos \alpha \cos \beta\) , \(H F = \cos \beta \cos \gamma\) , and \(H M^{\prime}M\sim H F D^{\prime}\) , therefore

\[F D^{\prime} = \frac{H F}{H M^{\prime}} M^{\prime}M = \frac{\cos\gamma\sin(\beta - \alpha)}{\cos\alpha}\]

and similarly

\[E^{\prime}F = \frac{\cos\beta\sin(\gamma - \alpha)}{\cos\alpha}.\]

We also have the standard \(FD = \sin \gamma \cos \beta\) and \(EF = \sin \beta \cos \gamma\) . We can finally compute

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

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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 \(D^{\prime}\neq D\) and \(E^{\prime}\neq E\) 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

\[f(A) = AM\cdot AD - AN\cdot AE = \frac{AD^2 - AE^2}{2} = \frac{FD^2 - FE^2}{2}\] \[\qquad = (\tau (E) - \tau (D))\left(\tau (F) - \tau \left(\frac{D + E}{2}\right)\right),\] \[f(D) = -DE\cdot DE^{\prime} = (\tau (E) - \tau (D))(\tau (D) - \tau (E^{\prime})),\] \[f(E) = ED\cdot ED^{\prime} = (\tau (E) - \tau (D))(\tau (E) - \tau (D^{\prime})).\]

Rearranging, it suffices to show that \(3\tau (F) = \tau (D^{\prime}) + \tau (E^{\prime}) + \tau ((D + E) / 2)\) . This can be rewritten as \(3F = D^{\prime} + E^{\prime} + (D + E) / 2\) . 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 \((A D E)\) 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 \(U M\perp A D\) and \(U H\perp D H\perp A E\) . 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

\[(d + e)u = d u + e(a d\overline{{u}}) = d(h + a e\overline{{h}}) \Rightarrow u = \frac{d h + a d e\overline{{h}}}{d + e}.\]

Computing \(v\) similarly, we get that

\[v:= H + 2A - (O + U + V) = h + 2a - \frac{(d + e)h + 2a d e\overline{{h}}}{d + e} = 2\left(a - \frac{a d e\overline{{h}}}{d + e}\right) = -\frac{2d e}{d + e}.\]

This displacement vector \(v\) satisfies \(v = d e\overline{{v}}\) and so it is orthogonal to line \(D E\) , as desired.