Circles $\omega_1$ and $\omega_2$ inscribed into equal angles $X_1OY$ and $Y OX_2$ touch lines $OX_1$ and $OX_2$ at points $A_1$ and $A_2$ respectively. Also they touch $OY$ at points $B_1$ and $B_2$. Let $C_1$ be the second common point of $A_1B_2$ and $\omega_1, C_2$ be the second common point of $A_2B_1$ and $\omega_2$. Prove that $C_1C_2$ is the common tangent of two circles.
Problem
Source: Sharygin 2010 Final 8.4
Tags: geometry, tangent, circles, equal angles, common tangents
GeoMetrix
25.12.2019 15:33
[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 = -4.28264905283688, xmax = 24.98493192714741, ymin = -9.432274274362085, ymax = 9.895732126056082; /* image dimensions */ pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw((-0.16,1.06)--(8.74,-5.92), linewidth(1.2) + wvvxds); draw((-0.16,1.06)--(13.905061984724782,0.4312806143468913), linewidth(1.2) + wvvxds); draw((-0.16,1.06)--(11.130343268440559,8.375338308914044), linewidth(1.2) + wvvxds); draw(circle((9.898792551478865,3.7911132384613015), 3.177576144966448), linewidth(1.2)); draw(circle((4.88344320105104,-0.8089030853842818), 1.641817699530772), linewidth(1.2)); draw((3.87024717678906,-2.100800594830072)--(9.756894083933526,0.6167069976080424), linewidth(1.2) + blue); draw((4.956760536316561,0.8312767604991137)--(8.170935298802075,6.457852721693972), linewidth(1.2) + blue); draw((4.88344320105104,-0.8089030853842818)--(6.523898647261402,-0.8757707718350884), linewidth(1.2) + dtsfsf); draw((9.898792551478865,3.7911132384613015)--(6.7240376796165595,3.9249850452258106), linewidth(1.2) + red); draw((9.898792551478865,3.7911132384613015)--(8.170935298802075,6.457852721693972), linewidth(1.2) + red); draw((4.88344320105104,-0.8089030853842818)--(3.87024717678906,-2.100800594830072), linewidth(1.2) + red); draw((9.898792551478865,3.7911132384613015)--(9.756894083933526,0.6167069976080424), linewidth(1.2) + red); draw((8.170935298802075,6.457852721693972)--(9.756894083933526,0.6167069976080424), linewidth(1.2) + blue); draw((4.956760536316561,0.8312767604991137)--(4.88344320105104,-0.8089030853842818), linewidth(1.2) + red); draw((4.956760536316561,0.8312767604991137)--(3.87024717678906,-2.100800594830072), linewidth(1.2) + blue); draw((6.7240376796165595,3.9249850452258106)--(9.756894083933526,0.6167069976080424), linewidth(1.2) + blue); draw((4.956760536316561,0.8312767604991137)--(6.523898647261402,-0.8757707718350884), linewidth(1.2) + blue); draw(circle((7.390474488221161,1.4926522029345493), 2.5219795945946455), linewidth(1.2) + green); draw((4.88344320105104,-0.8089030853842818)--(-0.16,1.06), linewidth(1.2) + wvvxds); draw((-0.16,1.06)--(6.523898647261402,-0.8757707718350884), linewidth(1.2) + wvvxds); draw((9.898792551478865,3.7911132384613015)--(-0.16,1.06), linewidth(1.2) + wvvxds); draw((-0.16,1.06)--(6.7240376796165595,3.9249850452258106), linewidth(1.2) + wvvxds); draw((4.88344320105104,-0.8089030853842818)--(9.898792551478865,3.7911132384613015), linewidth(1.2) + red); /* dots and labels */ dot((-0.16,1.06),dotstyle); label("$O$", (-0.08256113298199137,1.248492291060737), NE * labelscalefactor); dot((8.74,-5.92),dotstyle); label("$X_1$", (8.81174269729895,-5.726314345078365), NE * labelscalefactor); dot((13.905061984724782,0.4312806143468913),dotstyle); label("$Y$", (13.981081675581889,0.6213298414896461), NE * labelscalefactor); dot((11.130343268440559,8.375338308914044),dotstyle); label("$X_2$", (11.206362959297664,8.565387536056798), NE * labelscalefactor); dot((9.898792551478865,3.7911132384613015),dotstyle); label("$O_2$", (9.971042982869756,3.9852011619164065), NE * labelscalefactor); dot((8.170935298802075,6.457852721693972),linewidth(4pt) + dotstyle); label("$A_2$", (8.241595015870685,6.607880496486423), NE * labelscalefactor); dot((4.88344320105104,-0.8089030853842818),dotstyle); label("$O_1$", (4.95374338630102,-0.6139901349382603), NE * labelscalefactor); dot((4.956760536316561,0.8312767604991137),linewidth(4pt) + dotstyle); label("$B_1$", (5.029763077158122,0.9824233730608803), NE * labelscalefactor); dot((3.87024717678906,-2.100800594830072),linewidth(4pt) + dotstyle); label("$A_1$", (3.946482482444418,-1.944334724937544), NE * labelscalefactor); dot((9.756894083933526,0.6167069976080424),linewidth(4pt) + dotstyle); label("$B_2$", (9.838008523869828,0.77336922320385), NE * labelscalefactor); dot((6.523898647261402,-0.8757707718350884),linewidth(4pt) + dotstyle); label("$C_1$", (6.6071716624429895,-0.7280196712239132), NE * labelscalefactor); dot((6.7240376796165595,3.9249850452258106),linewidth(4pt) + dotstyle); label("$C_2$", (6.797220889585745,4.080225775487784), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Solution:(with amar_04) Let $O_1,O_2$ be the centres of $\omega_1,\omega_2$ . Notice that we just need to prove $B_1C_1\parallel B_2C_2$. Note that $\angle B_2OA_1=\angle B_1OA_2$ and $OB_2=OA_2$ and $OA_1=OB_1 \implies \Delta OA_1B_2 \equiv \Delta OB_1A_2 \implies B_1C_1\parallel B_2C_2$. Now note that one can easily prove $\Delta O_1B_1A_1 \sim \Delta O_2A_2B_2 \implies B_1C_1B_2C_2$ cyclic $\implies$ its an iscoseles trapezoid $\implies \angle C_1C_2B_2=\angle B_1B_2C_2$. Now since $90^\circ=\angle B_1B_2C_2+\angle O_2B_2C_2=\angle C_1C_2B_2+\angle BC_2B_2 \implies C_1C_2$ is tangent to $\omega _2$ . Similiarly for $\omega_1$. Hence done $\blacksquare$.