2023 Chinese IMO National Team Selection Test

Proof. Let $A^{\ast }$ be the midpoint of segment $AH$. In the given diagram, we have $\angle A_{1}RH=\angle A_{1}CR=\angle RAH$. Thus, $A_{1}R$ is a tangent to the circle $\Gamma$ with diameter $AH$, and similarly, $A_{1}Q$ is also a tangent to $\Gamma$ (R and Q lie on the circle $\Gamma$). Hence, we have

a. $$A_1{A}^{\ast }\perp RQ.$$

Considering the polar with respect to $\Gamma$, we know that point $D$ lies on the polar line of $A_1$ with respect to $\Gamma$. Therefore, $A_1$ lies on the polar line of $D$ with respect to $\Gamma$. However, $DA$ is a tangent to $\Gamma$, so $A{A}_1$ is the polar line of $D$. This implies

b. $$D{A}^{\ast }\perp A{A}_1.$$

Combining equations (a) and (b), we conclude that $K$ is the orthocenter of triangle $△A^{\ast }{A}_{1}D$. In particular, $A^{\ast }K\perp A_{1}D$, which implies that $\angle A^{\ast }{A}_2{A}_1=\frac{\pi }{2}$. Therefore, point $A_2$ lies on the nine-point circle of triangle $△ABC$. Consequently, $\omega$ is the nine-point circle of triangle $△ABC$.

In the given diagram, let $N$ be the center of $\omega$ and $I$ be the incenter of $△ABC$. Let $BC=a$, $CA=b$, $AB=c$, and $s=\frac{a+b+c}{2}$. Consider the inversion $f$ centered at $A$ that preserves the nine-point circle. Let $\odot A^{\prime }$ be the image of the incircle under $f$. It is clear that $\odot A^{\prime }$ is tangent to the sides $AB$ and $AC$. Similarly, we can define $\odot B^{\prime }$ and $\odot C^{\prime }$.

Let $M$ be the foot of the perpendicular from $I$ to side $AB$. We will show that $\odot A^{\prime }$ and $\omega$ do not coincide; otherwise, we would have $AR\cdot A{C}_1=A{M}^2$, which implies $b\cdot \cos A\cdot \frac{c}{2}=(\frac{b+c-a}{2}{)}^2$. This leads to either $a=b$ or $a=c$, contradicting the non-isosceles nature of $△ABC$.

Next, we will verify the collinearity of $A^{\prime }$, $B^{\prime }$, and $C^{\prime }$. Let $$\overrightarrow{I{A}^{\prime }}=p\cdot \overrightarrow{IA},\overrightarrow{I{B}^{\prime }}=q\cdot \overrightarrow{IB},\overrightarrow{I{C}^{\prime }}=r\cdot \overrightarrow{IC}.$$ Then we have: $$\frac{a}{p}\cdot \overrightarrow{I{A}^{\prime }}+\frac{b}{q}\cdot \overrightarrow{I{B}^{\prime }}+\frac{c}{r}\cdot \overrightarrow{I{C}^{\prime }}=a\cdot \overrightarrow{IA}+b\cdot \overrightarrow{IB}+c\cdot \overrightarrow{IC}=\overrightarrow{0}.$$ To prove that $A^{\prime }$, $B^{\prime }$, and $C^{\prime }$ are collinear, it suffices to show that: $$\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=0.$$ Let $L$ be the foot of the perpendicular from $A^{\prime }$ to side $AB$. Then we have $\frac{A{A}^{\prime }}{AI}=\frac{AL}{AM}$, $AM=s-a$. According to the properties of inversion $f$, we know that $f(M)=L$. Hence, $$AM\cdot AL=AR\cdot A{C}_1=b\cos A\cdot \frac{c}{2}=\frac{b^2+c^2-a^2}{4}.$$ $$\frac{AL}{AM}=\frac{AM\cdot AL}{A{M}^2}=\frac{b^2+c^2-a^2}{4(s-a{)}^2}.$$ Therefore, we have: $$p=\frac{I{A}^{\prime }}{IA}=\frac{2(a-b)(a-c)}{(b+c-a{)}^2}$$ Similarly, we have: $$q=\frac{2(b-c)(b-a)}{(c+a-b{)}^2},r=\frac{2(c-a)(c-b)}{(a+b-c{)}^2}.$$ Plugging into the computations gives $\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=0$ (Here, upon substitution, we obtain a cyclic summation of $a(b-c)(b+c-a{)}^2$. Let us consider this as a function of $a$, denoted by $g(a)$. It is observed that $g(a)$ is actually a quadratic function, and we have $g(b)=g(c)=g(b+c)=0$. Hence, we conclude that $g\equiv 0$. $\square$