Let $A,B$ be two fixed points on the unit circle $\omega$, satisfying $\sqrt{2} < AB < 2$. Let $P$ be a point that can move on the unit circle, and it can move to anywhere on the unit circle satisfying $\triangle ABP$ is acute and $AP>AB>BP$. Let $H$ be the orthocenter of $\triangle ABP$ and $S$ be a point on the minor arc $AP$ satisfying $SH=AH$. Let $T$ be a point on the minor arc $AB$ satisfying $TB || AP$. Let $ST\cap BP = Q$. Show that (recall $P$ varies) the circle with diameter $HQ$ passes through a fixed point.
Problem
Source: China TST 2023 Problem 19
Tags: geometry
HaO-R-Zhe
01.04.2023 00:08
The fixed point is the midpoint of AB, regardless of where P and B lie. @CANBANKAN I'll send you my solution again soon
HaO-R-Zhe
01.04.2023 06:26
Basically complex bash, with A, B, P, S, T lying on the unit circle. One may compute that q=p(b+p)/(a+p), and then compute angle QMH is a right angle
sami1618
29.06.2024 23:50
Let $E\neq A$ be the point along $AB$ such that $PA=PE$ (notice $HA=HS=HE$). Let $TB$ intersect $(PBE)$ again at $F$. Let $D$ be the foot of $A$ onto $BP$. Let $M$ be the midpoint of $BC$. We claim that $M$ is the desired fixed point. Claim: $\angle HEF=90^{\circ}$ $$\angle HEF=\angle HBF=180^{\circ}-\angle HBT=90^{\circ}$$Claim: $STEF$ is cyclic $$\angle FES=90^{\circ}-\angle HES=\angle EAS=\angle BAS=\angle BTS=\angle FTS$$Claim: $Q$ lies on $EF$ Follows from the Radical Axes Theorem on $w$, $(BPEF)$, and $(STEF)$. Claim: $D$ and $E$ lie on the circle with diameter $HQ$ $$\angle HDQ=\angle HEQ=90^{\circ}$$Claim: $MHDE$ is cyclic $$\angle HDM=\angle MAH=\angle HEM$$
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