AO.AB=AC^2=AM.AN so OMNB cyclic. PNO=PBO=90 so ONPB cyclic, so MOPNB cyclic.

Solution: $\angle OBP = \angle ONP= 90^{\circ} \Longrightarrow OBPN $ is cyclic Let $\angle MAO =\alpha$ Since $ABN$ is right triangle $\angle OBN=90^{\circ}- \alpha$ $OBPN$ is deltoid because $ON=OB$ and $NP=PB$ $ \Longrightarrow \angle POB = \alpha $ $ \Delta AOM \cong \Delta OBP \Longrightarrow MO=BP \Longrightarrow MOPB $ is rectangle $\angle OMP= \angle ONP= 90^{\circ} \Longrightarrow OMNP $ is cyclic q.e.d

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Dear Mathlinkers, 1. by a converse of the Reim's theorem, M, O, B, N concyclic 2. by Thales theorem, O, B, P, N concyclic and we are done... Sincerely Jean-Louis

By HL, $\triangle PNO \cong \triangle PBO.$ Hence, $\angle PON=\angle POB$ and let the angles equal to $x$. Thus, $\angle NPO=90^{\circ}-x$. Note that $\angle NAB=\frac{1}{2}\angle NOB=x$ and $\angle AMO=90^{\circ}-x$ so $\angle OMN=90^{\circ}+x$. Hence, $OMNP$ is cyclic as $\angle OMN+\angle OPN=(90^{\circ}+x)+(90^{\circ}-x)=180^{\circ}$, as desired. $\blacksquare$

$\angle MOB=90=180-90=180-\angle ANB=180-\angle MNB$. So $OBMN$ is cyclic. $\angle NMB=\angle NOB=2\angle BAN=180-\angle NPB$. So $BPMN$ is cyclic too and we are done.

Let $(A B C)$ be the unit circle and be in the complex plane $O=(0,0)$, Then $M=\frac{c}{2}, P=\frac{b n}{b+n}$ We have to prove $$ S=\frac{p-0}{p-b}: \frac{n-0}{n-b} \in \mathbb{R} $$but $S=-\frac{n}{b}: \frac{n}{n-b}=\frac{n-b}{b}$ and $\bar{S}=\frac{\frac{1}{n}-\frac{1}{b}}{\frac{1}{n}}=\frac{n-b}{b}$ which means $S=\bar{S} \Leftrightarrow S \in \mathbb{R}$ Ups I read wrong, thanks @below [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.209898506723724, xmax = 2.1831789717080152, ymin = -2.6729488991816752, ymax = 2.0186262360518135; /* image dimensions */ pen xfqqff = rgb(0.4980392156862745,0.,1.); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); /* draw grid of horizontal/vertical lines */ pen gridstyle = linewidth(0.7) + cqcqcq; real gridx = 0.5, gridy = 0.5; /* grid intervals */ for(real i = ceil(xmin/gridx)*gridx; i <= floor(xmax/gridx)*gridx; i += gridx) draw((i,ymin)--(i,ymax), gridstyle); for(real i = ceil(ymin/gridy)*gridy; i <= floor(ymax/gridy)*gridy; i += gridy) draw((xmin,i)--(xmax,i), gridstyle); /* end grid */ Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax,linewidth(1.2), Ticks(laxis, Step = 0.5, Size = 2, NoZero),EndArrow(9), above = true); yaxis(ymin, ymax,linewidth(1.2), Ticks(laxis, Step = 0.5, Size = 2, NoZero),EndArrow(9), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw(circle((0.,0.), 1.), linewidth(1.) + red); draw(circle((-0.0486093877675316,0.5568995667271306), 0.5590169943749475), linewidth(1.) + linetype("4 4") + xfqqff); draw((-0.523294673809755,0.8521517965493837)--(-0.09721877553506314,1.1137991334542612), linewidth(1.)); draw((-0.09721877553506314,1.1137991334542612)--(0.3677446329536539,0.9299268169774342), linewidth(1.)); draw((0.,0.)--(0.8521517965493837,0.523294673809755), linewidth(2.) + blue); draw((-0.523294673809755,0.8521517965493837)--(0.523294673809755,-0.8521517965493837), linewidth(1.) + blue); /* dots and labels */ dot((0.,0.),dotstyle); label("$O$", (0.025175999846809776,0.06219497035346265), NE * labelscalefactor); dot((0.523294673809755,-0.8521517965493837),dotstyle); label("$A$", (0.47574804891673167,-0.495091511390916), NE * labelscalefactor); dot((-0.523294673809755,0.8521517965493837),linewidth(4.pt) + dotstyle); label("$B$", (-0.5024675839324408,0.8981246929700306), NE * labelscalefactor); dot((0.8521517965493837,0.523294673809755),linewidth(4.pt) + dotstyle); label("$C$", (0.872962881649426,0.5720528153536388), NE * labelscalefactor); dot((0.42607589827469183,0.2616473369048775),linewidth(4.pt) + dotstyle); label("$M$", (0.45203373054463053,0.31119531326052546), NE * labelscalefactor); dot((0.3677446329536539,0.9299268169774342),linewidth(4.pt) + dotstyle); label("$N$", (0.3927479346143777,0.9751962276793597), NE * labelscalefactor); dot((-0.09721877553506314,1.1137991334542612),linewidth(4.pt) + dotstyle); label("$P$", (-0.07560985323462013,1.158982195063144), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Iora wrote: Let $(A B C)$ be the unit circle and be in the complex plane $O=(0,0)$, Then $M=\frac{c}{2}, P=\frac{b n}{b+n}$ We have to prove $$ S=\frac{p-0}{p-b}: \frac{n-0}{n-b} \in \mathbb{R} $$but $S=-\frac{n}{b}: \frac{n}{n-b}=\frac{n-b}{b}$ and $\bar{S}=\frac{\frac{1}{n}-\frac{1}{b}}{\frac{1}{n}}=\frac{n-b}{b}$ which means $S=\bar{S} \Leftrightarrow S \in \mathbb{R}$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.209898506723724, xmax = 2.1831789717080152, ymin = -2.6729488991816752, ymax = 2.0186262360518135; /* image dimensions */ pen xfqqff = rgb(0.4980392156862745,0.,1.); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); /* draw grid of horizontal/vertical lines */ pen gridstyle = linewidth(0.7) + cqcqcq; real gridx = 0.5, gridy = 0.5; /* grid intervals */ for(real i = ceil(xmin/gridx)*gridx; i <= floor(xmax/gridx)*gridx; i += gridx) draw((i,ymin)--(i,ymax), gridstyle); for(real i = ceil(ymin/gridy)*gridy; i <= floor(ymax/gridy)*gridy; i += gridy) draw((xmin,i)--(xmax,i), gridstyle); /* end grid */ Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax,linewidth(1.2), Ticks(laxis, Step = 0.5, Size = 2, NoZero),EndArrow(9), above = true); yaxis(ymin, ymax,linewidth(1.2), Ticks(laxis, Step = 0.5, Size = 2, NoZero),EndArrow(9), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw(circle((0.,0.), 1.), linewidth(1.) + red); draw(circle((-0.0486093877675316,0.5568995667271306), 0.5590169943749475), linewidth(1.) + linetype("4 4") + xfqqff); draw((-0.523294673809755,0.8521517965493837)--(-0.09721877553506314,1.1137991334542612), linewidth(1.)); draw((-0.09721877553506314,1.1137991334542612)--(0.3677446329536539,0.9299268169774342), linewidth(1.)); draw((0.,0.)--(0.8521517965493837,0.523294673809755), linewidth(2.) + blue); draw((-0.523294673809755,0.8521517965493837)--(0.523294673809755,-0.8521517965493837), linewidth(1.) + blue); /* dots and labels */ dot((0.,0.),dotstyle); label("$O$", (0.025175999846809776,0.06219497035346265), NE * labelscalefactor); dot((0.523294673809755,-0.8521517965493837),dotstyle); label("$A$", (0.47574804891673167,-0.495091511390916), NE * labelscalefactor); dot((-0.523294673809755,0.8521517965493837),linewidth(4.pt) + dotstyle); label("$B$", (-0.5024675839324408,0.8981246929700306), NE * labelscalefactor); dot((0.8521517965493837,0.523294673809755),linewidth(4.pt) + dotstyle); label("$C$", (0.872962881649426,0.5720528153536388), NE * labelscalefactor); dot((0.42607589827469183,0.2616473369048775),linewidth(4.pt) + dotstyle); label("$M$", (0.45203373054463053,0.31119531326052546), NE * labelscalefactor); dot((0.3677446329536539,0.9299268169774342),linewidth(4.pt) + dotstyle); label("$N$", (0.3927479346143777,0.9751962276793597), NE * labelscalefactor); dot((-0.09721877553506314,1.1137991334542612),linewidth(4.pt) + dotstyle); label("$P$", (-0.07560985323462013,1.158982195063144), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] But $M$ is not necessarily the midpoint of $OC$.

let <BAN=a <=> <PNA=90+a <=> <BPN=180-2a. since BP and PN are both tangents PO is bisector of <BPN, so <OPN=90-a.<OMA=180-<MOA-<OAM=90-a <=> <OMN=90+a. so <OPN+<OMN=180 and we are done.

$$\text{My Solution:}$$$$MOPN \text{cyclic}=OBPN \text{cyclic}+OMNB \text{cyclic}$$$$\angle OBP=\angle PNO=90$$$$(1)\implies{OBPN \text{cyclic}}$$$$\angle ANB=\angle MOB=90$$$$(2)\implies{OMNB \text{cyclic}}$$So,we are done

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