I had more fun constructing the diagram than trying to get anywhere with it, which I didn't. Plus, next time people post their solution to a geometry problem, I'd like to see at least one with a diagram.

awesomest solution ever: F is the isogonal conjugate of the point X along AM such that MA * MX = MB^2 = MC^2. everything else is angle chasing here

Let pt K is intersection of tangents to circumcircle of ABC at B and C. Draw AK. Let it intersect circumcircle of KBC at F'. This circumcircle also goes through O. Then after angle chasing get F' is the pt F (as angle BAK is equal to angle CAM). Then after a bit more angle chasing u r done.

I tried to use analytic geometry for this problem (which was a really stupid move), but didn't get the coordinates for F in time. I'd given up on angle chasing at this point.

Benjamin Hu wrote: I tried to use analytic geometry for this problem (which was a really stupid move), but didn't get the coordinates for F in time. I'd given up on angle chasing at this point. Haha, I did this as well. I actually got the coordinates of $ F$. They were really REALLY nasty. At that point, I was running out of time so I wrote "With the coordinates of the four points of interest, it is trivial to show the points lie on one circle." We'll see how the graders like that.

I used coordinates. I assigned the Origin to $ A$, $ (10,0)$ to $ B$ (I really should have used $ (4,0)$...), then $ (a,b)$ to $ C$, and $ a < 10$. I solved for lines and points and found the coordinates for $ F$ after a while. Then I found the equation of the circle containing $ A,$ $ N,$ and $ P$ (This was less messy and easier to do than if I tried $ A$, $ N$, $ F$ or any other way containing $ F$). At the end I said that if $ F$ was plugged into that equation, it would be true.

Outline of my solution: Median $ AM$ of a triangle $ ABC$ implies $ \frac{\sin{BAM}}{\sin{CAM}} = \frac{\sin{B}}{\sin{C}}$. Trig ceva for $ F$ shows that $ AF$ is a symmedian. Then $ FP$ is a median, use the lemma again to show that $ AFP = C$, and similarly $ AFN = B$, so you're done.

WLOG $ AB < AC$. The intersection of $ NE$ and $ PD$ is $ O$, the circumcenter of $ \triangle ABC$. Let $ \angle BAM = y$ and $ \angle CAM = z$. Note $ D$ lies on the perpendicular bisector of $ AB$, so $ AD = BD$. So $ \angle FBC = \angle B - \angle ABD = B - y$. Similarly, $ \angle FCE = C - z$, so $ \angle BFC = 180 - (B + C) + (y + z) = 2A$. Notice that $ \angle BOC$ intercepts the minor arc $ BC$ in the circumcircle of $ \triangle ABC$, which is double $ \angle A$. Hence $ \angle BFC = \angle BOC$, so $ FOBC$ is cyclic. Lemma 1: $ \triangle FEO$ is directly similar to $ \triangle NEM$ \[ \angle OFE = \angle OFC = \angle OBC = \frac {1}{2}\cdot (180 - 2A) = 90 - A \] since $ F$, $ E$, $ C$ are collinear, $ FOBC$ is cyclic, and $ OB = OC$. Also \[ \angle ENM = 90 - \angle MNC = 90 - A \] because $ NE\perp AC$, and $ MNP$ is the medial triangle of $ \triangle ABC$ so $ AB // MN$. Hence $ \angle OFE = \angle ENM$. Notice that $ \angle AEN = 90 - z = \angle CEN$ since $ NE\perp BC$. $ \angle FED = \angle MEC = 2z$. Then \[ \angle FEO = \angle FED + \angle AEN = \angle CEM + \angle CEN = \angle NEM \] Hence $ \angle FEO = \angle NEM$. Hence $ \triangle FEO$ is similar to $ \triangle NEM$ by AA similarity. It is easy to see that they are oriented such that they are directly similar. End Lemma 1. By the similarity in Lemma 1, $ FE: EO = NE: EM\implies FE: EN = OE: NM$. $ \angle FEN = \angle OEM$ so $ \triangle FEN\sim\triangle OEM$ by SAS similarity. Hence \[ \angle EMO = \angle ENF = \angle ONF \] Using essentially the same angle chasing, we can show that $ \triangle PDM$ is directly similar to $ \triangle FMO$. It follows that $ \triangle PDF$ is directly similar to $ MDO$. So \[ \angle EMO = \angle DMO = \angle DPF = \angle OPF \] Hence $ \angle OPF = \angle ONF$, so $ FONP$ is cyclic. In other words, $ F$ lies on the circumcircle of $ \triangle PON$. Note that $ \angle ONA = \angle OPA = 90$, so $ APON$ is cyclic. In other words, $ A$ lies on the circumcircle of $ \triangle PON$. $ A$, $ P$, $ N$, $ O$, and $ F$ all lie on the circumcircle of $ \triangle PON$. Hence $ A$, $ P$, $ F$, and $ N$ lie on a circle, as desired.

rem wrote: Let pt K is intersection of tangents to circumcircle of ABC at B and C. Draw AK. Let it intersect circumcircle of KBC at F'. This circumcircle also goes through O. Then after angle chasing get F' is the pt F (as angle BAK is equal to angle CAM). Then after a bit more angle chasing u r done. This is pretty much what I did. I know it's the day before the winner phone call. People are anxious about the result, so let's look at the intriguing background of symmedian from perspective drawing, while waiting patiently and calmly. In my Art I class, we learned perspective drawing, and my teacher claimed that one vanishing point is the midpoint of two others. I immediately wondered why. After I went home and did some calculation, the problem is equivalent to the following: Problem: Let $ \triangle AA'C$ be an isosceles triangle with $ AC = A'C$(C is a vanishing point), point $ E$ on $ AC$ and $ E'$ on $ A'C$ such that $ AA'//EE'$. $ X$ denotes the intersection of $ A'E$ and $ AE'$ and the gable(an art term) pass through $ X$ is parallel to $ AA'$. Let $ D$ be a point on the gable on the different side of $ AC$ as $ A'$, and $ AD$ intersect the line through $ C$ parallel to $ AA'$ at $ B$(This is a vanishing point). $ DE$ meets $ BC$ at $ B'$(third vanishing point). Prove that $ BC = B'C$. Solution: Let the gable meet $ AC$ at $ F$, since $ \frac {DF}{BC} = \frac {AF}{AC}$ and $ \frac {DF}{B'C} = \frac {FE}{ED}$ we want to show that $ AFEC$ is a harmonic range. This follows from \[ \frac {AF}{FE} = \frac {AA'}{EE'} = \frac {AC}{EC} \] Now we are going to look at a special case of the 1996 Chinese Math Olympiad Problem 1. Problem: With the same diagram as the previous problem, prove that the gable pass through the two points of tangency (namely, $ P$, $ Q$) from $ C$ to the circumcircle of $ AA'EE'$. Note: In the original problem, $ AA'EE'$ is any cyclic quadrilateral, and the problem asks to prove $ X$ and the two points of tangency are colliner. The proof is basically the same. Solution: Assume $ AC$ and $ PQ$ intersect at $ F'$, then it's well known that $ CEF'A$ is a harmonic range and by the problem above, we conclude that $ F = F'$. Since $ PQ//AA'//FX$, $ PQ$ must be the gable. Comment 1: The well know fact can be proven by \[ \frac {AF'}{F'E} = \frac {PA^2}{PE^2} = \frac {AC}{EC} \] Comment 2: However, we can avoid (or prove by proving the problem first) this "well known" fact and solve this problem using the following fact: if $ ABCDEF$ are six points on a circle then $ AD, BE, CF$ concur iff $ AB\cdot CD\cdot EF = BC\cdot DE\cdot FA.$ This result come from a slightly stronger version of the 1988 IMO Longlist Problem 88. Now let's see how all these relates symmedian. On the 2007 TST problem 5, I proved the following property of symmedian using Ptolemy's Theorem, but now we can see that it's a trivial consequence of the previous problem. Problem: In the same diagram as the previous problem, prove that $ \angle PAC = \angle XAQ$ where $ AC$ is the symmedian and $ AX$ is the median. Solution: This is equivalent to showing $ PE = QE'$ which is obviously true. Now we have proved the most import property of symmedian, and this is my solution to USAMO 2008 Problem 2: Let $ AS$ be the symmedian, and $ G$ be a point on the symmedian such that $ OG$ is perpendicular to $ AS$, then $ \angle ABG = \angle BGS - \angle BAG = \angle BAC - \angle BAG = \angle GAC = \angle BAM$, therefore, $ G$ lies on $ BF$ and similarly $ G$ lies on $ CF$, so $ G = F$, and we are done. It's amazing to see how one art technique can relate to many Olympiad geometry problems, and that's the beauty of the math.

nice... i spent half an hour banging my head before i noticed that FO and FA are the angle bisectors of DFE

What is a symmedian? Is it something similar to a median?

See http://mathworld.wolfram.com/Symmedian.html. Basically a symmedian is a line obtained by reflecting the median across the angle bisector. If $ AM$ is a median and $ AM'$ is a symmedian, then $ \angle BAM = \angle CAM'$. (Also see http://mathworld.wolfram.com/IsogonalConjugate.html.)

A bunch of diagram requests were made, the wiki has a number of them: 2008 USAMO Problems/Problem 2. In the meantime it'd be really nice if someone would look through the solutions users have assembled on the wiki.

I guess I'll post my full solution. It's essentially a trig bash.

Let's use the power of barycentric coordinates. Use $\bigtriangleup ABC$ as the reference triangle, with A = (1:0:0), B=(0:1:0) and C = (0:0:1). Calculate the midpoints of BC, AC and AB and we get (0:1/2:1/2), (1/2:0:1/2) and (1/2:1/2:0) the points are M, N and P. So let's calculate the perpendicular bissector of AB and AC. By the Evan's Forgotten Trick (Perpendicularity Criterion), two vectors $\vec{AB}$ with coordinates $(x_1,y_1,z_1)$ and $\vec{MN}$ with coordinates$(x_2, y_2, z_2)$ are perpendicular if and only if : $$a^2(y_1 z_2 + y_2 z_1) + b^2(x_1 z_2 + x_2 z_1) + c^2(x_1 y_2 + x_2 y_1) = 0$$. Plugging the values we get the perpendicular bissector of AB as : $$z(b^2-a^2) + c^2(y-x) = 0$$and AC : $$y(c^2-a^2)+ b^2(z-x) = 0 $$ Using the fact that the points on line $AM$ are in the form (t :1/2 : 1/2), plugging these we can calculate t and ,furthermore, calculate the coordinates o D and E : $$ D = (\frac{b^2 + c^2-a^2}{2c^2} ; 1/2 ; 1/2) $$$$ E = (\frac{b^2 + c^2-a^2}{2b^2} ; 1/2 ; 1/2) $$Now, let's calculate the intersection of lines BD and CE : First, let's use the fact that points on BD are in the form $(\frac{b^2+c^2-a^2}{2c^2} : t : 1/2)$ and points on CE are in the form : $(\frac{b^2 + c^2 -a^2}{2b^2} : 1/2 : t)$. Now, we're gonna multiply the coordinate in line BD by $2c^2$ and CE by $2b^2$, then, F gonna be (b^2+c^2-a^2 : b^2 : c^2). Now, let's use the circle of $(ANP)$ equation that is : $$-a^2yz -b^2xz -c^2xy + (ux + vy + wz)(x+y+z) = 0$$and find the constants u,v,w. A is immediate, $u=0$. Use this on N we get that $v = \frac{c^2}{2}$ and $w = \frac{b^2}{2}$. And this equation gonna be equal to zero if and only if F lies on this $(APN)$. $$-a^2yz - b^2xz - c^2xy + (ux + vy + wz)(x+y+z)$$ $$\iff$$ $$-a^2(b^2)(c^2) - b^2(b^2 + c^2 -a^2)(c^2) - c^2(b^2 + c^2 -a^2)(b^2) + (b^2c^2)(2b^2 + 2c^2 - a^2)$$$$\iff$$$$-a^2b^2c^2 - b^4c^2 -b^2c^4 + a^2b^2c^2 -b^4c^2 - b^2c^4 + a^2b^2c^2 + 2b^4c^2 + 2b^2c^4 -a^2b^2c^2$$which is clearly equal to zero, so, ANFP is cyclic.

Solution: Let $O$ be the circumcenter of $\triangle ABC$ and let $Q := AF \cap BC$. We wish to prove that $F$ lies on $A-$symmedian. [asy][asy] import olympiad; size(10cm); pair A = dir(120); pair B = dir(210); pair C = dir(330); pair M = (B+C)/2; pair N = (A+C)/2; pair P = (A+B)/2; pair O = circumcenter(A,B,C); pair D = extension(A,M,P,O); pair E = extension(N,O,A,M); pair F = extension(B,D,E,C); pair Q = extension(A,F,B,C); draw(A--B--C--A, darkmagenta); draw(P--N, orange); draw(N--M, orange); draw(P--M, orange); draw(A--M, orange); draw(P--O, orange); draw(E--N, orange); draw(B--D, orange); draw(C--F, orange); draw(O--F, orange); draw(A--Q, fuchsia); draw(circumcircle(A,P,N), purple + dashed); markscalefactor = 0.02; draw(anglemark(D,B,A), deepgreen); draw(anglemark(B,A,D), deepgreen); draw(anglemark(A,C,E), cyan); draw(anglemark(E,A,C), cyan); markscalefactor = 0.006; draw(rightanglemark(O,F,A), blue); markscalefactor = 0.008; draw(anglemark(B,F,Q)); draw(anglemark(Q,F,C)); markscalefactor = 0.01; draw(anglemark(B,F,Q)); draw(anglemark(Q,F,C)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$P$", P, dir(P)); dot("$O$", O, dir(O)); dot("$D$", D, NE); dot("$E$", E, dir(E)); dot("$F$", F, NW); dot("$Q$", Q, dir(Q)); [/asy][/asy] Claim: $FQ$ is the angle bisector of $\angle BFC$. Moreover, $\angle BFQ = \angle QFC = \angle BAC$. Proof: Start by applying Law of Sines in $\triangle FAB$ and $\triangle FAC$. This would yield \[\frac{\overline{FB}}{\sin(\angle BAF)} = \frac{\overline{AF}}{\sin(\angle ABF)} \qquad \text{and} \qquad \frac{\overline{AF}}{\sin(\angle ACF)} = \frac{\overline{CF}}{\sin(\angle FAC)}.\]Dividing both equations, eliminating $\overline{AF}$ and putting $\angle ACF = \angle MAC$ and $\angle ABF = \angle MAB$, we get \[\frac{\sin(\angle ACF)}{\sin(\angle ABF)} = \frac{\overline{BF}}{\overline{FC}} \cdot \frac{\sin(\angle FAC)}{\sin(\angle FAB)} = \frac{\sin(\angle MAC)}{\sin(\angle MAB)}.\]Applying ``Ratio Lemma'' on $\triangle ABC$ with cevian $AM$ and $AQ$ would yield us \[\frac{\overline{BF}}{\overline{FC}} \cdot \frac{\overline{QC}\cdot \overline{AB}}{\overline{BQ}\cdot \overline{AC}} = \frac{\overline{AB}}{\overline{AC}} \iff \frac{\overline{FB}}{\overline{FC}} = \frac{\overline{QB}}{\overline{QC}}\]By the converse of ``Angle Bisector Theorem'', we get that $FQ$ is the angle bisector of $\angle BFC$. For the claimed angle equality, observe that \begin{align*} \angle BFC & = \angle CFQ + \angle BFQ \\ & = 2\angle BAQ + 2\angle CAQ \\ & = 2\angle BAC \end{align*}and the claim is proven. $\square$ Clearly $O$ is the $F-$excenter of $\triangle FED$ since $EN$ and $DP$ are exterior angle bisectors by symmetry. Due to the above claim, $FD$ is exterior angle bisector of $\angle EFD$. With this, we conclude that $OF \perp AF$. Finally since \[\angle OFN = \angle ONA = \angle OPA = 90^\circ\]we get that $F \in \odot(APON)$ with diameter $\overline{AO}$ as desired. $\blacksquare$

The key claim is that $\overline{AF}$ is a symmedian. (Actually, it turns out $F$ is the A-dumpty point, but we will not need this.) Notice that $\angle BFC = 2\angle A$, so $B, F, O, C$ are concyclic, where $O$ is the circumcenter. Now, let $F'$ be the point on $(ANM)$ such that $\overline{AF'}$ is a symmedian. Then $\overline{AF'}$ passes through the midpoint of major arc $\widehat{BC}$ in $(BOC)$, and as $\angle OF'A = 90^\circ$, $F'$ lies on $(BOC)$. Finally, an angle chase reveals $\angle FBO = \angle FGO = \angle B - \angle BAD$, which is enough to characterize $F = F'$.

Barycentric coordinates finishes off in style

I claim that $F$ is the $A$-dumpty point. It is clear that the $A$-dumpty point lies on $\odot(ANP)$. Also, let $O$ denote the center of $\odot(ABC)$. Firstly, \begin{align*} \measuredangle BFC&=\measuredangle FBC+\measuredangle BCF\\ &=(\measuredangle ABC-\measuredangle ABF)+(\measuredangle BCA-\measuredangle FCA)\\ &=(\measuredangle ABC+\measuredangle BCA)+\measuredangle FBA+\measuredangle ACF\\ &=\measuredangle BAC+\measuredangle BAF+\measuredangle FCA\\ &=2\measuredangle A .\end{align*} So $F$ lies on $\odot(BOC)$. Now it is enough to prove that $AF$ is the $A$-symmedian. Firstly, redefine $F$ as the $A$-dumpty point. Now note that we have $\measuredangle FCA=\measuredangle FAB=\measuredangle CAM=\measuredangle CAE=\measuredangle ECA$. This gives us that $\overline{C-E-F}$ are collinear. Similarly we also get that $\overline{B-D-F}$ are collinear and we are done.

After drawing several diagrams, we conjecture that $AF$ is the A-symmedian. We would hope that then $F$ is the midpoint of $AT$, where $T$ is the intersection of the A-symmedian with the circumcircle. Note that $T=(-a^2:2b^2:2c^2)$, as the second two coordinates are in a $b^2$ to $c^2$ ratio and it satisfies the circumcircle equation. Thus, letting $A=(-a^2+2b^2+2c^2:0:0)$ (we make them the same sum so we can just average them), we have that the midpoint of $AT$ is $$(-a^2+b^2+c^2:b^2:c^2).$$Denote this point by $F'$, and we will try to show that $F'=F.$ Suppose that $BF'$ intersects $AM$ at $D'$. Then, $B=(0,1,0)$ and $F'=(-a^2+b^2+c^2:b^2:c^2)$, so the equation of line $BF'$ is $$x(c^2)=z(-a^2+b^2+c^2).$$The equation of $AM$ is just $y=z$, so we have $$D'=(-a^2+b^2+c^2:c^2:c^2).$$ Now, we check that $D'P\perp AB.$ Note that $$P=(1:1:0)=(-a^2+b^2+3c^2:-a^2+b^2+3c^2:0)$$and $$D'=(-2a^2+2b^2+2c^2:2c^2:2c^2).$$This is done so that $P$ and $D'$ have the same sum now. We have $$\overrightarrow{AB}=(-1,1,0)$$and $$\overrightarrow{D'P}=(a^2-b^2+c^2:-a^2+b^2+c^2:-2c^2)$$(strictly speaking this is not exactly $\overrightarrow{D'P}$, but it doesn't matter as it is a real multiple that does not affect perpendicularity). This, it suffices to check that $$a^2(-2c^2)+b^2(2c^2)+c^2(2a^2-2b^2)=0,$$which is clearly true. Hence, by EFFT, $D'P\perp AB,$ so $D'=D$. Similarly, $E'=E$. Since $F'$ lies on $BD'$ by definition (and thus $BD$), and similarly $F'$ lies on $CE$, we have $F'=F$, as desired.

As $\angle ABF = \angle BAM$ and $\angle ACF = \angle CAM$, $F$ is the $A-$ Dumpty point, which lies on $(ANP)$ since it is the midpoint of the $A-$ symmedian chord.

I have seem this point many times in the past few days Apply $\sqrt{bc}$ inversion, then a homothety of scale factor $\tfrac{1}{2}$ centered at $A$. The problem becomes: Restated problem wrote: In $\triangle ABC$, let $\ell$ be the $A$-symmedian and $B',C'$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively. Let $X$ and $Y$ be the feet from $B$ and $C$ to $\ell$ respectively. Prove that $(AB'X)$ and $(AC'Y)$ meet on $\overline{BC}$. I claim that this intersection point is in fact the midpoint $M$ of $\overline{BC}$. Consider $(AB'M)$ and suppose it intersects $\ell$ again at $X' \neq A$. Then $\measuredangle B'MX'=\measuredangle B'AX'=\measuredangle MAC=\measuredangle B'MA$, hence $B'$ is the midpoint of arc $AX'$, so $B'A=B'X'=B'B$. This implies that $\angle AX'B$ is right (since $B'$ is now its circumcenter), so $X'=X$; done. $\blacksquare$ Remark: Kind of got spoiled as to the identity of the point in another thread but I think this solution is more or less motivated independently of this fact. In some sense the diagram feels kind of random (at least to me) and a $\sqrt{bc}$ inversion cleans it up significantly. The other advantage of this $\sqrt{bc}$ inversion is that it makes it much easier to guess the mystery point since midpoints are easy (not that I had to do any guessing...)

IAmTheHazard wrote: I have seem this point many times in the past few days 2020 iran tst G3 or something i forget A-HM point of I_ABC :skull:

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.56, xmax = 5.32, ymin = -4.42, ymax = 7.74; /* image dimensions */ draw((-1.16,6.34)--(-2.88,-0.96)--(4.18,-0.9)--cycle, linewidth(0.4)); draw(arc((4.18,-0.9),0.6,158.05618236276638,180.48692124781314)--(4.18,-0.9)--cycle, linewidth(2) + blue); draw(arc((-1.16,6.34),0.6,-76.01940149306334,-53.58866260801658)--(-1.16,6.34)--cycle, linewidth(2) + blue); draw(arc((-1.16,6.34),0.6,-103.25801748053517,-76.01940149306334)--(-1.16,6.34)--cycle, linewidth(2) + red); draw(arc((-2.88,-0.96),0.6,0.48692124781311374,27.725537235284936)--(-2.88,-0.96)--cycle, linewidth(2) + red); draw(arc((-2.88,-0.96),0.6,49.50336653199302,76.74198251946484)--(-2.88,-0.96)--cycle, linewidth(2) + red); draw(arc((4.18,-0.9),0.6,126.41133739198344,148.8420762770302)--(4.18,-0.9)--cycle, linewidth(2) + blue); /* draw figures */ draw((-1.16,6.34)--(-2.88,-0.96), linewidth(0.4)); draw((-2.88,-0.96)--(4.18,-0.9), linewidth(0.4)); draw((4.18,-0.9)--(-1.16,6.34), linewidth(0.4)); draw((-1.16,6.34)--(0.65,-0.93), linewidth(0.4)); draw((-2.88,-0.96)--(0.2481425288175455,0.6840905057991408), linewidth(0.4)); draw((0.2481425288175455,0.6840905057991408)--(4.18,-0.9), linewidth(0.4)); draw((-2.88,-0.96)--(-0.14104396110317108,2.2472870702873227), linewidth(0.4)); draw((-0.4421832207949546,1.8946563452047962)--(4.18,-0.9), linewidth(0.4)); /* dots and labels */ dot((-1.16,6.34),linewidth(2pt) + dotstyle); label("$A$", (-1.08,6.42), NE * labelscalefactor); dot((-2.88,-0.96),linewidth(2pt) + dotstyle); label("$B$", (-3.28,-0.9), NE * labelscalefactor); dot((4.18,-0.9),linewidth(2pt) + dotstyle); label("$C$", (4.26,-0.82), NE * labelscalefactor); dot((0.65,-0.93),linewidth(1pt) + dotstyle); label("$M$", (0.74,-0.9), NE * labelscalefactor); dot((0.2481425288175455,0.6840905057991408),linewidth(2pt) + dotstyle); label("$Q$", (0.32,0.76), NE * labelscalefactor); dot((-0.14104396110317108,2.2472870702873227),linewidth(1pt) + dotstyle); label("$D$", (0.12,2.3), NE * labelscalefactor); dot((0.015671121441022638,1.6178292525545666),linewidth(1pt) + dotstyle); label("$E$", (0.16,1.52), NE * labelscalefactor); dot((-0.4421832207949546,1.8946563452047962),linewidth(2pt) + dotstyle); label("$F$", (-0.7,1.82), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] As $Q=A-$Humpty satisfies $\angle QBC=\angle QAB$ and $\angle QCB=\angle QAC$ then $F$ is the isogonal conjugate of $Q$ i.e the $A-$ Dumpty point, so taking a homotety with ratio $1/2$ sends $(ABXC)$ (where $X$ is the intersection of $(ABC)$ with $A-$symmedian) to $(ANFP)$

The angle conditions on $F$ uniquely determine it to be the $A$-Dumpty point, which is the midpoint of the $A$-symmedian chord, so a homothety of scale factor $\tfrac 12$ finishes. $\blacksquare$

We'll prove that $F$ is $A$-Dumpty point. Let $F'$ be $A$-Dumpty point and let $X = AF' \cap (ABC)$. Then it's well-known that $F'$ is the midpoint of $AX$. Thus $\angle ECA = \angle EAC = \angle MAC = \angle BAX = \angle BCX = \angle ACF'$, so $C, E, F'$ is collinear. Similarly $B, D, F'$ is collinear, hence $F' = F$. Then $F$ is the midpoint of $AX$, which means $F$ lies on $(APN)$, as needed. $\blacksquare$

We show that $F$ is the $A-$Dumpty Point. Define $F_B=BF\cap AC$ and $F_C=CF\cap AB$. Then, from Menelaus' Theorem, we have \[\frac{AD}{DM}\cdot\frac{MB}{BC}\cdot\frac{CF_B}{F_BA}=-1\]\[\frac{AE}{EM}\cdot\frac{MC}{CB}\cdot\frac{BF_C}{F_CA}=-1\]\[\Longrightarrow\frac{CF_B}{F_BA}\cdot\frac{AF_C}{F_CB}=\frac{DM}{AD}\cdot\frac{AE}{EM}\]\[\frac{BF_A}{F_AC}\cdot\frac{CF_B}{F_BA}\cdot\frac{AF_C}{F_CB}=1\]\[\Longrightarrow\frac{BF_A}{F_AC}=\frac{AD}{DM}\cdot\frac{EM}{AE}=\frac{BD}{DM}\cdot\frac{EM}{CE}=\frac{\sin(\angle FCM)}{\sin( \angle MBF)}=\frac{BF}{FC}\]Hence, $\angle BFF_A=\angle F_AFC$. Note that \[\angle BFC=\angle BAC+\angle FBA+\angle ACF=\angle BAF+\angle FAC+\angle FBA+\angle ACF=2\angle BAC.\]Hence, $B,F,O,C$ are concyclic. Moreover, \[\angle BAF+\angle FBA=\angle FAC+\angle ACF=\angle BAC=\angle BAF+\angle FAC\Longrightarrow \angle FBA=\angle FAC,\angle BAF=\angle ACF\]This means $F$ is the $A-$Dumpty by definition. So, $F\in (ANPO)$, as required. $\blacksquare$

We claim that $F$ is the dumpty point, which evidently finishes the problem. To do so, redefine $F$ as the dumpty point; we will show that $B,D,F$ are collinear, which would imply by symmetry that $C,E,F$ are collinear, solving the problem. Take a force-overlaid inversion at $A.$ This sends $B$ to $C$ and the dumpty point $F$ to the point $X$ such that $ABXC$ is a parallelogram. Additionally, the perpendicular bisector of $AB$ gets sent to the circle with center $C$ passing through $A.$ Thus $D$ gets sent to the point $Z$ on the $A$-symmedian such that $AC = CZ.$ We wish to show that $A,C,X,Z$ are concyclic. Indeed, $$\measuredangle AXC = \measuredangle XAB = \measuredangle CAZ = \measuredangle AZC,$$so we are done.

Let $X$ be the $A$-humpty point. Then, it's well known that $\angle XBM = \angle XAB$ and $\angle XCM = \angle XAC$. So, $X$ is the isogogonal conjugate of $F$, implying that $F$ is the $A$-dumpty point. Since the dumpty point is the foot from the circumcenter to the $A$-symmedian, it clearly lies on $(APN)$.