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Source: Sharygin 2018

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3042 posts

#1 h Apr 4, 2018, 7:30 PM • 4 Y

Y by mathematicsy, CoolJupiter, Adventure10, Mango247

Let the incircle of a nonisosceles triangle $ABC$ touch $AB$ , $AC$ and $BC$ at points $D$ , $E$ and $F$ respectively. The corresponding excircle touches the side $BC$ at point $N$ . Let $T$ be the common point of $AN$ and the incircle, closest to $N$ , and $K$ be the common point of $DE$ and $FT$ . Prove that $AK||BC$ .

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1600 posts

Vrangr

#2 h Apr 4, 2018, 7:35 PM • 6 Y

Y by anantmudgal09, Drunken_Master, Maths_Guy, Pluto1708, Adventure10, Mango247

$\omega$ be the incircle of $\triangle ABC$ .
Let $T' = AT\cap \omega\ (\neq T)$ , it's well-known that $T'$ is the point diametrically opposite $D$ w.r.t. $\omega$ . Therefore, $\angle ATD$ is $90^{\circ}$ .

Now consider the radical axes of $\odot ADT$ , $\odot AEFI$ and $\omega$ .

This post has been edited 2 times. Last edited by Vrangr, Apr 4, 2018, 7:51 PM
Reason: Angle typo

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66 posts

Fumiko

#3 h Apr 4, 2018, 7:38 PM • 2 Y

Y by Adventure10, Mango247

Let $\ell$ be the line through $A$ parallel to $BC$ it is well known that $\ell$ is the polar of midpoint of $DE$ and also this midpoint lies on $A-$ median then the rest is trivial

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2487 posts

#4 h Apr 5, 2018, 1:39 AM • 2 Y

Y by Adventure10, Mango247

Sharygin 2013 along with the fact that the antipode of the incircle touch point lies on the Nagel cevian.

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1600 posts

Vrangr

#5 h Apr 5, 2018, 5:52 AM • 5 Y

Y by AlastorMoody, Pluto04, Adventure10, Mango247, busy-beaver

Vrangr wrote:

$\omega$ be the incircle of $\triangle ABC$ .
Let $T' = AT\cap \omega\ (\neq T)$ , it's well-known that $T'$ is the point diametrically opposite $D$ w.r.t. $\omega$ . Therefore, $\angle ATD$ is $90^{\circ}$ .

Now consider the radical axes of $\odot ADT$ , $\odot AEFI$ and $\omega$ .

Expanding upon my previous sketch.

$\measuredangle$ refers to directed angles.

Let $I$ be the incentre and $\omega$ be the incircle of $\triangle ABC$ .
$[asy] import geometry; import olympiad; unitsize(4cm);pair A = dir(135), B = dir(210), C = -1/B; pair I = incenter(A, B, C); pair D = foot(I, B, C), E = foot(I, C, A), F = foot(I, A, B); pair T_ = 2I - D, N = B + C - D, T = intersectionpoints(incircle(A,B,C), A -- N)[1]; pair K = extension(E, F, D, T); pair L = extension(D, I, A, K); draw(circumcircle(A,E,F)^^circumcircle(D, T, A), linetype("0 2")); draw(D--L, linetype("2 4")); draw(A--B--C--cycle); draw(incircle(A,B,C)); draw(A--N); draw(D--K--F); draw(A--K, dashed); draw(rightanglemark(A, L, D, 1.5)^^rightanglemark(I, D, B, 1.5)); draw(rightanglemark(A, F, I, 1.5)^^rightanglemark(A, E, I, 1.5)); draw(rightanglemark(A, T, D, 1.5)); draw(I--F^^I--E, linetype("2 5")); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(T); dot(T_); dot(N); dot(K); dot(I); dot(L); label("A", A, dir(90)); label("B", B, dir(-90)); label("C", C, dir(-90)); label("D", D, dir(-90)); label("E", E, dir(90)); label("F", F, -dir(30)); label("K", K, dir(90)); label("T$'$", T_, dir(-135)); label("T", T, dir(-7)); label("I", I, dir(-45)); label("N", N, dir(-90)); label("L", L, dir(90));[/asy]$
Claim 1
Claim 2
Claim 3
And one more thing

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3042 posts

#6 h Apr 5, 2018, 6:54 AM • 4 Y

Y by Maths_Guy, MelonGirl, Adventure10, Mango247

Lemma:
Main problem

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307 posts

#7 h Apr 19, 2018, 9:11 AM • 2 Y

Y by Adventure10, Mango247

Let $F'$ be the point diametrically opposite $F$ in the incircle and let $X=FI \cap DE$ . $K'$ be the intersection of $DE$ and the line through $A$ parallel to $BC$ . Let $M$ be the midpoint of $BC$ . As above $A,F,T,N$ colinear.

It's well-known $X$ lies on the $A$ -median so:
$-1=(\infty_{BC},M;C,B) \stackrel{A}{=} (K',X;E,F)$ Also:
$-1=(T,F';E,F) \stackrel{F}{=}(K,X;E,F)$ So $K=K'$ as desired.

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214 posts

Arkmmq

#9 h Apr 19, 2018, 5:01 PM • 3 Y

Y by Mmqark, Adventure10, Mango247

Let AN intersect the incircle in F' and T and let FF' intersect DE at P ..
it is wellknown that FF' is a diameter in the incircle and AP is median in ABC .
we have $(D,E;F',T)=-1$ so $(D,E;P,K)=-1$ and project from A we get thd desired result.

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Synthetic_Potato

114 posts

Synthetic_Potato

#10 h May 29, 2018, 2:04 PM • 2 Y

Y by Adventure10, Mango247

I solved this in my sleep

My Solution:
We will prove that the line parallel to $BC$ through $A$ , $FT$ and $DE$ are concurrent.

Let $I$ be the incenter of $ABC$ . It is a well known result that $AN, FI$ meet on the incircle at the antipode of $F$ . Let the antipode of $F$ on the incircle be $F'$ . Then we can say $A,T,F'$ are collinear. Now, let $FF'$ meet the line parallel to $BC$ at $X$ . As $IF \perp BC$ , we get $\angle FXA=\angle IXA=90^\circ (\spadesuit)$ . Now see that as $FF'$ is a diameter, $\angle FTA =90^\circ$ . So, $FTXA$ is cyclic from $(\spadesuit)$ . Note that $\angle IDA = \angle IEA = \angle IXA = 90^\circ$ from $(\spadesuit)$ . So, $DEAX$ is cyclic and $I$ also lies on this circle. Now, Applying Radical Axis theorem on circles $DETF, AXTF, AXDE$ , we get $AX, DE, FT$ are concurrent. Hence $K$ lies on $AX$ , so $AK\parallel BC$ .

$\blacksquare$ .

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66 posts

neel02

#11 h Jun 2, 2018, 11:54 AM • 2 Y

Y by Adventure10, Mango247

Too easy for Sharygin !
Just note that , if $G$ is diametrically op. to $F$ then $DEGT$ is a harmonic quad. Since $A,G,T,N$ collinear
So projecting by $F$ we have $(D,E;X,K)=-1$ .Where $X$ is the intersection pt. of $DE$ & $A-median$ .
Again projecting by $A$ we have $AK,AM,AB,AC$ are harmonic pencils ! where $M$ is the intersection pt of $A-median$ & $BC$ . Since $M$ is the midpoint of $BC$ so $AK$ is parallel to $BC$

This post has been edited 1 time. Last edited by neel02, Jun 2, 2018, 11:55 AM
Reason: typo

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1218 posts

#12 h Sep 16, 2018, 3:05 PM • 2 Y

Y by Adventure10, Mango247

Here's the solution that I submitted during the actual exam.

Attachments:

Solution_Q 20.pdf (455kb)

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1618 posts

#13 h Feb 22, 2019, 10:11 AM • 1 Y

Y by Adventure10

Here is my solution for this problem
Solution
Let $G$ , $P$ be second intersections of $AN$ , $AF$ with the incircle of $\triangle$ $ABC$ , respectively; $Q$ $\equiv$ $FG$ $\cap$ $DE$
It's easy to see that: $AN$ $\perp$ $FK$ at $T$
Since: $DFEP$ , $DTEG$ are harmonic quadrilaterals, we have: $\dfrac{FD}{FE}$ = $\dfrac{PD}{PE}$ , $\dfrac{TD}{TE}$ = $\dfrac{GD}{GE}$
So: $\dfrac{KD}{KE}$ = $\dfrac{FD}{FE}$ . $\dfrac{TD}{TE}$ = $\dfrac{FD}{FE}$ . $\dfrac{TD}{TE}$ = $\dfrac{PD}{PE}$ . $\dfrac{GD}{GE}$ or $K$ , $G$ , $P$ are collinear
But: $GP$ $\perp$ $AF$ then: $G$ is orthocenter of $\triangle$ $AFK$ or $FG$ $\perp$ $AK$
Combine with: $FG$ $\perp$ $BC$ , we have: $AK$ $\parallel$ $BC$

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1051 posts

#14 h Oct 20, 2020, 8:33 PM • 2 Y

Y by starchan, sixoneeight

The hardest part of the problem is dealing with the strange point labels.

Let $\overline{AT}$ meet the incircle again at a point $L$ , and suppose that $\overline{DE}$ and $\overline{AM}$ meet at a point $X$ . Let $Y$ be the intersection of lines $AK$ and $BC$ , possibly at infinity. It is well-known that $\overline{FIXL}$ is collinear, so $-1 = (DE;LT) \stackrel{F}{=} (DE;FK) \stackrel{A}{=} (BC;MY)$ , so $\overline{AK} \parallel \overline{BC}$ , as desired.

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160 posts

#15 h Oct 20, 2020, 11:51 PM • 1 Y

Y by Mango247

Let $L = \overline{ET} \cap \overline{DF}$ . Pascal on $FFTEED$ $\implies$ $\overline{LAK}$ collinear. Since $TD$ is a diameter, $T$ is the orthocenter of $\triangle DKL$ . So, $\overline{TD} \perp \overline{AK}$ , as desired.

This post has been edited 1 time. Last edited by ppanther, Oct 20, 2020, 11:52 PM

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1162 posts

#16 h Aug 14, 2021, 3:06 AM

Y by

Relabel the points such that the incircle touches $BC, CA, AB$ at $D, E, F$ respectively.

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

The angle condition implies $AG$ 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. Also, I should've used projective... lol.

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8857 posts

#17 h Sep 5, 2021, 2:50 AM

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Let $K = \overline{AP_\infty} \cap \overline{DE}$ be a point such that $\overline{AK} \parallel \overline{BC}$ . EGMO 9.49 from Sharygin 2013 establishes that $\overline{AM}$ is the polar of $K$ with respect to the incircle, where $M$ is the midpoint of $BC$ . But now, $-1 = (FN; MP_\infty) \stackrel A= (F, T; \overline{AM} \cap \overline{FT}, \overline{AP_\infty} \cap \overline{FT}),$ so $\overline{AM} \cap \overline{FT}$ lies on the polar of $\overline{FT} \cap \overline{AP_\infty}$ . But $K$ also lies on the polar by La Hire's, and since the polar is a line, such $K$ is unique. From here $\overline{AP_\infty} \cap \overline{FT}=K$ so we are done.

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primesarespecial

364 posts

primesarespecial

#18 h Jan 15, 2022, 1:05 PM

Y by

Let $M$ be the midpoint of $BC$ .It is well known that $FT$ is the polar of $M$ .By La Hire , $AM$ is the polar of $K$ .Now,it is well known that $FI,AM,DE$ concur ,at $Q$ .By La Hire, $AK$ is the polar of $Q$ .Now $IQ \perp AK, IF \perp BC$ ,so done.

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1080 posts

#19 h Aug 13, 2022, 4:56 AM • 1 Y

Y by centslordm

Let $F'$ be the antipode of $F,$ let $M$ be the midpoint of $\overline{BC},$ and let $Y=\overline{AM}\cap\overline{DE}.$ It is well-known that $Y\in\overline{FF'}.$ Notice $-1=(DE;F'T)\stackrel{F}=(DE;YK)\stackrel{A}=(BC;M,\overline{AK}\cap\overline{BC}),$ so $\overline{AK}\cap\overline{BC}$ is the point at infinity along $\overline{BC}.$ $\square$

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49 posts

jampm

#20 h Aug 27, 2022, 8:30 PM

Y by

$M$ is the pole of $DT$ wrt. the incircle by symmetry. $A$ is the pole of $EF$ , and thus $K$ is the pole of $AM$ . Now $P_{\infty}$ , the point at infinity of line $BC$ is the pole of $DI$ , and since $EF$ , $AM$ and $DI$ concur, by the polar transformation, it's poles are colineal, which implies the problem.

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1472 posts

#21 h Aug 27, 2022, 10:57 PM

Y by

Let $FI \cap DE=G$ then its known that if u let $M$ the midpoint of $BC$ then $A,G,M$ are colinear, now taking polars w.r.t. $\omega$ we have that this is $\mathcal P_A, \mathcal P_G, \mathcal P_M$ concurrent and that means $DE,FT$ and the line through $A$ parallel to $BC$ concurrent thus we are done

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1353 posts

#22 h Aug 28, 2022, 2:54 PM • 1 Y

Y by Mango247

solution

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465 posts

#23 h Jun 3, 2023, 5:01 PM • 1 Y

Y by HoripodoKrishno

Alright, finally back from hospital after 6 straight days T__T Let's get back to work... Also woahhh!! This is so similar to the Sharygin 2023 P22. :no_mouth:

https://i.imgur.com/85yviDu.png

Firstly, let $M$ be the midpoint of $BC$ , $P=AM\cap DE$ and $Q=AM\cap FT$ . Also $\ell$ denote the line through $A$ parallel to $BC$ . Now it is well known that $M$ is also the midpoint of $FN$ . Also from the Diameter of the Incircle Lemma, we have that $\angle FTN=90^\circ$ which along with the fact that $M$ is the midpoint of $FN$ we have that $MF=MT$ and as $MF$ is tangent to the incircle, we have that $MT$ is tangent to the incircle.

Now $-1=(F,N;M,\infty_{BC})\overset{A}{=}(F,T;Q,FT\cap\ell)$ and $-1=(B,C;M,\infty_{BC})\overset{A}{=}(D,E;P,DE\cap\ell)$ . Now firstly note that the polar of $A$ w.r.t. the incircle is the line $DE$ so $DE\cap\ell$ lies on the polar of $A$ w.r.t. the incircle and by La Hire's Theorem, we thus have that $A$ lies on the polar of $DE\cap\ell$ and also as $(D,E;P,DE\cap\ell)=-1$ , we also have that $P$ also lies on the polar of $DE\cap\ell$ and thus the line $AP$ becomes that polar of $DE\cap\ell$ . Now using a similar logic, we see that the polar of $FT\cap\ell$ is the line $MQ\equiv AP$ . Now two points have the same polar means that both the points are identical and we thus have that $K=\ell\cap DE\cap FT$ and we are done.

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5001 posts

#24 h Aug 8, 2023, 6:18 PM • 1 Y

Y by centslordm

Relabel points so that $D$ is the $\overline{BC}$ -intouch point, etc. Here are two solutions.

Solution 1: Let $P$ be the $D$ -antipode with respect to the incircle and $Q$ the point on $\overline{DP}$ such that $\overline{AQ} \parallel \overline{BC}$ . It is well-known that $A,P,T,N$ are collinear, so $AEFIQ$ and $AQTD$ can easily be seen to be cyclic, and by radical center on the incircle, $(AEFIQ)$ , and $(AQTD)$ , we find that $\overline{EF}$ , $\overline{DT}$ , and $\overline{AQ}$ concur, hence $K$ lies on $\overline{AQ}$ . $\blacksquare$

Solution 2: Define $P$ as before and let $X=\overline{EF} \cap \overline{DP}$ . Since $\overline{TP}$ passes through $A$ ,
$-1=(F,E;T,P)\stackrel{D}{=}(F,E;K,X)\stackrel{A}{=}(B,C;\overline{AX} \cap \overline{BC},\overline{AK} \cap \overline{BC}).$ It is well-known that $\overline{AX} \cap \overline{BC}$ is just the midpoint of $\overline{BC}$ , so $\overline{AK} \cap \overline{BC}=P_{\infty \overline{BC}}$ which implies the desired result. $\blacksquare$

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598 posts

eibc

#25 h Aug 9, 2023, 8:01 PM

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Let $I$ be the incenter of $\triangle ABC$ , $X = \overline{AM} \cap \overline{DE},$ and $Y$ be the second intersection of $\overline{AN}$ and the incircle. Then from EGMO chapter 4, we know that $A$ , $I$ , $X$ , and $Y$ are collinear.

Claim: $AM$ is the polar of $K$ wrt the incircle

Proof: Since $DE$ is the polar of $A$ , by La Hire's we find that $A$ lies on the polar of $K$ . However, by taking a homothety at $A$ , we see that $Y$ is the $F$ -antipode wrt the incircle, so $\overline{YN} \perp \overline{FT}$ . By taking a homothety at $F$ we see that $\overline{IM} \perp \overline{FT}$ . But since $\overline{MF}$ is tangent to the incircle, this is enough to imply that line $FT$ is the polar of $M$ ; therefore, by La Hire's we see that $M$ lies on the polar of $K$ , too, which proves the claim.

Now, since $X$ lies on $\overline{AM}$ , we have
$-1 = (D, E; X, K) \overset{A}{=} (B, C; M, \overline{AK} \cap \overline{BC}).$ But since $(B, C; M, P_{\infty}) = -1$ , where $P_{\infty}$ is the point at infinity along line $BC$ , this implies that $\overline{AK} \parallel \overline{BC}$ , as needed.

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504 posts

kn07

#26 h Aug 9, 2023, 9:18 PM

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Relabel the points so $D=BC \cap \omega$ , $E=AC \cap \omega$ , and $F=AB \cap \omega$ .

Let the midpoint of $BC$ be $M$ and $AM \cap \omega=\{P,Q\}$ . Projecting the reciprocal pairs $\{B,C\}, \{D,N\}, \{M,M\}$ onto $\omega$ yields $DT, EF, PP,$ and $QQ$ are concurrent. Let $PQ \cap EF = X$ so that $K$ and $X$ are conjugated. Then, $\mathcal{H} (F,E;X,K) = \mathcal{H}(B,C;M,AK \cap BC)$ . The harmonic conjugate of $M$ is the point at infinity, thus $AK \parallel BC$ .

This post has been edited 2 times. Last edited by kn07, Aug 9, 2023, 9:20 PM
Reason: points are in wrong order

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1502 posts

#27 h Aug 30, 2023, 2:04 AM

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Let $M$ be the midpoint of $BC$ . It is well known that $\overline{AM}$ , $\overline{DE}$ , and $\overline{IF}$ concur at some point $P$ .
Let $S$ be the other intersection point of $AN$ and the incircle. It is well known that $SF$ is a diameter of the incircle.
As such, it follows that $(DE;ST) \overset{F}= (DE;PK) \overset{A}= (B,C;M,\overline{AK} \cap \overline{BC}) = -1$ which implies that $\overline{AK} \cap \overline{BC}$ intersect at $\infty$ , or that $\overline{AK} \parallel \overline{BC}$ .

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667 posts

#28 h Nov 19, 2023, 6:29 PM

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Let $M$ be the midpoint of $BC$ . Let $\Gamma$ denote the incircle. Define $P = AN \cap \Gamma$ . It is well known that $PX$ is a diameter of $\Gamma$ . Let $Q = PX \cap YZ$ .

Now we claim that $(ZY, QK) = -1$ . Indeed we find,
$\begin{align*} -1 = (ZY, PT) \overset{X}{=} (ZY, QK) \end{align*}$ Now projecting through $A$ we have,
$\begin{align*} -1 = (ZY,QK) \overset{A}{=} (BC,M\infty) \end{align*}$ so we indeed have $AK \parallel BC$ .

Attachments:

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793 posts

#29 h Jan 13, 2024, 5:46 PM

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We take note of the following:

$XK \perp AN$ through Diameter of the Incircle lemma.
$K = XT \cap YZ$ is the pole of $AM$ , so $IK \perp AM$ .
$XI$ , $AM$ , and $YZ$ concur by an incircle concurrence lemma.

Combining this with $AT \perp YZ$ , we get that the concurrence is the orthocenter of $\triangle AIK$ . Thus $XI \perp AK$ , which finishes. $\blacksquare$

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1880 posts

#30 h Feb 5, 2024, 2:06 PM

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Let $M$ be the midpoint of $BC$ (and therefore the midpoint of $XN$ , by a property I am too lazy to find the name of), let $P$ be the other intersection of $AT$ with $\omega$ , let $Q=AM\cap YZ$ , and let $\omega$ be the incircle.

Claim. $AN\perp XT$ .
Proof. Note that $P$ is the "top point" of $\omega$ and $X$ is the "bottom point" of $\omega$ (assuming that $BC$ is the "bottom tangent" of $\omega$ ), $PX$ is a diameter of $\omega$ , so $\angle PTX=90^{\circ}$ .

Claim. Line $XT$ is the polar of $M$ with respect to $\omega$ .
Proof. Since $MX$ is tangent to $\omega$ , we just need to prove that $MX=MT$ , which is true because triangle $XTN$ is a right triangle.

Claim. Line $AM$ is the polar of $K$ with respect to $\omega$ .
Proof. $YZ$ is the polar of $A$ wrt $\omega$ , and $XT$ is the polar of $M$ wrt $\omega$ , so we can finish by La Hire's.

Claim. $P,Q,X$ are collinear.
Proof. We employ barycentric coordinates with reference triangle $ABC$ . Then $I=(a:b:c)$ , $X=(0:a+b-c:a-b+c)$ , $Y=(a+b-c:0:-a+b+c)$ , and $Z=(a-b+c:-a+b+c:0)$ . Thus line $YZ$ is
$(-a+b+c)x+(-a+b-c)y+(-a-b+c)z=0,$ line $XI$ is
$(a-b-c)(c-b)x+a(a-b+c)y+a(-a-b+c)z=0,$ and line $AM$ is
$(0)x+(1)y+(-1)z=0.$ Now we plug everything into the concurrence lemma:
$\begin{align*} &\begin{vmatrix} 0 & 1 & -1 \\ -a+b+c & -a+b-c & -a-b+c \\ (a-b-c)(c-b) & a(a-b+c) & a(-a-b+c) \end{vmatrix} \\ =& \begin{vmatrix} 0 & 1 & 0 \\ -a+b+c & -a+b-c & -2a \\ (a-b-c)(c-b) & a(a-b+c) & 2a(c-b) \end{vmatrix} \\ =& -\begin{vmatrix} -a+b+c & -2a \\ (a-b-c)(c-b) & 2a(c-b) \end{vmatrix} \\ =& -2a(c-b)\begin{vmatrix} -a+b+c & -1 \\ a-b-c & 1 \end{vmatrix} \\ =& \; 0, \end{align*}$ as desired.

Therefore,
$-1=(Z,Y;P,T)\stackrel{X}{=}(Z,Y;Q,K).$ However, since
$-1=(B,C;M,\infty_{BC})\stackrel{A}{=}(Z,Y;Q,(A\infty_{BC}\cap ZY)),$ we indeed have $K=A\infty_{BC}\cap ZY$ , as desired. $\blacksquare$

This post has been edited 1 time. Last edited by cj13609517288, Feb 5, 2024, 2:10 PM

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Markas

#31 h Jun 16, 2024, 7:54 PM • 1 Y

Y by GeoKing

Let M be the midpoint of BC. Let l be the line which is parallel to BC trough A. Let $AM \cap DE = X$ and $AM \cap FT = Y$ . Now it follows that $(B,C;M,P\infty_{BC}) = -1$ and also $(B,C;M,P\infty_{BC})\stackrel{A}{=}(D,E;X,DE \cap l) = -1$ . Also its well known that BF = NC and since BM = MC, it follows that FM = MN $\Rightarrow$ $(F,N;M,P\infty_{BC}) = -1$ . Now projecting trough A we get $(F,N;M,P\infty_{BC})\stackrel{A}{=}(F,T;Y,FT \cap l) = -1$ . Now from the diameter of the incircle lemma, we have that $\angle NTF = 90^{\circ}$ and since FM = MN, then MT = MF and as MF is tangent to the incircle, we get that MT is also tangent to the incircle. Now its obvious that the polar of A is DE $\Rightarrow$ $DE \cap l \in DE$ $\Rightarrow$ $DE \cap l$ lies on the polar of A. Now by La Hire, A should lie on the polar of $DE \cap l$ . Now from $(D,E;X,DE \cap l) = -1$ and since D and E lie on a circle, we have that X lies on the polar of $DE \cap l$ . Since A should lie on the polar of $DE \cap l$ and X lies on the polar of $DE \cap l$ , then AX is the polar of $DE \cap l$ . Similarly the polar of $FT \cap l$ is the line $MY \equiv AX$ $\Rightarrow$ the polar of $DE \cap l$ and the polar of $FT \cap l$ is AX $\Rightarrow$ since the two points have the same polar, the points are identical $\Rightarrow$ we have that $K = DE \cap FT \cap l$ $\Rightarrow$ $K\in l$ $\Rightarrow$ $AK \parallel BC$ and we are ready.

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204 posts

Eka01

#32 h Aug 14, 2024, 10:03 AM • 1 Y

Y by Sammy27

Realise that $T$ is the same as $G$ in GOTEEM 2020 P1 and then the same solution follows.

This post has been edited 5 times. Last edited by Eka01, Aug 14, 2024, 10:07 AM

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985 posts

bebebe

#33 h Aug 14, 2024, 7:28 PM

Y by

Let $M$ be the midpoint of $BC$ and $XN,$ so $MT$ is tangent to incircle. Let $R=(AYIZ) \cap AM$ , so $90=\angle IRA=\angle IRM=\angle IXM,$ so $IRMX$ is cyclic. Thus $IRTMX$ is also cyclic, so $R=inverse(K)$ (follows by taking inverses of lines $XT$ and $YZ$ wrt incircle). All of this means $AM=polar(K).$

\textit{Lemma:} $XI, YZ, AM$ concur.

\textit{Proof:} Follows from Simson lines and homothethy.

Let $S=XI \cap YZ \cap AM.$ Since (where $P_{\infty}$ is point at infinity wrt $BC$ ) $XI=polar(P_{\infty})$ and $YZ=polar(A)$ and $AM=polar(K),$ we know $polar(S) = \overline{AKP_{\infty}}$ so $A, K, P_{\infty}$ are collinear, and we are done.

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onyqz

#34 h Aug 15, 2024, 4:19 PM

Y by

Basically the same as the above ones. Nonetheless a fun projective geometry exercise.
Click to reveal hidden text

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#35 h Aug 21, 2024, 5:31 AM • 1 Y

Y by dolphinday

Suppose that $X'$ is the antipode of $X$ , which lies on $\overline{AN}$ , and let $P = \overline{IX} \cap \overline{YZ}$ . It is well-known that $P$ lies on the $A$ -median, which we will call $\overline{AM}$ . Finally, let $Q = \overline{AK} \cap \overline{BC}$ .

Since $T'YTZ$ is harmonic, we have

$-1 = (Z,Y;T',T) \overset{X}{=} (Z,Y;P,K) \overset{A}{=} (B,C;M,Q).$
However, we also know that $(B,C,M,\infty) = -1$ , so $Q$ is the point at infinity. This implies that $\overline{AK} \parallel \overline{BC}$ . $\square$

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N3bula

#36 h Dec 7, 2024, 5:49 AM

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Let $M$ be the midpoint of $BC$ , I will prove that if $R=XY\cap XI$ , $A-R-M$ , clearly $R$ lies on the polar of $A$ , as well as this $R$ lies on the polar of the intersection of $BC$ and
the line tangent to the incircle and passing through the antipode. Thus the polar of $R$ is the line through $A$ tangent to $BC$ . Thus if we let the second
intersection of $XI$ with the incircle be $P$ and the intersection of $XI$ with the polar of $R$ be $Q$ , we get that $-1=(XP;RQ)\overset{\mathrm{A}}{=}(XN;M'\infty)$ ,
Thus $M'$ is the midpoint of $XN$ so $M'=M$ and $A-R-M$ now let $AK\cap BC$ be $J$ . Note that $-1=(PT;YZ)\overset{\mathrm{X}}{=}(YZ;RK)\overset{\mathrm{A}}{=}(BC;MJ)$
Thus as $M$ is the midpoint of $BC$ $J$ a point at infinity so $AK\parallel BC$ .

This post has been edited 1 time. Last edited by N3bula, Dec 7, 2024, 5:49 AM

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Ritwin

#37 h Dec 22, 2024, 7:52 PM • 1 Y

Y by OronSH

did someone say projective nuke?

Rename the intouch triangle to $XYZ$ as per the OTIS problem. Let $D$ be the antipode of $X$ on the incircle, and recall that $\overline{ADTN}$ are collinear (proof by a homothety sending the incircle to the $A$ -excircle).

Now let $PQR$ be the $D$ -cevian triangle in $XYZ$ .

Brocard on $XYDZ$ implies $QR \perp \overline{XPD} \perp BC$ .
Pascal on $XYYDZZ$ implies $\overline{QAR}$ collinear.

So, the goal is now to show that $K \in \overline{QAR}$ . Pascal on $XTDZYY$ says $K$ , $A$ , $Q$ are collinear, done. $\blacksquare$

This post has been edited 2 times. Last edited by Ritwin, Dec 22, 2024, 7:59 PM

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#38 h Mar 29, 2025, 3:43 AM

Y by

Let $R$ be the antipode of $D$ in the incircle. It is well known $A, R, N$ are collinear.
Let $M$ be midpoint of $BC.$ Clearly, $M$ is also midpoint of $DN.$ We have that $\angle DTN = 180 - \angle DTA = 90,$ so $DM = MT = MN,$ so $MT$ is tangent to the incircle.
Let $G$ denote the concurrency point of $RD, EF, AM.$
With respect to incircle: $G$ lies on the polar of $A.$
$A$ is the pole of $EF,$ so $A$ lies on the polar of $K.$ $DT$ is the polar of $M,$ so $M$ lies on the polar of $K.$ Hence, $AM$ is the polar of $K.$
However, both $A, K$ lie on the polar of $G.$ Thus, $AK$ is the polar of $G.$ Thus, $AK\perp IG=RD\perp BC,$ which finishes

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E50

#39 h Mar 29, 2025, 3:20 PM

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Let $I$ be the incenter of $\triangle ABC$ , $AN$ intersects the incircle of $\triangle ABC$ again at $L \ne T$ , $DF$ intersects $LE$ at $O$ and $DL$ intersects $EF$ at $M$ . Pascal on $DDELTF$ and $EEDLTF$ implies that $O,A,M,K$ collinear. It is well-known that $LF$ passes through $I$ . Applying Brocard's Theorem on $(DLEF)$ implies that $IL$ $\bot$ $OM$ and since $IL$ $\bot$ $BC$ we obtain that $OM$ $\parallel$ $BC$ , thus $AK$ $\parallel$ $BC$ .

This post has been edited 1 time. Last edited by E50, Mar 29, 2025, 3:21 PM
Reason: wrong typing

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