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#1 h Mar 31, 2016, 1:07 PM • 2 Y

Y by Adventure10, Mango247

Let $I$ be an incenter of a triangle $\triangle ABC$ . Points $A_1, B_1, C_1$ are the tangent points of the inscribed circle on sides $BC$ , $CA$ and $AB$ respectively. Circumcircle of $\triangle BC_1B_1$ intersects line $BC$ at points $B$ and $K$ and Circumcircle of $\triangle CB_1C_1$ intersects line $BC$ at points $C$ and $L$ . Prove that lines $LC_1$ , $KB_1$ and $IA_1$ are concurrent.

This post has been edited 1 time. Last edited by proximo, Mar 31, 2016, 1:07 PM

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#2 h Mar 31, 2016, 1:54 PM • 2 Y

Y by Adventure10, Mango247

Let $X=LC_1\cap KB_1$ . $\angle XLK=\angle C_1LC=\angle AB_1C_1=90-\tfrac{\angle A}{2}$ , $\angle XKL=\angle AC_1B_1=90-\tfrac{\angle A}{2}$ $\rightarrow \bigtriangleup XKL$ is $X$ -isosceles with $\angle KXL=\angle A\rightarrow AC_1IB_1X$ is cyclic $\rightarrow \angle LXI=\tfrac{\angle A}{2}$ .
From sine law in $\bigtriangleup A_1B_1K$ , $\bigtriangleup A_1B_1C_1$ and $\bigtriangleup A_1C_1L$ :
$A_1K=\frac{A_1B_1\cdot \sin(90-\tfrac{\angle B}{2})}{\sin(90-\tfrac{\angle A}{2})}=\frac{A_1B_1\cdot A_1C_1}{B_1C_1}=\frac{A_1C_1\cdot \sin(90-\tfrac{\angle C}{2})}{\sin(90-\tfrac{\angle A}{2})}=A_1L$
So $XA_1$ is angle bisector of $\angle LXK\rightarrow \angle LXA_1=\tfrac{\angle A}{2}\rightarrow A_1, I, X$ are collinear.

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#3 h Mar 29, 2019, 8:07 PM • 4 Y

Y by XbenX, Pluto1708, lilavati_2005, Adventure10

Nice Problem

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -9.458926695058976, xmax = 8.324797056975646, ymin = -5.364384529820596, ymax = 5.20381196216717; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-4.038972461057737,4.5456684151144175)--(-7.261258400278971,-3.978189777622795)--(4.578741599721029,-3.938189777622795)--cycle, linewidth(0.8) + rvwvcq); /* draw figures */ draw((-4.038972461057737,4.5456684151144175)--(-7.261258400278971,-3.978189777622795), linewidth(0.8) + rvwvcq); draw((-7.261258400278971,-3.978189777622795)--(4.578741599721029,-3.938189777622795), linewidth(0.8) + rvwvcq); draw((4.578741599721029,-3.938189777622795)--(-4.038972461057737,4.5456684151144175), linewidth(0.8) + rvwvcq); draw(circle((-2.8417637049549698,-0.9131111870423753), 3.050130459066838), linewidth(0.4)); draw(circle((-2.2754930710017005,-3.495072411571849), 5.009117517887086), linewidth(0.4) + linetype("4 4") + sexdts); draw(circle((-1.8945019851244536,-5.232235088998642), 6.6013207600174235), linewidth(0.4) + linetype("4 4") + dtsfsf); draw((xmin, 1.5466676069433636*xmin + 8.97345873793856)--(xmax, 1.5466676069433636*xmax + 8.97345873793856), linewidth(0.4)); /* line */ draw((xmin, -296.0000000000161*xmin-842.0751678537591)--(xmax, -296.0000000000161*xmax-842.0751678537591), linewidth(0.8)); /* line */ draw(circle((-3.440368083006353,1.8162786140360208), 2.7942612418406854), linewidth(1.2)); draw((-2.8417637049549698,-0.9131111870423753)--(-0.701942576874821,1.2604714217280129), linewidth(0.4)); draw((-2.8417637049549698,-0.9131111870423753)--(-5.6948363526110555,0.16543982451154654), linewidth(0.4)); draw((-5.6948363526110555,0.16543982451154654)--(-0.701942576874821,1.2604714217280129), linewidth(0.4)); draw((-8.376341257304185,-3.981956949437069)--(-7.261258400278971,-3.978189777622795), linewidth(0.8) + rvwvcq); draw((-2.860218981567893,4.549650690382964)--(2.7134227193737988,-3.9444915305969404), linewidth(0.4)); /* dots and labels */ dot((-4.038972461057737,4.5456684151144175),dotstyle); label("$A$", (-5.012995757050321,4.751930456992521), NE * labelscalefactor); dot((-7.261258400278971,-3.978189777622795),dotstyle); label("$B$", (-7.155788700943017,-4.970810315636224), NE * labelscalefactor); dot((4.578741599721029,-3.938189777622795),dotstyle); label("$C$", (5.074165584267678,-4.3294301147431735), NE * labelscalefactor); dot((-2.8417637049549698,-0.9131111870423753),dotstyle); label("$I$", (-3.4678525458079683,-1.5161033244622921), NE * labelscalefactor); dot((-2.831459268965191,-3.9632242400170044),dotstyle); label("$A_1$", (-3.6573512415263703,-4.577236101451852), NE * labelscalefactor); dot((-0.701942576874821,1.2604714217280129),dotstyle); label("$B_1$", (0.0014312681135396832,1.6324903890126838), NE * labelscalefactor); dot((-5.6948363526110555,0.16543982451154654),dotstyle); label("$C_1$", (-6.587292613787812,0.07277035502276516), NE * labelscalefactor); dot((-8.376341257304185,-3.981956949437069),dotstyle); label("$X_1$", (-9.123659771864881,-3.6880499138501226), NE * labelscalefactor); dot((-2.860218981567893,4.549650690382964),dotstyle); label("$T$", (-2.2579762577597116,4.649892697759536), NE * labelscalefactor); dot((2.7134227193737988,-3.9444915305969404),dotstyle); label("$X_2$", (2.1150705665110974,-4.766734797170253), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let $\odot (AC_1IB_1)$ be $\omega_1$ , define, $IA_1 \cap \omega_1=T$ , Let $TC_1 \cap BC=X_1$ , then, $\angle TC_1B_1=\angle TIB_1=\angle ACB$ $\implies$ $X_1$ $,$ $C_1$ $,$ $B_1$ $,$ $C$ concyclic $\implies$ $X_1 \equiv L$ and similarly, Let $TB_1 \cap BC=X_2$ and in the similar manner show that, $B$ $,$ $C_1$ $,$ $B_1$ $,$ $X_2$ concyclic $\implies$ $X_2 \equiv K$ $\implies$ $KB_1,LC_1,IA_1$ concurrent at $T \in \omega_1$

This post has been edited 3 times. Last edited by AlastorMoody, Mar 29, 2019, 8:16 PM

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