China Team Selection Test for IMO, 2022

(1) This problem can be solved by computing the angles, but it will (seriously) depend on the relative positions of points. To avoid a case-by-case discussion, we use complex numbers. Assume that the circumcircle of $\triangle ABC$ is the unit circle on the complex plane. We use the corresponding lowercase letters to denote the complex number corresponding to the points on the plane, e.g. point $A$ corresponds to the complex number $a$.

For a point $P$ on the plane, its projection image on the line $BC$ is the midpoint of $P$ and the symmetry point $P'$ of $P$ with respect to $BC$. Note that $\angle PBC = \angle P'BC$ and they are on opposite sides of the line $BC$. We have
$$\frac{(p' - b)(p - b)}{|p - b|^2} = \frac{(c - b)^2}{|c - b|^2},$$
i.e.
$$p' = b + \frac{(c - b)(\bar{p} - \bar{b})}{\bar{c} - \bar{b}},$$
For a complex number on the unit circle, $\bar{b} = \frac{1}{b}, \bar{c} = \frac{1}{c}$. We have
$$p' = b - bc\left(\bar{p} - \frac{1}{b}\right) = b + c - bc\bar{p}.$$
So
$$d = \frac{1}{2}(p - bc\bar{p} + b + c).$$
Similarly, we have
$$e = \frac{1}{2}(p - ca\bar{p} + c + a), \quad f = \frac{1}{2}(p - ab\bar{p} + a + b).$$
Then
$$\angle EDF = 90^\circ \text{ and } ED = DF \Leftrightarrow \frac{e-d}{f-d} = \pm i \Leftrightarrow \left(\frac{e-d}{f-d}\right)^2 = -1.$$
i.e.
$$\left[ \frac{(bc - ca)\bar{p} + (a - b)}{(bc - ab)\bar{p} + (a - c)} \right]^2 = -1.$$
we obtain an equation on $\bar{p}$:
$$[c^2(a-b)^2 + b^2(c-a)^2]\bar{p}^2 - 2[c(a-b)^2 + b(c-a)^2]\bar{p} + [(a-b)^2 + (c-a)^2] = 0, \quad (*)$$

The leading coefficient of this quadratic equation is zero if and only if $|a-b| = |c-a|$ (i.e. $\triangle ABC$ is an isosceles triangle with $AB = AC$) and $\frac{b}{c} = \pm i\frac{a-b}{c-a}$ (i.e. the central angle corresponding to $\widehat{BC}$ is $\frac{\pi}{2}$ larger than the inscribed angle corresponding to $\widehat{BC}$), then $\triangle ABC$ is isosceles triangle with a right angle; this contradicts with our assumption. So (*) is a quadratic equation.
The discriminant of (*) is
$$\begin{aligned} \Delta &= 4 \left\{[c(a-b)^2 + b(c-a)^2]^2 - [c^2(a-b)^2 + b^2(c-a)^2]^2[(a-b)^2 + (c-a)^2]\right\} \\ &= 4\left\{2bc(a-b)^2(c-a)^2 - (b^2+c^2)(a-b)^2(c-a)^2\right\} \\ &= -4(a-b)^2(b-c)^2(c-a)^2 \neq 0. \end{aligned}$$
So the equation (*) has two roots, this proves the existence of the two points $P_1$ and $P_2$.

We next prove that $W$ lies on the nine-point circle of $\triangle ABC$.

Step 1: show that $P_1P_2$ passes through the circumcenter $O$ of $\triangle ABC$. This is equivalent to
$$\frac{c(a-b)^2 + b(c-a)^2}{(a-b)(b-c)(c-a)} \in i\mathbb{R} \Leftrightarrow \frac{c(a-b)^2 + b(c-a)^2}{(a-b)(b-c)(c-a)} + \frac{\overline{c(a-b)^2 + b(c-a)^2}}{(a-b)(b-c)(c-a)} = 0.$$
But
$$\begin{aligned} \frac{c(a-b)^2 + b(c-a)^2}{(a-b)(b-c)(c-a)} &= \frac{c(a-b)}{(b-c)(c-a)} + \frac{b(c-a)}{(a-b)(b-c)} \\ &= -\frac{c}{b-c} - \frac{c}{c-a} - \frac{b}{a-b} - \frac{b}{b-c}, \end{aligned}$$
Yet $\frac{\overline{b+c}}{b-c} = \frac{b+c}{c-b}$,
$$\frac{\bar{c}}{\bar{c}-\bar{a}} + \frac{\bar{b}}{\bar{a}-\bar{b}} = \frac{a}{a-c} + \frac{a}{b-a} = \left(1-\frac{c}{c-a}\right) + \left(-1-\frac{b}{a-b}\right),$$
This proves Step 1.

For a point $Q$ on the circumcircle of $\triangle ABC$, its feets on the three sides of $\triangle ABC$ are collinear (on the well-known Simson line), and this line passes through the midpoint of $Q$ and orthocenter $H$ of $\triangle ABC$. Let $X$ and $Y$ denote the projections of a point $Q$ on the circumcircle onto $BC$ and $CA$, respectively. Then, we know
$$x = \frac{1}{2}(q - bc\bar{q} + b + c), \quad y = \frac{1}{2}(q - ca\bar{q} + c + a).$$
In order to prove that these two points are colinear with $\frac{1}{2}(q + b + b + c)$, we only need to show that
$$\frac{a + \frac{bc}{q}}{b + \frac{ca}{q}} = \frac{aq + bc}{bq + ac} \in \mathbb{R},$$

But this is obvious. Since $H$ is the homothetic center of the circumcircle and the nine-point circle, $X_1Y_1$ and $X_2Y_2$ each passes through a pair of opposite points on the nine-point circle of $\triangle ABC$.

Step 3: prove that $X_1Y_1 \perp X_2Y_2$.
We have
$$\begin{align*} \angle CY_1X_1 &= \angle CQ_1X_1 = 90^\circ - \angle X_1CQ_1 \\ \angle BAQ_2 &= \angle Q_2Y_2X_2 = 90^\circ - \angle CY_2X_2, \end{align*}$$
So $X_1Y_1 \perp X_2Y_2$.

Combine these three steps, we see that the intersect point of $X_1Y_1$ and $X_2Y_2$ is on the nine-point circle of $\triangle ABC$.