, 2011

Since $\angle ABC$ and $\angle ACB$ are acute, $E$ lies on the opposite ray to ray $AB$ on the edge $AB$, at the same time, $F$ lies on the edge $AC$ or on the opposite ray to ray $AC$. Hence, it follows from the definition of the points $M, N, P$ that $E, M, N$ are collinear and $M, F, P$ are collinear.
Hence $\angle NMP = \frac{1}{2}(\angle AEF + \angle AFE) = \frac{1}{2}\angle BAC$.
Consequently: $A, M, N, P$ lie on a circle if and only if $\angle NAP = \frac{1}{2}\angle BAC$. (1)

Further we will show
$\angle NAP = \frac{1}{2}\angle BAC$ if and only if $d$ passes through the center $I$ of the inscribed circle of triangle $ABC$. (2)

Without loss of generality, assume that $AB < AC$. (3)

* The necessary condition: Assume that $I \in d$. Then, it follows from (3) that $E$ lies on the opposite ray to ray $AB$ and $F$ lies on edge $AC$.
Draw line $Ax$ (distinct from $AC$) tangent to $(P)$. We will show that $Ax$ is tangent to $(N)$.
Indeed, let $T, T_1, T_2, T_3$ be the tangent points of $(P)$ with $Ax, CD, DF, FC$. Let $S$ be the intersection of $Ax$ and $DF$. We have: $AT = AT_3, CT_3 = CT_1, DT_1 = DT_2$ and $ST_2 = ST$.
$$\text{Hence } AS - SD = (AT - ST) - (DT_2 - ST_2) = AT_3 - DT_1 = AC - CD. \quad (4)$$
Since $I \in d$, $D$ is the tangent point of $(I)$ and $BC$. Consequently $AC - CD = AB - BD$. (5)

It follows from (4) and (5) that $AS + BD = AB + SD$. Thus $ABDS$ is a cyclic quadrilateral.
Consequently $Ax$ is tangent to $(N)$.
Hence, we have $\overline{NAP} = \overline{NAx} + \overline{xAP} = \frac{1}{2}\overline{BAx} + \frac{1}{2}\overline{xAC} = \frac{1}{2}\overline{BAC}$.

* Sufficient condition: Assume $\overline{NAP} = \frac{1}{2}\overline{BAC}$. Consider the following two cases:

- Case 1: $E$ lies in the opposite ray to ray $AB$ and $F$ lies on edge $AC$.
Draw the tangent $Ax$ (distinct from $AC$) to $(P)$, which intersects $DF$ at $S$. We have
$$\overline{NAx} = \overline{NAP} - \overline{xAP} = \frac{1}{2}\overline{BAC} - \frac{1}{2}\overline{xAC} = \frac{1}{2}\overline{BAx}.$$
Consequently, $Ax$ is tangent to $(N)$. Hence $ABDS$ is a tangent quadrilateral. Consequently
$$AS + BD = AB + SD.$$
Moreover, according to the proof of the previous part, we have $AS - SD = AC - CD$. (See (4)).
Hence we obtain $BD = AB + CD - AC$. Consequently $2BD = AB + BC - AC$.
Hence $BD = p - b$, where $p = \frac{AB + BC + CA}{2}$ and $b = AC$.
Consequently $BD = BK$, where $K$ is the tangent point of $(I)$ and edge $BC$.
Hence $D \equiv K$, as $D$ and $K$ both lie on edge $BC$. Thus $I \in d$.

- Case 2: $E$ lies on edge $AB$ and $F$ lies on the opposite ray to ray $AC$.
Then, by means of (3), $CD > CK$. (*)
On the other hand, in this case $B$ plays the role of $C$ and $C$ plays the role of $B$, $E$ plays the role of $F$ and $F$ plays the role of $E$, $(N)$ plays the role of $(P)$ and $(P)$ plays the role of $(N)$ of the previous case. Thus, according to the above proof, we have $CD = CK$, contradicting (*). The obtained contradiction shows that this case cannot happen.

Thus (2) is verified. It follows from (1) and (2) the claim of the problem