Quite hard problem. I unfortunately did not solve this synthetically, and I think few of the graders who attempted it were able to either. Nonetheless, I did hear there were some correct solutions during the contest (including someone who solved only this problem!?), so it is doable and I would like to see such a solution. Here is my rather boring one. [asy][asy] /* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(12cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair M_A = (-1.790212311643596, 6.877060317202126); pair M_B = (-4.0, -3.0); pair M_C = (8.0, -3.0); pair C = M_A+M_B-M_C; pair B = M_C+M_A-M_B; pair A = M_B+M_C-M_A; pair O = circumcenter(A,B,C); pair H = orthocenter(A,B,C); pair N_9 = circumcenter(M_A,M_B,M_C); pair S = (-0.49630963792819704, -0.2453435658119223); pair X = (2)*(foot(N_9,M_A,S))-M_A; pair Y = (2)*(foot(N_9,M_B,S))-M_B; pair Z = (2)*(foot(N_9,M_C,S))-M_C; pair P = IntersectionPoint(Line(A,X,lisf),Line(B,Y,lisf)); pair D = IntersectionPoint(Line(A,X,lisf),Line(M_B,M_C,lisf)); pair E = IntersectionPoint(Line(B,Y,lisf),Line(M_A,M_C,lisf)); pair F = IntersectionPoint(Line(M_A,M_B,lisf),Line(C,P,lisf)); pair X_prime = M_B+M_C-X; pair D_prime = M_B+M_C-D; pair E_prime = M_A+M_C-E; pair F_prime = M_A+M_B-F; /* Draw objects */ draw(M_A--M_B, rgb(0.0,0.2,0.8)); draw(M_B--M_C, rgb(0.0,0.2,0.8)); draw(M_C--M_A, rgb(0.0,0.2,0.8)); draw(circumcircle(M_A,M_B,M_C)); draw(M_B--Y, dotted); draw(M_C--Z, dotted); draw(D--M_A, rgb(1.0,0.5,0.0)); draw(E--M_B, rgb(1.0,0.5,0.0)); draw(F--M_C, rgb(1.0,0.5,0.0)); draw(M_A--X, dotted); draw(A--D, rgb(1.0,0.2,0.8)); draw(M_A--D_prime, rgb(1.0,0.2,0.8) + linewidth(1.0)); draw(M_C--F_prime, rgb(1.0,0.2,0.8) + linewidth(1.0)); draw(M_B--E_prime, rgb(1.0,0.2,0.8) + linewidth(1.0)); draw(A--M_B, rgb(0.2,0.6,0.0) + linetype("4 4")); draw(A--M_C, rgb(0.2,0.6,0.0) + linetype("4 4")); draw(A--M_A, rgb(0.2,0.6,0.0) + linetype("4 4")); /* Place dots on each point */ dot(M_A); dot(M_B); dot(M_C); dot(A); dot(O); dot(S); dot(X); dot(Y); dot(Z); dot(D); dot(E); dot(F); dot(X_prime); dot(D_prime); dot(E_prime); dot(F_prime); /* Label points */ label("$M_A$", M_A, lsf * dir(135)); label("$M_B$", M_B, lsf * dir(225)); label("$M_C$", M_C, lsf * dir(-45)); label("$A$", A, lsf * dir(45)); label("$O$", O, lsf * dir(45)); label("$S$", S, lsf * dir(45)); label("$X$", X, lsf * dir(225)); label("$Y$", Y, lsf * dir(45)); label("$Z$", Z, lsf * dir(170)); label("$D$", D, lsf * dir(225)); label("$E$", E, lsf * dir(45)); label("$F$", F, lsf * dir(160)); label("$X'$", X_prime, lsf * dir(45)); label("$D'$", D_prime, lsf * dir(-45)); label("$E'$", E_prime, lsf * dir(45)); label("$F'$", F_prime, lsf * dir(160)); [/asy][/asy] Let $AX$ meet $M_BM_C$ at $D$, and let $X$ reflected over $M_BM_C$'s midpoint be $X'$. Let $Y'$, $Z'$, $E$, $F$ be similarly defined. By Cevian Nest Theorem it suffices to prove that $M_AD$, $M_BE$, $M_CF$ are concurrent. Taking the isotomic conjugate and recalling that $A$ $M_AM_BAM_C$ is a parallelogram, we see that it suffices to prove $M_AX'$, $M_BY'$, $M_CZ'$ are concurrent. We now use barycentric coordinates on $\triangle M_AM_BM_C$. Let $S = \left( a^2S_A + t : b^2S_B + t : c^2S_C + t \right)$ (possibly $t=\infty$ if $S$ is the centroid). Let $v = b^2S_B+t$, $w=c^2S_C+t$. Hence \[ X = \left( -a^2 vw : (b^2w+c^2v)v : (b^2w+c^2v)w \right). \] Consequently, \[ X' = \left( a^2vw : -a^2vw + (b^2w+c^2v)w : -a^2vw + (b^2w+c^2v)v \right) \] We can compute \[ b^2w+c^2v = (bc)^2(S_B+S_C)+ (b^2+c^2)t = (abc)^2 + 2t. \] so \[ -a^2v+b^2w+c^2v = (b^2+c^2) + (abc)^2 - (ab)^2S_B - a^2t = S_A(ab+t). \] Thus \[ X' = \left( a^2vw : S_A (b^2S_B+t) (ab+t) : S_A (c^2S_C+t) (ac+t) \right) \] and from this it's evident that $AX'$, $BY'$, $CZ'$ are concurrent.

This problem is a special case of the problem Concurrent 14 (b)

Here is a quick "almost synthetic" solution. The non-synthetic part is showing that the $AX,BY,CZ$ concurring is equivalent to $H_aX,H_bY,H_cZ$ concurring, which follows by trig Ceva in a way similar to the Cevian Nest Theorem. ($H_a,H_b,H_c$ are the feet of the altitudes.) Let $H,N$ denote the orthocenter, and nine-point center as usual, and let $T_a,T_b,T_c$ (not sure what the standard notation for this is) be the midpoints of $AH,BH,CH$. By Pascal on $M_bT_bH_bM_cT_cH_c$, $M_cH_b$ and $M_bH_c$ meet on the Euler line $NH$. By Pascal on $M_bYH_bM_cZH_c$, $H_bY$ and $H_cZ$ meet on the Euler line. Thus $H_aX,H_bY,H_cZ$ concur and we are done. Unfortunately getting up to saying Pascal but not saying which hexagons was only worth 1 point QQ

I kind of find this more straightforward then the other two problems, especially using a ceva approach. A sketch: Applying Ceva's directly on $ABC$ and $X,Y,Z$ we need to show that (directed angles clockwise ($mod 180$)) $$\frac{sin\angle(XA,AB)}{sin\angle(XA,AC)}\cdot\frac{sin\angle(ZC,AC)}{sin\angle(ZC,BC)}\cdot\frac{sin\angle(YB,BC)}{sin\angle(YB,AB)}=1$$$$\Leftrightarrow\frac{sin\angle(XM_C,AB)}{sin\angle(XM_B,AC)}\cdot\frac{sin\angle(ZM_B,AC)}{sin\angle(ZM_A,BC)}\cdot\frac{sin\angle(YM_A,BC)}{sin\angle(YM_C,AB)}=1$$(As $\frac{M_CX}{M_BX}\cdot\frac{M_BZ}{M_AZ}\cdot\frac{M_AY}{M_CY}=1$). We want to bring the angles into the circle. Now from some pascals we know that $M_AH_B,M_BH_A$ intersect on the Euler line, and similarly for other pairs. Let those intersections be $P,Q,R$ ($P$ corresponds to $M_BH_C,M_CH_B$ and etc.). Thus $$\frac{sin\angle(XM_C,AB)}{sin\angle(ZM_A,BC)}=\frac{sin\angle(QM_A,M_AS)}{sin\angle(QM_C,M_CS)}=\frac{sin\angle(M_AQ,OH)}{sin\angle(M_CQ,OH)}\cdot\frac{M_CS}{M_AS}$$ By multiplying the equations obtained from the remaining two pairs of angles we are left to show: $$\frac{sin\angle(M_AQ,OH)}{sin\angle(M_CQ,OH)}\cdot\frac{sin\angle(M_CP,OH)}{sin\angle(M_BP,OH)}\cdot\frac{sin\angle(M_BR,OH)}{sin\angle(M_AR,OH)}=1$$$$\Leftrightarrow\frac{H_AQ}{H_AR}\cdot\frac{H_BR}{H_BP}\cdot\frac{H_CP}{H_CQ}=1$$(Pairing $sin\angle(M_AQ,OH),sin\angle(M_BP,OH)$ and etc) $$\Leftrightarrow \frac{sin\angle H_AH_CM_A}{sin\angle H_CH_AM_C}\cdot\frac{sin\angle H_CH_BM_C}{sin\angle H_BH_CM_B}\cdot\frac{sin\angle H_BH_AM_B}{sin\angle H_AH_BM_A}=1$$(Pairing $H_AQ,H_CQ$ and etc) which is easy to show.

I found a generalisation a while ago and posted it on my blog: Let $ABC$ be a triangle, $P, Q$ be points with cevian triangles $P_AP_BP_C$ and $Q_AQ_BQ_C$ respectively. Denote $\mathcal{C}$ as the conic through $P_AP_BP_CQ_AQ_BQ_C$. Consider point $R$ on $PQ$. Let $Q_AR \cap \mathcal{C} = X$ and define $Y, Z$ similarly. Then, $ABC$ and $XYZ$ are perspective. Proof: Let $Q_AX \cap AP = R_A$ and define $R_B$ similarly. Then by cross ratio $P_BR_A \cap P_AR_B = W$ lies on a conic as $R$ moves through $P_A$. But of course $CR \cap P_BR_A$ moves along a conic also through $P_A$. If we take the $R$ for the four positions $P, Q, \{PQ \cap \mathcal{C}\}$ then we have the conics coincide, hence $W \in CR$. Now, because \[A(X,P,R,C)=P_A(X,R_A,R,C) = P_B(Y,R_B,R,C) = B(Y,P,R,C)\] $\implies AX \cap BY$ lies on a conic through $ABCPR$ so done. Indeed, from this one gets that the perspector of $ABC$ and $XYZ$ lies on the rectangular circumhyperbola through $S$. Telv Cohl has a nice solutoin to this generalisation using Jerabek theorem and pascal theorem.

Thanks for IDMasterz for remind me about that . Here (Post #6) is the link of my proof ( take $ P, Q $ as the Orthocenter, Centroid of $ \triangle ABC $ we get the original problem ) .

Sine rule in $\triangle{AXH_c}$ yeilds $\frac{H_cX}{sin\angle{BAX}}=\frac{AX}{sin\angle{XH_cA}}=\frac{AX}{sin\angle{M_cM_aX}}$ Apply sine rule in $\triangle{AXH_b}$ similarly and divide to obtain $\frac{sin{\angle{BAX}}}{sin{\angle{CAX}}}=\frac{sin{\angle{M_cM_aX}}}{sin{\angle{M_bM_aX}}} \cdot \frac{H_cX}{H_bX}$.Using these facts for points $Y$ and $Z$ and the fact that $M_aX,M_bY,M_cZ$ concur we see that we need to prove $\frac{H_bX}{H_cX} \cdot \frac{H_cY}{H_aY} \cdot \frac{H_aZ}{H_bZ}=1$. In other words we need to prove that $H_aX,H_bY,H_aZ$ are concurrent.Let $J_a,J_b,J_c$ be the midpoints of $AH,BH,CH$ respectively.Apply Pascal's theorem with $M_aXJ_bM_bYJ_a$ we get $J_aY \cap J_bX$ lies on the euler line.Now Pascal with $J_aH_aXJ_bH_bY$ gives that $H_aX \cap H_bY$ lies on the euler line.Similarly $H_bY \cap H_cZ$ lies on the euler line.Thus they concur on the euler line as desired.

IDMasterz wrote: I found a generalisation a while ago and posted it on my blog: Let $ABC$ be a triangle, $P, Q$ be points with cevian triangles $P_AP_BP_C$ and $Q_AQ_BQ_C$ respectively. Denote $\mathcal{C}$ as the conic through $P_AP_BP_CQ_AQ_BQ_C$. Consider point $R$ on $PQ$. Let $Q_AR \cap \mathcal{C} = X$ and define $Y, Z$ similarly. Then, $ABC$ and $XYZ$ are perspective. Another proof:(long ) $AP_A,BP_B,CP_C$ cut again to $\mathcal{C}$ at $A^*,B^*,C^*$ respectively. From Pascal's theorem on $Q_A,Q_C,P_A,P_C,A^*,C^*$ we get $AC,Q_AQ_C,A^*C^*$ are concurrent. i.e. $\triangle A^*B^*C^*$ is perspective to $\triangle Q_AQ_BQ_C$, with the same perspective axis that the triangles $ABC$ and $Q_AQ_BQ_C$ are perspectives. i.e. the perspective centers are collinear. Let $K$ be the perspective center between $\triangle A^*B^*C^*$ and $\triangle Q_AQ_BQ_C$, hence $K \in PQ$, since the composition of involutions that fixed $\mathcal{C}$ with centers $P,K,R$ changes $P_A$ and $X$, hence is a involution too, with pole on $PQ$ i.e. $\triangle XYZ$ is perspective to $\triangle P_AP_BP_C$ with center on $PQ$. From post-4-part (b) we get $ABC$ is perspective to $XYZ$.

Delightful configuration!

Solution Let $H,O$ be the orthocenter and circumcenter of $\triangle ABC$, $AD,BE,CF$ are the altitudes of $\triangle ABC$. $AU,BV,CW$ are concurrent if and only if $\prod \frac{sin \angle UAC}{sin \angle UAB} = 1$ $\leftrightarrow \prod \frac{UE}{UF}$ $.\prod \frac{sin \angle UEA}{sin \angle UFA} = 1$ $\leftrightarrow$ $\prod$ $\frac{sin \angle UFE}{sin \angle UEF}$. $\prod$ $\frac{\sin \angle UM_aM_b}{sin \angle UM_aM_c}= 1 \leftrightarrow \prod \frac{sin \angle UFE}{sin \angle UEF} \leftrightarrow $ $DU,EV,FW$ are concurrent So we need to prove $DU,EV,FW$ are concurrent on $OH$ Let $AH,BH,CH$ cuts Euler circle at $M,N,P$ Applying Pascal $\begin{pmatrix} M_a & P & W\\ M_c & M & U \end{pmatrix}$ and $\begin{pmatrix} F & U & M\\ D & W & P \end{pmatrix}$ we have $FW,DU, OH$ are concurrent Similary, we have $DU,EV,FW,OH$ are concurrent

Let $H_a, H_b, H_c$ be the feet of the altitudes from $A, B, C$ in $\triangle ABC$. From Pappus we know that $H_cM_b\cap M_cH_b\in OH$, so by Pascal on $ZH_cM_bYH_bM_c$ we get that $ZH_c\cap YH_b \in OH$ as well. Then we necessarily have that $H_aX, H_bY, H_cZ, OH$ are concurrent at a point $T$. So by Jerabek's theorem on $\triangle XYZ$ with respect to $\{S, T\}$, we have that $\triangle ABC$ and $\triangle XYZ$ are perspective, implying the result.

Can anyone say what Cevian Nest theorem and Jerabek's theorem is?

Murad.Aghazade wrote: Can anyone say what Cevian Nest theorem and Jerabek's theorem is? Cevian Nest Theorem: Let $\triangle UVW$ be the cevian triangle of $P$ with respect to $\triangle ABC$, and let $\triangle XYZ$ be the cevian triangle of $Q$ with respect to $\triangle UVW$. Then $AX, BY, CZ$ are concurrent. Jerabek's Theorem: Let $\triangle UVW$ and $\triangle XYZ$ be the circumcevian triangles of $P$ and $Q$ with respect to $\triangle ABC$. Then the triangle formed by the lines $\{XU, YV, ZW\}$ is perspective with $\triangle ABC$.

So here is an "algebraic geometry" solution: We proceed by "barycentric coordinates" with respect to $ABC$. Suppose $X=(1-t)M_a + tS$. Since $X$ is on the nine point circle, we have \[(X-N)^2 = R^2/4\]where a vector squared is it dotted with itself, and here $R/2$ is the radius of the nine point circle. Note then that we have \[(t(S-M_a)+M_a-N)^2 = R^2/4,\]or \[t^2(S-M_a)^2 + 2t(S-M_a)\cdot(M_a-N)+(M_a-N)^2-R^2/4=0.\]But $M_a$ is on the nine point circle so that constant term is $0$. Thus, we have that \[t_A = -\frac{2(M_a-N)\cdot(S-M_a)}{(S-M_a)^2},\]where we label $t$ as $t_A$ to distinguish the three cases. Now $S$ varies along the Euler line, so there exist some linear functions $\alpha,\beta,\gamma$ such that $S=(\alpha(s),\beta(s),\gamma(s))$. Then, we see that \[t_A = \frac{f_A(s)}{g_A(s)}\]where $f_A$ is linear and $g_A$ is quadratic, so \[X=\frac{1}{g_A(s)}(\alpha_1(s),\beta_1(s),\gamma_1(s)) = (\alpha_1(s):\beta_1(s):\gamma_1(s))\]and similarly for $Y,Z$ where all the $\alpha_i,\beta_i,\gamma_i$ are some explicit quadratic functions of $s$. Now, the statement that $AX,BY,CZ$ concur is by Ceva's theorem equivalent to \[\frac{\gamma_1(s)}{\beta_1(s)}\cdot\frac{\alpha_2(s)}{\gamma_2(s)}\cdot\frac{\beta_3(s)}{\alpha_3(s)}=1,\]which when clearing the denominator is equivalent some sixth degree polynomial in $s$ equal to $0$. Thus, it suffices to verify the statement for $7$ distinct values of $s$, so it suffices to show the problem for $7$ different points $S$ on the Euler line. But one can easily check that the nine point center is sent to the orthocenter, the centroid and the two intersections of the euler line with the nine point circle are sent to themselves, and the intersections of the euler line with sides of the medial triangle are sent to $A,B,C$. Thus the problem has been verified for $7$ points, so it must be true for all points.

Here's a sort of mixed solution: let $H$ be the orthocenter of $ABC$ and let $H_a,H_b,H_c$ be such that $H_a = AH\cap BC$, etc. we have \begin{align*} \frac{\sin\angle BAX}{\sin\angle CAX} &= \frac{M_cX}{M_bX} \cdot \frac{M_bX}{\sin\angle M_bAX} \bigg/ \frac{M_cX}{\sin\angle M_cAX} \\ &= \frac{\sin\angle M_cM_aX}{\sin\angle M_bM_aX} \cdot \frac{\sin\angle AM_cX}{\sin\angle AM_bX} \\ &= \frac{\sin\angle M_cM_aX}{\sin\angle M_bM_aX} \cdot \frac{\sin\angle H_cH_aX}{\sin\angle H_bH_cX}\end{align*}Taking the product cyclically, it suffices to prove that $H_aX, H_bY, H_cZ$ concur. In fact, we prove that they concur at a point on the Euler line. Let $\phi_a$ denote the composition of maps taking \[S \overset{M_a}{\longrightarrow} X \overset{H_a}{\longrightarrow} T_a\]where $T_a$ is the intersection between the Euler line and $XH_a$. Similarly define $\phi_b$, we will prove that $\phi_a = \phi_b$. Because the maps $\phi_a, \phi_b$ are projective, it suffices to verify this for three distinct points: When $S$ is the circumcenter of $ABC$, $T_a = T_b$ is the nine-point center. When $S$ is the nine-point center, $T_a=T_b=H$. When $S$ is on the nine-point circle, $T_a=T_b=S$. Therefore the two maps are indeed the same as desired.

Is it possible to complex bash? I couldn't do it.

Pretty similar to other solutions. Let $D,E,F$ be the orthic triangle of $\triangle ABC$. By the law of sines in $\triangle AM_CX$ and $\triangle AM_BX$, we have \[\frac{\sin \angle M_CAX}{\sin\angle AM_CX}=\frac{M_CX}{AX}\qquad \text{and}\qquad \frac{\sin \angle XAM_B}{\sin \angle AM_BX}=\frac{XM_B}{AX}\]Dividing them shows that \[\frac{\sin \angle M_CAX}{\sin \angle XAM_B}=\frac{\sin \angle AM_CX}{\sin \angle AM_BX}\cdot \frac{M_CX}{XM_B}\]By the trigonometric form of Ceva's theorem, it suffices to show that \[\frac{\sin\angle AM_CX}{\sin \angle AM_BX}\cdot \frac{\sin \angle CM_BZ}{\sin\angle CM_AZ}\cdot \frac{\sin\angle BM_AY}{\sin \angle BM_CY}\cdot \frac{M_CX}{XM_B}\cdot \frac{M_BZ}{ZM_A}\cdot \frac{M_AY}{YM_C}=1\]Since $M_AX,M_BY,M_CZ$ are concurrent at $S$, the latter three terms multiply to one, so by easy angle chasing, we just have to show that $DX,EY,FZ$ are concurrent. Now let $N$ be the nine-point center, and let $D',E',F'$ be the antipodes of $D,E,F$ in the nine-point circle. Then $D'M_A$ is perpendicular to $BC$, so it passes through $O$. Similary, $E'M_B$ and $F'M_C$ also pass through $O$. Then the desired concurrency follows by the Circumcevian Pingpong theorem (see https://artofproblemsolving.com/community/c6h66285p391497) on triangles $\triangle XYZ,\triangle M_AM_BM_C, \triangle D'E'F', \triangle DEF$, and points $S,O,N$. In addition, this concurrency point also lies on the Euler line as well.

First fix the triangle $ABC$ and animate $S$ on the Euler line. Now notice how $X,Y,Z$ have a degree of $2$, this hold ssince $S \rightarrow X$ is projective. Thus we must have that $AX,BY,CZ$ all have a degree of $2$. The hypothetical intersection point must have at most $2+2+2=6$ degree. Thus we need to cover $7$ special cases. The first case is when $S$ is the center of the nine-point circle. We easily see that the intersection of $AX,BY,CZ$ is the orthocenter of $ABC$. The second case is when $S$ is the centroid $G$ of $ABC$, again easily seen that $AX,BY,CZ$ are concurrent at $G$. The third and fourth case is when $S$ is on the nine-point circle. It is easily see that $AX,BY,CZ$ are concurrent on the nine-point circle. The fifth,sixth and seventh cases are when $S$ is the intersection of the Euler line with $M_aM_b,M_bM_c,M_cM_a$, respectively. It is easily seen by going case to case that the intersections are $C,A$ and $B$, respectively. Thus by the moving points we have that $AX,BY$ and $CZ$ are concurrent.

Wait what Let $D, E,F$ be the feet of the altitudes from $A,B,C$ respectively. Let $\ell$ be the euler line, $O,G,N,H$ the circumcenter, centroid, nine-point center, orthocenter of $\triangle ABC$, respectively. Observe that \[\frac{\sin \angle M_CAX}{M_CX} = \frac{\sin \angle AM_C X}{AX} \text{ and } \frac{\sin \angle M_B AX}{M_B X} = \frac{\sin \angle AM_B X}{AX}\]Dividing these two gives \[\frac{\sin \angle M_CAX}{\sin \angle M_B AX} = \frac{\sin \angle AM_CX}{\sin \angle AM_BX}\cdot \frac{M_CX}{M_BX}\]Multiplying this cyclically over $A,B,C$, we need to show \[1 = \frac{\sin \angle M_C AX}{M_B AX}\cdot \frac{\sin M_B CZ}{\sin M_A CZ} \cdot \frac{\sin M_A BY}{\sin M_C BY} = \left(\frac{M_CX}{M_BX} \cdot \frac{M_B Z}{M_A Z}\cdot \frac{M_AY}{M_CY}\right) \cdot \left(\frac{\sin \angle AM_C X}{\sin \angle AM_B X}\cdot \frac{\sin \angle CM_B Z}{\sin \angle CM_A Z}\cdot \frac{\sin \angle BM_AY}{\sin \angle BM_CY}\right)\]Since $M_AX, M_BY, M_CZ$ are collinear, this means $\frac{M_CX}{M_BX} \cdot \frac{M_BZ}{M_A Z}\cdot \frac{M_AY}{M_CY} = 1$. To show the other term is $1$, observe that \[\angle AM_C X = 180 - \angle FM_CX = \angle FM_AX = \angle FDX\]Therefore, it suffices to prove that $DX, EY, FZ$ are concurrent. Let $T$ be one of the intersections with the nine point circle, and $M_AO, M_AN\cap (M_AM_BM_C) = A_1, A_2$ respectively. Observe that $DA_1\cap \ell = N$ since $\angle DM_A A_1 = 90$, and $DA_2\cap \ell = H$ since $\angle M_ADA_3 = 90$.Then, \[(ON;ST) \overset{M_A}{=} (A_1A_2; XT) \overset{D}{=} (NH; DX\cap \ell T)\]Now, we get the same thing with respect to $E,F$. Therefore, $DX\cap \ell = EY\cap \ell = FZ \cap \ell$ so $DX, EY, FZ$ are concurrent. We conclude that $AX,BY,CZ$ concur.

Fix $\triangle ABC$ and animate $S$ along the Euler line. Since $S \mapsto X$ is a projective map, $X, Y, Z$ all have degree $2$, from which the desired concurrence of $\overline{AX}, \overline{BY}, \overline{CZ}$ has degree $\deg X + \deg Y + \deg Z = 6$. Thus, it suffices to prove the problem for $7$ distinct positions of $S$. Take $S$ to be the nine-point center. Then $\overline{AX}, \overline{BY}, \overline{CZ}$ all contain the orthocenter of $\triangle ABC$. Take $S$ to be the centroid $G$ of $ABC$. Then $\overline{AX}, \overline{BY}, \overline{CZ}$ all contain $G$. Take $S$ to be either of the two intersections of the Euler line and nine-point circle. Then $\overline{AX}, \overline{BY}, \overline{CZ}$ all contain $S$. Take $S$ to be the intersection of $M_bM_c$ and the Euler line. Then $\overline{AX}, \overline{BY}, \overline{CZ}$ all contain $A$. We can achieve two other analogous configurations by considering $M_aM_c$ and $M_bM_c$. Therefore, the above configurations finish.

Let $T_a, T_b, T_c$ be the feet of the altitudes from $A, B, C$ to $BC, CA, AB$ respectively. Let $K_a, K_b, K_c$ be the midpoints of $AH, BH, CH$ respectively. Then the nine-point circle is $\Gamma = (M_aM_bM_cT_aT_bT_cK_aK_bK_c)$. Now, note that $\angle T_aM_aO = 90^{\circ}$ and $K_a$ is the antipode of $M_a$, so by Pascal on $T_aK_aM_aT_bK_bM_b$, we have that $M_aT_b \cap M_bT_a$ lies on the Euler line. Call this point $S_c$, and define $S_a, S_b$ similarly. Now define $f(P)$ on the Euler line to equal $\frac{PS}{PN}$. Then it follows that $\frac{f(S_b)}{f(S_c)} = (T_cT_b; XK_a)_{\Gamma}$ by projection through $M_a$, so taking a cyclic product gives that $1 = \prod \frac{T_cX}{T_bX} \cdot \prod \frac{T_bK_a}{T_cK_a}$. However the latter product is equal to $1$ by Trig Ceva within $T_aT_bT_c$, so it follows that $1 = \prod \frac{T_cX}{T_bX} = \prod \frac{\sin \angle AM_cX}{\sin \angle AM_bX}$. Now, we will prove the desired result via Trig Ceva. By LoS, it is equivalent to showing that $\prod \frac{XM_c}{XM_b} \cdot \prod \frac{\sin \angle AM_cX}{\sin \angle AM_bX} = 1$. However, we know the latter product is equal to $1$, so it suffices to show that $\prod \frac{XM_c}{XM_b}$. By the inversion length formula on the inversion at $S$ fixing $\Gamma$, it follows that: $$\prod \frac{XM_c}{XM_b} = \prod \frac{XM_c}{ZM_a} = \prod \frac{r^2}{SZ \cdot SM_a} = \prod \frac{r^2}{SX \cdot SM_a} = \prod \frac{r^2}{r^2} = 1$$ This finishes.

mmp is a drug Animate $S$ linearly on the Euler line $\ell$. Then $X,Y,Z$ move with degree $2$, so the intersection condition has degree $6$. It thus suffices to check $7$ cases: When $S$ is either of the two intersections with the $\ell$ and, $X=Y=Z=S$ and the concurrency point is $S$ When $S=N_9$, the center of $(M_AM_BM_C)$, $X$ is the midpoint of $\overline{AH}$ (where $H$ is the orthocenter of $\triangle ABC$) and symmetric statements are true for $Y,Z$, so the concurrency point is $H$. When $S=G$, the centroid of $\triangle ABC$, the concurrency point is clearly the centroid. When $S=\ell \cap \overline{M_bM_c}$, $Y=M_c$ and $Z=M_b$, so the concurrency point is $A$. Likewise, the concurrency points when $S$ is the intersection of $\overline{M_aM_c}$ and $\overline{M_aM_b}$ are $B$ and $C$ respectively. This gives a total of $2+1+1+3=7$ cases, as desired. Note that this solution assumes all seven choices are distinct. If they aren't distinct, we can move $A$ around a bit to make them distinct and use continuity. This finishes the problem. $\blacksquare$

IDMasterz wrote: I found a generalisation a while ago and posted it on my blog: Let $ABC$ be a triangle, $P, Q$ be points with cevian triangles $P_AP_BP_C$ and $Q_AQ_BQ_C$ respectively. Denote $\mathcal{C}$ as the conic through $P_AP_BP_CQ_AQ_BQ_C$. Consider point $R$ on $PQ$. Let $Q_AR \cap \mathcal{C} = X$ and define $Y, Z$ similarly. Then, $ABC$ and $XYZ$ are perspective. Proof: Let $Q_AX \cap AP = R_A$ and define $R_B$ similarly. Then by cross ratio $P_BR_A \cap P_AR_B = W$ lies on a conic as $R$ moves through $P_A$. But of course $CR \cap P_BR_A$ moves along a conic also through $P_A$. If we take the $R$ for the four positions $P, Q, \{PQ \cap \mathcal{C}\}$ then we have the conics coincide, hence $W \in CR$. Now, because \[A(X,P,R,C)=P_A(X,R_A,R,C) = P_B(Y,R_B,R,C) = B(Y,P,R,C)\] $\implies AX \cap BY$ lies on a conic through $ABCPR$ so done. Indeed, from this one gets that the perspector of $ABC$ and $XYZ$ lies on the rectangular circumhyperbola through $S$. Telv Cohl has a nice solutoin to this generalisation using Jerabek theorem and pascal theorem. I don't understand this finish. Here's a different one as such, or maybe it's the same one with more details. Note that $\ell_{P_BQ_B} \cup \ell{P_AQ_A}, (XYP_BQ_BP_AQ_A), (R_AR_BP_BQ_BP_AQ_A)$ are coconic, where the last one follows by $W - R - C$. As such, there exists a projective map $t \colon RR_A \to RR_B$ by the two intersections of the pencil of conics through $P_BQ_BP_AQ_A$. Then, \[ A(X,P,R,C) = (X, R_A, R, AC \cap RR_A) \overset{steiner}= (Y, R_A, R, BC \cap RR_B) = B(Y,P,R,C) \]where the $steiner$ step uses the fact that mapping on hte mentioned conics.

Ratio bash :p Let $H, G, N$ be the orthocenter, centroid and nine-point center respectively. Let $\Delta L_aL_bL_c$ be an orthic triangle. Let $T_a = L_bL_c \cap M_bM_c$ and so on, and $AL_a, AX$ meet $\omega$ again at $R_a, X'$ respectively and so on. It is well-known that $M_aR_a \cap M_bR_c \cap M_cR_c = N$. All poles and polars mentioned below are with respect to the nine-point circle. Claim 1. Line $AT_bT_c$ is the polar of $T_a$ and so on. In particular, $\Delta T_aT_bT_c$ is self-polar. Proof. Firstly, $A, T_b, T_c$ are collinear because $L_a(T_b,T_c;A,M_a) = M_a(T_b,T_c;A,L_a)=-1$. Also, by Pascal on $L_aL_cM_bM_aM_cL_b$, we get $T_b, T_c, L_cM_b \cap M_cL_b$ collinear, but since $A$ and $L_cM_b \cap M_cL_b$ both lie on the polar of $T_a$ by Brocard, the desired claim is implied. Claim 2. $T_a$ lies on $Y'Z'$ and so on. Proof. Note that \[ (L_c,L_b;Y',M_b) = (M_c, R_b; Y, BM_b \cap \omega) = (M_bM_c \cap HG, N; S, G) = (M_b, R_c; Z, CM_c \cap \omega) = (L_b, L_c; Z', M_c) \]so $L_bL_c, Y'Z', M_bM_c$ concur. Claim 3. $X'T_a, Y'T_b, Z'T_c$ concur at some $P \in \omega$. Proof. It follows directly from Claims 1 and 2: if $X'T_a$ meets $\omega$ again at $P$, then by Brocard $Y'P \cap X'Z'$ lies on $T_bT_c$, so it must be actually $T_b$. Similarly, $Z'P \cap Z'Y' = T_c$. Claim 4. $AT_a, BT_b, CT_c$ concur projectively. (They are actually parallel, but we won't need that.) Proof. Note that \[ \prod_{\text{cyc}} \frac{\sin\angle M_bAT_a}{\sin\angle T_aAM_c} = \prod_{\text{cyc}} \frac{M_bT_a \cdot AM_c}{T_aM_c \cdot AM_b} = \prod_{\text{cyc}} \frac{M_bL_b \cdot AL_c \cdot AM_c}{L_bA \cdot M_cL_c \cdot AM_b } = \prod_{\text{cyc}} \frac{AL_c}{L_bA} = 1 \]by four applications of Ceva, so the conclusion follows by Trig Ceva. Finally note that \[ \prod_{\text{cyc}} \frac{\sin \angle T_cX'A}{\sin \angle AX'T_b} = \prod_{\text{cyc}} \frac{T_cA \cdot X'T_b}{AT_b \cdot X'T_c} = 1 \]by two more applications of Ceva, so the problem is proved by Trig Ceva.

EulerMacaroni wrote: Murad.Aghazade wrote: Can anyone say what Cevian Nest theorem and Jerabek's theorem is? Cevian Nest Theorem: Let $\triangle UVW$ be the cevian triangle of $P$ with respect to $\triangle ABC$, and let $\triangle XYZ$ be the cevian triangle of $Q$ with respect to $\triangle UVW$. Then $AX, BY, CZ$ are concurrent. Jerabek's Theorem: Let $\triangle UVW$ and $\triangle XYZ$ be the circumcevian triangles of $P$ and $Q$ with respect to $\triangle ABC$. Then the triangle formed by the lines $\{XU, YV, ZW\}$ is perspective with $\triangle ABC$. Can you provide a proof of jerabek's theorem?

$ $