Let $ABC$ be an acute-angled triangle with incircle $\omega$ and incenter $I$. Let $\omega$ touch $AB, BC$ and $CA $ at points $D, E, F$ respectively. The circles $\omega_1$ and $\omega_2$ centered at $J_1$ and $J_2$ respectively are inscribed into A$DIF$ and $BDIE$. Let $J_1J_2$ intersect $AB$ at point $M$. Prove that $CD$ is perpendicular to $IM$.
Problem
Source: Sharygin Geometry Olympiad 2017 First Round P15 grades 9-11
Tags: geometry, circumscribed quadrilateral, perpendicular, incircle
GeoMetrix
09.01.2020 15:39
Easy problem. Heres my solution. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); 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 = -13.733297405303937, xmax = 24.78110432088361, ymin = -11.88603698325395, ymax = 13.548473767091943; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw((-1.675320020216008,-5.563177958383545)--(21.22192784198372,-5.5104193227563565), linewidth(0.4) + rvwvcq); draw((21.22192784198372,-5.5104193227563565)--(-3.2970048276789528,12.200355789078085), linewidth(0.4) + rvwvcq); draw((-3.2970048276789528,12.200355789078085)--(-1.675320020216008,-5.563177958383545), linewidth(0.4) + rvwvcq); draw(circle((3.5555760732831905,0.18028969895944455), 5.731399687113113), linewidth(0.4) + sexdts); draw((3.568782100429702,-5.5510947737737615)--(3.5555760732831905,0.18028969895944455), linewidth(0.4)); draw((3.5555760732831905,0.18028969895944455)--(6.91158733471615,4.826377385021283), linewidth(0.4) + wvvxds); draw((3.5555760732831905,0.18028969895944455)--(-2.1520879952393717,-0.34077963840052844), linewidth(0.4) + wvvxds); draw((3.568782100429702,-5.5510947737737615)--(-3.2970048276789528,12.200355789078085), linewidth(0.4) + wrwrwr); draw((-1.675320020216008,-5.563177958383545)--(3.5555760732831905,0.18028969895944455), linewidth(0.4) + dtsfsf); draw((3.5555760732831905,0.18028969895944455)--(21.22192784198372,-5.5104193227563565), linewidth(0.4) + dtsfsf); draw((3.568782100429702,-5.5510947737737615)--(7.885475105740837,-1.214463178056403), linewidth(0.4) + dtsfsf); draw((0.824009077036602,-2.818941487510733)--(3.568782100429702,-5.5510947737737615), linewidth(0.4) + dtsfsf); draw((-11.351770883035856,-5.585473928113544)--(3.5555760732831905,0.18028969895944455), linewidth(0.4)); draw((-11.351770883035856,-5.585473928113544)--(6.91158733471615,4.826377385021283), linewidth(0.4)); draw((-11.351770883035856,-5.585473928113544)--(-0.2912289614950867,4.428940472775149), linewidth(0.4)); draw((-11.351770883035856,-5.585473928113544)--(7.885475105740837,-1.214463178056403), linewidth(0.4)); draw((-11.351770883035856,-5.585473928113544)--(-1.675320020216008,-5.563177958383545), linewidth(0.4) + rvwvcq); /* dots and labels */ dot((-1.675320020216008,-5.563177958383545),dotstyle); label("$A$", (-1.578752444961634,-5.308577467677774), NE * labelscalefactor); dot((21.22192784198372,-5.5104193227563565),dotstyle); label("$B$", (21.329813776588882,-5.2585587641372715), NE * labelscalefactor); dot((-3.2970048276789528,12.200355789078085),dotstyle); label("$C$", (-3.204360310027991,12.448062289200871), NE * labelscalefactor); dot((3.5555760732831905,0.18028969895944455),linewidth(4pt) + dotstyle); label("$I$", (3.648202075020962,0.368545384169342), NE * labelscalefactor); dot((6.91158733471615,4.826377385021283),linewidth(4pt) + dotstyle); label("$E$", (6.999455212234683,5.020284813436143), NE * labelscalefactor); dot((3.568782100429702,-5.5510947737737615),linewidth(4pt) + dotstyle); label("$D$", (3.6732114267912133,-5.358596171218278), NE * labelscalefactor); dot((-2.1520879952393717,-0.34077963840052844),linewidth(4pt) + dotstyle); label("$F$", (-2.053930128596415,-0.13164165123569027), NE * labelscalefactor); dot((7.885475105740837,-1.214463178056403),linewidth(4pt) + dotstyle); label("$J_1$", (7.974819931274498,-1.006968963194497), NE * labelscalefactor); dot((0.824009077036602,-2.818941487510733),linewidth(4pt) + dotstyle); label("$J_2$", (0.9221827320635315,-2.6075674764906003), NE * labelscalefactor); dot((-11.351770883035856,-5.585473928113544),linewidth(4pt) + dotstyle); label("M", (-11.257371580049023,-5.383605522988529), NE * labelscalefactor); dot((-0.2912289614950867,4.428940472775149),linewidth(4pt) + dotstyle); label("$H$", (-0.2032380975977929,4.620135185112116), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof: Let $EF \cap AB=M'$ and let $CD \cap \omega=H$. Notice that $(H,D;F,E)=-1 \implies HD$ is the polar of $M'$ w.r.t $\omega \implies IM' \perp CD$ . So following a phantom approach we just need to show $M=M' \iff J_1J_2,AB,EF$ concurrent. Notice that $(M',D;A,B)=-1$ . So we just need to show $BJ_2,AJ_1,ID$ concurrent which is true by cevas $\blacksquare$.
Pluto1708
09.01.2020 16:24
An alternate finish though pretty much same- It suffices to show $-1=(M,ID\cap J_1J_2,J_1,J_2)$.But since $\angle{J_2DJ_1}=90^{\circ}$ and $\angle{J_2DA}=\angle{J_2DI}$ so this is clear.$\square$