Inside an angle $\angle BOC$ there are three disjoint circles: $k_1,k_2$ and $k_3$, which are, each one, tangent to its sides $BO$ and $OC$. Let $r_1, r_2$ and $r_3$, respectively, be the radii of these circles, with $r_1<r_2<r_3$. The circles $k_1$ and $k_3$ are tangent to the side $OB$ at $A$ and $B$, respectively, and $k_2$ is tangent to the side $OC$ at $C$. Let $K=AC\cap k_1,L=AC\cap k_2,M=BC\cap k_2$ and $N=BC\cap k_3$. Besides that, let $P=AM\cap BK,Q=AM\cap BL,R=AN\cap BK$ and $S=AN\cap BL$. If the intersections of $CP,CQ,CR$ and $CS$ with $AB$ are $X,Y,Z$ and $T$, respectively, prove that $XZ = YT$.
Problem
Source: Brazil Cono Sur TST 2024 - T2/P2
Tags: geometry, power of a point, Isotomic conjugate
Thelink_20
05.11.2024 02:37
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(55cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(20); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -22.108669623383125, xmax = 34.466462623730884, ymin = -9.925205373501543, ymax = 15.669715857541119; /* image dimensions */ pen ffdxqq = rgb(1,0.8431372549019608,0); pen ffffww = rgb(1,1,0.4); pen qqzzqq = rgb(0,0.6,0); pen ffqqff = rgb(1,0,1); pen ffxfqq = rgb(1,0.4980392156862745,0); /* draw figures */ draw(circle((-6.731762098415997,0.2225214273728322), 2.130023381915785), linewidth(2) + linetype("2 2") + red); draw(circle((0.7744766577468636,2.309621806441035), 4.198860323075723), linewidth(2) + linetype("2 2") + red); draw(circle((13.149428773482526,5.750461898357554), 7.609590905166444), linewidth(2) + linetype("2 2") + red); draw((-6.7265812880175355,-1.90749565394637)--(4.968100728737103,2.519252824925454), linewidth(2) + ffdxqq); draw((-4.745507571800929,0.9918019763334265)--(13.167937418985776,-1.859106497683937), linewidth(2) + ffdxqq); draw((13.167937418985776,-1.859106497683937)--(-3.365254518533487,3.011799704720432), linewidth(2) + ffdxqq); draw((6.815655887931793,1.5327380123908365)--(-6.7265812880175355,-1.90749565394637), linewidth(2) + ffdxqq); draw((-1.3841808023168805,5.911097335000227)--(-3.365254518533487,3.011799704720432), linewidth(2) + qqzzqq); draw((-2.228132505138145,4.361287710120124)--(-2.5213028157122213,4.561609329600533), linewidth(2) + qqzzqq); draw((-3.365254518533487,3.011799704720432)--(-4.745507571800929,0.9918019763334265), linewidth(2) + qqzzqq); draw((-4.745507571800929,0.9918019763334265)--(-6.7265812880175355,-1.90749565394637), linewidth(2) + qqzzqq); draw((-5.589459274622195,-0.5580076485466771)--(-5.882629585196271,-0.3576860290662683), linewidth(2) + qqzzqq); draw((-1.3841808023168805,5.911097335000227)--(4.968100728737103,2.519252824925454), linewidth(2) + qqzzqq); draw((1.7450746938924153,4.44147054217338)--(1.5778289833717907,4.12825104318616), linewidth(2) + qqzzqq); draw((1.8755828184704237,4.371784829456451)--(1.708337107949799,4.058565330469231), linewidth(2) + qqzzqq); draw((2.006090943048432,4.302099116739523)--(1.8388452325278075,3.988879617752303), linewidth(2) + qqzzqq); draw((4.968100728737103,2.519252824925454)--(6.815655887931793,1.5327380123908365), linewidth(2) + qqzzqq); draw((6.815655887931793,1.5327380123908365)--(13.167937418985776,-1.859106497683937), linewidth(2) + qqzzqq); draw((9.944911384141086,0.06311121956398406)--(9.777665673620461,-0.25010827942323577), linewidth(2) + qqzzqq); draw((10.075419508719094,-0.00657449315294279)--(9.908173798198469,-0.31979399214016263), linewidth(2) + qqzzqq); draw((10.205927633297103,-0.07626020586986965)--(10.038681922776478,-0.38947970485708944), linewidth(2) + qqzzqq); draw((-6.7265812880175355,-1.90749565394637)--(-0.4904636629445183,-1.8923276333134136), linewidth(2) + ffqqff); draw((-3.682927836968292,-1.7225550677024042)--(-3.682064197932589,-2.0776280691555873), linewidth(2) + ffqqff); draw((-3.534980753029466,-1.7221952181041977)--(-3.534117113993763,-2.077268219557381), linewidth(2) + ffqqff); draw((6.93181979391276,-1.8742745183168936)--(13.167937418985776,-1.859106497683937), linewidth(2) + ffqqff); draw((9.975473244962002,-1.689333932072929)--(9.97633688399771,-2.0444069335261124), linewidth(2) + ffqqff); draw((10.123420328900828,-1.6889740824747224)--(10.124283967936536,-2.0440470839279055), linewidth(2) + ffqqff); draw((-14.46,-1.926305538820097)--(xmax, 0.5993814043566019*xmax + 6.740749568176367), linewidth(2) + blue); /* ray */ draw((-14.46,-1.9263055388200971)--(xmax, 0.0024322858459198972*xmax-1.8911346854880955), linewidth(2) + blue); /* ray */ draw((-1.3841808023168805,5.911097335000227)--(-0.4904636629445183,-1.8923276333134136), linewidth(2) + ffxfqq); draw((-1.3841808023168805,5.911097335000227)--(4.572044551745734,-1.880014166237969), linewidth(2) + ffxfqq); draw((-1.3841808023168805,5.911097335000227)--(1.869311579222509,-1.8865879853923389), linewidth(2) + ffxfqq); draw((-1.3841808023168805,5.911097335000227)--(6.93181979391276,-1.8742745183168936), linewidth(2) + ffxfqq); /* dots and labels */ dot((13.167937418985776,-1.859106497683937),dotstyle); label("$B$", (13.280377535376473,-1.551375652825157), NE * labelscalefactor); dot((-14.46,-1.9263055388200971),dotstyle); label("$O$", (-14.356219493286961,-1.6401441657651894), NE * labelscalefactor); dot((-1.3841808023168805,5.911097335000227),dotstyle); label("$C$", (-1.2776585867888475,6.201074477271002), NE * labelscalefactor); dot((-6.7265812880175355,-1.90749565394637),linewidth(4pt) + dotstyle); label("$A$", (-6.6037693631907946,-1.6697336700785335), NE * labelscalefactor); dot((-4.745507571800929,0.9918019763334265),linewidth(4pt) + dotstyle); label("$K$", (-4.621272574196737,1.23003775262919), NE * labelscalefactor); dot((-3.365254518533487,3.011799704720432),linewidth(4pt) + dotstyle); label("$L$", (-3.2601553757829054,3.24212404593659), NE * labelscalefactor); dot((4.968100728737103,2.519252824925454),linewidth(4pt) + dotstyle); label("$M$", (5.202442857833521,2.6207444553563635), NE * labelscalefactor); dot((6.815655887931793,1.5327380123908365),linewidth(4pt) + dotstyle); label("$N$", (7.1553501425142345,1.5555223000759753), NE * labelscalefactor); dot((2.0525508466716533,1.4156395768681307),linewidth(4pt) + dotstyle); label("$Q$", (2.184313417872418,1.6442908130160077), NE * labelscalefactor); dot((-0.7479148972241164,0.3555887388750598),linewidth(4pt) + dotstyle); label("$P$", (-0.6266894918952763,0.5790686577356193), NE * labelscalefactor); dot((1.0534027925696388,0.06891066724385501),linewidth(4pt) + dotstyle); label("$R$", (1.1782702712187167,0.31276311891552233), NE * labelscalefactor); dot((4.044541760374791,0.8287716004145997),linewidth(4pt) + dotstyle); label("$S$", (4.166810206866476,1.0525007267491253), NE * labelscalefactor); dot((-0.4904636629445183,-1.8923276333134136),linewidth(4pt) + dotstyle); label("$X$", (-0.360383953075179,-1.6697336700785335), NE * labelscalefactor); dot((1.869311579222509,-1.8865879853923389),linewidth(4pt) + dotstyle); label("$Z$", (1.9771868876790086,-1.6401441657651894), NE * labelscalefactor); dot((4.572044551745734,-1.880014166237969),linewidth(4pt) + dotstyle); label("$Y$", (4.699421284506671,-1.6401441657651894), NE * labelscalefactor); dot((6.93181979391276,-1.8742745183168936),linewidth(4pt) + dotstyle); label("$T$", (7.036992125260858,-1.6401441657651894), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Lemma: $AK=CL$ and $BN=CM$. Proof: Because of the tangency points: $Pow_{k_2}B=Pow_{k_3}C\implies BM\cdot BC= CN\cdot BC\implies BM=CN\implies \boxed{BN=CM}$. Analogously $\boxed{AK=CL}$.$\ _{\blacksquare}$ But, that means that $(P, S)$ and $(Q, R)$ are pairs of isotomic conjugates, and the result follows directly. $_{\blacksquare}$ Remark: You can find info. about isotomic conjugates in Chapter 4 of EGMO.