Solution 1 (synthetic)
Without Loss of Generality, assume $AB >AC$. It is sufficient to prove that $\angle OFA = 90^{\circ}$, as this would immediately prove that $A,P,O,F,N$ are concyclic.
By applying the Menelaus' Theorem in the Triangle $\triangle BFC$ for the transversal $E,M,D$, we have (in magnitude)
$$\frac{FE}{EC} \cdot \frac{CM}{MB} \cdot \frac{BD}{DF} = 1 \iff \frac{FE}{EC} = \frac{DF}{BD}$$
Here, we used that $BM=MC$, as $M$ is the midpoint of $BC$. Now, since $EC =EA$ and $BD=DA$, we have
$$\frac{FE}{EA} = \frac{DF}{DA} \iff \frac{DA}{AE} = \frac{DF}{FE} \iff AF \text{ bisects exterior } \angle EFD$$
Now, note that $OE$ bisects the exterior $\angle FED$ and $OD$ bisects exterior $\angle FDE$, making $O$ the $F$-excentre of $\triangle FED$. This implies that $OF$ bisects interior $\angle EFD$, making $OF \perp AF$, as was required.

Solution 2 (complex)
Let $A=1,B=b,C=c$ where $b,c$ all lie on the unit circle. Then $O$ is 0. As noted earlier, $(FOBC)$ is cyclic. We will find the ghost point $F',$ the second intersection of $OBC$ and $ANP$.
We know that these two circles already intersect at $O$ so we can reflect over the line between their centers. The center of $ANPO$ is the midpoint of $AO$ namely $\frac12$. With the tangent formula and then taking the midpoint, we find that the center of $OBC$ is $\frac{bc}{b+c}.$ Then we want to find the reflection of 0 over the line through $\frac12$ and $\frac{bc}{b+c}.$ Then we get
$$\begin{align*} f' &= \frac{\left(\frac{bc}{b+c}-\overline{\frac{bc}{b+c}}\right)\div2}{(1/2)-\overline{\frac{bc}{b+c}}}\\ &= \frac{\frac{bc-1}{2(b+c)}}{\frac{b+c-2}{2(b+c)}}\\ &= \frac{bc-1}{b+c-2}. \end{align*}$$
Now it remains to show $\angle F'BA=\angle ABM;$ the other angle equality would follow by symmetry.
Then we get:
$$\begin{align*} \frac{f'-b}{b-a}\div\frac{b-a}{a-m} &=\frac{\frac{bc-1}{b+c-2}-b}{b-1}\div\frac{b-1}{1-\frac{b+c}2}\\ &=\frac{\frac{bc-1-b(b+c-2)}{b+c-2}(2-b+c)}{2(b-1)^2}\\ &=\frac{b^2-2b+1}{2(b-1)^2}\\ &=\frac12. \end{align*}$$
Thus $\measuredangle F'BA=\measuredangle BAM,$ so $F'=F$ and we're done.
~cocohearts

Solution 3 (synthetic)

Without loss of generality $AB 180^\circ$ so $\angle AFC = \angle AFB$.
Then $\angle FAB + \angle ABF = 180^\circ - \angle AFB = 180^\circ - \angle AFC = \angle FAC + \angle ACF$ and $\angle ABF + \angle ACF = \angle A = \angle FAB + \angle FAC$. Subtracting these two equations, $\angle FAB - \angle FCA = \angle FCA - \angle FAB$ so $\angle BAF = \angle ACF$. Therefore, $\triangle ABF\sim\triangle CAF$ (by AA similarity), so a spiral similarity centered at $F$ takes $B$ to $A$ and $A$ to $C$. Therefore, it takes the midpoint of $\overline{BA}$ to the midpoint of $\overline{AC}$, or $P$ to $N$. So $\angle APF = \angle CNF = 180^\circ - \angle ANF$ and $APFN$ is cyclic.

Solution 6 (isogonal conjugates)

/* setup and variables */ size(280); pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8); pair B=(0,0),C=(5,0),A=(4,4); /* A.x > C.x/2 */   /* construction and drawing */ pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C); D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s)); D(C--D(MP("E",E,NW,s))--MP("N",N,(1,0),s)--D(MP("O",O,SW,s))); D(D(MP("D",D,SE,s))--MP("P",P,W,s)); D(B--D(MP("F",F,s))); D(O--A--F,linetype("4 4")+linewidth(0.7)); D(MP("O'",circumcenter(A,P,N),NW,s)); D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); picture p = new picture; draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7)); draw(p,circumcircle(A,B,C),linetype("1 4")+linewidth(0.7)); clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p);

Solution 7 (symmedians)
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.

Solution 8 (inversion) (Official Solution #2)
Invert the figure about a circle centered at $A$, and let $X'$ denote the image of the point $X$ under this inversion. Find point $F_1'$ so that $AB'F_1'C'$ is a parallelogram and let $Z'$ denote the center of this parallelogram. Note that $\triangle BAC\sim\triangle C'AB'$ and $\triangle BAD\sim\triangle D'AB'$. Because $M$ is the midpoint of $BC$ and $Z'$ is the midpoint of $B'C'$, we also have $\triangle BAM\sim\triangle C'AZ'$. Thus
$$\angle AF_1'B' = \angle F_1'AC' = \angle Z'AC' = \angle MAB = \angle DAB = \angle DBA = \angle AD'B'.$$
Hence quadrilateral $AB'D'F_1'$ is cyclic and, by a similar argument, quadrilateral $AC'E'F_1'$ is also cyclic. Because the images under the inversion of lines $BDF$ and $CFE$ are circles that intersect in $A$ and $F'$, it follows that $F_1' = F'$.
Next note that $B'$, $Z'$, and $C'$ are collinear and are the images of $P'$, $F'$, and $N'$, respectively, under a homothety centered at $A$ and with ratio $1/2$. It follows that $P'$, $F'$, and $N'$ are collinear, and then that the points $A$, $P$, $F$, and $N$ lie on a circle.

2008usamo2-sol8.png
Solution 9 (Official Solution #1)
Let $O$ be the circumcenter of triangle $ABC$. We prove that
$$\angle APO = \angle ANO = \angle AFO = 90^\circ.\qquad\qquad (1)$$
It will then follow that $A, P, O, F, N$ lie on the circle with diameter $\overline{AO}$. Indeed, the fact that the first two angles in $(1)$ are right is immediate because $\overline{OP}$ and $\overline{ON}$ are the perpendicular bisectors of $\overline{AB}$ and $\overline{AC}$, respectively. Thus we need only prove that $\angle AFO = 90^\circ$.

2008usamo2-sol9.png
We may assume, without loss of generality, that $AB > AC$. This leads to configurations similar to the ones shown above. The proof can be adapted to other configurations. Because $\overline{PO}$ is the perpendicular bisector of $\overline{AB}$, it follows that triangle $ADB$ is an isosceles triangle with $AD = BD$. Likewise, triangle $AEC$ is isosceles with $AE = CE$. Let $x = \angle ABD = \angle BAD$ and $y = \angle CAE = \angle ACE$, so $x + y = \angle BAC$.
Applying the Law of Sines to triangles $ABM$ and $ACM$ gives
$$\frac{BM}{\sin x} = \frac{AB}{\sin\angle BMA}\quad\text{and}\quad\frac{CM}{\sin y} = \frac{AC}{\sin\angle CMA}.$$
Taking the quotient of the two equations and noting that $\sin\angle BMA = \sin\angle CMA$, we find
$$\frac{BM}{CM}\frac{\sin y}{\sin x} = \frac{AB}{AC}\frac{\sin\angle CMA}{\sin\angle BMA} = \frac{AB}{AC}.$$
Because $BM = MC$, we have
$$\frac{\sin x}{\sin y} = \frac{AC}{AB}.\qquad\qquad (2)$$
Applying the Law of Sines to triangles $ABF$ and $ACF$, we find
$$\frac{AF}{\sin x} = \frac{AB}{\sin\angle AFB}\quad\text{and}\quad\frac{AF}{\sin y} = \frac{AC}{\sin\angle AFC}.$$
Taking the quotient of the two equations yields
$$\frac{\sin x}{\sin y} = \frac{AC}{AB}\frac{\sin\angle AFB}{\sin\angle AFC},$$
so by $(2)$,
$$\sin\angle AFB = \sin\angle AFC.\qquad\qquad (3)$$
Because $\angle ADF$ is an exterior angle to triangle $ADB$, we have $\angle EDF = 2x$. Similarly, $\angle DEF = 2y$. Hence
$$\angle EFD = 180^\circ - 2x - 2y = 180^\circ - 2\angle BAC.$$
Thus $\angle BFC = 2\angle BAC = \angle BOC$, so $BOFC$ is cyclic. In addition,
$$\angle AFB + \angle AFC = 360^\circ - 2\angle BAC > 180^\circ,$$
and hence, from $(3)$, $\angle AFB = \angle AFC = 180^\circ - \angle BAC$. Because $BOFC$ is cyclic and $\triangle BOC$ is isosceles with vertex angle $\angle BOC = 2\angle BAC$, we have $\angle OFB = \angle OCB = 90^\circ - \angle BAC$. Therefore,
$$\angle AFO = \angle AFB - \angle OFB = (180^\circ - \angle BAC) - (90^\circ - \angle BAC) = 90^\circ.$$
This completes the proof.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.