Vijetnam 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: $AT=A{T}_3,C{T}_3=C{T}_1,D{T}_1=D{T}_2$ and $S{T}_2=ST$. $$\text{Hence }AS-SD=(AT-ST)-(D{T}_2-S{T}_2)=A{T}_3-D{T}_1=AC-CD.(4)$$ 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