Let $ABCD$ be a cyclic quadrilateral. A point $P$ moves along the arc $AD$ which does not contain $B$ and $C$. A fixed line $l$, perpendicular to $BC$, meets the rays $BP$, $CP$ at points $B_0$, $C_0$ respectively. Prove that the tangent at $P$ to the circumcircle of triangle $PB_0C_0$ passes through some fixed point.
Problem
Source: Sharygin 2018
Tags: geometry
TheDarkPrince
05.04.2018 09:38
We claim that the fixed point is the center of the circle circumscribing the quadrilateral $ABCD$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -12.780312388489444, xmax = 19.494797944691786, ymin = -6.29015279504083, ymax = 11.593432930557963; /* image dimensions */ pair A = (0.,4.), B = (-2.,0.), C = (4.,0.), D = (4.,2.), P = (2.,4.), B_0 = (0.7381969395889266,2.7381969395889265), C_0 = (0.7381969395889266,6.523606120822147), O = (1.,1.), Q = (0.,-2.); /* draw figures */ draw(A--B, linewidth(1.)); draw(B--C, linewidth(1.)); draw(circle(O, 3.1622776601683795), linewidth(1.)); draw(A--D, linewidth(1.)); draw(D--C, linewidth(1.)); draw(C_0--(0.7381969395889266,0.), linewidth(1.)); draw(P--B, linewidth(1.)); draw(C_0--C, linewidth(1.)); draw(P--Q, linewidth(1.)); /* dots and labels */ dot(A,dotstyle); label("$A$", A, dir(90)); dot(B,dotstyle); label("$B$", B, dir(-135)); dot(C,dotstyle); label("$C$", C, dir(0)); dot(D,dotstyle); label("$D$", D, dir(0)); dot(P,dotstyle); label("$P$", P, dir(60)); dot(B_0,linewidth(4.pt) + dotstyle); label("$B_0$", B_0, dir(-30)); dot(C_0,linewidth(4.pt) + dotstyle); label("$C_0$", C_0, dir(90)); dot(O,linewidth(4.pt) + dotstyle); label("$O$", O, dir(0)); dot(Q,linewidth(4.pt) + dotstyle); label("$Q$", Q, dir(-135)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $O$ be the center of the circle circumscribing $ABCD$. Suppose $PO$ meet the circle again at $Q$. Let $B_0C_0\cap BC=X$. As $B_0C_0\perp BC$. we have, \begin{align*} \angle OPC&=\angle QPC=\angle QBC\\ &=\angle PBQ-\angle PBC\\ &=90^{\circ}-\angle PBC\\ &=90^{\circ}-\angle B_0BX\\ &=\angle BB_0X\\ &=\angle PB_0C_0. \end{align*} So, from the converse of tangent chord theorem, $OP$ is tangent to circle $(PB_0C_0)$. So we have proved our original claim.