USAMO $2015/2$ (Doubt in complex numbers solution)
By Sebastian Wright
Quadrilateral $APBQ$ is inscribed in circle $\omega$ with $\angle P = \angle Q = 90^{\circ}$ and $AP = AQ < BP$. Let $X$ be a variable point on segment $\overline{PQ}$. Line $AX$ meets $\omega$ again at $S$ (other than $A$). Point $T$ lies on arc $AQB$ of $\omega$ such that $\overline{XT}$ is perpendicular to $\overline{AX}$. Let $M$ denote the midpoint of chord $\overline{ST}$. As $X$ varies on segment $\overline{PQ}$, show that $M$ moves along a circle.
In This ($4$th post)
Why $4\operatorname{Re} x + 2 = s + t + \frac 1s + \frac 1t + \frac st + \frac ts$ depends only on $P$ and $Q$, and not on $X$ ?
and also where we have used the fact that $AP=AQ$ and $\angle P = \angle Q = 90$ ?
Please someone help me clear this doubts .
thankyou very much
$\endgroup$ 11 Answer
$\begingroup$$∠P=∠Q=90^\circ$ is used to show $AB$ is the circumference. Then $AP=AQ$ shows that $PQ\perp AB$, which $AB$ is the real line, so $P$ and $Q$ have the same real component. Since $X$ lies on $PQ$, then its real component only depends on $P$ and $Q$.
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