orz otl fr

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Hip1zzzil

24 posts

Hip1zzzil

#1 h Mar 29, 2025, 10:23 AM

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An acute triangle $\bigtriangleup ABC$ is given which $BC>CA>AB$ .
$I$ is the interior and the incircle of $\bigtriangleup ABC$ meets $BC, CA, AB$ at $D,E,F$ . $AD$ and $BE$ meet at $P$ . Let $l_{1}$ be a tangent from D to the circumcircle of $\bigtriangleup DIP$ , and define $l_{2}$ and $l_{3}$ on $E$ and $F$ , respectively.
Prove $l_{1},l_{2},l_{3}$ meet at one point.

This post has been edited 4 times. Last edited by Hip1zzzil, Mar 29, 2025, 10:12 PM
Reason: Mistake

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whwlqkd

187 posts

whwlqkd

#2 h Mar 29, 2025, 10:32 AM

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Why is geometry in p3/6
It is very unexpected and soooooo hard

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EthanWYX2009

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EthanWYX2009

#3 h Mar 29, 2025, 11:35 AM

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Complex bash should be working, ig 30 minute calculation

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kmh1

105 posts

kmh1

#4 h Mar 29, 2025, 12:47 PM • 1 Y

Y by Retemoeg

I assume you mean $I$ is the incenter, not just any point in the interior.

solution

This post has been edited 2 times. Last edited by kmh1, Mar 29, 2025, 12:57 PM

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Acorn-SJ

59 posts

Acorn-SJ

#5 h Mar 29, 2025, 12:50 PM

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Yes, $I$ is supposed to be the incenter. Also, the original statement contained the fact that $BC > CA > AB$ .

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LoloChen

482 posts

LoloChen

#6 h Mar 29, 2025, 1:25 PM • 5 Y

Y by qwerty123456asdfgzxcvb, whwlqkd, Resolut1on07, Funcshun840, sebasl195

But this can be solved instantly using cubic curve. :maybe:

Generalization (using basically the same method): $I$ is the Miquel point of $\bigtriangleup ABC$ and $D,E,F$ which lies on $BC, CA, AB$ respectively. Furthermore, $AD$ , $BE$ and $CF$ meet at $P$ . Let $l_{1}$ be a tangent from D to the circumcircle of $\bigtriangleup DIP$ , and define $l_{2}$ and $l_{3}$ on $E$ and $F$ , respectively. Then $l_{1},l_{2},l_{3}$ meet at one point.
Sketch of proof: Consider the Qa-Cu1 cubic curve of $ABCDEFP$ . $l_{1},l_{2},l_{3}$ are the tangents of Qa-Cu1. Consider its Abelian group, we only need $D+D=E+E=F+F$ . But that's pretty obvious because $A+P+D=B+P+E=C+P+F$ and $A+A=B+B=C+C$ . See here for details.

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space10

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space10

#7 h Mar 30, 2025, 11:29 AM • 10 Y

Y by qwedsazxc, ehuseyinyigit, segment, GuvercinciHoca, Mop2018, Kingsbane2139, Lufin, Gaussss, X.Luser, k12byda5h

Very good problem. This is my solution.

Attachments:

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sansae

120 posts

sansae

#8 h Mar 30, 2025, 3:24 PM

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Complex bash should definitely work in contest hour.

Let the incircle be the unit circle, with $D=a, E=b, F=c$ . Then $P=K$ is the symmedian point, whose complex coordinate is given in Post #3 by
$k=\frac{2(\sum a^{2}b^{2} - \sum a^{2}bc)}{\sum_{sym}a^{2}b-6abc}$ (All the summations without index will denote a cyclic sum.)

Let $Q=q$ be the intersection of $l_{1}$ and $l_{2}$ . It is enough to show that $q$ is symmetric w.r.t. $a, b, c$ .

Since $\angle QDO = \angle OKD$ ,
$\frac{k(q-a)}{a(k-a)} = \frac{a\overline{k}(\overline{q}-\overline{a})}{\overline{k}-\overline{a}}, \quad \overline{q}-\overline{a} = \frac{k(q-a)(\overline{k}-\overline{a})}{a^{2}\overline{k}(k-a)}$
Note that by definition of $K=k$ , $D=a$ , and $A=\frac{2bc}{b+c}$ are collinear, hence
$\frac{\overline{k}-\overline{a}}{k-a} = \frac{b+c-2a}{a(ab+ac-2bc)}$
Using the formula for $k$ , one can easily see that
$\frac{k}{\overline{k}} = \frac{\sum a^{2}b^{2} - \sum a^{2} bc}{\sum a^{2} - \sum ab}$ .

Plugging these into the equation for $q$ , and subtracting with the similar equation for the $E=b$ -side gives a linear equation on $q$ :
$\overline{b}-\overline{a} = \frac{\sum a^{2}b^{2} - \sum a^{2} bc}{\sum a^{2} - \sum ab}\left ( \frac{(b+c-2a)(q-a)}{a^{3}(ab+ac-2bc)} - \frac{(c+a-2b)(q-b)}{b^{3}(bc+ba-2ca)}\right )$
Solving this equation gives
$q=\frac{abc(\sum ab) (\sum_{sym}a^{2}b - 6abc)}{(\sum_{sym}a^{2}b) (\sum a^{2}b^{2} - \sum a^{2}bc)},$ which is symmetric. Hence the claim.

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jerrome2685

47 posts

jerrome2685

#9 h Mar 31, 2025, 7:05 AM

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Same problem with https://artofproblemsolving.com/community/q2h2407674p19737244 (Note that $P$ is a symmedian point of $DEF$ )

This post has been edited 1 time. Last edited by jerrome2685, Mar 31, 2025, 7:05 AM

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hectorleo123

368 posts

hectorleo123

#10 h Jun 6, 2025, 4:13 AM

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Hip1zzzil wrote:

An acute triangle $\bigtriangleup ABC$ is given which $BC>CA>AB$ .
$I$ is the interior and the incircle of $\bigtriangleup ABC$ meets $BC, CA, AB$ at $D,E,F$ . $AD$ and $BE$ meet at $P$ . Let $l_{1}$ be a tangent from D to the circumcircle of $\bigtriangleup DIP$ , and define $l_{2}$ and $l_{3}$ on $E$ and $F$ , respectively.
Prove $l_{1},l_{2},l_{3}$ meet at one point.

Let $G'$ be the antigonal conjugate of $G$ with respect to triangle $\triangle DEF$ , and let $L$ be the inverse of $G'$ with respect to $I$

Since the isogonal conjugate of $G$ with respect to $\triangle DEF$ lies on the Euler line of $\triangle DEF$ , it follows that
$\angle AGI = \angle AG'I = \angle LIA.$
Therefore, $L$ lies on $l_{1},l_{2},l_{3}$

This post has been edited 2 times. Last edited by hectorleo123, Jun 6, 2025, 4:15 AM

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