2022 China Team Selection Test for IMO

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

The leading coefficient of this quadratic equation is zero if and only if $|a-b|=|c-a|$ (i.e. $△ABC$ 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 $△ABC$ is isosceles triangle with a right angle; this contradicts with our assumption. So () is a quadratic equation. The discriminant of () is $$\begin{array}{rl}\Delta & =4\left\{[c(a-b{)}^2+b(c-a{)}^2{]}^2-[c^2(a-b{)}^2+b^2(c-a{)}^2{]}^2[(a-b{)}^2+(c-a{)}^2]\right\}\\ & =4\left\{2bc(a-b{)}^2(c-a{)}^2-(b^2+c^2)(a-b{)}^2(c-a{)}^2\right\}\\ & =-4(a-b{)}^2(b-c{)}^2(c-a{)}^2\ne 0.\end{array}$$ 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 $△ABC$.

Step 1: show that $P_1{P}_2$ passes through the circumcenter $O$ of $△ABC$. This is equivalent to $$\frac{c(a-b{)}^2+b(c-a{)}^2}{(a-b)(b-c)(c-a)}\in i\mathbb{R}\iff \frac{c(a-b{)}^2+b(c-a{)}^2}{(a-b)(b-c)(c-a)}+\frac{\overline{c(a-b{)}^2+b(c-a{)}^2}}{(a-b)(b-c)(c-a)}=0.$$ But $$\begin{array}{rl}\frac{c(a-b{)}^2+b(c-a{)}^2}{(a-b)(b-c)(c-a)}& =\frac{c(a-b)}{(b-c)(c-a)}+\frac{b(c-a)}{(a-b)(b-c)}\\ & =-\frac{c}{b-c}-\frac{c}{c-a}-\frac{b}{a-b}-\frac{b}{b-c},\end{array}$$ 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 $△ABC$, its feets on the three sides of $△ABC$ are collinear (on the well-known Simson line), and this line passes through the midpoint of $Q$ and orthocenter $H$ of $△ABC$. Let $X$ and $Y$ denote the projections of a point $Q$ on the circumcircle onto $BC$ and $CA$, respectively. Then, we know $$x=\frac{1}{2}(q-bc\bar{q}+b+c),y=\frac{1}{2}(q-ca\bar{q}+c+a).$$ 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, $X_1{Y}_1$ and $X_2{Y}_2$ each passes through a pair of opposite points on the nine-point circle of $△ABC$.

Step 3: prove that $X_1{Y}_1\perp X_2{Y}_2$. We have $$\begin{array}{rl}\angle C{Y}_1{X}_1& =\angle C{Q}_1{X}_1=90^{\circ }-\angle X_{1}C{Q}_1\\ \angle BA{Q}_2& =\angle Q_2{Y}_2{X}_2=90^{\circ }-\angle C{Y}_2{X}_2,\end{array}$$ So $X_1{Y}_1\perp X_2{Y}_2$.

Combine these three steps, we see that the intersect point of $X_1{Y}_1$ and $X_2{Y}_2$ is on the nine-point circle of $△ABC$.