China National Team Selection Test

Let $\omega$ be the circumcircle of $\triangle BCG$. Let $E'$ be the other intersection of the line $BE$ and $\omega$, $F'$ be the other intersection of the line $CF$ and $\omega$, as shown in Fig. 1.2. Note that $\angle BE'F' = \angle BCF' = \angle BCF = \angle BEF$, and hence $EF \parallel E'F'$; similarly, $\angle CF'G = \angle CBG = \angle CFA$, and hence $GF' \parallel HF$; moreover, $GE' \parallel HE$. It follows that $\triangle E'F'G$ and $\triangle EFH$ are homothetic triangles. Let $P$ be the homothetic centre. Then $EE'$ and $FF'$ pass through $P$; the line connecting the circumcentres of $\triangle E'F'G$ and $\triangle EFH$, which is $MN$, passes through $P$ as well. Therefore, the lines $BE$, $CF$, and $MN$ meet at $P$. $\square$

Fig. 1.2

(Edited from Wang Zihan's proof.) As in Fig. 1.3, we first apply Pascal's theorem to the cyclic hexagon $ABCDEF$ to deduce that $G$, $P$, $H$ are collinear. Let $A'$ and $D'$ be the antipodal points of $A$ and $D$, respectively. Let $BA'$ and $CD'$ meet at $G'$, $D'E$ and $A'F$ meet at $H'$. Apply Pascal's theorem to the cyclic hexagon $BA'FCD'E$ to deduce that $G'$, $H'$, $P$ are collinear.

Since $A$ and $A'$, $D$ and $D'$ are antipodal points, we have $A'B \perp AB$, $A'F \perp AF$, $CD' \perp CD$, $D'E \perp DE$. Notice that $\angle AHH' = 90^\circ - \angle FH'H = 90^\circ - \angle FAD$. Thus, the angle between $AH$ and $GG'$ is given by
$$\begin{align*} 180^\circ - \angle AGG' - \angle BAF &= 90^\circ - \angle BAF + \angle BG'G \\ &= 90^\circ - \angle BAF + \angle DAB \\ &= 90^\circ - \angle DAF. \end{align*}$$
This implies $HH' \parallel GG'$. As $M$ is the midpoint of $GG'$, $N$ is the midpoint of $HH'$, and furthermore it is shown that $GH$, $G'H'$ pass through $P$. We infer that $MN$ passes through $P$, too. Therefore, $BE$, $CF$, and $MN$ are concurrent. $\square$

Fig. 1.3

As shown in Fig. 1.4, let $P$ be the intersection of $BE$ and $CF$. It suffices to show that $M$, $P$, $N$ are collinear, or equivalently,
$$\frac{\sin \angle MPB}{\sin \angle MPC} = \frac{\sin \angle NPE}{\sin \angle NPF}.$$
Denote $\alpha = \angle MPB$, $\beta = \angle NPE$, $\gamma = \angle BPC = \angle FPE$. The above relation is equivalent to
$$\sin \alpha \sin(\gamma - \beta) = \sin \beta \sin(\gamma - \alpha),$$
or rewritten as $\sin \gamma \sin(\alpha - \beta) = 0$, or just $\alpha = \beta$ (for the collinearity of $M$, $P$, $N$).

In $\triangle MPC$ and $\triangle MPB$, apply the law of sines to obtain
$$\frac{BM}{\sin \angle MPB} = \frac{PM}{\sin \angle PBM}, \quad \frac{CM}{\sin \angle MPC} = \frac{PM}{\sin \angle PCM}.$$
As $BM = MC$, it follows that
$$\frac{\sin \angle MPB}{\sin \angle MPC} = \frac{\sin \angle PBM}{\sin \angle PCM}.$$
Similarly, in $\triangle NPF$ and $\triangle NPE$, we have
$$\frac{\sin \angle NPE}{\sin \angle NPF} = \frac{\sin \angle NEP}{\sin \angle NFP}.$$
Now it suffices to show
$$\frac{\sin \angle PBM}{\sin \angle PCM} = \frac{\sin \angle NEP}{\sin \angle NFP}. \qquad \textcircled{1}$$

The angles appearing in the numerators are supplementary, as
$$\begin{align*} \angle PBM + \angle NEP &= \angle PBC + \angle MBC + \angle PEF + \angle NFE \\ &= (180^\circ - \angle CDE) + (90^\circ - \angle BGC) \\ &\quad + (180^\circ - \angle BAF) + (90^\circ - \angle AHD) \\ &= 540^\circ - (\angle CDE + \angle BGC + \angle BAF + \angle AHD) \\ &= 540^\circ - 360^\circ = 180^\circ, \end{align*}$$
and thus $\sin \angle PBM = \sin \angle NEP$. Analogously, $\sin \angle PCM = \sin \angle NFP$. We have verified (1), and the statement. $\square$

Fig. 1.4

Assume that the circumcircle of $\triangle ABC$ is the unit circle in the complex plane. We use lowercase letters to represent the complex numbers corresponding to the points in the plane (for instance, $a$ is the complex number for the point $A$, and so on). Let $BE \cap CF = X$. We have
$$\begin{align*} x \in CF &\Leftrightarrow \frac{x-c}{x-f} = \overline{\left(\frac{x-c}{x-f}\right)} \\ &\Leftrightarrow x\bar{x} - \bar{f}x - c\bar{x} + c\bar{f} = x\bar{x} - c\bar{x} - f\bar{x} + c\bar{f} \\ &\Leftrightarrow (c-f)\bar{x} - (\bar{c} - \bar{f})x = c\bar{f} - c\bar{f} \\ &\Leftrightarrow \bar{x} + c\bar{f}\bar{x} = \bar{c} + \bar{f}. \end{align*}$$
In the same manner, $\bar{x} + \bar{b}\bar{e}\bar{x} = \bar{b} + \bar{e}$. Thus,
$$x = \frac{\bar{b} + \bar{e} - \bar{c} - \bar{f}}{\bar{b}\bar{e} - \bar{c}\bar{f}}. \qquad \textcircled{1}$$
On the other hand, $\triangle BMC$ is an isosceles triangle, $\angle BMC$ is twice the angle between the lines $CD$ and $AB$, and this implies that
$$\frac{c-m}{b-m} = \left(\frac{d-c}{a-b}\right)^2 / \left|\frac{d-c}{a-b}\right|^2 = \frac{d-c}{a-b} / \frac{\bar{d}-\bar{c}}{\bar{a}-\bar{b}} = \frac{cd}{ab}.$$
Consequently,
$$m = \frac{bc(a-d)}{ad-bc} = \frac{\bar{d}-\bar{a}}{\bar{c}\bar{d}-\bar{a}\bar{b}}. \qquad \textcircled{2}$$
In the same way, we can derive
$$n = \frac{(a-d)ef}{af-be} = \frac{\bar{d}-\bar{a}}{\bar{d}\bar{e}-\bar{a}\bar{f}}. \qquad \textcircled{3}$$
To prove that $BE$, $CF$, and $MN$ are concurrent, it is equivalent to prove that $M$, $X$, $N$ are collinear, namely
$$\frac{x-n}{x-m} = \left( \frac{x-n}{x-m} \right) \in \mathbb{R}. \qquad (4)$$
From (1)–(3), it follows that
$$\begin{align*} x - n &= \frac{1}{(\bar{b}\bar{e} - \bar{c}\bar{f})(\bar{d}\bar{e} - \bar{a}\bar{f})} [ (\bar{b} + \bar{e} - \bar{c} - \bar{f})(\bar{d}\bar{e} - \bar{a}\bar{f}) - (\bar{d} - \bar{a})(\bar{b}\bar{e} - \bar{c}\bar{f}) ] \\ &= \frac{1}{(\bar{b}\bar{e} - \bar{c}\bar{f})(\bar{d}\bar{e} - \bar{c}\bar{f})} [ \bar{b}\bar{d}\bar{e} + \bar{d}\bar{e}^2 - \bar{d}\bar{e}\bar{f} - \bar{c}\bar{d}\bar{e} - \bar{a}\bar{b}\bar{f} - \bar{a}\bar{e}\bar{f} + \bar{a}\bar{f}^2 \\ &\quad + \bar{a}\bar{c}\bar{f} - \bar{b}\bar{d}\bar{e} + \bar{a}\bar{b}\bar{e} + \bar{c}\bar{d}\bar{f} - \bar{a}\bar{c}\bar{f} ] \\ &= \frac{1}{(\bar{b}\bar{e} - \bar{c}\bar{f})(\bar{d}\bar{e} - \bar{a}\bar{f})} (\bar{e} - \bar{f})(\bar{a}\bar{b} - \bar{a}\bar{f} - \bar{c}\bar{d} + \bar{d}\bar{e}), \end{align*}$$
while
$$m - n = \frac{\bar{d} - \bar{a}}{(\bar{c}\bar{d} - \bar{a}\bar{b})(\bar{d}\bar{e} - \bar{a}\bar{f})}(\bar{a}\bar{b} - \bar{a}\bar{f} - \bar{c}\bar{d} + \bar{d}\bar{e}).$$
Hence,
$$\frac{x - n}{x - m} = \frac{(\bar{e} - \bar{f})(\bar{c}\bar{d} - \bar{a}\bar{b})}{(\bar{d} - \bar{a})(\bar{b}\bar{e} - \bar{c}\bar{f})}.$$
In the last fraction, the unit complex numbers $a, b, c, d, e$, and $f$ all appear exactly once in the numerator and in the denominator; moreover, in each parenthesis, the coefficients of the homogeneous terms sum to zero (one minus one). So, the fraction is real, which justifies (4), and the argument as well.