China National Team Selection Test

Fig. 2.1

First, we need a lemma.

Lemma Let $R'$ be the midpoint of $BC$. Then $\angle AMI = \angle IR'B$, $IR' // AE$.

Proof of lemma As illustrated in Fig. 2.2, let $I_b$ and $I_c$ be the escentres of $\triangle ABC$ relative to the vertices $B$ and $C$, respectively. Then $A, M, I_b$, and $I_c$ are collinear. As $\angle I_bBI_c = \angle I_bCI_c = 90^\circ$, the points $B, C, I_b$, and $I_c$ all lie on the circle with diameter $I_bI_c$; since $MB = MC$, $M$ is the centre of this circle, $MI_b = MI_c$. Since $\triangle I_bI_c \sim \triangle ICB$, $M, R'$ are midpoints of $I_bI_c$ and $BC$, respectively, it follows that $\angle AMI = \angle IR'B$.

Let the incircle $\odot I$ of $\triangle ABC$ touch $BC$ at $Z$ and $ZW$ be a diameter. Through $W$, draw a line parallel to $BC$ that crosses $AB$, $AC$ at $B_1$, $C_1$, respectively. Then $B_1C_1 // BC$. Since $\triangle A_1B_1C_1$ and $\triangle ABC$ are homothetic with centre $A$, $W$ and $E$ are correspondent points, it follows that $A, W, E$ are collinear. Finally, from properties of escribed circles, we infer that $BZ = CE$, $R'$ is the midpoint of $ZE$, as $I$ is the midpoint of $ZW$, $IR' // AE$. The lemma is verified.

Return to the original problem. Let $S$ be the intersection of $AI$ and $BC$. Then
$$\begin{align*} \angle AIF &= \angle ASB = \angle SAC + \angle ACS \\ &= \angle BCN + \angle ACS \\ &= \angle ACN = 180^\circ - \angle ARN, \end{align*}$$
indicating that $A, I, F$, and $R$ lie on a circle, say $\omega$, as shown in Fig. 2.3.

Suppose the lines $AF$ and $MI$ intersect at $X$. We show that $X$ is on $\odot O$. By the lemma,
$$\begin{align*} \angle AMI &= \angle IRB = \angle AEB, \\ \text{also } \angle AMN &= \angle ACN = \angle ASB, \text{ thus} \\ \angle IMN &= \angle AMN - \angle AMI \\ &= \angle ASB - \angle AEB \\ &= \angle IAE. \end{align*}$$
It is given that $\angle IAE = \angle FAI$, hence $\angle IMN = \angle FAI$, $A, M, N$, and $X$ are concyclic, that is, $X$ lies on $\odot O$.

Let $Y$ be the other intersection of the line $DI$ and $\odot O$. Clearly,
$$\angle IYX = \angle DYX = \angle DAX = \angle IFX,$$
and $I, X, Y, F$ lie on a circle, say $\odot T$.

Through $M$, draw the tangent line $MG$ of $\odot O$ ($G$ and $Y$ are on the same side of $MN$), $MG // IF$. Then
$$\angle YMG = \angle MXY = \angle IXY = \angle YFL,$$
yielding the collinearity of $Y, F, M$. Apply the radical axis theorem to $\odot O, \odot \omega$, and $\odot T$, to derive the collinearity of $X, Y$, and $L$. In the cyclic quadrilateral $IXYF$, apply Brocard's theorem to derive the orthocentre $T$ of $\triangle MKL$, $TK \perp ML$.

Finally, in $\triangle TFI$ and $\triangle NAD$, $TF = TI, NA = ND$, and
$$\angle FTI = 2\angle FXI = 2\angle AXM = \angle AND.$$
Observing $FI // BC // AD$, we conclude that $\triangle TFI$ and $\triangle NAD$ are homothetic with centre $K$, and moreover $K, T, N$ are collinear.

From $TK \perp ML$ and the collinearity of $K, T, N, NK \perp ML$ follows. $\square$