Dual concurrence of cevians in symmedian picture

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

w

18 users

TOPICS

No tags match your search

M

No tags match your search

M

Dual concurrence of cevians in symmedian picture

G H J

Reveal topic tags

geometry solved

L

G H BBookmark kLocked kLocked NReply

Source: USA December TST for IMO 2017, Problem 2, by Evan Chen

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

140 posts

#47 h Sep 30, 2023, 5:32 PM

Y by

YAYY THE POWER OF INVERSION AND I CAN FINALLY POST LATEX

Invert about (ABC), s.t. $A_1\mapsto(AT)\cap BC=T,A_2\mapsto(AT)\cap BC=D$ due to orthogonality. Part a) follows immediately since $OA_1O_A=OMT=90=OMO_A$ where $O_A$ is the antipode of O wrt (BOC) and M is the midpoint of BC, so $A_1\in AO_A$ implies $AA_1$ is a symmedian in ABC (since $AO_A$ is a symmedian since $O_A$ is the intersection of tangents at B and C), whence the lines concur at the symmedian point. Part b) follows since $AD$ etc. concur at the orthocenter, so their preimage of concurrence is a point on line OH.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

everythingpi3141592

85 posts

everythingpi3141592

#48 h Oct 3, 2023, 4:02 PM

Y by

My solution was kinda funny lol

Invert about the circumcircle of $(ABC)$ . Then, $A_1$ goes to $T_A$ (we use $T_A$ , $T_B$ , $T_C$ ). So, $AA_1$ goes to circumcircle of $AOT_A$ , i.e. circle with diameter $OT_A$ .

Claim: $T_A, T_B, T_C$ are collinear
Proof.
Apply Pascal's on $ABBCCAA$ .

Now, the circle with diameters $OT_A$ , passes through the feet from $O$ onto $T_A$ , $T_B$ , $T_C$ , hence done.

Now, $A_2$ goes to $D$ , the feet from $A$ to $BC$ . Thus, $AA_2$ goes to circumcircle of $AOD$ . Note that defining $BOE$ and $COF$ similarly, Now, $O$ has equal power with respect to all the circles, and similarly, $HA\cdot HD = HB \cdot HE = HC \cdot HF$ , since $ABDE$ , and analogous quadrilaterals for $BC$ and $CA$ are cyclic. So, $OH$ is the common radical axis of all the circles and done.

Edit: $A_1$ and $A_2$ , both lie on the circumcircle of $BOC$ , so after inversion go to $BC$ , and the circumcircle of $AOT$ is orthogonal to $ABC$ , so it inverts to itself, thus, $A_1$ goes to $T_A$ and $A_2$ to $D$ .

This post has been edited 1 time. Last edited by everythingpi3141592, Oct 4, 2023, 12:27 PM
Reason: Written

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

194 posts

#49 h Oct 11, 2023, 3:12 PM

Y by

Let $T_A$ be point on $BC$ such that $\angle{T_AAO} = 90^{\circ}$ .Define points $T_B, T_C$ analogously. Let $O_A$ be midpoint of $AT_A$ . Define points $O_B, O_C$ analogously.

Now consider inversion around $(ABC)$ . Points $A, B, C$ are fixed under inversion. Since $\angle{OAO_A} = 90^{\circ}$ , so circle with diameter $AT_A$ is fixed under inversion. Note that $(BOC)^* = BC$ , therefore $A_1^* = BC \cap (AT_AA_1)$ . So $A_1^*$ is either foot of altitude $A$ to $BC$ or $T_A$ . Let $D$ be foot of altitude $A$ to $BC$ , then since $OD < OT_A$ and $OA_1 < OA_2$ , so $A_1^* = T_A$ . Therefore $AA_1, BB_1, CC_1$ concurrent $\iff$ the circles $(OAT_A), (OBT_B), (OCT_C)$ are coaxial. By Pascal's theorem on cyclic hexagon $AABBCC$ yields $T_A, T_B, T_C$ are collinear. Let $P_{AB} = (OAT_A) \cap (OBT_B)$ . Then it's not hard to see that $T_A, T_B, P_{AB}$ are collinear. Thus $P_{AB}, P_{BC}, P_{CA}$ are collinear, thus $P_{AB} = P_{BC} = P_{CA}$ . Hence the circles $(OAT_A), (OBT_B), (OCT_C)$ are coaxial.

Now since $A_2^* = D$ , therefore line $AA_2$ becomes $OAD$ under the inversion. Let $E, F$ be feet of the perpendicular from $B, C$ to $AC, AB$ , respectively. Then lines $BB_2, CC_2$ become $(OBE), (OCF)$ , respectively. Let $H$ be orthocenter of $ABC$ . Then since $AH \cdot HD = BH \cdot HE = CH \cdot HF$ , the circles $(OAD), (OBE), (OCF)$ are coaxial and radical axis of these circles is $OH$ . Thus $AA_2, BB_2, CC_2, OH$ are concurrent. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1729 posts

OronSH

#50 h Oct 27, 2023, 6:06 PM

Y by

Invert about the circumcircle. The circle with diameter $AT$ is orthogonal to the circumcircle, so it is fixed, and circle $(BOC)$ is sent to $BC.$ Thus, $A_1$ is sent to $T$ and $A_2$ is sent to the foot $D$ from $A$ to $BC.$

a. We get that $AA_1$ inverts to the circle with diameter $OT.$ First, by Pascal's on $AABBCC$ we find that the three $T$ s are collinear. Then, all three of these circles pass through the foot from $O$ to the three $T$ line, so inverting back we get our desired result.

b. We get $AA_2$ inverts to $(AOD).$ First, let $E,F$ be the feet from $B,C$ to $AC,AB$ respectively. Letting $P$ be the point on the Euler line such that $H$ is between $P$ and $O$ and $OH \cdot PH=AH \cdot DH=BH \cdot EH=CH \cdot FH.$ Then $(AOD),(BOE),(COF)$ pass through $P$ and inverting back we get our desired result.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

5001 posts

#51 h Nov 13, 2023, 3:32 PM

Y by

It is morally correct to view the problem with $\triangle ABC$ as the intouch triangle most of the time:

Restated problem wrote:

Let $\triangle ABC$ have incenter $I$ and intouch triangle $DEF$ (with $D$ on $\overline{BC}$ , etc.). Let $T=\overline{BC} \cap \overline{EF}$ , and let the circle with diameter $\overline{DT}$ intersect $(AEFI)$ at points $A_1,A_2$ with $IA_1<IA_2$ , and define $B_1,B_2,C_1,C_2$ similarly. Then prove that $\overline{DA_1}$ , etc. concur, and $\overline{DA_2}$ , etc. concur on the Euler line of $\triangle DEF$ .

For the first part, I claim that $A_1$ is the foot of $I$ onto $\overline{AD}$ . This point obviously lies on $(AEFI)$ , which has diameter $\overline{AI}$ . Furthermore, since $T$ lies on the polar of $A$ wrt the incircle, $A$ lies on the polar of $T$ . Since $D$ does as well, we conclude $\overline{AD}$ is the polar of $T$ and thus $T,A_1,I$ are collinear (since $A_1,T$ are now inverses) and $\angle TA_1D=\angle AA_1I=90^\circ$ as desired. The conclusion then follows from the well-known fact (proof by Ceva) that $\overline{AD},\overline{BE},\overline{CF}$ concur.

For the second part, I claim that $A_2$ is the inverse of the foot of the altitude from $D$ to $\overline{EF}$ wrt the incircle. Note that since $\triangle DEF$ is always acute, $A_2$ will always lie outside the incircle, but since $T$ lies outside the incircle its inverse $A_1$ lies inside and we indeed have $IA_1<IA_2$ . To see this, let $H$ be the foot of the altitude. An inversion about the incircle fixes the circle with diameter $\overline{DT}$ , since it's clearly orthogonal, but swaps $(AEFI)$ with $\overline{EF}$ , so clearly $A_1$ gets sent to $H$ and the conclusion follows.

Now return to the original problem statement. We want to prove that if $D,E,F$ are the feet of the altitudes of a triangle $ABC$ and $O$ is its circumcenter, then $(AOD),(BOE),(COF)$ concur again at some point on the Euler line. This follows because $O$ lies on all three points, and the orthocenter $H$ of $ABC$ has equal power to all of these circles, since $HB\cdot HE=HC\cdot HF$ from $(BCEF)$ cyclic, and cyclic equalities hold as well. This establishes coaxiality, and since the common radical center is $\overline{OH}$ their intersection lies on the Euler line as well. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

819 posts

#52 h Nov 26, 2023, 6:57 PM • 3 Y

Y by mathmax12, OronSH, GeoKing

For part a, we make the following claim.

Claim: $A_1$ is the $A$ -dumpty point.
Proof. We prove that the Dumpty point is $(AT)\cap(BOC)$ . To do so, let $K$ be the intersection of the tangent from $T$ to the circumcircle. Then, as $D_A$ is the midpoint of $AK$ , and $TA=TK$ , $D_A$ lies on $AT$ . As it is well known that $D_A$ lies on $(BOC)$ we conclude. $\blacksquare$

Now, for part b, note that $(AT)$ and $(ABC)$ are orthogonal. Hence $A_2$ inverts to $(AT)\cap BC=D$ , the foot from $A$ to $BC$ . It thus suffices to prove that the euler line is the radical axis of coaxial circles $(AOD)$ , $(BOE)$ , $(COF)$ which is clear since $O$ lies on all circles and $AH\cdot HD=CH\cdot HF=BH\cdot HE$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

662 posts

#53 h Dec 12, 2023, 9:36 AM • 1 Y

Y by GeoKing

Amazing Problem.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = 2.134378529921642, xmax = 15.574498558294096, ymin = -8.566245459549759, ymax = 3.5151558254614903; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen ccqqqq = rgb(0.8,0,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); draw((6.3640414700338654,3.1219211393949418)--(8.146705533050916,-4.58053301967872)--(14.96,-5.17)--cycle, linewidth(0.4) + zzttqq); /* draw figures */ draw((6.3640414700338654,3.1219211393949418)--(8.146705533050916,-4.58053301967872), linewidth(0.4) + zzttqq); draw((8.146705533050916,-4.58053301967872)--(14.96,-5.17), linewidth(0.4) + zzttqq); draw((14.96,-5.17)--(6.3640414700338654,3.1219211393949418), linewidth(0.4) + zzttqq); draw(circle((12.007198316715282,0.3704613751601351), 6.27819639948054), linewidth(0.4)); draw((xmin, -0.08651717362000891*xmin-3.875703082644667)--(xmax, -0.08651717362000891*xmax-3.875703082644667), linewidth(0.4)); /* line */ draw((xmin, 2.050968333258838*xmin-9.930526387190534)--(xmax, 2.050968333258838*xmax-9.930526387190534), linewidth(0.4)); /* line */ draw(circle((11.684573567473066,-3.3585641551838576), 3.7429558018738223), linewidth(0.4) + ccqqqq); draw(circle((4.598363275132471,-0.4994289250734425), 4.028870285498829), linewidth(0.4) + ccqqqq); draw((xmin, -2.0427571208589823*xmin + 16.122112169748487)--(xmax, -2.0427571208589823*xmax + 16.122112169748487), linewidth(0.4)); /* line */ draw((xmin, -0.7797408915850768*xmin + 1.7717864161434749)--(xmax, -0.7797408915850768*xmax + 1.7717864161434749), linewidth(0.4)); /* line */ draw((xmin, -0.9646301934203966*xmin + 9.260867693569134)--(xmax, -0.9646301934203966*xmax + 9.260867693569134), linewidth(0.4)); /* line */ draw((11.361948818230825,-7.0875896855278375)--(14.96,-5.17), linewidth(0.4)); draw((2.832685080231075,-4.120778989541827)--(12.007198316715282,0.3704613751601351), linewidth(0.4)); draw((8.011104920945979,-2.6406879557153524)--(6.3640414700338654,3.1219211393949418), linewidth(0.4)); draw((8.011104920945979,-2.6406879557153524)--(2.832685080231075,-4.120778989541827), linewidth(0.4)); draw((8.011104920945979,-2.6406879557153524)--(11.361948818230825,-7.0875896855278375), linewidth(0.4)); draw((8.011104920945979,-2.6406879557153524)--(12.007198316715282,0.3704613751601351), linewidth(0.4)); draw((6.3640414700338654,3.1219211393949418)--(5.7158424544224875,-4.370221616658555), linewidth(0.4)); draw((12.007198316715282,0.3704613751601351)--(11.361948818230825,-7.0875896855278375), linewidth(0.4)); draw((5.7158424544224875,-4.370221616658555)--(8.011104920945979,-2.6406879557153524), linewidth(0.4)); draw((xmin, 0.5329523090844281*xmin-13.142966543903047)--(xmax, 0.5329523090844281*xmax-13.142966543903047), linewidth(0.4)); /* line */ draw((xmin, -4.320754716981127*xmin + 30.619383340107266)--(xmax, -4.320754716981127*xmax + 30.619383340107266), linewidth(0.4)); /* line */ draw(circle((11.988136734355425,-6.753861389708466), 3.367578977766797), linewidth(0.4) + fuqqzz); draw(circle((8.070627573114596,-0.5406226834024689), 4.040626607487973), linewidth(0.4) + fuqqzz); draw((xmin, -0.1899834899593175*xmin-2.327846990208609)--(xmax, -0.1899834899593175*xmax-2.327846990208609), linewidth(0.4)); /* line */ draw((xmin, -3.498717121018664*xmin + 25.387901989475214)--(xmax, -3.498717121018664*xmax + 25.387901989475214), linewidth(0.4)); /* line */ draw((xmin, 1.1815258220482348*xmin-13.816353486493071)--(xmax, 1.1815258220482348*xmax-13.816353486493071), linewidth(0.4)); /* line */ draw(circle((13.668914754254354,2.093107471947692), 7.376966264044081), linewidth(0.8) + linetype("4 4") + blue); draw((xmin, 0.23144104803493185*xmin-8.63235807860258)--(xmax, 0.23144104803493185*xmax-8.63235807860258), linewidth(0.4)); /* line */ draw((5.7158424544224875,-4.370221616658555)--(5.456350369654387,-7.369534630603981), linewidth(0.4)); draw((8.622595214459407,-6.636735605387107)--(12.007198316715282,0.3704613751601351), linewidth(0.4)); draw((9.016273468710851,-8.337722779416932)--(12.007198316715282,0.3704613751601351), linewidth(0.4)); /* dots and labels */ dot((6.3640414700338654,3.1219211393949418),dotstyle); label("$A$", (6.428118258274618,3.096147688409077), NE * labelscalefactor); dot((8.146705533050916,-4.58053301967872),dotstyle); label("$B$", (8.208902840747376,-4.411081433779994), NE * labelscalefactor); dot((14.96,-5.17),dotstyle); label("$C$", (15.03524374022628,-4.987217622227062), NE * labelscalefactor); dot((2.832685080231075,-4.120778989541827),linewidth(4pt) + dotstyle); label("$T$", (2.90146643808347,-3.974614624350397), NE * labelscalefactor); dot((12.007198316715282,0.3704613751601351),linewidth(4pt) + dotstyle); label("$O$", (12.084728108482201,0.5122641765858619), NE * labelscalefactor); dot((4.598363275132471,-0.4994289250734425),linewidth(4pt) + dotstyle); label("$D$", (4.664792348179044,-0.3606694422733324), NE * labelscalefactor); dot((8.541507737600647,-1.3261135841073304),linewidth(4pt) + dotstyle); label("$A_1$", (8.610452305422605,-1.1812270440009751), NE * labelscalefactor); dot((8.011104920945979,-2.6406879557153524),linewidth(4pt) + dotstyle); label("$A_2$", (8.086692134107087,-2.5080861446669505), NE * labelscalefactor); dot((11.361948818230825,-7.0875896855278375),linewidth(4pt) + dotstyle); label("$E$", (11.438757230526397,-6.9425889284716575), NE * labelscalefactor); dot((5.7158424544224875,-4.370221616658555),linewidth(4pt) + dotstyle); label("$F$", (5.782147380318814,-4.236494710008155), NE * labelscalefactor); dot((9.016273468710851,-8.337722779416932),linewidth(4pt) + dotstyle); label("$U$", (9.08183645960657,-8.199613339628897), NE * labelscalefactor); dot((11.988136734355425,-6.753861389708466),linewidth(4pt) + dotstyle); label("$K$", (12.049810763727834,-6.610874153305164), NE * labelscalefactor); dot((9.804181210614669,-4.190479552794751),linewidth(4pt) + dotstyle); label("$C_2$", (9.867476716579846,-4.044449313859133), NE * labelscalefactor); dot((8.376542831829639,-3.9192518311933076),linewidth(4pt) + dotstyle); label("$I$", (8.453324254027951,-3.7825692282013743), NE * labelscalefactor); dot((10.616632328666311,-3.678220995144719),linewidth(4pt) + dotstyle); label("$C_1$", (10.688034318307489,-3.5381478149207997), NE * labelscalefactor); dot((8.622595214459407,-6.636735605387107),linewidth(4pt) + dotstyle); label("$G$", (8.697745667308524,-6.4886634466648765), NE * labelscalefactor); dot((5.456350369654387,-7.369534630603981),linewidth(4pt) + dotstyle); label("$H$", (5.520267294661055,-7.2219276865066), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
For part (a), just notice $A_1,B_1,C_1$ are dumpty points, so $AA_1,BB_1,CC_1$ concurr at Lemoine Point.
For part(b),
Let $F$ be the feet of perpendicular from $A$ to $BC$
Let $D$ be the intersection of tangents at $B$ and $C$
We start off with a claim
Claim 1: $O-A_2-F$
$[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 = -3.196667360633523, xmax = 14.023736069784894, ymin = -6.000531614861945, ymax = 2.774519323171599; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); pen zzttqq = rgb(0.6,0.2,0); draw((4.909534718710192,2.300199406267056)--(5.756625777182814,-2.8681799510800343)--(11.676625777182814,-2.968179951080035)--cycle, linewidth(0.4)); /* draw figures */ draw((4.909534718710192,2.300199406267056)--(5.756625777182814,-2.8681799510800343), linewidth(0.4)); draw((5.756625777182814,-2.8681799510800343)--(11.676625777182814,-2.968179951080035), linewidth(0.4)); draw((11.676625777182814,-2.968179951080035)--(4.909534718710192,2.300199406267056), linewidth(0.4)); draw(circle((8.770639314469276,0.27942145627853776), 4.357943577256562), linewidth(0.4)); draw((xmin, 1.9107020619366035*xmin-7.080458703921846)--(xmax, 1.9107020619366035*xmax-7.080458703921846), linewidth(0.4)); /* line */ draw((xmin, -0.016891891891891983*xmin-2.770939650789783)--(xmax, -0.016891891891891983*xmax-2.770939650789783), linewidth(0.4)); /* line */ draw((xmin, 59.19999999999968*xmin-288.34425594137474)--(xmax, 59.19999999999968*xmax-288.34425594137474), linewidth(0.4)); /* line */ draw(circle((3.572616646175254,-0.2542527115657585), 2.8831537515316596), linewidth(0.8) + linetype("4 4") + ccqqqq); draw((4.822497553771272,-2.8524007581170014)--(8.770639314469276,0.27942145627853776), linewidth(0.8) + linetype("4 4") + zzttqq); draw(circle((8.720490199244079,-2.689406165051738), 2.9692511478370176), linewidth(0.8) + linetype("4 4") + blue); draw((5.756625777182814,-2.8681799510800343)--(8.670341084018874,-5.658233786382011), linewidth(0.4)); draw((8.670341084018874,-5.658233786382011)--(11.676625777182814,-2.968179951080035), linewidth(0.4)); draw((4.909534718710192,2.300199406267056)--(5.818149972595824,-2.062609876205987), linewidth(0.4)); draw((xmin, -0.5233678342223821*xmin + 4.869691959038001)--(xmax, -0.5233678342223821*xmax + 4.869691959038001), linewidth(0.4)); /* line */ draw((2.235698573640315,-2.808704829398572)--(6.394973887038401,-0.8432137967984308), linewidth(0.4)); draw((6.394973887038401,-0.8432137967984308)--(11.676625777182814,-2.968179951080035), linewidth(0.4)); draw((5.818149972595824,-2.062609876205987)--(6.394973887038401,-0.8432137967984308), linewidth(0.4)); draw((6.394973887038401,-0.8432137967984308)--(8.770639314469276,0.27942145627853776), linewidth(0.4)); draw((4.909534718710192,2.300199406267056)--(8.670341084018874,-5.658233786382011), linewidth(0.4)); draw((4.822497553771272,-2.8524007581170014)--(6.394973887038401,-0.8432137967984308), linewidth(0.4)); draw((8.770639314469276,0.27942145627853776)--(11.676625777182814,-2.968179951080035), linewidth(0.4)); draw((8.770639314469276,0.27942145627853776)--(8.670341084018874,-5.658233786382011), linewidth(0.4)); /* dots and labels */ dot((4.909534718710192,2.300199406267056),dotstyle); label("$A$", (4.957028814446773,2.4321401680749144), NE * labelscalefactor); dot((5.756625777182814,-2.8681799510800343),dotstyle); label("$B$", (5.8066363474644715,-2.74158928671943), NE * labelscalefactor); dot((11.676625777182814,-2.968179951080035),dotstyle); label("$C$", (11.728527659692308,-2.8430349623036326), NE * labelscalefactor); dot((2.235698573640315,-2.808704829398572),linewidth(4pt) + dotstyle); label("$T$", (2.281399120913426,-2.7035471583753536), NE * labelscalefactor); dot((4.822497553771272,-2.8524007581170014),linewidth(4pt) + dotstyle); label("$F$", (4.868263848310596,-2.754269996167455), NE * labelscalefactor); dot((8.770639314469276,0.27942145627853776),linewidth(4pt) + dotstyle); label("$O$", (8.824645196094503,0.3778652374948072), NE * labelscalefactor); dot((5.818149972595824,-2.062609876205987),linewidth(4pt) + dotstyle); label("$A_2$", (5.870039894704598,-1.9553853009418578), NE * labelscalefactor); dot((8.670341084018874,-5.658233786382011),linewidth(4pt) + dotstyle); label("$D$", (8.7231995205103,-5.556706784181058), NE * labelscalefactor); dot((6.394973887038401,-0.8432137967984308),linewidth(4pt) + dotstyle); label("$A_1$", (6.440671819865738,-0.7380371939314239), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
Proof
Claim 2: $AA_2C_2C$ is cyclic
Proof
So by Claim 2 and radax, we have $AA_2,BB_2,CC_2$ concurr.
Let $H$ be the orthocenter of $\triangle ABC$
Let $U = CC \cap AB$
Let $G$ be the foot of altitude from $C$ to $AB$
It suffices to prove $AA_2,CC_2,OH$ concurrent.
Claim 3: $OH$ is the radical axis of $(CU),(AT)$
Proof
By Claim 3, radax on $(CU),(AT),(AA_2C_2C)$ , we're done

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

794 posts

#54 h Dec 27, 2023, 6:28 AM

Y by

Clearly $T = AA \cap BC$ . We begin by assuming $A_1$ lies inside the circle and $A_2$ lies outside, which we will soon prove. Inverting about $(ABC)$ tells us
$A_1^*, A_2^* = (AT)^* \cap (BOC)^* = (AT) \cap BC,$
so our inverted points correspond to $T$ and the foot from $A$ to $BC$ , essentially proving our earlier statement. Solving (b) first, we need the radical axis of $(AOD)$ , $(BOE)$ , and $(COF)$ to be the Euler Line, where $\triangle DEF$ is the orthic triangle. This is well known, with a proof being to consider the power of $H$ .

We tackle (a) by simply noting that $A_1$ is the $A$ -Dumpty point as it lies on $(AT)$ , $(BOC)$ , and is inside the circle. Hence the concurrence is the symmedian point. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

278 posts

#55 h Apr 23, 2024, 2:54 PM

Y by

Let $A'$ be the foot of the altitude from $A$ to $BC$ , and define $B',C'$ analogously.
Inverting about $(ABC)$ , $(AT)$ is unchanged and $(BOC)$ goes to line $BC$ , so $\{ A_1,A_2 \}$ goes to $\{T,A'\}$ . Since $OA_1<OA_2$ and $OT>OA'$ , $T$ is the inverse of $A_1$ and $A'$ is the inverse of $A_2$ .

Note that $A_1$ is the $A$ -Dumpty point of $\Delta ABC$ . Thus, $AA_1, BB_1, CC_1$ all pass through the symmedian point of $\Delta ABC$ .

Since line $AA_2$ is inverted to $(AOA')$ , it suffices to show that the circles $(AOA'),(BOB'),(COC')$ have radical axis $OH$ , where $H$ is the orthocentre of $\Delta ABC$ .
Trivially $O$ has equal power to all three circles, and $HA \cdot HA' = HB \cdot HB' = HC \cdot HC'$ , so $OH$ is the radical axis of these three circles. $\square$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

297 posts

#56 h Jul 5, 2024, 8:07 PM

Y by

makes no sense how this isnt the canonical solution imo. maybe im just not inversionpilled enough
also i might be fakesolving bc phantom point stuff

Pass to the tangential triangle - the problem rephrases as the following;

Rephrased Problem wrote:

Let $\triangle ABC$ be a triangle with intouch triangle $\triangle DEF$ , circumcenter $O$ , and incenter $I$ . Let $T = \overline{EF} \cap \overline{BC}$ , and suppose that $(AI) \cap (DT)$ = $D_1$ and $X$ with $ID_1 < ID_2$ . Define $E_1$ , $Y$ , $F_1$ , and $Z$ similarly.

Prove that $\overline{DD_1}$ , $\overline{EE_1}$ , and $\overline{FF_1}$ concur.
Prove that $\overline{DX}$ , $\overline{EY}$ , and $\overline{FZ}$ concur on line $\overline{OI}$ ,

Now, we can begin solving the problem.

a) I claim that $\overline{AD_1D}$ collinear, which finishes (as then the requested concurrence point is the Gergonne point of $\triangle ABC$ ). Indeed, note that $\overline{IT} \perp \overline{AD}$ for pole-polar reasons, so if $\overline{AD} \cap (AI) = D'_1$ , then $I$ , $D'_1$ , and $T$ are collinear, so $D'_1$ lies on $(AT)$ and $D_1 = D'_1$ .

b) In fact, the following characterization of $X$ holds:

Claim: $X$ is the $A$ -Sharkydevil point in $\triangle ABC$ .

Proof. Let $X'$ be the $A$ -Sharkydevil point of $\triangle ABC$ - we wish to show that $X = X'$ . If we let $N$ be the midpoint of arc $\widehat{BAC}$ and $M$ be the midpoint of arc $\widehat{BC}$ . It is well-known that $\overline{X'DM}$ collinear. Note that
$(N, M; B, C) \overset{X'}{=} (\overline{NX'} \cap \overline{BC}, D; B, C)$ so $\overline{TNX'}$ collinear. Hence $\measuredangle NX'M = \measuredangle TX'M = \measuredangle TX'D = 90$ , so $X'$ lies on $(TD)$ . By the definition of the Sharykdevil Point, $X'$ lies on $(AI)$ , so $X = X'$ and we're done. $\square$

Now this reduces to Canada 2007/5, so the three lines mentioned concur at the exsimilicenter of the incircle and the circumcircle. Clearly this exsimilicenter lies on $\overline{IO}$ .

This post has been edited 1 time. Last edited by pikapika007, Jul 5, 2024, 8:07 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

750 posts

#57 h Sep 26, 2024, 5:02 AM

Y by

Subjective Rating (MOHs) $\text{[math]}$

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1892 posts

#58 h Nov 26, 2024, 9:03 PM

Y by

what ok

Note that $(AT)$ is fixed by power of a point (since it remains tangent to $AO$ ). Thus $A_1,A_2$ are the intersections of $(AT)$ with $BC$ . Let the foot from $A$ to $BC$ be $D$ ; then $A_1$ maps to $T$ and $A_2$ maps to $D$ .

(a) We claim that the concurrency point is the symmedian point $K$ of triangle $ABC$ . If we invert, we want to prove that $K'TOA$ is concyclic. But $(TOA)$ passes through $M$ , the midpoint of $BC$ , and the inverse of $(AOM)$ is the line passing through $A$ and the pole of $BC$ , which is the $A$ -symmedian.

(b) The inverse of $AA_2$ is $(AOD)$ . Similarly define $E$ and $F$ , then we want to prove that $(AOD),(BOE),(COF)$ are coaxial with their radical axis at the Euler line (which is fixed under our inversion). But this is just a simple power of a point calculation at $H$ . $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

204 posts

Eka01

#59 h Dec 22, 2024, 1:32 PM • 1 Y

Y by L13832

mfw when it is not orthocentric config

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

359 posts

#60 h Feb 21, 2025, 1:48 AM

Y by

Sharkydevil point config. cool!

Let $X, Y, Z$ be the poles of $AB, AC, BC$ with respect to $(ABC).$ We use reference triangle $XYZ.$ Let $T_A, T_B, T_C$ denote the $T$ given in the problem.

Claim: $A_2$ is the $X$ -sharkydevil point
Proof: Invert with respect to the incircle. We get that $(XBOC) \leftrightarrow BC$ , and so $A_2$ goes to the 2nd intersection between $BC$ and $(T_AA)$ . However, this is also the foot from $A$ to $BC.$ This finishes.

Claim: $A,A_1,X$ are collinear
Proof: This is ELMO problem. However, we will still prove this result. Note that $T_A$ lies on the polar of point $X$ , so by La Hire’s, $X$ lies on the polar of point $T_A$ . However, we also know that $AA_1$ is the polar of point $T_A$ . This finishes.

(I) they concur on the symmedian point
(ii) first, invert about incircle so that we can rephrase the problems in terms of reference triangle $ABC$ :

Quote:

Let $ABC$ be a triangle with $O$ as circumcenter, $D, E, F$ the feet of respective sides. Show that $(ADO), (BEO), (COF)$ are concurrent on the Euler line.

I'm 99% sure this was a past TST problem, but we will still prove this result. $HD \cdot HA = HB \cdot HE = HC \cdot HF$ by considering the proper circumcircles. This finishes.

remark: they concur at the $X_56$ point of $XYZ$ by sharkydevil point properties

This post has been edited 2 times. Last edited by Ilikeminecraft, Feb 21, 2025, 1:50 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1014 posts

bjump

#61 h Apr 8, 2025, 12:25 AM

Y by

Invert with respect to $(ABC)$ . Let $X'$ denote the image of $X$ inverted. We have $A_1' =T$ , $A_2'$ is the foot from $A$ . We can get $B_1'$ , $B_2'$ , $C_1'$ , and $C_2'$ by symmetry.

Part (a): By pascals theorem on $(AABBCC)$ we have $A_1'$ , $B_1'$ , and $C_1'$ collinear. Let $A \neq X = (A_1'OA) \cap (B_1' O B)$ . We have $\measuredangle A_1' XO = 90^\circ = \measuredangle B_1' XO$ which is true only if $A_1'$ , $B_1'$ , and $X$ are collinear. By the previous collinearity $\measuredangle C_1' XP = 90^\circ$ . Therefore the three circles intersect at $X$ . Inverting back gives $AA_1\cap BB_1 \cap CC_1 = X'$ therefore Part (a) is finished.

Part (b): Let $H$ be the orthocenter of $\triangle ABC$
$HA\cdot HA_2' = HB \cdot HB_2' = HC \cdot HC_2'$ Therefore $HO$ is the radical axis of all $3$ pairs of circle $(AOA_2')$ , $(BOB_2')$ , $(COC_2')$ . Which means the three previously mentioned circles meet at the Euler line at a point besides $O$ . Since $O$ lies on the euler line inverting back gives that $AA_2 \cap BB_2 \cap CC_2 \in OH$ .

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a