Let $ABC$ be an acute triangle with $AC> AB$. be $\Gamma$ the circumcircle circumscribed to the triangle $ABC$ and $D$ the midpoint of the smallest arc $BC$ of this circle. Let $E$ and $F$ points of the segments $AB$ and $AC$ respectively such that $AE = AF$. Let $P \neq A$ be the second intersection point of the circumcircle circumscribed to $AEF$ with $\Gamma$. Let $G$ and $H$ be the intersections of lines $PE$ and $PF$ with $\Gamma$ other than $P$, respectively. Let $J$ and $K$ be the intersection points of lines $DG$ and $DH$ with lines $AB$ and $AC$ respectively. Show that the $JK$ line passes through the midpoint of $BC$
Problem
Source: Rioplatense Olympiad 2018 level 3 p4
Tags: geometry, midpoint, intersections, circumcircle, arc midpoint
lxhoanghsgs
12.12.2018 05:14
My solution: Let $M$ be the midpoint of $BC$, so $DM\perp BC$. Because $\angle JGE+\angle JEG=\angle AEP+\angle PAD=\angle AEF+\angle BAD=90$, so $DJ\perp AB$. By the same proof like this, we have $DK\perp AC$. But by Simson line we have $J,M,K$ are collinear, so we complete the proof. Sincerely, XH
DNCT1
04.01.2021 22:15
[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 = 4.133073170731708, xmax = 21.41112195121952, ymin = -11.07141463414634, ymax = 4.725658536585367; /* image dimensions */ pen ffwwqq = rgb(1.,0.4,0.); pen xfqqff = rgb(0.4980392156862745,0.,1.); pen qqccqq = rgb(0.,0.8,0.); /* draw figures */ draw(circle((13.56,-3.16), 7.200277772419616), blue); draw((6.819066778880511,-5.690576872653271)--(20.15326240969683,-6.053594787954724), red); draw((20.15326240969683,-6.053594787954724)--(10.38,3.3), red); draw(circle((10.917470246108792,0.8400665846574656), 2.517964787952965), ffwwqq); draw((8.86566019507541,2.2995946549080473)--(8.77438163698692,-8.539763645699686), xfqqff); draw((8.86566019507541,2.2995946549080473)--(20.742089371683175,-2.648539095183681), xfqqff); draw((13.364048044551708,-10.357610911348011)--(20.742089371683175,-2.648539095183681)); draw((10.38,3.3)--(6.108748230209608,-7.483975714850472), red); draw((13.364048044551708,-10.357610911348011)--(6.108748230209608,-7.483975714850472)); draw((6.108748230209608,-7.483975714850472)--(18.75969991296587,-4.719872397243202)); draw((13.56,-3.16)--(13.364048044551708,-10.357610911348011)); draw((6.819066778880511,-5.690576872653271)--(8.86566019507541,2.2995946549080473), qqccqq); draw((8.86566019507541,2.2995946549080473)--(20.15326240969683,-6.053594787954724), qqccqq); draw((8.841371462102767,-0.5847002427285903)--(13.398610496411207,0.4110097915632345)); draw((10.38,3.3)--(8.86566019507541,2.2995946549080473)); draw((10.38,3.3)--(13.364048044551708,-10.357610911348011)); draw((8.841371462102767,-0.5847002427285903)--(11.454940492217577,-1.6198668306850688)); /* dots and labels */ dot((13.56,-3.16),linewidth(3.pt) + dotstyle); label("$O$", (13.598439024390249,-2.8294634146341444), NE * labelscalefactor); dot((10.38,3.3),linewidth(3.pt) + dotstyle); label("$A$", (10.507707317073173,3.8241951219512207), NE * labelscalefactor); dot((6.819066778880511,-5.690576872653271),linewidth(3.pt) + dotstyle); label("$B$", (6.172097560975611,-5.426536585365851), NE * labelscalefactor); dot((20.15326240969683,-6.053594787954724),linewidth(3.pt) + dotstyle); label("$C$", (20.23063414634147,-6.950439024390242), NE * labelscalefactor); dot((8.841371462102767,-0.5847002427285903),linewidth(3.pt) + dotstyle); label("$E$", (8.168195121951221,-1.0050731707317058), NE * labelscalefactor); dot((13.398610496411207,0.4110097915632345),linewidth(3.pt) + dotstyle); label("$F$", (13.727219512195127,0.5832195121951234), NE * labelscalefactor); dot((13.364048044551708,-10.357610911348011),linewidth(3.pt) + dotstyle); label("$D$", (13.684292682926834,-10.985560975609754), NE * labelscalefactor); dot((8.86566019507541,2.2995946549080473),linewidth(3.pt) + dotstyle); label("$P$", (8.254048780487807,2.643707317073172), NE * labelscalefactor); dot((8.77438163698692,-8.539763645699686),linewidth(3.pt) + dotstyle); label("$G$", (8.103804878048782,-9.118243902439023), NE * labelscalefactor); dot((20.742089371683175,-2.648539095183681),linewidth(3.pt) + dotstyle); label("$H$", (20.981853658536593,-2.486048780487803), NE * labelscalefactor); dot((6.108748230209608,-7.483975714850472),linewidth(3.pt) + dotstyle); label("$J$", (5.571121951219514,-7.0362926829268275), NE * labelscalefactor); dot((18.75969991296587,-4.719872397243202),linewidth(3.pt) + dotstyle); label("$K$", (18.663804878048786,-4.160195121951218), NE * labelscalefactor); dot((13.486164594288645,-5.8720858303039964),linewidth(3.pt) + dotstyle); label("$M$", (12.933073170731712,-5.598243902439022), NE * labelscalefactor); dot((11.119990979256984,-0.08684522558267845),linewidth(3.pt) + dotstyle); label("$N$", (10.507707317073173,0.06809756097561126), NE * labelscalefactor); dot((11.454940492217577,-1.6198668306850688),linewidth(3.pt) + dotstyle); label("$I$", (11.130146341463417,-2.2928780487804863), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] $\textit{Proof}$ Let $M,N$ be the midpoint of $BC,EF$ respectively. Let $AD\cap(AEF)=I\neq A$ it's easy to show that $I$ is the midpoint of arc $EF$ of $(AEF)$ So that $EI\parallel GD$ since $\angle PEI=\angle PGI=180^o-\angle PAI$ Since $AEF$ is isoceles so $AI\bot EF$ and $AI$ is also diameter of $(AEF)$ we have $\angle AEN=\angle AIE=\angle NDJ$ implies that $ENDJ$ cyclic since $EN\bot ND$ so $DG\bot AB$ at $J$ Similarly we have $BK\bot AC$ and since we also have $DM\bot BC$ By Simson line we have $JK$ passing through $M$ (Q,E,D).$\quad\blacksquare$