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Source: FKMO 2013 #4

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202 posts

#1 h Mar 24, 2013, 6:19 AM • 1 Y

Y by Adventure10

For a triangle $ABC$ , let $B_1 ,C_1$ be the excenters of $B, C$ . Line $B_1 C_1$ meets with the circumcircle of $\triangle ABC$ at point $D (\ne A)$ . $E$ is the point which satisfies $B_1 E \bot CA$ and $C_1 E \bot AB$ . Let $w$ be the circumcircle of $\triangle ADE$ . The tangent to the circle $w$ at $D$ meets $AE$ at $F$ . $G , H$ are the points on $AE, w$ such that $DGH \bot AE$ . The circumcircle of $\triangle HGF$ meets $w$ at point $I ( \ne H )$ , and $J$ be the foot of perpendicular from $D$ to $AH$ . Prove that $AI$ passes the midpoint of $DJ$ .

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#2 h Mar 24, 2013, 9:59 AM • 2 Y

Y by Adventure10, Mango247

first of all we angle chase that $EB_{1}=EC_{1}$ then by projecting $B_{1},C_{1}$ to $BC$ in $B',C'$ we have $B'C=C'B$ (well known for excircles) but $D$ is the midpoint of arc $BAC$ so $D$ project on to $BC$ in the midpoint of $BC$ therefore $D$ is the midpoint of $B_{1}C_{1}$ so $\angle EDA=90$ now $H$ is the symmetric point of $G$ wrt $AE$ and $G$ is the midpoint of $DH$ now let $M$ be the midpoint of $DJ$ we have $\angle DMG=\angle DJH=90$ let $K$ be the second intersection point of circle $DMG$ and line $AM$ we may assume that $M-I-A$ because it's done similarly in other cases. we are goint to prove that $K=I$ since $180-\angle AKD=\angle MKD=\angle MGD=\angle AHD=\angle AED$ so $K$ is on $w$ . since $\angle DKG=\angle DMG=90=\angle AKM$ we have $\angle GKE=\angle MKD=\angle AED=x$ and $\angle AKH=\angle ADH=\angle AED=x$ so $\angle GKH=\angle EKA-2x=90-2x$ but since $\angle FHA=\angle ADH=x$ and $\angle FGH=90$ and $\angle AHD=x$ we have $\angle GFH=90-2x=\angle GHK$ so $K$ is on the circle $HGF$ therefore $K=I$ and clearly $AI$ bisects $DJ$ .

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49 posts

#3 h Jun 6, 2013, 8:43 AM • 2 Y

Y by Adventure10, Mango247

Using the same assumption as leader (until $M$ ), we first note that $\angle FIH=\angle FGH=90$ . Now suppose that $N$ is the second intersection point of $FI$ and $w$ , hence $\angle HIN=90$ . It implies that $HN$ is the diameter of $w$ . Since $AE$ is a diameter too, then $AN||HE$ . We know that $HE$ is perpendicular to $HA$ , then $AN\bot HA\rightarrow AN||DJ$ . Note also that $HF$ is tangent to $w$ since $HF=HD$ . By the well known lemma, $(I,N;H,F)$ is harmonic. Using pencil $(AI,AN;AH,AF)$ , it intersects $DJ$ at $(M,\infty ;J,D)$ . Hence $M$ is the midpoint of $DJ$ and we are done.

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6872 posts

#4 h May 26, 2014, 12:19 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Well, first let's get rid of the concrete...

Let $A_1$ be the $A$ -excenter. Then $\triangle ABC$ is the orthic triangle of $A_1B_1C_1$ , and in particular its circumcircle is the nine-point circle of $A_1B_1C_1$ . So $D$ is just the midpoint of $B_1C_1$ and $E$ is the circumcenter.

In other words, $\overline{AE}$ is a diameter of $\omega$ . Now we can ignore $B_1$ , $C_1$ , $B$ , $C$ . Oops.

$[asy]size(8cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair O = (0.0, 0.0); pair A = (-1.0, 0.0); pair E = (1.0, 0.0); pair D = (-0.5999999999999999, -0.8000000000000002); pair H = (2)*(foot(D,A,E))-D; pair F = (2)/(H+D); pair G = (H+D)/(2); pair J = foot(D,A,H); pair M = (D+J)/(2); pair I = (2)*(foot(O,A,M))-A; pair D_1 = IntersectionPoint(Line(D,A,lisf),Line(H,E,lisf)); pair N = midpoint(D_1--H); /* Draw objects */ draw(CirclebyRadius(O,1)); draw(E--F); draw(F--D); draw(F--H); draw(H--J); draw(J--D); draw(A--M); draw(D--H); draw(circumcircle(F,G,H)); draw(D--D_1); draw(E--D_1); /* Place dots on each point */ dot(O); dot(A); dot(E); dot(D); dot(H); dot(F); dot(G); dot(J); dot(M); dot(I); dot(D_1); dot(N); /* Label points */ label("$O$", O, lsf * dir(45)); label("$A$", A, lsf * dir(45)); label("$E$", E, lsf * dir(45)); label("$D$", D, lsf * dir(225)); label("$H$", H, lsf * dir(90)); label("$F$", F, lsf * dir(225)); label("$G$", G, lsf * dir(-45)); label("$J$", J, lsf * dir(225)); label("$M$", M, lsf * dir(225)); label("$I$", I, lsf * dir(225)); label("$D_1$", D_1, lsf * dir(45)); label("$N$", N, lsf * dir(45)); [/asy]$

Now to do the actual problem... let $D_1$ denote the intersection of $\overline{AD}$ and $\overline{EH}$ , and $N$ the midpoint of $\overline{D_1H}$ . It suffices to prove that $A$ , $I$ , $N$ are collinear as homothety will then do the rest. (Now we can erase $J$ and $M$ too!)

By Brokard's Theorem on $AHED$ we find that $\overline{D_1F}$ is perpendicular to $\overline{AE}$ . Then angle chasing gives $FHD_1$ isosceles, whence $\overline{FN} \perp \overline{D_1H}$ , and hence $N$ lies on $(FGHI)$ , and moreover $\overline{FN}$ is parallel to $\overline{AH}$ .

Finally, spiral similarity $\triangle FIH \sim \triangle GID$ gives $\angle GID = 90^{\circ}$ . So $\angle AIG = \angle AID - 90^{\circ} = \angle EAD = \angle HAE = \angle NFE = \angle NIG$ implying $A$ , $I$ , $N$ are collinear.

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1721 posts

#5 h Mar 12, 2016, 9:01 AM • 2 Y

Y by Eray, Adventure10

Let $A_1$ be the $A$ -excenter.
$D$ is the midpoint of $B_1C_1$ and angle chasing gives us that $E$ is the circumcenter of $\triangle A_1B_1C_1$ .

Now delete $B_1, C_1, B, C$ . All we need is that $AE$ is the diameter of $\omega$ , true since $\angle ADE = 90$ .

Extend $FI$ to hit $\omega$ again at $N$ . Note that $HN$ is a diameter of $\omega$ as $\angle HIF=\angle HIN=90$ .
Since $DH$ is the polar of $F$ , it follows that $(I, N, D, H)$ form a harmonic bundle.
Note that $\angle HAN = 90$ , so $AN \parallel JD$ . Now taking perspectivity at $A$ , we know that $AI \cap DJ$ is the midpoint of $DJ$ .

This post has been edited 3 times. Last edited by rkm0959, Mar 12, 2016, 9:05 AM

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125 posts

#6 h Feb 20, 2019, 9:51 PM • 3 Y

Y by amar_04, Adventure10, Mango247

I wasted 15 minutes drawing the whole configuration xd. My solution: get that
$GIMD$ is cyclic through simple angle chasing and then $MGD=JHD=180-AID =MID$ thus $A, I, M are colinear.$

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308 posts

#7 h Oct 13, 2020, 3:42 AM

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$[asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -25.800306410292965, xmax = 18.69755750735967, ymin = -26.36714817636629, ymax = 11.432112570887025; /* image dimensions */ /* draw figures */draw(circle((1.51,-5.91), 10.121744656995443), linewidth(0.8)); draw(circle((-9.704152976988905,-9.594995367805822), 6.0734424300098295), linewidth(0.8)); draw((-14.298672709415895,-5.622956674768868)--(8.129633244561914,1.747034060842779), linewidth(0.8)); draw((-10.850919602737585,-0.7068894497390896)--(-5.1096332445619135,-13.56703406084278), linewidth(0.8)); draw((-10.850919602737585,-0.7068894497390896)--(-4.827298864876047,1.9822973840736995), linewidth(0.8)); draw((-14.298672709415895,-5.622956674768868)--(11.63007655692946,-6.093753593985436), linewidth(0.8)); draw((-4.827298864876047,1.9822973840736995)--(-5.1096332445619135,-13.56703406084278), linewidth(0.8)); draw((-8.61007655692946,-5.726246406014564)--(8.129633244561914,1.747034060842779), linewidth(0.8)); draw((-5.1096332445619135,-13.56703406084278)--(8.129633244561914,1.747034060842779), linewidth(0.8)); /* dots and labels */dot((1.51,-5.91),dotstyle); label("$O$", (0.9542336333852767,-5.394140630949368), NE * labelscalefactor); dot((-4.827298864876047,1.9822973840736995),dotstyle); label("$D$", (-6.661440209152047,2.261405849507), NE * labelscalefactor); dot((-8.61007655692946,-5.726246406014564),dotstyle); label("$A$", (-8.455708915509009,-5.31439535511128), NE * labelscalefactor); dot((11.63007655692946,-6.093753593985436),dotstyle); label("$E$", (11.799591147365131,-5.7131217343017155), NE * labelscalefactor); dot((-14.298672709415895,-5.622956674768868),linewidth(4pt) + dotstyle); label("$F$", (-15.39354791342259,-4.716305786325626), NE * labelscalefactor); dot((-5.1096332445619135,-13.56703406084278),dotstyle); label("$H$", (-6.58169493331396,-13.687649318110433), NE * labelscalefactor); dot((-4.96846605471898,-5.7923683383845415),linewidth(4pt) + dotstyle); label("$G$", (-4.308954571928476,-5.473885906787454), NE * labelscalefactor); dot((-8.361199418772483,-3.6718898739692025),linewidth(4pt) + dotstyle); label("$I$", (-7.777874070885267,-3.440381372916232), NE * labelscalefactor); dot((8.129633244561914,1.747034060842779),dotstyle); label("$H'$", (8.290799010489295,2.141787935749869), NE * labelscalefactor); dot((-10.850919602737585,-0.7068894497390896),linewidth(4pt) + dotstyle); label("$J$", (-10.688576638975448,-0.3701882531498763), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$
Basically the excenter set-up is useless...
Firstly, let $A_1$ be the A-excenter. Notice that $\triangle ABC$ is the orthic triangle of $\triangle A_1B_1C_1$ , hence $E$ is the circumcenter of $\triangle A_1B_1C_1$ , since $D$ is the midpoint of $B_1C_1$ , $ED\perp DA$ . Let $H'$ be the refoection of $H$ with resepct to $O$ , then $\angle FIH=\angle FGH=90^{\circ}=\angle H'IH$ , hence $F,I,H'$ are collinear. Moreover, $AH'\|DJ$ . Let $AI\cap JD=X$
Now since $FD,FH$ are tangent to $(DAE)$ , $(H',I;D,H)=-1$ . Projecting at $A$ we have $(J,D;X,\infty_{JD})=-1$ , this completes the proof.

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235 posts

DNCT1

#8 h Jan 3, 2021, 2:17 PM

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$[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 = -20.60931532349808, xmax = 57.94594262920248, ymin = -11.923382857502832, ymax = 31.52942715176661; /* image dimensions */ pen ffqqff = rgb(1.,0.,1.); pen qqccqq = rgb(0.,0.8,0.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); /* draw figures */ draw((5.946964785775411,18.127449325978752)--(-4.753145523657345,-6.638460566542636), ffqqff); draw((-4.753145523657345,-6.638460566542636)--(39.123037581004304,-8.57338376786798), ffqqff); draw((39.123037581004304,-8.57338376786798)--(5.946964785775411,18.127449325978752), ffqqff); draw((-16.744108013774635,12.510848548587553)--(53.65676215412123,29.93680216510684)); draw((-16.744108013774635,12.510848548587553)--(25.086495536042218,-5.562061925351868)); draw((25.086495536042218,-5.562061925351868)--(53.65676215412123,29.93680216510684)); draw(circle((15.516730160908809,6.282693700313439), 15.227562023082754), red); draw((18.45632707017328,21.22382535684719)--(25.086495536042218,-5.562061925351868)); draw((0.6082211705700606,24.735355684321082)--(25.086495536042218,-5.562061925351868)); draw((0.2914768664954937,6.547848914836248)--(9.577219998049351,25.560389745724507)); draw((9.577219998049351,25.560389745724507)--(18.45632707017328,21.22382535684719)); draw((0.6082211705700606,24.735355684321082)--(18.45632707017328,21.22382535684719)); draw((18.45632707017328,21.22382535684719)--(0.2914768664954937,6.547848914836248)); draw(circle((0.44984901853277715,15.641602299578667), 9.095132344339751), qqccqq); draw((5.946964785775411,18.127449325978752)--(14.016773534111325,23.39210755128585)); draw((0.6082211705700606,24.735355684321082)--(0.2914768664954937,6.547848914836248)); draw((9.37390196833439,13.88583713584172)--(14.016773534111325,23.39210755128585)); draw((5.946964785775411,18.127449325978752)--(30.741983455322124,6.017538485790626)); draw((0.6082211705700606,24.735355684321082)--(30.741983455322124,6.017538485790626), ffxfqq); draw((0.2914768664954937,6.547848914836248)--(8.531639366171959,19.813663796951545)); draw((-16.744108013774635,12.510848548587553)--(-4.753145523657345,-6.638460566542636)); draw((39.123037581004304,-8.57338376786798)--(53.65676215412123,29.93680216510684)); /* dots and labels */ dot((5.946964785775411,18.127449325978752),linewidth(3.pt) + dotstyle); label("$A$", (3.2819983648724627,18.539975049351547), NE * labelscalefactor); dot((-4.753145523657345,-6.638460566542636),linewidth(3.pt) + dotstyle); label("$B$", (-6.537408879215107,-9.526519671938146), NE * labelscalefactor); dot((39.123037581004304,-8.57338376786798),linewidth(3.pt) + dotstyle); label("$C$", (40.626673159315736,-10.91824668291119), NE * labelscalefactor); dot((-16.744108013774635,12.510848548587553),linewidth(3.pt) + dotstyle); label("$C_1$", (-19.14027014524876,13.823566845498457), NE * labelscalefactor); dot((53.65676215412123,29.93680216510684),linewidth(3.pt) + dotstyle); label("$B_1$", (55.626397610914076,27.6635187879526), NE * labelscalefactor); dot((18.45632707017328,21.22382535684719),linewidth(3.pt) + dotstyle); label("$D$", (19.286858991062438,22.947110584099512), NE * labelscalefactor); dot((25.086495536042218,-5.562061925351868),linewidth(3.pt) + dotstyle); label("$E$", (26.168175878651365,-6.743065649992062), NE * labelscalefactor); dot((0.6082211705700606,24.735355684321082),linewidth(3.pt) + dotstyle); label("$F$", (-1.7436825080857423,26.03983727515072), NE * labelscalefactor); dot((9.37390196833439,13.88583713584172),linewidth(3.pt) + dotstyle); label("$G$", (9.080860910593469,10.49888565261841), NE * labelscalefactor); dot((0.2914768664954937,6.547848914836248),linewidth(3.pt) + dotstyle); label("$H$", (-1.4344098389806221,7.483477128843485), NE * labelscalefactor); dot((9.577219998049351,25.560389745724507),linewidth(3.pt) + dotstyle); label("$J$", (7.998406568725547,26.96765528246608), NE * labelscalefactor); dot((8.531639366171959,19.813663796951545),linewidth(3.pt) + dotstyle); label("$I$", (8.616951906935787,21.09147456946879), NE * labelscalefactor); dot((14.016773534111325,23.39210755128585),linewidth(3.pt) + dotstyle); label("$P$", (14.338496285380513,23.874928591414875), NE * labelscalefactor); dot((30.741983455322124,6.017538485790626),linewidth(3.pt) + dotstyle); label("$K$", (31.03922041705701,6.478340954251843), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
$\textit{Proof}$
It's easy to see that $E$ lies on the perpendicular bisector of $\overline{B_1C_1}$
Since $90^o-\angle C_1AB=90^o-\angle B_1AC\implies \angle EC_1B_1=\angle EB_1C_1$
Well known $(ABC)$ passes the midpoint of $B_1C_1$ (the Euler circle of the triangle which has three vertices be the excenters of $\bigtriangleup ABC$ )
This implies that $G$ is the midpoint of $HD$ here $\implies F=\overline {HH}\cap\overline{DD} (\star)$ of the circle $\omega$
Let $\overline{AI}\cap\overline{JD}=P$ and $K\in\omega$ such that $\overline{AK}\parallel\overline{JD}$
Since $(FGH)$ is the circle with diameter $FH$ so $\angle FIH=90^o (1)$
We have $\angle HIK=\angle HAK=\angle HJD=90^o (2)$
So by $(1),(2)\implies F,I,K\quad\text{collinear}$ and by $(\star)$ we have $HIDK$ is harmonic and since $\overline{AK}\parallel\overline{JD}$ so
$-1=A(KI,DH)\overset{JD}{=}A(P\infty_{JD},JD)\implies P\quad\text{is the midpoint of}\quad\overline{JD}$ The end the proof. $\quad\blacksquare$

This post has been edited 1 time. Last edited by DNCT1, Jan 3, 2021, 2:19 PM

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