GOTEEM #1: Incircle Concurrency

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GOTEEM #1: Incircle Concurrency

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Source: GOTEEM: Mock Geometry Olympiad

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#1 h Jan 2, 2020, 8:49 PM • 4 Y

Y by jhu08, Adventure10, Mango247, ismayilzadei1387

Let $ABC$ be a scalene triangle. The incircle of $\triangle ABC$ is tangent to sides $\overline{BC}$ , $\overline{CA}$ , $\overline{AB}$ at $D, E, F$ , respectively. Let $G$ be a point on the incircle of $\triangle ABC$ such that $\angle AGD = 90^\circ$ . If lines $DG$ and $EF$ intersect at $P$ , prove that $AP$ is parallel to $BC$ .

Proposed by bluek

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#2 h Jan 2, 2020, 10:40 PM • 3 Y

Y by jhu08, Adventure10, Mango247

A bunch of point definitions:

$D_1$ is the point diametrically opposite $D$ wrt the incircle
$D_2$ is the second intersection of $AD$ with the incircle
$T$ is the intersection of $DD_1$ with $AP$
$X$ is the intersection of $EF$ and $DD_1$
$E_1$ is the intersection of $DE$ with $AP$ and define $F_1$ analogously
$M$ is the midpoint of $BC$
$Q$ is the extouch point on $BC$

Since $DD_1$ is the diameter of the incircle, it is clear from the angle condition that $A,G, D_1$ are collinear. I will now show that $(T, X; D_1,D)=-1$ , which follows from the following claim:

Claim: $F, D_1, E_1$ are collinear, and similarly for $E, D_1, F_1$ .
Proof: Apply Pascal's on hexagon $FD_1GDEE$ . We find that $FD_1\cap DE$ , $A$ , and $P$ are collinear. However, the intersection of $DE$ with $AP$ is $E_1$ , so we must have $FD_1\cap DE=E_1$ , or $F, D_1, E_1$ collinear. A similar argument with Pascal's on hexagon $ED_1GDFF$ yields $E, D_1, F_1$ collinear.

Now, applying Ceva-Menalaus to $\triangle DE_1F_1$ , we easily get $(T, X; D_1, D)=-1$ . Projecting through $A$ , we obtain $(T, X; D_1,D)=(AT\cap BC, AX\cap BC; AD_1\cap BC, D)=-1$ . However, it is well known that $AX\cap BC$ is the midpoint of $BC$ , and that $AD_1\cap BC$ is the $A$ extouch point. Furthermore, we know that the intouch point and the extouch point are symmetric about the midpoint, so $(AT\cap BC, M; Q, D)=-1$ implies that $AT\cap BC$ is the point of infinity wrt $BC$ , or $BC$ is parallel to $AT$ . Since $A, T, P$ are collinear by definition, $AP$ is parallel to $BC$ .

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#3 h Jan 3, 2020, 2:41 AM • 5 Y

Y by Lol_man000, Tekashi6ix9ine, YerMom, jhu08, Adventure10

Let $M$ denote the midpoint of $\overline{BC}$ and $\overline{AM}\cap\overline{EF}=X$ . Let $Y$ denote the antipode of $D$ with respect to the incircle, $\omega$ . Let $\overline{GX}\cap \omega = Z.$ Note that $\angle YGD=\angle AGD=90^{\circ},$ so $A, Y, G$ are collinear.

It is well known that $\overline{AM}, \overline{EF}, \overline{DY}$ concur at $X.$

Claim: Quadrilateral $DZFE$ is harmonic.

Proof. Note that it suffices to prove that $A,Z,D$ are collinear.

Note $X$ lies on $\overline{EF}$ , the polar of $A$ . By La Hire's $A$ lies on the polar of $X$ . Furthermore, by Brocard's on quadrilateral $DGYZ$ , $\overline{GY}\cap\overline{DZ}$ lies on the polar of $X$ as well. Moreover, since $A\in\overline{GY}$ and on the polar of $X$ , then $\overline{GY}\cap\overline{DZ}=A.$ Hence, $A,Z,D$ must be collinear. $\blacksquare$

Projecting $\omega$ onto line $\overline{EF}$ and then onto $\overline{BC}$ gives,
$(D,Z;F,E)\stackrel{G}{=} (P,X;F,E)\stackrel{A}{=}(\overline{AP}\cap\overline{BC}, M;B,C)=-1.$ For which it is well known then that $\overline{AP}\cap\overline{BC} = \infty,$ so $\overline{AP}\parallel \overline{BC}.$

This post has been edited 3 times. Last edited by cmsgr8er, Nov 16, 2020, 2:42 PM

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#4 h Jan 3, 2020, 2:55 AM • 7 Y

Y by rzlng, Pluto1708, cmsgr8er, tree_3, jhu08, Adventure10, ehuseyinyigit

Let $I$ be the incenter, and let $D'$ be the antipode of $D$ on the incircle so that $A, D', G$ collinear. Let $X$ be the foot from $A$ to $DD'$ , and let $P = AX \cap GD$ be the orthocenter of $\triangle AD'D$ . Right angles imply $AXGD$ cyclic, so $PX \cdot PA = PG \cdot PD$ . Then $P$ lies on the radical axis of the incircle and $(AIEFX)$ , which is $EF$ , meaning $P = EF \cap GD \cap AX$ and we are done.

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#5 h Jan 3, 2020, 3:25 AM • 9 Y

Y by Aryan-23, amar_04, Kagebaka, tree_3, mijail, jhu08, MrOreoJuice, Adventure10, Mango247

Here is my solution for p1 i submitted.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 2.3613380392430985, xmax = 23.887380706934806, ymin = -9.061409326786983, ymax = 5.15416560376006; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); /* draw figures */ draw((7.06,2.8)--(3.46,-4.48), linewidth(1.2) + dtsfsf); draw((3.46,-4.48)--(16.02,-4.46), linewidth(1.2) + dtsfsf); draw((16.02,-4.46)--(7.06,2.8), linewidth(1.2) + dtsfsf); draw(circle((8.03017757107319,-1.6364997487962782), 2.8362193026272435), linewidth(1.2)); draw((8.034693838122354,-4.472715455671781)--(10.537308327920737,-2.9625598155596276), linewidth(1.2)); draw((7.06,2.8)--(20.121430537802492,2.820798456270373), linewidth(1.2) + dtsfsf); draw((7.06,2.8)--(11.445306161877651,-4.467284544328221), linewidth(1.2) + rvwvcq); draw((7.06,2.8)--(9.74,-4.47), linewidth(1.2) + rvwvcq); draw((10.537308327920737,-2.9625598155596276)--(20.121430537802492,2.820798456270373), linewidth(1.2)); draw((5.487822929371477,-0.37929140949323353)--(20.121430537802492,2.820798456270373), linewidth(1.2)); draw((8.025661304024027,1.199715958079226)--(8.034693838122354,-4.472715455671781), linewidth(1.2) + rvwvcq); /* dots and labels */ dot((7.06,2.8),dotstyle); label("$A$", (7.113840965876333,2.9456495378540297), NE * labelscalefactor); dot((3.46,-4.48),dotstyle); label("$B$", (3.521507871332976,-4.336862299722183), NE * labelscalefactor); dot((16.02,-4.46),dotstyle); label("$C$", (16.073706777558403,-4.322884349937968), NE * labelscalefactor); dot((8.03017757107319,-1.6364997487962782),linewidth(4pt) + dotstyle); label("$I$", (8.09229745077141,-1.527294393094892), NE * labelscalefactor); dot((5.487822929371477,-0.37929140949323353),linewidth(4pt) + dotstyle); label("$F$", (5.548310590044209,-0.2692789125155078), NE * labelscalefactor); dot((8.034693838122354,-4.472715455671781),linewidth(4pt) + dotstyle); label("$D$", (8.09229745077141,-4.364818199290614), NE * labelscalefactor); dot((9.815712055935128,0.567135097534706),linewidth(4pt) + dotstyle); label("$E$", (9.867497073366765,0.6812216728111381), NE * labelscalefactor); dot((8.025661304024027,1.199715958079226),linewidth(4pt) + dotstyle); label("$Z$", (8.078319500987195,1.31022941310083), NE * labelscalefactor); dot((10.537308327920737,-2.9625598155596276),linewidth(4pt) + dotstyle); label("G", (10.594350462145966,-2.8551996225953533), NE * labelscalefactor); dot((20.121430537802492,2.820798456270373),linewidth(4pt) + dotstyle); label("$P$", (20.183224014117727,2.9316715880698143), NE * labelscalefactor); dot((11.445306161877651,-4.467284544328221),linewidth(4pt) + dotstyle); label("$K$", (11.502917198119967,-4.3508402495063985), NE * labelscalefactor); dot((8.027291358535031,0.17604172516803054),linewidth(4pt) + dotstyle); label("$L$", (8.078319500987195,0.2898390788531074), NE * labelscalefactor); dot((9.74,-4.47),linewidth(4pt) + dotstyle); label("$M$", (9.79760732444569,-4.364818199290614), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Let $Z$ be the $D-$ antipode w.r.t $\odot(I)$ . Clearly $G=AZ \cap \odot(I)$ . Now let $ZD \cap EF=L$ . Its well known (also egmo lemma 4.17) that $AL \cap BC=M$ is the midpoint of $BC$ . Now note that $-1=(Z,G;F,E)\overset{D}{=}(F,E;L,P)$ . Now let $AP \cap BC =T$ . Note that $-1=(F,E;L,P)\overset{A}{=}(B,C;M,T)$ but its well known that $(B,C,M;P_{\infty})$ is harmonic $\implies T=P_{\infty} \implies AP \parallel BC$ . Done $\blacksquare$

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1218 posts

#6 h Jan 3, 2020, 4:23 AM • 5 Y

Y by AlastorMoody, amar_04, Pluto04, jhu08, Adventure10

Sharygin 2018 CR P20

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#7 h Jan 3, 2020, 4:58 AM • 2 Y

Y by jhu08, Adventure10

It is well known that $AM$ , $EF$ and $DD'$ concur where $D'$ is the point diametrically opposite $D$ wrt the incircle and $M$ is the midpoint of $BC$ . Note that $A,G,D$ are collinear due to obvious reasons . Call that concurrency point $Z$ . Let the parallel from $A$ to $BC$ intersect $EF$ at $P'$ .
Now we have
$-1 = (BC;M\infty_{BC})\stackrel{A}{=}(FE;ZP')$ But since $(F,E;D',G) = -1$ (the quadrilateral is harmonic )
Hence we have
$-1 = (F,E;D',G)\stackrel{D}{=}(FE;ZP)$ So $P'=P$ , done

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579 posts

#8 h Jan 3, 2020, 5:05 PM • 3 Y

Y by jhu08, Adventure10, Mango247

Simple solution:

We let Q be the intersection of the line through A parallel to BC and the circle with diameter of AD.

We know that Q is on circle with diameter AD or (AGD), so AQD=AQI=90 degrees where I is the incenter. Thus, Q lies on the circle with diameter AI. This is the same circle as AEF as AEIF is obviously cyclic. Thus, AGEF is cyclic. Taking the radical axis theorem on the circle with (AQGD), (AQEIF), (FEGD) which gives us that AQ, EF, GD are concurrent, and thus we have achieved the result that we wanted.

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1119 posts

#9 h Mar 16, 2020, 1:33 AM • 1 Y

Y by jhu08

AwesomeYRY wrote:

Simple solution:

We let Q be the intersection of the line through A parallel to BC and the circle with diameter of AD.

We know that Q is on circle with diameter AD or (AGD), so AQD=AQI=90 degrees where I is the incenter. Thus, Q lies on the circle with diameter AI. This is the same circle as AEF as AEIF is obviously cyclic. Thus, AGEF is cyclic. Taking the radical axis theorem on the circle with (AQGD), (AQEIF), (FEGD) which gives us that AQ, EF, GD are concurrent, and thus we have achieved the result that we wanted.

Why $\angle AQI=90^{\circ}$ ? Can someone explain pls? :huh:

:huh:

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jayme

#10 h Mar 16, 2020, 7:33 AM • 1 Y

Y by jhu08

Dear Mathlinkers,
a way of a proof can be seen on

https://artofproblemsolving.com/community/c6t48f6h1621417_sharygin_cr_p20

Sincerely
Jean-Louis

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pad

#11 h May 23, 2020, 3:38 AM • 4 Y

Y by mathlogician, anonman, jhu08, Mango247

Diagram
Let $Q=(AEFI) \cap (AD)$ .

Claim: $AQ\parallel BC$ .
Proof: We know $\angle AQD=90$ and $\angle AQI=90$ since $(AEFI)$ has diameter $AI$ . Therefore, $Q,I,D$ are collinear. But $ID\perp BC$ , so $QD\perp BC$ . Finally, since $\angle AQD=90$ , we conclude that $AQ\parallel BC$ . $\square$

Now, radical axis on $(AEFIQ), (DEF), (AD)$ tells us that $AQ, EF, DG$ concur. Since $AQ$ is the line through $A$ parallel to $BC$ , we are done.

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#12 h Jun 19, 2020, 12:01 AM • 2 Y

Y by jhu08, Mango247

wait can someone check if this works? seems really sketchy

Let $M$ be the midpoint of $BC$ , let $AG$ intersect the incircle at point $X$ , and let $Q = XG \cap EF$ . Note that $EF, AM, DX$ concur, and call the concurrence point $T$ . Note that $A,X,G$ are collinear, so by tangents from $A$ note that $EFXG$ is harmonic. Now let $P'$ be the point on $EF$ such that $AP' \parallel BC$ . It is obvious that $(B,C;M, P_{\infty}) = (F,E;T,P) = -1$ by projecting through $A$ . But also by projecting through point $D$ , $(E,F;X,G) = (E,F;T,P) = -1$ thus $P = P'$ and we win.

wait yay seems identical to something else in this thread

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#13 h Aug 23, 2020, 3:36 PM • 1 Y

Y by jhu08

Let $H$ be where $AG$ hits the incircle of $\triangle ABC$ . Since $\angle AGD = \angle HGD = 90^{\circ}$ , then $H$ is the other point where the perpendicular line to $BC$ passing through $D$ hits the incircle of $\triangle ABC$ . Letting $M$ be the midpoint of line $BC$ and $H'$ be where $EF$ hits $DH$ , we have $A, H', M$ are collinear. Since $AE$ and $AF$ are tangents and $A, G, H$ are collinear, then $(HG; EF) = -1$ so taking projectivity with respect to point $D$ and line $EF$ we get $(H'P; EF) = -1$ and now with respect to $A$ and line $BC$ , letting $AP$ hit $BC$ at $Q$ , we have $(MQ; CB) = -1$ so since $BM=CM$ then $Q = \infty$ meaning that $AP$ and $BC$ are parallel.

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#14 h Sep 21, 2020, 8:54 PM • 3 Y

Y by A-Thought-Of-God, jhu08, Mango247

More storage

tworigami wrote:

Let $ABC$ be a scalene triangle. The incircle of $\triangle ABC$ is tangent to sides $\overline{BC}$ , $\overline{CA}$ , $\overline{AB}$ at $D, E, F$ , respectively. Let $G$ be a point on the incircle of $\triangle ABC$ such that $\angle AGD = 90^\circ$ . If lines $DG$ and $EF$ intersect at $P$ , prove that $AP$ is parallel to $BC$ .

Proposed by bluek

Solution: Let $D'$ be the antipode of $D$ w.r.t $\odot(I), \overline{AP} \cap \overline{BC} = X$ and $\overline{DD'} \cap \overline{EF} = K.$ Now, by Incenter Concurrency Lemma we have $\overline{AK}$ passing through the midpoint of $BC$ , let's say $M.$

Since $\angle AGD = 90^{\circ}$ , so $\overline{AG}$ passes through $D'$ . Next, as $AE, AF$ are tangent to $\odot(I)$ and $A,D',G$ are collinear, hence, $D'GEF$ is harmonic due to Harmonic Quad Lemma. Thus, $-1=(D',G;F,E)\stackrel{D}= (K,P;F,E)\stackrel{A}=(M,X;B,C).$ But by Midpoints and Parallel Lines Lemma $(B,C;M,P_{\infty})=-1,$ where $P_{\infty}$ is the point at infinity along $\overline{BC}.$ Hence, $X=P_{\infty} \implies AP \parallel BC. \quad\square$

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#15 h Feb 5, 2021, 12:45 AM • 1 Y

Y by jhu08

Let $Q$ be the antipode of $D$ with respect to the incircle and set $R = DM\cap EF$ ; clearly, $A$ , $Q$ , and $G$ are collinear. Then $GEMF$ is a harmonic quadrilateral, so
$-1 = (G,M;E,F) \stackrel{D}=(P,R;E,F) \stackrel{A}=(PA\cap BC, AR\cap BC; B,C).$ But it's known that $AR$ is actually the $A$ -median of $\triangle ABC$ , so this last equality is only possible when $PA\cap BC = \infty_{BC}$ -- that is, when $PA\parallel BC$ .

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#16 h Jun 26, 2021, 2:50 AM • 1 Y

Y by jhu08

Let $l$ be the line through $A$ parallel to $BC$ , $D'$ be the antipode of $D$ (with respect to the incircle), and $N = l \cap DD'$ . The desired conclusion is equivalent to showing $l$ , $EF$ , and $DG$ are concurrent.

The angle condition implies that $AG$ passes through $D'$ . It's also easy to see $l \perp DD'$ by parallel lines.

Claim: $ANEIF$ and $ANGD$ are cyclic.

Proof. Because $N, D', I, D$ are collinear, $\angle AND = \angle ANI = 90^{\circ} = \angle AFI = \angle AEI$ proving the first claim.

Observe $\angle AGD = \angle AND = 90^{\circ}$ which proves our second claim. $\square$

Now, it follows that $AN = l$ , $EF$ , and $DG$ are concurrent the Radical Center of $(ANEIF)$ , $(ANGD)$ , and $(DGEF)$ . $\blacksquare$

Note: If the centers of the $3$ circles are collinear, then the $3$ lines concur at infinity.

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Mathscienceclass

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Mathscienceclass

#17 h Aug 18, 2021, 9:11 PM • 1 Y

Y by jhu08

All antipodes, poles, and polars are taken w.r.t. incircle. With $M$ as the midpoint of $BC$ , first establish the following:

Lemma, $M$ , $I$ , midpoint of $AD$ are collinear.
Proof. Let $D'$ be the antipode of $D$ w.r.t. incircle and $K$ the $A$ -extouch. $A$ , $D'$ , $K$ are collinear by a common lemma. Since $M$ is the midpoint of $D$ and $K$ (isotomic), we are done by a $\frac{1}{2}$ homothety at $D$ . $\square$

Since the midpoint of $AD$ is the center of $(AD)$ , we know that $M$ lies on the perpendicular bisector of $DG$ . Thus $MG$ is also tangent, so $P$ lies on the polar of $M$ . By La Hire's $M$ lies on the polar of $P$ .

Also we know that $P$ lies on $EF$ , the polar of $A$ , so $A$ lies on the polar of $P$ and thus line $AM$ is really the polar. Then $AM$ contains $P'$ , the harmonic conjugate of $P$ w.r.t $EF$ . This means $(B, C; M, P_{\infty}) \stackrel{A}{=} (E, F; P', AP_{\infty} \cap EF)$ , implying $AP \parallel BC$ . $\blacksquare$

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#18 h Aug 18, 2021, 9:19 PM • 2 Y

Y by jhu08, pad

The key observation is that the polar of $P$ with respect to the incircle is in fact $AM$ , where $M$ is the midpoint of $BC$ .

First, note that $A$ must lie on the polar of $P$ , since the polar of $A$ is $EF$ , so by La Hire, we get what we want. It's well known that $MG$ is tangent to the incircle, so the polar of $M$ is $DG$ , but as $P$ lies on $DG$ , then $M$ must lie on the polar of $P$ by La Hire.

Now, if we let $N=EF\cap AM$ , we have
$-1=(F,E;N,P) \stackrel{A} = (B,C;M,AP\cap BC)$ which implies $AP$ is parallel to $BC$ , as desired.

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#19 h Oct 17, 2021, 11:59 AM

Y by

Harmonics = scam. Call the incircle $\omega$ , let $I$ be the incenter, $D'$ be the anitpode of $D$ on the incircle, since $\measuredangle AGD = \measuredangle D'GD$ so $A-D'-G$ , now let $L=D'D \cap (AFIE)$ . Note that $\measuredangle ALD = \measuredangle ALI = 90^\circ = \measuredangle AGD$ so $L \in (AGD)$ . Also because of $90^\circ$ we have $AL \parallel BC$ , finish by radical axis on $\{(ADG) , (AFIE) , \omega\}$ .
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#21 h Aug 11, 2023, 12:48 PM

Y by

Let $M$ be the midpoint of $BC$ . Furthermore, let $D'$ be the antipode of $D$ . Clearly, as $\angle AGD=90^\circ$ , the intersection of $AG$ with the incircle must be $D'$ . Now, it is well known that $A$ , $D'$ and the $A$ -excirlce touch point $X$ are collinear. Furthermore, we know that $BD=CX$ , so $DM=MX$ . As $\angle DGX=90^\circ$ , we have $MG=MD$ , so $MG$ is tangent to the incircle. Therefore, $P$ lies on the polar of $M$ , so $M$ lies on the polar of $P$ . Also, $P$ lies on the polar of $A$ , so $A$ lies on the polar of $P$ . It follow that $AM$ is the polar of $P$ , so if $K\coloneq EF\cap AM$ . By projecting from $A$ onto $BC$ , $-1=(F,E;K,P)=(A,C;M,P'=BC\cap AP)$ , so $P'$ is the point at infinity, and thus $AP\parallel BC$ , as desired.

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#22 h Feb 28, 2024, 1:23 AM

Y by

Let $H$ be the $D$ -antipode and $K = \overline{EF} \cap \overline{DI}$ . By incenter concurrency lemma $\overline{AK}$ bisects $\overline{BC}$ . Hence $-1 = (EF;GH) \stackrel D= (EF; PK) \stackrel A= (CB; \overline{AP} \cap \overline{BC}, \overline{AK} \cap \overline{BC}).$ The previous observation implies $\overline{AP} \cap \overline{BC} = \infty_{BC}$ , as needed.

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#23 h Feb 29, 2024, 5:26 AM

Y by

Complex numbers can also put this one away very quickly. Take the incircle as the unit circle, so that
$|d|=|e|=|f|=1$ $a=\frac{2ef}{e+f}$ $b=\frac{2df}{d+f}$ $c=\frac{2de}{d+e}$ $g=\frac{d+a}{d\overline{a}+1} = \frac{de+df+2ef}{2d+e+f}$ $p = \frac{dg(e+f) - ef(d+g)}{dg-ef} = \frac{d^2e^2+d^2f^2-2e^2f^2}{(e+f)(d^2-ef)}$ Then
$p-a = \frac{d^2(e-f)^2}{(e+f)(d^2-ef)}$ $b-c = -\frac{2d^2(e-f)}{(d+e)(d+f)}$ so that
$\frac{a-p}{b-c} = \frac{(d+e)(d+f)(e-f)}{(e+f)(d^2-ef)}$ which is equal to its conjugate and thus real. $\blacksquare$

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Om245

#24 h Mar 10, 2024, 5:45 PM • 1 Y

Y by GeoKing

If $D'$ is antipode of $D$ in incircle and $M$ is midpoint of $BC$ then it well know that $A-D'-G$ and $MQ=MD$ .

Consider pole of $DG$ and $EF$ with respect to incircle which are $A$ and $M$ respectively and hence $AM$ is polar of $P$ .
By well know fact we know $EF,AM,ID$ are concurrent at one point $X$ .

Polar of $X$ is parallel to $BC$ as $ID \perp BC$ . From fact that $X$ lie on $AM$ we get $AP \parallel BC$ .

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eibc

#25 h Mar 10, 2024, 8:22 PM

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By radical center lemma on $(AD), (AEIF)$ , and the incircle, we find that $PA$ , $(AEIF)$ , and $(AD)$ meet again at some point $X \neq A$ . Since $\overline{AX} \perp \overline{XI}$ and $\overline{AX} \perp \overline{XD}$ , we have $X$ , $I$ , and $D$ collinear. But because $\overline{XID} \perp \overline{BC}$ and $\overline{AP} \perp \overline{XID}$ , we have $\overline{AP} \parallel \overline{BC}$ , as desired.

This post has been edited 1 time. Last edited by eibc, Mar 10, 2024, 8:22 PM

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ismayilzadei1387

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ismayilzadei1387

#26 h Mar 12, 2024, 6:32 PM

Y by

It is very intersting

first $DI \cap EF=J$
$DI \cap AG=K$
so $LFDE$ is harmonic if we take pencil from $A$ it quickly implies
$(F,E;J,P)=-1$
So we can take some intersection point which will be infinity point soon
from easy lemma $AJ \cap BC$ at midpoint of $BC$
$AP \cap BC = R$
$(B,C;M,R)=-1$
also we know that
$(B,C;M,Q_{\infty})=-1$
so if 3 points in parantheses are coincide
then the forth ones will be coincide
$R \equiv Q_{\infty}$
and we are done

This post has been edited 2 times. Last edited by ismayilzadei1387, Mar 12, 2024, 6:33 PM

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bjump

#27 h Aug 2, 2024, 12:31 AM

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XIOO 10 min solve let $\overline{AG}$ intersect the incircle of $\triangle ABC$ again at $H$ then since $90^\circ = \angle AGD = \angle AHD$ . $H$ is the $D$ -antipode on the incircle. Now pascal on $HHFDDE$ gives $IJ \parallel BC$ , and pascal on $FFHEED$ gives $AIJ$ are collinear. Now brokard gives that $\overline{AIJ}$ is the pole of $K=\overline{EF} \cap \overline{HD}$ . Now $-1=(FE; HG) \stackrel{D}= (FE; KP)$ so $P$ lies on the polar of $K$ and as $\overline{AIJP} \parallel \overline{BC}$ we are finished.

Heavily rushed cuz my brother is making me do stuff

This post has been edited 1 time. Last edited by bjump, Aug 2, 2024, 12:32 AM

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Eka01

#28 h Aug 14, 2024, 9:45 AM • 1 Y

Y by Sammy27

We define:- $M$ as midpoint of $BC$ , $I$ as the incenter and $DI \cap EF=Q$ , .
All poles and polars with respect to $(DEF)$ .
It is easy to see that $EF$ is polar of $A$ , $AP$ is polar of $Q$ since the required polar is perpendicular to $IQ$ and passes through $A$ due to La Hire's [Since $P$ lies on polar of $A$ , $A$ must lie on polar of $P$ .]
and it is well known that $DG$ is polar of $M$ because $M$ is circumcenter of $\Delta DGX$ where $X$ is the $A$ extouch point and $MD$ is tangent to the incircle. So taking the polar duality, it suffices to show that $A, Q,M$ are collinear but this is well known and hence we are done.

This post has been edited 6 times. Last edited by Eka01, Aug 14, 2024, 10:10 AM

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#29 h Aug 26, 2024, 2:45 AM

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Let $M$ is the midpoint of $BC$ .
Let $AM \cap EF = X$ . It is well known that $X \in DI$ .
Let the $D$ antipode on the incircle be $D'$ and notice that $\angle DGA = \angle DGD'$ so $G-D'-A$ .
Since $(B, C; N, \infty_{BC}) \overset{A}= (F, E; X, A\infty_{BC} \cap EF) = -1$ , it suffices to show that $(F, E; X, P) = -1$ .
This is equivalent to $AN$ being the polar of $P$ wrt the incircle.
Notice that the polar of $A$ wrt the incircle is $EF$ so by La Hire's, $A$ lies on the polar of $P$ . Then since $AD' \cap BC = T$ which is the extouch point, and $N$ is the circumcenter of $(DGT)$ we have $ND = NG$ which implies $NG$ is a tangent to the incircle. So this implies $P$ lies on the polar of $N$ so by La Hire's we have that the polar of $P$ is $AN$ as desired.

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#30 h Dec 14, 2024, 8:43 PM

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Let $M$ be the midpoint of $BC$ . With respect to the incircle, notice $A$ is the pole of $EF$ and $M$ is the pole of $DG$ , so $P = EF \cap DG$ is the pole of $AM$ . Thus
$-1 = (E, F; EF \cap AM, P) \overset{A}{=} (B, C; M, AP \cap BC) \implies AP \parallel BC. \quad \blacksquare$

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#31 h Dec 17, 2024, 11:48 AM

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First note that $EF$ is the polar of $A$ .
Let $I$ be the incentre, $D'$ be the antipode of $D$ , and $U$ be the $A$ -extouch point. Then $A,D',G,U$ collinear.
Let $M$ be the midpoint of $BC$ . Then by a homothety at $D$ with scale factor $2$ we have that $D'G // IM$ , so $DG // IM$ . Since $MD$ is tangent to the incircle we have that $DG$ is the polar of $M$ .
It follows that the polar of $P=EF \cap DG$ is $AM$ , so $-1=(E,F;P,EF \cap AM) \stackrel{A}{=} (C,B;AP \cap BC,M)$ , from which it follows that $AP//BC$ .

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#32 h Mar 30, 2025, 4:35 PM

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Let $D'$ be reflection $D$ in $I$ , then it is wellknown that $A, D, G$ are collinear. Let $X=DE\cap D'F$ and $Y=DF\cap D'E$ and $Z=DD'\cap EF$ . By Pascal on $DGD'EFF$ and $DGD'FEE$ we see that line $AP$ is the same line as $XY$ . Brocard gives $ZI\perp XY$ , but since $ZI$ is the same line as $DI$ we have $ZI\perp BC$ and combining gives $BC\parallel XY$ or $BC\parallel AP$ and we are done.

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