WakeUp wrote: Let $AB$ and $CD$ be two diameters of the circle $C$. For an arbitrary point $P$ on $C$, let $R$ and $S$ be the feet of the perpendiculars from $P$ to $AB$ and $CD$, respectively. Show that the length of $RS$ is independent of the choice of $P$. Denote circle $C$ is centered at $O$ and $r$ is radius of this. Let $I$ is the midpoint of $OP$ then $PRSO$ is cyclic centered at $I$ and this circle has radius $ \frac{r}{2} $ Setting $ \angle{AOD} = p $ so that $ RS = 2 . \frac{r}{2} . sinp = rsinp $.

Remark: The problem has a correspondent in mechanics, like this Given a fixed point $O$, two fixed directions $Ox$ and $Oy$ and a bar $RS$ of fixed length with the ends $R$ and $S$ sliding along $Ox$ and $Oy$ respectively, then the locus of its center of rotation is a circle of center $O$ and radius $\frac{RS}{sin (\widehat{xOy})}$. Best regards, sunken rock

Could complex numbers work? edit: yes, see below

Solution 1. [Synthetic] Let $O$ be the center of $\mathcal C$. Firstly, $\angle PSO+\angle PRO=90^\circ+90^\circ=180^\circ$, thus $PROS$ is cyclic quadrilateral. Also note that $OP$ is the diameter of $\mathcal C$, which is fixed value. Let $E$ be the center $(PROS)$, thus $ER=ES$ is fixed value, as if diameter is fixed, so is the radius. Also, $\measuredangle SER=2\measuredangle SPR=2\measuredangle SOR=\measuredangle COB$, where the last angle is fixed, hence $\triangle SER$ is always with fixed size, therefore the length of $RS$ does not depend on $P$. Solution 2. [Complex bash] Throw everything onto complex plane, let $\mathcal C$ be the unit circle, thus $b=-a$ ja $d=-c$, also since $R$ and $S$ foot from $AB$ and $CD$, respectively, then \begin{align*} r=\frac{a+b+p-\overline{p}ab}{2}=\frac{p+\overline{p}a^2}{2}=\frac{p^2+a^2}{2p}\end{align*}and \begin{align*}s=\frac{c+d+p-\overline{p}cd}{2}=\frac{p+\overline{p}c^2}{2}=\frac{p^2+c^2}{2p}.\end{align*}Conjugates of those are \begin{align*} \overline{r}&=\frac{\frac{1}{p}+pa^2}{2}=\frac{1+p^2a^2}{2p},\\ \overline{s}&=\frac{\frac{1}{p}+pc^2}{2}=\frac{1+p^2c^2}{2p}.\end{align*}Therefore, the length of $RS$ is, \begin{align*} \lvert r-s\rvert^2=(r-s)(\overline{r}-\overline{s})=\frac{a^2-c^2}{2p}\cdot \frac{p^2(a^2-c^2)}{2p}=\frac{(a^2-c^2)^2}{4}.\end{align*}Now it's easy to see that the length of $RS$ does not depend on $p$. [asy][asy] import olympiad;import geometry; defaultpen(fontsize(12pt)); size(7cm); pair A=dir(180); pair B=dir(0); pair C=dir(125); pair D=dir(-55); pair O=(0,0); pair P=dir(70); pair E=midpoint(O--P); path w = circle(O, 1); path q = circle(E,0.5); pair S=intersectionpoints(q,C--D)[0]; pair R=intersectionpoints(q,A--B)[1]; dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$P$",P,dir(P)); dot("$D$",D,dir(D)); dot("$E$",E,dir(100)); dot("$O$",O,dir(250)); dot("$R$",R,dir(-40)); dot("$S$",S,dir(200)); draw(rightanglemark(P,S,O,3)); draw(rightanglemark(P,R,O,3)); draw(w); draw(q); draw(O--P); draw(R--P); draw(S--P); draw(A--B); draw(C--D); draw(S--E); draw(R--E); draw(R--S,blue); [/asy][/asy]