Perpendicularity with Incircle Chord

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Perpendicularity with Incircle Chord

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Source: 2019 ELMO Shortlist G3

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#1 h Jun 27, 2019, 9:40 PM • 8 Y

Y by amar_04, GeoMetrix, itslumi, tiendung2006, Adventure10, Mango247, cubres, Rounak_iitr

Let $\triangle ABC$ be an acute triangle with incenter $I$ and circumcenter $O$ . The incircle touches sides $BC,CA,$ and $AB$ at $D,E,$ and $F$ respectively, and $A'$ is the reflection of $A$ over $O$ . The circumcircles of $ABC$ and $A'EF$ meet at $G$ , and the circumcircles of $AMG$ and $A'EF$ meet at a point $H\neq G$ , where $M$ is the midpoint of $EF$ . Prove that if $GH$ and $EF$ meet at $T$ , then $DT\perp EF$ .

Proposed by Ankit Bisain

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#2 h Jun 28, 2019, 1:49 AM • 6 Y

Y by XianYing-Li, Muaaz.SY, Adventure10, Mango247, MS_asdfgzxcvb, cubres

Define $T$ instead as the foot to $EF$ from $D$ ; we wish to show $T \in GH$ . Let $(AI)$ meet $(ABC)$ a second time at a point $T'$ so that $I, T, T'$ are collinear, say by inversion about the incircle. By radical axis on $(AI), (ABC), (A'EFG)$ we get a point $X = AT' \cap EF \cap A'G$ . Since $\angle XGA = \angle XMA = 90^{\circ}$ , point $X$ lies on $(AMG)$ .

Now note that
$-1 = (A, I; E, F) \stackrel{T'}{=} (X, T; E, F),$ so by properties of harmonic divisions we have $TM \cdot TX = TE \cdot TF$ . This implies that $T$ lies on the radical axis of $(AMG)$ and $(A'EFG)$ , as desired.

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#3 h Jun 28, 2019, 10:40 AM • 4 Y

Y by Ramisoka, Adventure10, Mango247, cubres

This is a really rich configuration!

tastymath75025 wrote:

Let $\triangle ABC$ be an acute triangle with incenter $I$ and circumcenter $O$ . The incircle touches sides $BC,CA,$ and $AB$ at $D,E,$ and $F$ respectively, and $A'$ is the reflection of $A$ over $O$ . The circumcircles of $ABC$ and $A'EF$ meet at $G$ , and the circumcircles of $AMG$ and $A'EF$ meet at a point $H\neq G$ , where $M$ is the midpoint of $EF$ . Prove that if $GH$ and $EF$ meet at $T$ , then $DT\perp EF$ .

Proposed by Ankit Bisain

Let $X=(AMG) \cap AT.$ Since $T$ lies on the radical axis of $(AMG),(EFG),$ hence power of a point gives $X \in (AFE).$ Define $Y=(AEF) \cap (ABC).$ Clearly $\measuredangle IYA=\pi/2=\measuredangle A'YA,$ and so $Y \in A'I.$

Now define $P$ to be the radical center of $(AEF), (FEG), (ABC).$ Hence $P$ lies on $AY,EF$ and $GA'.$

Key Claim: $I,T$ and $Y$ are collinear.
Proof: We have $\measuredangle PMA=\pi/2=\measuredangle A'GA=\measuredangle PGA$ so $P \in (AMG).$
Further, we get $\measuredangle PXA=\pi/2=\measuredangle IXA$ and so $I,X,P$ are also collinear. $\square$

Since $AT \perp PI, PT \perp AI,$ hence $T$ is the orthocenter of $\triangle API.$ Hence $IT \perp AP$ which implies that $T, I, Y$ are collinear.

Notice that the power of $I$ with respect to $(AMP)$ is $\text{[math]}$ $IM \cdot IA=r^2,$ where $r$ is the inradius of $ABC.$

So inverting about the incircle of $\triangle ABC,$ we find that $T=AX \cap FE \mapsto (IMP) \cap (IEF)=Y.$ But $Y \in (ABC),$ which is the image of the nine-point circle of $DEF$ under this inversion. So $T$ must be the foot of the perpendicular from $D,$ and so we are done. $\blacksquare$
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This post has been edited 1 time. Last edited by Wizard_32, Jul 4, 2019, 3:26 AM
Reason: Undefined variable.

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#4 h Jun 28, 2019, 10:50 PM • 4 Y

Y by Adventure10, Mango247, Radin.AmirAslani, cubres

Let $N$ be the middle of the arc $BC$ .
Claim1: The intersection of $A'I$ and $ND$ lies on $\odot O$ .
Proof: Let $J = ND \cap \odot O$ , then $ND \cdot NJ = NC^2 = NI^2$ , thus $\triangle NID \sim \triangle NJI$ . Hence
$\angle IJN = \angle NID = 90^\circ - (\angle B + \frac{1}{2} \angle A) =90^\circ - \angle NBA = 90^\circ - \angle NA'A = \angle A'AN = \angle A'JN$ Therefore, $A',I,J$ are collinear.
Claim 2: Let $T = A'J \cap EF$ , then $DT \perp EF$ .
Proof: Since $90^\circ = \angle IJA = \angle IMT$ , we have $IT \cdot IJ = IM \cdot IA = r^2 = ID^2$ . Therefore,
$\angle TDI = \angle IJD =\angle NID$ implying that $DT \parallel NI$ , hence $DT \perp EF$ .
Claim 3: $AJ$ , $EF$ , $A'G$ are concurrent, denote the intersection by $L$ .
Proof: Application of radical axis theorem to $\odot O$ , $\odot (AEFIJ)$ and $\odot(A'EFG)$ .
Claim 4: $L,G,M,A$ are concyclic; $L,I,M,J$ are concyclic.
Proof: Since $\angle AML =\angle AGL=90^\circ$ and $\angle IML=\angle IJL =90^\circ$ as well.
Finally, $MT \cdot TL = IT \cdot TJ = FT \cdot TE$ , meaning that $T$ lies on the radical axis of $\odot(AMLG)$ and $\odot(A'EFG)$ , which is $HG$ .

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jbaca

#5 h Jun 29, 2019, 6:56 PM • 4 Y

Y by translate, Adventure10, Mango247, cubres

Solution. Redefine $T$ as the $D$ -foot of altitude in $\bigtriangleup DEF$ . It's not hard to show that $T,\ I$ and $A'$ are collinear. Redefine also $H$ as $\overline{GT}\cap (A'EF)$ , $G\neq H$ , so it suffices to show that $A,\ H,\ M$ and $G$ are concyclic.
Let $R=\overline{A'I}\cap (ABC),\ R\neq A'$ . Clearly, it lies on $(AEF)$ . By the radical axis theorem, $AR,\ EF$ and $GA'$ concur at a point, say $P$ . Moreover, being $\angle AMP=\angle AGP=90^\circ$ , we infer that $AMGP$ is cyclic. Because $\angle TMI=\angle ART=90^\circ$ , we get
$PT\cdot PM=PR\cdot PA=PF\cdot PE$ which gives us that $(P,T;F,E)=-1$ , implying the following equality
$PT\cdot TM=FT\cdot TE=GT\cdot TH$ thus $H$ lies on $(PGM)$ and then it lies on $(AMG)$ as well, as required. $\blacksquare$

This post has been edited 1 time. Last edited by jbaca, Jun 30, 2019, 4:32 AM
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#6 h Feb 26, 2020, 11:47 AM • 4 Y

Y by AlastorMoody, mueller.25, amar_04, cubres

Here i present a solution that I,mueller.25,amar_04 found.
$[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 = -24.620625249953825, xmax = 27.856252010786847, ymin = -16.24019088506529, ymax = 18.41499364361862; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen ffqqff = rgb(1,0,1); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen qqffff = rgb(0,1,1); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); /* draw figures */ draw((-3.527646662682087,13.712520226747058)--(-7.26,-4.11), linewidth(0.4) + rvwvcq); draw((-7.26,-4.11)--(13.395645203182287,-4.419894946997829), linewidth(0.4) + rvwvcq); draw((13.395645203182287,-4.419894946997829)--(-3.527646662682087,13.712520226747058), linewidth(0.4) + rvwvcq); draw(circle((-0.14166328775828607,1.5839579345872608), 5.800101026607598), linewidth(0.4) + wvvxds); draw((-5.818616292086269,2.7728130584379342)--(-0.14166328775828446,1.5839579345872612), linewidth(0.4) + ffqqff); draw((-0.14166328775828446,1.5839579345872612)--(4.098565608016035,5.541435771580792), linewidth(0.4) + green); draw((-0.14166328775828446,1.5839579345872612)--(-0.22867193493617494,-4.2154904369538455), linewidth(0.4) + wrwrwr); draw(circle((3.176913954878332,3.006394992923919), 12.632191044978963), linewidth(0.4)); draw((-3.527646662682087,13.712520226747058)--(9.88147457243875,-7.699730240899219), linewidth(0.4) + qqffff); draw((-5.818616292086269,2.7728130584379342)--(4.098565608016035,5.541435771580792), linewidth(0.4) + linetype("2 2") + wvvxds); draw(circle((1.2960492996280217,-3.5659157486317214), 9.528795798604248), linewidth(0.4) + dtsfsf); draw(circle((-9.961397440505575,6.766318481782098), 9.467991748670812), linewidth(0.4) + dtsfsf); draw((-0.5445625005599162,5.78342099642709)--(-8.17647769874491,-2.531903351900492), linewidth(0.4) + ffqqff); draw((-2.442737265572527,3.7152718581028)--(-0.22867193493617494,-4.2154904369538455), linewidth(0.4) + qqffff); draw((-0.14166328775828446,1.5839579345872612)--(-5.12964304175612,-1.3759795466042293), linewidth(0.4) + green); draw((-0.14166328775828446,1.5839579345872612)--(-0.5128097549578929,7.372172013104032), linewidth(0.4) + wrwrwr); draw((-5.12964304175612,-1.3759795466042293)--(-0.5128097549578929,7.372172013104032), linewidth(0.4) + ffqqff); draw((-3.527646662682087,13.712520226747058)--(-0.14166328775828446,1.5839579345872612), linewidth(0.4) + wrwrwr); draw((9.88147457243875,-7.699730240899219)--(-2.442737265572527,3.7152718581028), linewidth(0.4) + blue); draw((-16.395148218329062,-0.17988326318285885)--(-3.527646662682087,13.712520226747058), linewidth(0.4) + linetype("4 4") + wvvxds); 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label("$T$", (-2.3009144669764607,4.000890344077527), NE * labelscalefactor); dot((-5.12964304175612,-1.3759795466042293),linewidth(4pt) + dotstyle); label("$K$", (-4.992910118663807,-1.1104938046959105), NE * labelscalefactor); dot((-0.5128097549578929,7.372172013104032),linewidth(4pt) + dotstyle); label("$L$", (-0.3926643847677092,7.647011036869246), NE * labelscalefactor); dot((-16.395148218329062,-0.17988326318285885),linewidth(4pt) + dotstyle); label("$T'$", (-16.272031140290537,0.08216249668455824), NE * labelscalefactor); dot((-8.010663903216715,8.872428769327545),linewidth(4pt) + dotstyle); label("$J$", (-7.889361136302091,9.14635038717612), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Proof: Let $J=A'I \cap \odot(ABC)$ . Notice that it is sufficient to show that if $T$ is the foot of the altitude from $D$ onto $EF$ then $T \in$ radical axis of $\odot(AMG)$ and $\odot(A'EF)$ . Now we state a lemma.

Claim: If $AJ \cap EF=T'$ then $T'$ is the harmonic conjugate of $T$ w.r.t $EF$ .

Proof: Firstly it's a well known fact that $\overline{(I,T,J)}$ is a collinear triple (see here ) Notice that since $\overline{IE}=\overline{IF} \implies \angle FJI=\angle IJE$ . But also notice that $\angle TJT'=90^\circ$ $\implies$ $(T,T';F,E)=-1$ . Done $\square$ .

Now back to the main problem. Firstly notice that by radical axis theorem on $\odot(ABC),\odot(AEF),\odot(A'EF) \implies AJ,EF,A'G$ are concurrent. So we could define $T'=EF \cap A'G$ . But notice that $\angle AMT'=90^\circ$ and also $\angle AGT'=90^\circ$ $\implies$ $T' \in \odot(AMG)$ . But now finally notice that $\text{Pow}_{\odot(A'FE)}{T}=\overline{TF} \cdot \overline{TE}=\overline{TT'} \cdot \overline{TM}=\text{Pow}_{\odot(AMG)}{T}$ where the last part follows from the claim. This immediately implies $T \in$ radical axis of $\odot(A'FE)$ and $\odot(AMG)$ as desired $\blacksquare$ .

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#7 h May 13, 2020, 10:55 AM • 1 Y

Y by cubres

Comparing to imo problems ,what level is this.Can anyone give his opinion.

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#8 h May 13, 2020, 11:00 AM • 1 Y

Y by cubres

Maxito12345 wrote:

Comparing to imo problems ,what level is this.Can anyone give his opinion.

I would say $\leq$ G4. Actually this is quite a well known configuration now (just a mix of well known lemmas), so it is easy to most of the people.

@below Hmm maybe.

This post has been edited 2 times. Last edited by AmirKhusrau, May 13, 2020, 11:58 AM

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#9 h May 13, 2020, 11:04 AM • 2 Y

Y by Mango247, cubres

AmirKhusrau wrote:

Maxito12345 wrote:

Comparing to imo problems ,what level is this.Can anyone give his opinion.

I would say $\leq$ G4. Actually this is quite a well known configuration now (just a mix of well known lemmas), so it is easy to most of the people.

Could it be a p2 (like the one of imo 2019)

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Mathematicsislovely

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Mathematicsislovely

#10 h May 13, 2020, 12:58 PM • 1 Y

Y by cubres

Let circumcircle of $AEF$ cut $(ABC)$ at $R$ .

Claim: $AR,BG,EF$ concur at a point.
proof: Radical axis theorem on $(ARFE),(EFA'),(ABC)$ shows that these 3 lines are concurrent.Let this point of concurrency be $S$ . $\blacksquare$

Claim: $S$ lies on $AMG$
proof: $\angle AMS= \angle AGS=90^\circ$ $\blacksquare$

Now observe that, $ST.TM=GT.TH=FT.TE$ . As $M$ is the midpoint of $EF$ we have $(S,T;F,,E)=-1$ .[It can be seen considering a circle with diameter $EF$ and centre $M$ then under inversion in this circle $S,T$ swaps, as $ST.TM=FT.TE$ ].So we have $ST.SM=SF.SE$ .From this we get the ninepoint circle of $DEF$ cut $EF$ in $T$ except $M$ .So $DT\perp EF$ $\blacksquare$

This post has been edited 1 time. Last edited by Mathematicsislovely, May 13, 2020, 12:59 PM

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#11 h May 13, 2020, 1:06 PM • 1 Y

Y by cubres

Maxito12345 wrote:

AmirKhusrau wrote:

Maxito12345 wrote:

Comparing to imo problems ,what level is this.Can anyone give his opinion.

I would say $\leq$ G4. Actually this is quite a well known configuration now (just a mix of well known lemmas), so it is easy to most of the people.

Could it be a p2 (like the one of imo 2019)

?

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khina

#12 h May 13, 2020, 1:25 PM • 1 Y

Y by cubres

i think its a medium problem, so sure.

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#13 h May 18, 2020, 5:28 AM • 1 Y

Y by cubres

Mr Evan chen ,how many mohs?

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#14 h May 20, 2020, 9:17 AM • 1 Y

Y by cubres

bump????

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#15 h Jun 25, 2020, 11:55 PM • 1 Y

Y by cubres

My solution is the same as probably half of this thread, but whatever.

Let $T'$ be the foot of the perpendicular from $D$ to $EF$ . Let $R = EF \cap (AMG)$ , and let $Q = (AEIF) \cap (ABC)$ . It suffices to show that $G,T',H$ are collinear, or by radical axes and harmonic bundles it suffices to show that $(E,F;R,T') = -1$ .

Claim: $Q,T',I$ collinear.

Proof: We invert around the incircle. let $Q'$ be the intersection of $T'I$ with $(ABC)$ . Note that after the inversion, $Q'$ gets sent to the intersection of line $EF$ with the nine-point circle of $(DEF)$ , so $T'$ and $Q'$ are inverses. Now $\angle AQ'I = \angle AMT' = 90$ , so $Q=Q'$ .

Now, note that $\angle RGA + \angle AGA' = 90+90=180$ , so $R,G,A'$ are collinear. Moreover, by radical axes on $(AEIF), (A'GFE), (ABA'C)$ we find that $A,Q,R$ are collinear. Now, $(E,F;R,T') \stackrel{Q}{=} (E,F;A,I) = -1$ , which is what we wanted.

This post has been edited 2 times. Last edited by mathlogician, Feb 22, 2021, 5:09 AM

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#17 h Jun 26, 2020, 12:01 AM • 1 Y

Y by cubres

check this https://artofproblemsolving.com/community/c946900h1911664_properties_of_the_sharkydevil_point

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Dr_Vex

#18 h Jun 26, 2020, 8:57 AM • 1 Y

Y by cubres

LeTs SpAm
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This post has been edited 1 time. Last edited by Dr_Vex, Jun 26, 2020, 9:29 AM

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MP8148

#19 h Jun 26, 2020, 11:19 PM • 2 Y

Y by Mango247, cubres

$[asy] size(8cm); defaultpen(fontsize(8.5pt)); pair A = dir(125), B = dir(210), C = dir(330), O = origin, A1 = 2O-A, I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,C,A), F = foot(I,A,B), T = foot(D,E,F), M = (E+F)/2, G = intersectionpoints(unitcircle,circumcircle(A1,E,F))[0], H = T+dir(G--T)*abs(E-T)*abs(F-T)/abs(G-T), J = extension(A1,G,E,F), N = dir(270), K = intersectionpoints(unitcircle,circumcircle(A,E,F))[1], L = intersectionpoint(unitcircle,A+dir(A--T)*0.0069--A+dir(A--T)*69); draw(circumcircle(G,E,F)^^circumcircle(A,M,G)); draw(A--B--C--A--L^^unitcircle); draw(H--G, dashed); draw(A--J--E^^A1--J, blue); draw(circumcircle(A,E,F)); draw(E--D--F^^T--D, gray); dot("$A$", A, dir(90)); dot("$B$", B, dir(250)); dot("$C$", C, dir(330)); dot("$E$", E, dir(60)); dot("$F$", F, dir(135)); dot("$T$", T, dir(270)); dot("$M$", M, dir(315)); dot("$G$", G, dir(225)); dot("$H$", H, dir(45)); dot("$J$", J, dir(225)); dot("$A'$", A1, dir(315)); dot("$K$", K, dir(150)); dot("$L$", L, dir(240)); dot("$D$", D, dir(45)); [/asy]$
Redefine $T$ be the point on $\overline{EF}$ such that $\overline{DT} \perp \overline{EF}$ . Let $L = \overline{AT} \cap (ABC)$ and $K = (AEF) \cap (ABC)$ .

By radical axis $\overline{AK}$ , $\overline{EF}$ , and $\overline{A'G}$ concur at a point $J$ , which lies on $(AMG)$ from $\angle AGJ = \angle AMJ = 90^\circ$ . It is well-known that $KBLC$ is harmonic, so $-1 = (K,L;B,C) \overset{A}{=} (J,T;F,E).$ This implies $\text{Pow}(T,(A'EF)) = ET \cdot FE = MT \cdot JT = \text{Pow}(T,(AMG)),$ so $T$ lies on the radical axis $\overline{HG}$ of $(A'EF)$ and $(AMG)$ as desired. $\blacksquare$

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#20 h Dec 14, 2020, 8:05 PM • 2 Y

Y by Didier, cubres

Redefine $T$ to be the foot of the perpendicular from $D$ to $\overline{EF}$ . We will prove that it lies on the radical axis of $(A'EF)$ and $(AMG)$ . Let $R$ be the second intersection of $(ABC)$ and $(AEF)$ . Then it's well-known that $R,T,I,A'$ are collinear. Notice that by radical center $A'G$ , $AR$ , $EF$ are concurrent, say at $S$ . Then $\angle AGS=180^{\circ}-\angle AGA'=90^{\circ}=\angle SMA \implies S \in (AMG)$ . Also $\angle SRI-180^{\circ}-\angle ARI=90^{\circ}=\angle SMI \implies SRMI$ is cyclic. Then $\text{Pow}(T,(AMG))=ST\cdot TM=RT\cdot TI=FT \cdot TE=\text{Pow}(T,(A'EF))$ , as desired.

This post has been edited 1 time. Last edited by snakeaid, Dec 14, 2020, 8:55 PM

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#21 h Mar 8, 2021, 7:11 AM • 1 Y

Y by cubres

Funny problem.
Redefine $T$ to be the foot of perpendicular from $D$ to $EF$ . We will prove $G,T,H$ are collinear instead, i.e.
$\text{Pow}_T (AMG) = \text{Pow}_T (EFA')$ Apparently, $\text{Pow}_T (EFA') = TE \cdot TF$ , and by letting $EF \cap (AMG) = J$ , we have $\text{Pow}_T (AMG) = TM \cdot TJ$ .
Therefore, we need to prove
$TE \cdot TF = TM \cdot TJ$ which is equivalent to proving $(E,F;T,J) = -1$ . Let $AJ \cap (ABC) = K$ .

Claim 01. $J,G,A'$ collinear.
Proof. Let $A'G \cap (AMG) = J'$ . Since $A'$ is the antipode of $A$ , we have $\measuredangle AGA' = 90^{\circ}$ , and hence $\measuredangle AGJ' = 90^{\circ} = \measuredangle AMJ' = \measuredangle AMJ$ , proving $J' \equiv J$ .

Claim 02. $K,D,Y$ collinear.
Proof. By our previous claim, $J$ lies on the radical axis of $(ABC)$ and $(EFA')$ , and therefore,
$JK \cdot JA = JF \cdot JE$ which means $K = (AEF) \cap (ABC)$ . Therefore, we know that $K$ is the incenter Miquel Point. Therefore, if $X$ and $Y$ are the midpoint of arcs $BC$ containing $A$ and not containing $A$ respectively, we have $K,D,Y$ collinear. By letting $AT \cap (ABC) = L$ , we have $L,D,X$ by a well known lemma.
Thus,
$-1 = (X,Y;B,C) \overset{D}{=} (L,K;C,B) \overset{A}{=} (T,J;E,F)$ which is what we wanted.

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#22 h Apr 6, 2021, 3:31 PM • 1 Y

Y by cubres

Redefine $T$ to be the foot from $D$ to $EF$ and $H$ to be the second intersection of $GT$ with $(A'EF)$ , and we will show $AGMH$ cyclic. Add in the point $S=(AEF) \cap (ABC)$ and let $L= AS \cap EF$ . We will in fact show $G,M,H$ lie on the circle with diameter $AL$ .

First note $M$ lies on that circle since $AM \perp ML$ for obvious reasons. By radical axis on $(ABC),(A'EF),(AEF)$ we get $A'GL$ collinear hence $AG \perp GL$ . It remains to show $AH \perp HL$ . Indeed, letting $LH$ intersect $(A'EF)$ again at $K$ and noting $KH$ and $AS$ intersect on the radical axis of $(AEF)$ and $(A'EF)$ , we have $ASKH$ cyclic and thus $AH \perp HK$ as desired.

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GeronimoStilton

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GeronimoStilton

#23 h Apr 6, 2021, 9:25 PM • 4 Y

Y by Mango247, Mango247, Mango247, cubres

Solution with hint from @above.

It is well-known that $T,I,A'$ are collinear along with $(AEF)\cap (ABC)=K\ne A$ . Let line $EF$ intersect $(AGM)$ again at point $J$ . Observe that $AJ$ is the diameter of $(AGM)$ . Moreover, since $\angle A'GA=90^\circ=\angle AGJ$ , $A',G,J$ are collinear. So by radical axis theorem on $(AEF)$ , $(ABC)$ , $(A'EF)$ , $K$ lies on $AJ$ .

Now $JT\cdot JM = JK\cdot JA=JE\cdot JF$ , implying $(JT;EF)$ harmonic. It is well-known that $TF\cdot TE=TJ\cdot TM$ then. This implies the desired, since $T$ must lie on the radical axis of $(AHMGJ)$ and $(A'GFHE)$ .

Sketch for second well-known part

This post has been edited 1 time. Last edited by GeronimoStilton, Apr 6, 2021, 9:27 PM

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#25 h Aug 12, 2021, 9:10 AM • 1 Y

Y by cubres

deleted as required

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#27 h Oct 16, 2021, 3:26 AM • 1 Y

Y by cubres

$[asy] size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.664267992854413, xmax = 8.910818228494016, ymin = -5.8993890210618805, ymax = 5.012432929595713; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen zzttff = rgb(0.6,0.2,1); pen ffvvqq = rgb(1,0.3333333333333333,0); /* draw figures */ draw(circle((-1.7768158352132182,-0.7516822995618229), 1.7652367603555659), linewidth(0.8) + qqwuqq); draw(circle((-0.5651633926932255,-0.2764813199926565), 3.9584030633428173), linewidth(0.8) + fuqqzz); draw(circle((-2.3384079414166115,1.08415885457241), 1.919817270343408), linewidth(0.8) + qqwuqq); draw(circle((-1.2225909183064403,-2.5634401570264176), 3.1274391741754353), linewidth(0.8) + qqwuqq); draw(circle((-4.48752331183388,0.8510717143897553), 2.6078141261627823), linewidth(0.8) + qqwuqq); draw((-4.257045907673084,1.1514396538248008)--(-1.7768158352132182,-0.7516822995618229), linewidth(0.8) + linetype("4 4") + zzttff); draw((-1.8423521740436113,-2.5157020864132407)--(-2.5675883939229163,-0.1449093096982009), linewidth(0.8) + zzttff); draw((-6.075046564922025,-1.2178566162970728)--(-0.5151604411450491,0.4829376770825319), linewidth(0.8) + linetype("4 4") + ffvvqq); draw((-6.075046564922025,-1.2178566162970728)--(1.769673214613549,-3.472962639985313), linewidth(0.8) + linetype("4 4") + ffvvqq); draw((-6.075046564922025,-1.2178566162970728)--(-2.9,2.92), linewidth(0.8) + linetype("4 4") + ffvvqq); draw((-2.9,2.92)--(-3.88,-2.44), linewidth(1.2) + blue); draw((-3.88,-2.44)--(2.58,-2.68), linewidth(1.6) + blue); draw((2.58,-2.68)--(-2.9,2.92), linewidth(1.6) + blue); /* dots and labels */ dot((-2.9,2.92),dotstyle); label("$A$", (-2.8131431421552735,3.138265037307195), NE * labelscalefactor); dot((-3.88,-2.44),dotstyle); label("$B$", (-3.791875263683722,-2.234349587253223), NE * labelscalefactor); dot((2.58,-2.68),dotstyle); label("$C$", (2.6635919208656196,-2.463414551866264), NE * labelscalefactor); dot((-1.8423521740436113,-2.5157020864132407),linewidth(4pt) + dotstyle); label("$D$", (-1.7511146698584463,-2.3592941134057908), NE * labelscalefactor); dot((-0.5151604411450491,0.4829376770825319),linewidth(4pt) + dotstyle); label("$E$", (-0.4391971452564833,0.639374514255838), NE * labelscalefactor); dot((-3.513267319342443,-0.4341967670158078),linewidth(4pt) + dotstyle); label("$F$", (-3.437865772918113,-0.27688534419632627), NE * labelscalefactor); dot((-0.5651633926932255,-0.2764813199926565),linewidth(4pt) + dotstyle); label("$O$", (-0.4808453206406726,-0.11029264265956915), NE * labelscalefactor); dot((1.769673214613549,-3.472962639985313),dotstyle); label("$A'$", (1.8514525008739282,-3.2547298841658603), NE * labelscalefactor); dot((-4.241058303247535,-1.7450695599165347),linewidth(4pt) + dotstyle); label("$G$", (-4.166708842141426,-1.588802868798289), NE * labelscalefactor); dot((-2.014213880243746,0.02437045503336205),linewidth(4pt) + dotstyle); label("$M$", (-1.8135869329347303,0.2853650234902291), NE * labelscalefactor); dot((-1.7768158352132182,-0.7516822995618229),linewidth(4pt) + dotstyle); label("$I$", (-1.6886424067821624,-0.5892466595777459), NE * labelscalefactor); dot((-6.075046564922025,-1.2178566162970728),linewidth(4pt) + dotstyle); label("$K$", (-5.999228559045755,-1.047376588803828), NE * labelscalefactor); dot((-4.257045907673084,1.1514396538248008),linewidth(4pt) + dotstyle); label("$J$", (-4.166708842141426,1.3265694080949613), NE * labelscalefactor); dot((-2.5675883939229163,-0.1449093096982009),linewidth(4pt) + dotstyle); label("$T$", (-2.479957739081759,0.01465188349299872), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

Let $EF$ meet $(AMG)$ at $K$ . Notice that
$\angle AGK=\angle AMK=90^{\circ}=\angle AGA'$ hence $K,G,A'$ are collinear. Let $AK$ meet $(ABC)$ at $J$ , then $KJ\times KA=KG\times KA'=KF\times KE$ Hence $J$ lies on $(AEF)$ . Redefine $T$ as the projection of $D$ on $EF$ , then
$\frac{FT}{TE}=\frac{\tan\angle FDT}{\tan\angle TDE}=\frac{\tan\angle BID}{\tan\angle DIC}=\frac{BD}{DC}$ Therefore, $J$ is the center of spiral sim. sending $\overline{FTE}$ to $\overline{BDC}$ . So
$\frac{JF}{JE}=\frac{FB}{EC}=\frac{BD}{DC}=\frac{FT}{TE}$ whichh implies $JT$ is the internal angle bisector of $\angle FJE$ , meanwhile since $AF=AE$ , $JK$ is the external angle bisector of $\angle FJE$ , so $(T,H;F,E)=-1$ . Therefore,
$HF\times HE=HT\times HM\hspace{20pt}(1)$ $MT\times MH=ME^2\hspace{20pt}(2)$ We now show that $T$ lies on the radical axis of $\Omega_1=(HMG)$ and $\Omega_2=(EFA')$ . Indeed, for each point $X$ on the plane define
$f(X)=Pow(X,\Omega_1)-Pow(X,\Omega_2)$ Then by linearity of PoP,
$MHf(T)=MTf(H)+HTf(M)=MT\cdot HF\cdot HE-HT\cdot ME^2=MT\cdot HT\cdot HM-HT\cdot MT\cdot MH=0$ as desired.

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91 posts

#28 h Jul 7, 2022, 1:41 PM • 1 Y

Y by cubres

hint 1
hint 2
solution

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1032 posts

#29 h Jul 8, 2022, 12:10 AM • 2 Y

Y by channing421, cubres

Great problem. I believe this works.

$[asy] import olympiad; unitsize(45); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -3.4556855291888393, xmax = 7.161644883742675, ymin = -1.975717796255949, ymax = 4.150731345409496; /* image dimensions */ pen zzwwff = rgb(0.6,0.4,1); pen qqzzff = rgb(0,0.6,1); draw((0.7751464073673895,3.3086911070471117)--(0,0)--(4,0)--cycle, linewidth(0.65) + zzwwff); /* draw figures */ draw((0.7751464073673895,3.3086911070471117)--(0,0), linewidth(0.65) + zzwwff); draw((0,0)--(4,0), linewidth(0.65) + zzwwff); draw((4,0)--(0.7751464073673895,3.3086911070471117), linewidth(0.65) + zzwwff); draw(circle((2,1.2765929021365985), 2.3726966594542893), linewidth(0.65) + qqzzff); draw(circle((1.3889916156368791,1.1011928091575605), 1.1011928091575605), linewidth(0.65) + qqzzff); draw(circle((1.6561974031409439,0.14027251010636455), 1.8064053063692798), linewidth(0.65) + qqzzff); draw((0.31682872142499685,1.3523746779608636)--(2.177579263719154,1.8697987707740351), linewidth(0.65)); draw((2.177579263719154,1.8697987707740351)--(1.3889916156368791,0), linewidth(0.65)); draw((1.3889916156368791,0)--(0.31682872142499685,1.3523746779608636), linewidth(0.65)); draw(circle((-0.5115130232317311,2.0364707547989784), 1.8094300525369909), linewidth(0.65) + qqzzff); draw((-0.14594123637954498,0.26435496183235674)--(1.2934839092391737,1.909888019820456), linewidth(0.65)); draw((1.3889916156368791,0)--(0.9629704469345858,1.5320491096223667), linewidth(0.65)); draw((1.3889916156368791,1.1011928091575605)--(-0.0181461969054606,2.524300947194244), linewidth(0.65)); draw(circle((1.0820690115021343,2.2049419581023364), 1.1456280673609427), linewidth(0.65) + qqzzff); draw((1.3889916156368791,1.1011928091575605)--(3.2248535926326105,-0.7555053027739147), linewidth(0.65)); draw((2,-1.0961037573176908)--(-0.0181461969054606,2.524300947194244), linewidth(0.65)); draw((0.7751464073673895,3.3086911070471117)--(3.2248535926326105,-0.7555053027739147), linewidth(0.65)); draw((0.7751464073673895,3.3086911070471117)--(2,-1.0961037573176908), linewidth(0.65)); draw((3.2248535926326105,-0.7555053027739147)--(-1.7981724538308512,0.7642504025508446), linewidth(0.65)); draw((-1.7981724538308512,0.7642504025508446)--(0.7751464073673895,3.3086911070471117), linewidth(0.65)); draw((-1.7981724538308512,0.7642504025508446)--(0.31682872142499685,1.3523746779608636), linewidth(0.65)); draw((0.7751464073673895,3.3086911070471117)--(-0.14594123637954498,0.26435496183235674), linewidth(0.65)); draw((-1.7981724538308512,0.7642504025508446)--(1.3889916156368791,1.1011928091575605), linewidth(0.65)); draw((0.7751464073673895,3.3086911070471117)--(1.0127252647845169,1.0614144711059215), linewidth(0.65)); /* dots and labels */ dot((0.7751464073673895,3.3086911070471117),dotstyle); label("$A$", (0.7995630827824076,3.381460016535215), N * labelscalefactor); dot((0,0),dotstyle); label("$B$", (0.10138532040368696,-0.1253083835583541), W * labelscalefactor); dot((4,0),dotstyle); label("$C$", (4.042977334252348,-0.11144763889395265), NE * labelscalefactor); dot((1.3889916156368791,1.1011928091575605),linewidth(4pt) + dotstyle); label("$I$", (1.4163662203482723,1.1568104978987808), NE * labelscalefactor); dot((2,1.2765929021365985),linewidth(4pt) + dotstyle); label("$O$", (2.0262389855819363,1.330069806203799), NE * labelscalefactor); dot((1.3889916156368791,0),linewidth(4pt) + dotstyle); label("$D$", (1.4424106353652614,-0.11837801122615338), E * labelscalefactor); dot((2.177579263719154,1.8697987707740351),linewidth(4pt) + dotstyle); label("$E$", (2.234150155547958,1.891429965112058), NE * labelscalefactor); dot((0.31682872142499685,1.3523746779608636),linewidth(4pt) + dotstyle); label("$F$", (0.2659244132029516,1.4478861358512114), NE * labelscalefactor); dot((3.2248535926326105,-0.7555053027739147),dotstyle); label("$A'$", (3.259845260713666,-0.8737885954360328), NE * labelscalefactor); dot((-0.14594123637954498,0.26435496183235674),linewidth(4pt) + dotstyle); label("$G$", (-0.2885053733731066,0.1380457650652736), SW * labelscalefactor); dot((1.2472039925720753,1.6110867243674494),linewidth(4pt) + dotstyle); label("$M$", (1.29854989070086,1.6835187951460362), NE * labelscalefactor); dot((1.2934839092391737,1.909888019820456),linewidth(4pt) + dotstyle); label("$H$", (1.319341007697462,1.967664060766266), NE * labelscalefactor); dot((0.9629704469345858,1.5320491096223667),linewidth(4pt) + dotstyle); label("$T$", (0.91737941242982,1.6211454441562296), NE * labelscalefactor); dot((-0.0181461969054606,2.524300947194244),linewidth(4pt) + dotstyle); label("$R$", (-0.12910680973248986,2.5498153366711276), N * labelscalefactor); dot((2,-1.0961037573176908),linewidth(4pt) + dotstyle); label("$J$", (1.9915871239209328,-1.1341679567104709), S * labelscalefactor); dot((-1.7981724538308512,0.7642504025508446),linewidth(4pt) + dotstyle); label("$K$", (-1.8379339884368798,0.6647540623125291), W * labelscalefactor); dot((1.0127252647845169,1.0614144711059215),linewidth(4pt) + dotstyle); label("$L$", (1.019389313553884,1.1338614601131567), NW * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Define $R$ the $A$ -Sharky Devil point of $\triangle ABC.$ Let $J$ be the midpoint of $\widehat{BC}$ not containing $A,$ and $K$ is the concurrence point of radical axes on $(AFE), (GFE),$ and $(ABC).$

Note $AMGK$ is then a cyclic quadrilateral with diameter $AK$ since $\angle AMK = \angle AGK = 90^\circ.$ Extend $AT$ to meet $(AMG)$ at $L.$ By Thales Theorem $\angle ALK = 90^\circ.$ From the problem statement, $T$ lies on the radical axis of $(AMG)$ and $(A'EF).$ Therefore $TL \cdot TA = TF \cdot TE,$ and by the converse of Power of a Point, $AFLE$ is cyclic. Since $AI$ is a diameter of $(AFE)$ it follows $\angle ALI = 90^\circ,$ so $K,L,I$ are collinear. It follows $T$ is the orthocenter of $\triangle AIK.$

It follows since $IR \perp AK$ that $I,T,R$ are collinear. By the Sharky-Devil Lemma, $DT \perp EF,$ as required.

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1391 posts

#30 h Oct 3, 2022, 6:53 PM • 1 Y

Y by cubres

Quite nice config geo.
We begin by applying radical axis to $(A'EF),(AEF),(ABC)$ . Let $TI \cap (ABC)=R$ , so $AR,EF,GA'$ concur at $P$ . Since $\angle AGA'= \angle AMF =90$ , we have that $P \in (AMG)$ (and it has diameter $AP$ ). We want $T\in GH$ , which the radical axis of $(A'EF)$ and $(AMG)$ , so we want $TF \cdot TE=TM \cdot TP \iff (P,T,F,E)=-1$ . But note that $PT \cdot PM=PR \cdot PA= PF \cdot PE$ , which is sufficient.

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116 posts

#31 h Jan 16, 2024, 6:57 PM • 1 Y

Y by cubres

Solution in Spanish

This post has been edited 1 time. Last edited by UI_MathZ_25, Jan 16, 2024, 7:03 PM

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419 posts

#32 h Jul 15, 2024, 6:41 PM • 1 Y

Y by cubres

Define $T$ to be the foot of $D$ onto $EF$ . We need to show that $T$ has same power w.r.t. circles $(DEF),(AMG),(A'EF)$ .

Define $K=MF\cap (AMG)$ . Since $AM\perp EF$ , we have $\angle AMK=90^\circ$ , which means that $K$ is the antipode of $A$ in $(AMG)$ . Since $A'$ is also the antipode of $A$ in $(ABC)$ , we have $\angle AGK=\angle AGA'=90^\circ$ . Hence, $A',K,G$ are collinear.

Claim 1: $(K,T;E,F)=-1\Leftrightarrow T\in HG$ .
Proof. $(K,T;E,F)=-1\Leftrightarrow MT\cdot MK=ME^2\Leftrightarrow TM\cdot TK=TE\cdot TF$ So, $T$ has equal power from circles $(AMG),(DEF)$ . However, $T$ also has equal power from circles $(A'EF),(DEF)$ as $T\in EF$ by definition. Hence, $T$ has equal power from all three circles (as required), which means $T\in HG$ . $\blacksquare$

Define $S=(AEF)\cap (ABC)$ to be the Sharkydevil Point in $ABC$ .

Claim 2: $K,S,A$ are collinear and $A',I,S,T$ collinear.
Proof. Simply note that $K$ lies on the radical axes of circles $(AEF),(A'EF)$ and $(A'EF),(ABC)$ as established. Hence, $K$ lies on the radical axis of $(ABC),(AEF)$ , which is line $AS$ . So, $K\in AS$ .
Note that $\angle ASI=90^\circ=\angle ASA'$ , which means $S,I,A'$ are collinear. Invert about the incircle to see that $(ABC)$ goes to ninepoint circle of the intouch triangle and $(AEF)$ goes to line $EF$ , which means $S$ goes to $T$ . So, $I,S,T$ are collinear as well. $\blacksquare$

Finally, note that taking perspective at $S$ gives $(K,T;E,F)=(A,I;E,F)=-1$ , which finishes the problem by Claim 1.

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1541 posts

#33 h Mar 19, 2025, 4:27 AM • 2 Y

Y by MS_asdfgzxcvb, cubres

the fish are dying

Let $S$ be the Sharkey-Devil point, so $(ASEFI), (DEF), (GFEA'), (ABC)$ share a radical center $T'$ . Since $\measuredangle AGT' = \measuredangle AMT' = 90^\circ$ , $T'$ lies on $(AMM')$ . We want to show that $T$ lies on the radical axis of $(AMG)$ and $(AEIF)$ , or that $TF \cdot TE = TM \cdot TM'$ , or that $M'$ is harmonic conjugate of $M$ in $EF$ . Then, since $S$ lies on $TI$ , and $AS \perp TI, AM \perp MM'$ , $T$ is the orthocenter of $\triangle AM'I$ . As such, $AT \perp MI$ , so the polar of $M'$ wrt the incircle is $AT$ and we are done.

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678 posts

#34 h Apr 24, 2025, 6:42 PM • 1 Y

Y by cubres

Define $T$ to be the foot from $D$ to $EF.$
Draw in $K,$ the $A$ -sharkydevil point.
By Radax on $(AEFI), (EFGA’), (ABC),$ we have that $AK, EF, A’G$ concur at a point $X.$
Since $\angle AGX’ = 180-\angle AGA’ = 90 = \angle AMF = \angle AMX’,$ we have $AMGX’$ is cyclic.
Observe that $-1= (AI;EF) \stackrel K= (X’T;EF).$
It is well known that this implies $TM\cdot TX’ = TE\cdot TF.$
Thus, $T$ is the radical center of $(AEFI), (AMGX’), (DEF),$ which implies $T,G,H$ are collinear.

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1930 posts

#35 h Apr 24, 2025, 7:56 PM • 1 Y

Y by cubres

Full diagram https://www.geogebra.org/calculator/dyrupagm
Diagram without the fluff https://www.geogebra.org/calculator/dszqgp3x

Let $K$ be the $A$ -Sharkydevil point. Then radax on $(AGM),(AEF),(A'EF)$ gives that $T$ lies on the line through $A$ and $(AMG)\cap(AEF)$ . The inverse of the latter point around the incircle is $(AMG)\cap EF$ , let's call it $X$ . Then it suffices to show that $XIMK$ are concyclic (since $K$ and $T$ are well known to be inverses). This is equivalent to $\angle XKI=90^\circ$ , which is equivalent to $AKX$ collinear. Now redefine $X=AK\cap EF$ , we will show that it lies on $(AMG)$ . But by radax on $(ABC),(AEF),(A'EF)$ we get that $X$ lies on $GA'$ too. So then $\angle AGX=90^\circ=\angle AMX$ , done. $\blacksquare$

This post has been edited 1 time. Last edited by cj13609517288, Apr 24, 2025, 7:57 PM

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