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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
inequality with roots
luci1337   2
N 8 minutes ago by Nguyenhuyen_AG
Source: an exam in vietnam
let $a,b,c\geq0: a+b+c=1$
prove that $\Sigma\sqrt{a+(b-c)^2}\geq\sqrt{3}$
2 replies
luci1337
May 18, 2025
Nguyenhuyen_AG
8 minutes ago
Very hard
steven_zhang123   0
26 minutes ago
Source: An article
Given a positive integer \(n\) and a positive real number \(m\), non-negative real numbers \(a_1, a_2, \cdots, a_n\) satisfy \(\sum_{i=1}^n a_i = m\). Define \(N\) as the number of elements in the following collection of subsets:

$$
\left\{ I \subseteq \{1, 2, \cdots, n\} : \prod_{i \in I} a_i \geq 1 \right\}.
$$
Find the maximum possible value of \(N\).
0 replies
steven_zhang123
26 minutes ago
0 replies
Interesting Inequality
lbh_qys   1
N an hour ago by nexu
Let $ a, b, c$ be real numbers such that $ (3a-2b-c)(3b-2c-a)(3c-2a-b)\neq 0 $ and $ a + b + c = 3 . $ Prove that
$$ \left( \frac{1}{3a - 2b - c} + \frac{1}{3b - 2c - a} + \frac{1}{3c - 2a - b} \right)^2 + a^2 + b^2 + c^2 \geq 3 + 3\sqrt{\frac 27}$$
1 reply
lbh_qys
2 hours ago
nexu
an hour ago
Goofy geometry
giangtruong13   1
N an hour ago by nabodorbuco2
Source: A Specialized School's Math Entrance Exam
Given the circle $(O)$, from $A$ outside the circle, draw tangents $AE,AF$ ($E,F$ are tangential points) and secant $ABC$ ($B,C$ lie on circle $O$, $B$ is between $A$ and $C$). $OA$ intersects $EF$ at $H$; $I$ is midpoint of $BC$. The line crossing $I$, paralleling with $CE$, intersects $EF$ at $D$. $CD$ intersects $AE$ at $K$. Let $N$ lie inside the triangle $FBC$ such that: $AF$=$AN$. From $N$ draw chords $BQ$, $RC$, $FP$ on circle $(O)$. Prove that: $PRQ$ is a isosceles triangle
1 reply
giangtruong13
Yesterday at 4:23 PM
nabodorbuco2
an hour ago
Rainbow vertices
goodar2006   3
N an hour ago by Sina_Sa
Source: Iran 3rd round 2012-Combinatorics exam-P1
We've colored edges of $K_n$ with $n-1$ colors. We call a vertex rainbow if it's connected to all of the colors. At most how many rainbows can exist?

Proposed by Morteza Saghafian
3 replies
goodar2006
Sep 20, 2012
Sina_Sa
an hour ago
Geometry hard problem.
noneofyou34   1
N an hour ago by User21837561
In a circle of radius R, three chords of length R are given. Their ends are joined with segments to
obtain a hexagon inscribed in the circle. Show that the midpoints of the new chords are the vertices of
an equilateral triang
1 reply
noneofyou34
2 hours ago
User21837561
an hour ago
Prove that two different boards can be obtained
hectorleo123   2
N an hour ago by alpha31415
Source: 2014 Peru Ibero TST P2
Let $n\ge 4$ be an integer. You have two $n\times n$ boards. Each board contains the numbers $1$ to $n^2$ inclusive, one number per square, arbitrarily arranged on each board. A move consists of exchanging two rows or two columns on the first board (no moves can be made on the second board). Show that it is possible to make a sequence of moves such that for all $1 \le i \le n$ and $1 \le j \le n$, the number that is in the $i-th$ row and $j-th$ column of the first board is different from the number that is in the $i-th$ row and $j-th$ column of the second board.
2 replies
hectorleo123
Sep 15, 2023
alpha31415
an hour ago
Inspired by lbh_qys
sqing   2
N 2 hours ago by JARP091
Source: Own
Let $ a, b $ be real numbers such that $ (a-3)(b-3)(a-b)\neq 0 $ and $ a + b =6 . $ Prove that
$$ \left( \frac{a}{b - 3} + \frac{b}{3 - a} + \frac{3}{a - b} \right)^2 + 2(a^2 + b^2 )\geq54$$Equality holds when $ (a,b)=\left(\frac{3}{2},\frac{9}{2}\right)$
$$\left( \frac{a+1}{b - 3} + \frac{b+1}{3 - a} + \frac{4}{a - b} \right)^2 + 2(a^2 + b^2 )\geq 60$$Equality holds when $ (a,b)=\left(3-\sqrt 3,3+\sqrt 3\right)$
$$  \left( \frac{a+3}{b - 3} + \frac{b+3}{3 - a} + \frac{6}{a - b} \right)^2 + 2\left(a^2 + b^2\right)\geq 72$$Equality holds when $ (a,b)=\left(3-\frac{3}{\sqrt 2},3+\frac{3}{\sqrt 2}\right)$
2 replies
sqing
3 hours ago
JARP091
2 hours ago
Interesting Inequality
lbh_qys   4
N 2 hours ago by lbh_qys
Given that \( a, b, c \) are pairwise distinct $\mathbf{real}$ numbers and \( a + b + c = 9 \), find the minimum value of

\[
\left( \frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b} \right)^2  + a^2 + b^2 + c^2.
\]
4 replies
lbh_qys
Today at 5:42 AM
lbh_qys
2 hours ago
Sum of complex fractions is an integer
Miquel-point   2
N 3 hours ago by cazanova19921
Source: Romanian IMO TST 1981, Day 3 P4
Let $n\geqslant 3$ be a fixed integer and $\omega=\cos\dfrac{2\pi}n+i\sin\dfrac{2\pi}n$.
Show that for every $a\in\mathbb{C}$ and $r>0$, the number
\[\sum\limits_{k=1}^n \dfrac{|a-r\omega^k|^2}{|a|^2+r^2}\]is an integer. Interpet this result geometrically.

Octavian Stănășilă
2 replies
Miquel-point
Apr 6, 2025
cazanova19921
3 hours ago
mexico 1999
cuenca   1
N 3 hours ago by FrancoGiosefAG
A polygon has each side integral and each pair of adjacent sides perpendicular (it is not necessarily convex). Show that if it can be covered by non-overlapping $2 x 1$ dominos, then at least one of its sides has even length.
1 reply
cuenca
Nov 18, 2006
FrancoGiosefAG
3 hours ago
segment of projections is half as sidelength, right triangle inscribed in right
parmenides51   4
N 3 hours ago by sunken rock
Source: 2020 Austrian Federal Competition For Advanced Students, Part 1, p2
Let $ABC$ be a right triangle with a right angle in $C$ and a circumcenter $U$. On the sides $AC$ and $BC$, the points $D$ and $E$ lie in such a way that $\angle EUD = 90 ^o$. Let $F$ and $G$ be the projection of $D$ and $E$ on $AB$, respectively. Prove that $FG$ is half as long as $AB$.

(Walther Janous)
4 replies
parmenides51
Nov 22, 2020
sunken rock
3 hours ago
Hard Inequality
danilorj   2
N 3 hours ago by sqing
Let $a, b, c > 0$ with $a + b + c = 1$. Prove that:
\[
\sqrt{a + (b - c)^2} + \sqrt{b + (c - a)^2} + \sqrt{c + (a - b)^2} \geq \sqrt{3},
\]with equality if and only if $a = b = c = \frac{1}{3}$.
2 replies
danilorj
Today at 5:17 AM
sqing
3 hours ago
Interesting inequalities
sqing   6
N 3 hours ago by sqing
Source: Own
Let $a,b,c \geq 0 $ and $ab+bc+ca- abc =3.$ Show that
$$a+k(b+c)\geq 2\sqrt{3 k}$$Where $ k\geq 1. $
Let $a,b,c \geq 0 $ and $2(ab+bc+ca)- abc =31.$ Show that
$$a+k(b+c)\geq \sqrt{62k}$$Where $ k\geq 1. $
6 replies
sqing
May 16, 2025
sqing
3 hours ago
Concurrency
Omid Hatami   14
N May 16, 2025 by Ilikeminecraft
Source: Iran TST 2008
Suppose that $ I$ is incenter of triangle $ ABC$ and $ l'$ is a line tangent to the incircle. Let $ l$ be another line such that intersects $ AB,AC,BC$ respectively at $ C',B',A'$. We draw a tangent from $ A'$ to the incircle other than $ BC$, and this line intersects with $ l'$ at $ A_1$. $ B_1,C_1$ are similarly defined. Prove that $ AA_1,BB_1,CC_1$ are concurrent.
14 replies
Omid Hatami
May 20, 2008
Ilikeminecraft
May 16, 2025
Concurrency
G H J
Source: Iran TST 2008
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Omid Hatami
1275 posts
#1 • 7 Y
Y by TRYTOSOLVE, itslumi, cadaeibf, Adventure10, Mango247, Mango247, and 1 other user
Suppose that $ I$ is incenter of triangle $ ABC$ and $ l'$ is a line tangent to the incircle. Let $ l$ be another line such that intersects $ AB,AC,BC$ respectively at $ C',B',A'$. We draw a tangent from $ A'$ to the incircle other than $ BC$, and this line intersects with $ l'$ at $ A_1$. $ B_1,C_1$ are similarly defined. Prove that $ AA_1,BB_1,CC_1$ are concurrent.
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mathangel
316 posts
#2 • 1 Y
Y by Adventure10
I think that we could reduce the problem to the following:

Given triangle $ ABC$ and $ (I)$ is its incircle. Let $ A', B'$ be points on $ BC, CA$ and $ d$ is a tangent of $ (I)$. Tangents of $ (I)$ from $ A', B'$ (different from $ BC, CA$) intersect $ d$ at $ A_1, B_1$ respectively. Then $ AA_1, BB_1, A'B'$ are concurrent.
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Shishkin
99 posts
#3 • 2 Y
Y by Adventure10, Mango247
Lets incircle tachs $ AB$, $ BC$, $ CA$ in $ F$, $ E$, $ D$ respectively.
Points $ F'$, $ E'$, $ D'$ lie on incircle. $ D' \in A_{1}A'$, $ E' \in B_{1}B'$, $ F' \in C_{1}C'$.
$ X = EF \cap LD'$ $ Y = DF \cap LE'$ $ Z = DE \cap LF'$
Let $ P$ be a pole of line $ l$. Then $ P = DD' \cap EE' \cap FF'$
$ AA_{1}$, $ BB_{1}$, $ CC_{1}$ are concurent. $ \Longleftrightarrow$ $ X$, $ Y$, $ Z$ are colinear. ($ X$ is a pole of $ AA_{1}$)
From Pascal teorem for $ DEFF'LD'$ we get that $ X$, $ Y$, $ P$ are colinear. Similarly $ X$, $ Z$, $ P$ are colinear. So $ X$, $ Y$, $ Z$, $ P$ are colinear.
Attachments:
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mathangel
316 posts
#4 • 2 Y
Y by Adventure10, ehuseyinyigit
My solution is similar to yours.

Let $ D, E$ be intersections of $ B'B_1, AB$ and $ l, AC$. We easily note that the hexagon $ B'EA_1A'BD$ circumscribes the incircle of triangle $ ABC$. It follows from Brianchon's Theorem that $ A'B', BE, A_1D$ are concurrent at $ K$. Hence, applying Pappus' Theorem to the 2 triples $ (B_1, A_1, E)$ and $ (A, B, D)$, we have the 3 points $ B', K$ and the intersection $ H$ of $ AA_1, BB_1$ are collinear. Hence, $ H$ is on $ A'B'$.
This means $ AA_1, BB_1, A'B'$ are concurrent.

The problem then follows from this.
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thecmd999
2860 posts
#5 • 2 Y
Y by Adventure10, Mango247
Solution
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TelvCohl
2312 posts
#6 • 15 Y
Y by navi_09220114, Eray, Fermat_Theorem, AlastorMoody, amar_04, Kamran011, mijail, SenatorPauline, Nymoldin, parola, mathleticguyyy, Adventure10, Mango247, EpicBird08, and 1 other user
My solution:

From Brianchon theorem ( for $ A'CAC'C_1A_1 $ and $ B'ABA'A_1B_1 $ )
we get $ AA_1, BB_1, CC_1, l $ are concurrent .

Q.E.D
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sayantanchakraborty
505 posts
#7 • 2 Y
Y by Adventure10, Mango247
Before proving the statement I need a prerequisite lemma:

Lemma:With respect to a given circle $\omega$,the polars are concurrent if and only if their respective poles are collinear.

Proof:Consider the intersection points of the polars and use La Hire's theorem.

Now back to our problem.I shall first name the points:$D,E,F$ are the tangency points of the incircle with $BC,CA,AB$ respectively.Let $X$ be the point on $(I)$ such that $B'X$ is tangent to $(I)$ (other than $E$),$Y$ be the point where $l'$ touches $(I)$,$Z$ be the point where $A'A_1$ touches $(I)$ and $L$ be the point where $C'C_1$ touches $(I)$.In the rest of my proof the poles and polars,etc will be considered with respect to the incircle.

Note that the polars of $A',B',C'$ are $DZ,EX,FL$ and as $A',B',C'$ are collinear,by our lemma $DZ,EX,FL$ concur.Set $YX \cap DF=P,EF \cap YZ=Q,YL \cap DE=R$.Applying Pascal's theorem on the cyclic hexagon $XYZDFE$ we see that $P,Q,EX \cap DZ$ are collinear.Similarly Pascal's theorem in $XYLFDE$ yeilds that $P,R,XE \cap LF$ are collinear.But $XE \cap DZ=XE \cap LF$ as $DZ,EX,FL$ concur so $P,Q,R$ are collinear.

Now see that $DF$ is the polar of $B$,while $XY$ is the polar of $B_1$.Hence by La Hire's theorem $P=DF \cap XY$ is the pole of $BB_1$.Similarly $Q$ is the pole of $AA_1$ and $R$ is the pole of $CC_1$.Finally by our lemma,as $P,Q,R$ are collinear,$AA_1,BB_1,CC_1$ concur,as desired.
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amar_04
1916 posts
#8 • 7 Y
Y by GeoMetrix, mueller.25, Hexagrammum16, BinomialMoriarty, Bumblebee60, A-Thought-Of-God, Mathematicsislovely
Wonderful Problem!!! I guess there is a solution with Dual of Desargues Involution too, can anyone provide that solution too. :)
Iran TST 2008 P2 wrote:
Suppose that $ I$ is incenter of triangle $ ABC$ and $ l'$ is a line tangent to the incircle. Let $ l$ be another line such that intersects $ AB,AC,BC$ respectively at $ C',B',A'$. We draw a tangent from $ A'$ to the incircle other than $ BC$, and this line intersects with $ l'$ at $ A_1$. $ B_1,C_1$ are similarly defined. Prove that $ AA_1,BB_1,CC_1$ are concurrent.

Let the Tangency Points made by $\odot(I)$ with $\{BC,C,A,AB\}$ be $\{D,E,F\}$ respectively. Let $\ell'$ be tangent to $\odot(I)$ at a point $X$ andlet $\{A'P,B'Q,C'R\}$ be the other tangents where $\{P,Q,R\}\in\odot(I)$.

Notice that the Polars of $\{A',B',C'\}$ are $\{PD,QE,RF\}$ respectively WRT $\odot(I)$ and as $\overline{A'-B'-C'}$. Hence, $PD,QE,RF$ are concurrent. Now by Pascal on $XQPDFR$ we get that $XQ\cap FD,QP\cap FR,PD\cap RX$ are collinear. Now Pascal on $XPDEFR$ we get that $XP\cap EF,PD\cap FR,DE\cap RX$ are collinear. So combining both the Collinearities we get that $XP\cap EF, DE\cap XR,FD\cap XQ$ are collinear. Now the Polar of $XP\cap EF$ is $AA_1$, the Polar of $XQ\cap FD$ is $BB_1$ and the Polar of $DE\cap RX$ is $CC_1$. So, $AA_1,BB_1,CC_1$ are concurrent.
This post has been edited 3 times. Last edited by amar_04, Mar 6, 2020, 6:25 AM
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rcorreaa
238 posts
#9
Y by
Suppose that $\ell$ touches the incircle at $P$.

We will prove that $AA_1,B_1,CC_1,\ell$ are concurrent.

Firstly, we will show that $BB_1,CC_1,\ell$ are concurrent, and similarly, we would have that $\{AA_1,CC_1,\ell\}; \{AA_1,BB_1,\ell\}$ are concurrent, implying the desired result.

Now, we use Moving Points. Fix $A_1,B_1,C_1$ and move $P$ projectively on the incircle. Let $T_A=C'C_1 \cap B'B_1$.

$\implies I$ is the incenter of $\Delta C_1T_AB_1$ $\implies \angle ZIY=90º+\frac{\angle FT_AE}{2}$ (fixed).

Hence, the composed map $$C'C_1 \mapsto \mathcal{C}_I \mapsto \mathcal{C}_I \mapsto B'B_1$$given by $$C_1 \mapsto IC_1 \mapsto IB_1 \mapsto B_1$$is projective $(*)$, since it's a projection followed by a rotation with fixed angle through $I$, followed by another projection.

Now, define $CC_1 \cap \ell= Q_C, BB_1 \cap \ell= Q_B$.

Thus, the composed map $$C_1C' \mapsto \mathcal{C}_C \mapsto \mathcal{C}_C \mapsto \ell$$given by $$ A_1 \mapsto CA_1 \mapsto CQ_C \mapsto Q_C$$is projective, since all maps are projections, which are projective maps.

Similarly, $Y \mapsto Q_B$ is projective $\implies$ from $(*)$, the map $$Q_B \mapsto B_1 \mapsto C_1 \mapsto Q_C$$is projective. Hence, since $Q_B,Q_C \in \ell$, which is a fixed line, by the Moving Points Lemma, in order to prove that $Q_B=Q_C$, it is sufficient to verify the problem for $3$ choices of $P$.

Choosing $P$ to be the points such that the incircle touches $AB,BC,CA$, we get the desired result easily.

Hence, we are done.

$\blacksquare$
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starchan
1610 posts
#10
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We will henceforth work in the projective plane. First of all send $\ell'$ to the line at infinity, so that $A', B', C'$ are the points at infinity for $BC, CA, AB$ respectively. Under this transformation, however the incircle becomes an ellipse. In order to fix this; we take an affine homography converting the ellipse into a circle again. Since affine transforms preserve the line at infinity, $\ell'$ remains the line at infinity. Now considering the projective dual of the problem, it can be seen to be equivalent to the following:

Reduced Problem:
Given a triangle $ABC$ with circumcircle $\gamma$, an arbitrary point $P$ on $\gamma$ and $A', B', C'$ as the respective $A, B, C$ antipodes with respect to $\gamma$ prove that $A'P \cap BC, B'P \cap CA$ and $C'P \cap AB$ are collinear.

Now this is easy to prove by spamming Pascal's Hexagram Theorem. Using Pascal on $APCA'BC'$ yields that centre of $\gamma$, $A'P \cap BC$, $C'P \cap AB$ are collinear. Reiterating this result for other pairs of intersections yields that $A'P \cap BC, B'P \cap CA$ and $C'P \cap AB$ are collinear and thus we are done.
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TheHazard
93 posts
#11 • 3 Y
Y by popop614, dolphinday, ohiorizzler1434
Polar reciprocate the entire diagram to get the equivalent problem statement.
New Problem Statement wrote:
Let $ABC$ be a triangle with point $X$ in the plane and point $P$ on circumcircle $\Gamma$. Let $XA, XB, XC$ intersect $\Gamma$ at $D, E, F$. Then let $DY, EY, FY$ intersect $BC, AC, AB$ at $G, H, I$. Show that $G, H, I$ are collinear.

Now, take a homography in $\mathbb{CP}^2$ that maps $X$ to a point of infinity. Let $L$ be one of the midpoint of arcs $AD, BE, CF$. Oh wait, this is just USAMO 2012/5 with $\gamma = PL$.
This post has been edited 1 time. Last edited by TheHazard, Nov 7, 2023, 6:52 PM
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OronSH
1747 posts
#12 • 6 Y
Y by centslordm, ihatemath123, GeoKing, ohiorizzler1434, ehuseyinyigit, EpicBird08
brianchon on $ABDXYE,ACDXZF$ gives $AX,BY,CZ,DEF$ concurrent done
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matematica007
17 posts
#13
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We apply Brianchon on $A_1A'BAB'B_1$ so we obtained that $AA_1,BB_1,l$ are concurrent.
Analogous $BB_1,CC_1,l$ are concurrent so $AA_1,BB_1,CC_1$ are concurrent. So the problem is proved.
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cursed_tangent1434
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#14
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Denote by $\omega$ the incircle of $\triangle ABC$. We claim that in particular $ AA_1,BB_1,CC_1$ are concurrent on line $\ell$.

Now, note that each side of $AC'C_1A_1A'C$ and $BC'C_1B_1B'C$ are tangent to $\omega$. Thus both these hexagons are tangential, which implies that by Brianchon's Theorem, $AA_1$ and $CC_1$ intersect on $\ell$ and also $BB_1$ and $CC_1$ intersect on $\ell$, which finishes the problem.
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Ilikeminecraft
658 posts
#15
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Brianchon applied on $AFZXDB$ implies that $AX\cap FD\cap BX$ are concurrent. Symmetry finishes.
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