China-TST-2023B

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 $Q{P}_1=Q{H}_1$.

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, $△PTQ\sim △BHO$, 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 $△OPQ\sim △HBA$. Therefore, $\angle OQP=\angle HAB=\angle H_{1}AB=\angle H_1{P}_{1}B$. Since $PQ\perp P_{1}B$, it follows that $OQ\perp P_1{H}_1$, and thus $OQ$ bisects $P_1{H}_1$ perpendicularly, leading to $Q{P}_1=Q{H}_1$.

Since $H_1$ and $H$ are symmetric with respect to $BP$, we have $QH=Q{H}_1=Q{P}_1$. Also, it is well-known that $M$ is the midpoint of $H{P}_1$, so $QM\perp MH$. Consequently, the circle with diameter $HQ$ passes through the midpoint $M$ of $AB$. $\square$



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Alternative solution.

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

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

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

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. $\square$