Problem 1850

Proof. Let $A^*$ be the midpoint of segment $AH$. In the given diagram, we have $\angle A_1RH = \angle A_1CR = \angle RAH$. Thus, $A_1R$ is a tangent to the circle $\Gamma$ with diameter $AH$, and similarly, $A_1Q$ is also a tangent to $\Gamma$ (R and Q lie on the circle $\Gamma$). Hence, we have

a.
$$A_1A^* \perp RQ.$$

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 $AA_1$ is the polar line of $D$. This implies

b.
$$DA^* \perp AA_1.$$

Combining equations (a) and (b), we conclude that $K$ is the orthocenter of triangle $\triangle A^*A_1D$. In particular, $A^*K \perp A_1D$, which implies that $\angle A^*A_2A_1 = \frac{\pi}{2}$. Therefore, point $A_2$ lies on the nine-point circle of triangle $\triangle ABC$. Consequently, $\omega$ is the nine-point circle of triangle $\triangle ABC$.

In the given diagram, let $N$ be the center of $\omega$ and $I$ be the incenter of $\triangle ABC$. 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 $\odot A'$ be the image of the incircle under $f$. It is clear that $\odot A'$ is tangent to the sides $AB$ and $AC$. Similarly, we can define $\odot B'$ and $\odot C'$.

Let $M$ be the foot of the perpendicular from $I$ to side $AB$. We will show that $\odot A'$ and $\omega$ do not coincide; otherwise, we would have $AR \cdot AC_1 = AM^2$, 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 $\triangle ABC$.

Next, we will verify the collinearity of $A'$, $B'$, and $C'$. Let
$$\overrightarrow{IA'} = p \cdot \overrightarrow{IA}, \quad \overrightarrow{IB'} = q \cdot \overrightarrow{IB}, \quad \overrightarrow{IC'} = r \cdot \overrightarrow{IC}.$$
Then we have:
$$\frac{a}{p} \cdot \overrightarrow{IA'} + \frac{b}{q} \cdot \overrightarrow{IB'} + \frac{c}{r} \cdot \overrightarrow{IC'} = a \cdot \overrightarrow{IA} + b \cdot \overrightarrow{IB} + c \cdot \overrightarrow{IC} = \overrightarrow{0}.$$
To prove that $A'$, $B'$, and $C'$ 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 $A'$ to side $AB$. Then we have $\frac{AA'}{AI} = \frac{AL}{AM}$, $AM = s-a$. According to the properties of inversion $f$, we know that $f(M) = L$. Hence,
$$AM \cdot AL = AR \cdot AC_1 = b \cos A \cdot \frac{c}{2} = \frac{b^2 + c^2 - a^2}{4}.$$
$$\frac{AL}{AM} = \frac{AM \cdot AL}{AM^2} = \frac{b^2 + c^2 - a^2}{4(s-a)^2}.$$
Therefore, we have:
$$p = \frac{IA'}{IA} = \frac{2(a-b)(a-c)}{(b+c-a)^2}$$
Similarly, we have:
$$q = \frac{2(b-c)(b-a)}{(c+a-b)^2}, \quad r = \frac{2(c-a)(c-b)}{(a+b-c)^2}.$$
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 $g \equiv 0$. $\Box$