China-TST-B

We prove that the midpoint $M$ of $AB$ satisfies the given condition.

Let $P_1$ be the antipodal point of $P$ on the circle $\omega$, and let $H_1$ be the intersection of the extension of $AH$ with $\omega$. We will show that $QP_1 = QH_1$.

Denote $O$ as the center of $\omega$. Since $SH = AH$, we have that $S$ and $A$ are symmetric with respect to $OH$, implying $OH \perp SA$. Also, $HB \perp AP$, so $\angle BHO - 180^\circ - \angle SAP = 180^\circ - \angle STP = \angle PTQ$. By noting that $TB \parallel AP$, we have $\angle TPQ = \angle APB - \angle APT = \angle APB - \angle PAB = \angle HBO$. Thus, $\triangle PTQ \sim \triangle BHO$, which gives $\frac{PQ}{PT} = \frac{BO}{BH}$. Since $PT = AB$ and $BO = PO$, we obtain $\frac{PQ}{AB} = \frac{PO}{BH}$. Furthermore, $\angle OPQ = 90^\circ - \angle PAB = \angle ABH$, implying $\triangle OPQ \sim \triangle HBA$. Therefore, $\angle OQP = \angle HAB = \angle H_1AB = \angle H_1P_1B$. Since $PQ \perp P_1B$, it follows that $OQ \perp P_1H_1$, and thus $OQ$ bisects $P_1H_1$ perpendicularly, leading to $QP_1 = QH_1$.

Since $H_1$ and $H$ are symmetric with respect to $BP$, we have $QH = QH_1 = QP_1$. Also, it is well-known that $M$ is the midpoint of $HP_1$, so $QM \perp MH$. Consequently, the circle with diameter $HQ$ passes through the midpoint $M$ of $AB$. $\square$

Let lowercase letters represent complex numbers corresponding to the respective points on the complex plane.

Firstly,
$$\begin{align*} Q \in PB & \\ \Leftrightarrow \frac{q-p}{q-b} \in \mathbb{R} & \Leftrightarrow \frac{q-p}{q-b} = \frac{\bar{q}-\bar{p}}{\bar{q}-\bar{b}} \\ & \Leftrightarrow q\bar{q} - p\bar{q} - q\bar{b} + p\bar{b} = q\bar{q} - b\bar{q} - q\bar{p} + b\bar{p} \\ & \Leftrightarrow (p-b)\bar{q} + \frac{p-b}{pb}q = \frac{p^2-b^2}{pb} \\ & \Leftrightarrow q + pb\bar{q} = p + b. \end{align*}$$

Similarly,
$$Q \in ST \Leftrightarrow q + st\bar{q} = s + t.$$
Therefore,
$$\bar{q} = \frac{s + t - p - b}{st - pb}.$$
According to the given conditions, $t = \frac{ap}{b}$, and $s$ satisfies
$$\begin{align*} (s-h)\left(\frac{1}{s}-\bar{h}\right) &= (a-h)\left(\frac{1}{a}-\bar{h}\right) \\ \Leftrightarrow 1-\frac{h}{s}-\bar{h}s+h\bar{h} &= 1-\frac{h}{a}-\bar{h}a+h\bar{h} \\ \Leftrightarrow \bar{h}s^2-\left(\frac{h}{a}+\bar{h}a\right)s+h = 0. \end{align*}$$
Hence, by Vieta's formulas, we have
$$s = \frac{h}{a\bar{h}} = bp \frac{a+b+p}{ab+bp+pa}.$$
Furthermore, we have
$$\begin{align*} s + t - p - b &= p \left[ \frac{a}{b} + \frac{b(a+b+p)}{ab+bp+pa} - 1 \right] - b \\ &= p \frac{a(ab + bp + pa) + b(b^2 - pa)}{b(ab + bp + pa)} - b \\ &= \frac{ap(ab + bp + pa) + bp(b^2 - pa) - b^2(ab + bp + pa)}{b(ab + bp + pa)} \\ &= \frac{(ab + pa)(ap - b^2)}{b(ab + bp + pa)} = \frac{a(b + p)(ap - b^2)}{b(ab + bp + pa)}. \end{align*}$$
$$st - pb = \frac{ap^2(a + b + p) - pb(ab + bp + pa)}{ab + bp + pa} = \frac{p[ap(a + p) - b^2(a + p)]}{ab + bp + pa} = \frac{p(ap - b^2)(a + p)}{ab + bp + pa}.$$
so
$$\bar{q} = \frac{a(p+b)}{bp(p+a)}, \quad q = \frac{\frac{1}{abp}(b+p)}{\frac{1}{abp^2}(p+a)} = \frac{p(p+b)}{p+a}.$$

We know that $h = a + b + p$. The center of the circle with diameter $HQ$ is
$$\frac{1}{2}(h+q) = \frac{a+b}{2} + \frac{p(p+a)+(p+b)}{p+a},$$
and its radius is
$$\left| \frac{1}{2}(h+q) \right| = \left| \frac{1}{2(p+a)} \left[ (a+b+p)(p+a) - p(p+b) \right] \right| = \left| \frac{a}{2} \frac{(p+a)+(p+b)}{p+a} \right|.$$
Since $|a| = |p| = 1$, we can conclude that the circle passes through the midpoint of $AB$, which is $\frac{a+b}{2}$. Thus, the proof is complete. $\square$