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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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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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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
Bosnia and Herzegovina 2022 IMO TST P1
Steve12345   3
N 2 minutes ago by waterbottle432
Let $ABC$ be a triangle such that $AB=AC$ and $\angle BAC$ is obtuse. Point $O$ is the circumcenter of triangle $ABC$, and $M$ is the reflection of $A$ in $BC$. Let $D$ be an arbitrary point on line $BC$, such that $B$ is in between $D$ and $C$. Line $DM$ cuts the circumcircle of $ABC$ in $E,F$. Circumcircles of triangles $ADE$ and $ADF$ cut $BC$ in $P,Q$ respectively. Prove that $DA$ is tangent to the circumcircle of triangle $OPQ$.
3 replies
Steve12345
May 22, 2022
waterbottle432
2 minutes ago
Functional equation of nonzero reals
proglote   6
N 3 minutes ago by pco
Source: Brazil MO 2013, problem #3
Find all injective functions $f\colon \mathbb{R}^* \to \mathbb{R}^* $ from the non-zero reals to the non-zero reals, such that \[f(x+y) \left(f(x) + f(y)\right) = f(xy)\] for all non-zero reals $x, y$ such that $x+y \neq 0$.
6 replies
proglote
Oct 24, 2013
pco
3 minutes ago
k Funcional equation problem
khongphaiwminh   1
N 7 minutes ago by jasperE3
Determine all functions $f \colon \mathbb R^+ \to \mathbb R^+$ that satisfy the equation
$$f(x+f(y))=f(x+y)+f(y)$$for any $x, y \in \mathbb R^+$. Note that $\mathbb R^+ \stackrel{\text{def}}{=} \{x \in \mathbb R \mid x > 0\}$.
1 reply
khongphaiwminh
16 minutes ago
jasperE3
7 minutes ago
AO and KI meet on $\Gamma$
Kayak   29
N 9 minutes ago by Mathandski
Source: Indian TST 3 P2
Let $ABC$ be an acute-angled scalene triangle with circumcircle $\Gamma$ and circumcenter $O$. Suppose $AB < AC$. Let $H$ be the orthocenter and $I$ be the incenter of triangle $ABC$. Let $F$ be the midpoint of the arc $BC$ of the circumcircle of triangle $BHC$, containing $H$.

Let $X$ be a point on the arc $AB$ of $\Gamma$ not containing $C$, such that $\angle AXH = \angle AFH$. Let $K$ be the circumcenter of triangle $XIA$. Prove that the lines $AO$ and $KI$ meet on $\Gamma$.

Proposed by Anant Mudgal
29 replies
Kayak
Jul 17, 2019
Mathandski
9 minutes ago
4 Variables Cyclic Ineq
nataliaonline75   1
N 9 minutes ago by NO_SQUARES
Prove that for every $x,y,z,w$ non-negative real numbers, then we have:
$\frac{x-y}{xy+2y+1}+\frac{y-z}{yz+2z+1} + \frac{z-w}{zw+2w+1} + \frac{w-x}{wx+2x+1} \geq 0$
1 reply
nataliaonline75
16 minutes ago
NO_SQUARES
9 minutes ago
IMO Genre Predictions
ohiorizzler1434   56
N 10 minutes ago by Theoryman007
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
56 replies
ohiorizzler1434
May 3, 2025
Theoryman007
10 minutes ago
Number theory
MathsII-enjoy   2
N 35 minutes ago by SimplisticFormulas
Prove that when $x^p+y^p$ | $(p^2-1)^n$ with $x,y$ are positive integers and $p$ is prime ($p>3$), we get: $x=y$
2 replies
MathsII-enjoy
Yesterday at 3:22 PM
SimplisticFormulas
35 minutes ago
National diophantine equation
KAME06   2
N 37 minutes ago by damyan
Source: OMEC Ecuador National Olympiad Final Round 2024 N3 P5 day 2
Find all triples of non-negative integer numbers $(E, C, U)$ such that $EC \ge 1$ and:
$$2^{3^E}+3^{2^C}=593 \cdot 5^U$$
2 replies
KAME06
Feb 28, 2025
damyan
37 minutes ago
IMO ShortList 1998, number theory problem 1
orl   55
N an hour ago by reni_wee
Source: IMO ShortList 1998, number theory problem 1
Determine all pairs $(x,y)$ of positive integers such that $x^{2}y+x+y$ is divisible by $xy^{2}+y+7$.
55 replies
orl
Oct 22, 2004
reni_wee
an hour ago
Number of sets S
Jackson0423   2
N an hour ago by Jackson0423
Let \( S \) be a set consisting of non-negative integers such that:

1. \( 0 \in S \),
2. For any \( k \in S \), both \( k + 9 \in S \) and \( k + 10 \in S \).

Find the number of such sets \( S \).
2 replies
Jackson0423
an hour ago
Jackson0423
an hour ago
F.E....can you solve it?
Jackson0423   16
N an hour ago by jasperE3
Find all functions \( f : \mathbb{R} \to \mathbb{R} \) such that
\[
f\left(\frac{x^2 - f(x)}{f(x) - 1}\right) = x
\]for all real numbers \( x \) satisfying \( f(x) \neq 1 \).
16 replies
Jackson0423
Yesterday at 1:27 PM
jasperE3
an hour ago
Find all positive a,b
shobber   14
N an hour ago by reni_wee
Source: APMO 2002
Find all positive integers $a$ and $b$ such that
\[ {a^2+b\over b^2-a}\quad\mbox{and}\quad{b^2+a\over a^2-b} \]
are both integers.
14 replies
shobber
Apr 8, 2006
reni_wee
an hour ago
Geo metry
TUAN2k8   2
N an hour ago by TUAN2k8
Help me plss!
Given an acute triangle $ABC$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively. Lines $BD$ and $CE$ intersect at point $F$. The circumcircles of triangles $BDF$ and $CEF$ intersect at a second point $P$. The circumcircles of triangles $ABC$ and $ADE$ intersect at a second point $Q$. Point $K$ lies on segment $AP$ such that $KQ \perp AQ$. Prove that triangles $\triangle BKD$ and $\triangle CKE$ are similar.
2 replies
TUAN2k8
Today at 10:33 AM
TUAN2k8
an hour ago
(not so) small set of residues generates all of F_p upon applying Q many times
62861   14
N 2 hours ago by john0512
Source: RMM 2019 Problem 6
Find all pairs of integers $(c, d)$, both greater than 1, such that the following holds:

For any monic polynomial $Q$ of degree $d$ with integer coefficients and for any prime $p > c(2c+1)$, there exists a set $S$ of at most $\big(\tfrac{2c-1}{2c+1}\big)p$ integers, such that
\[\bigcup_{s \in S} \{s,\; Q(s),\; Q(Q(s)),\; Q(Q(Q(s))),\; \dots\}\]contains a complete residue system modulo $p$ (i.e., intersects with every residue class modulo $p$).
14 replies
62861
Feb 24, 2019
john0512
2 hours ago
Parallel lines lead to similar triangles
a1267ab   29
N Dec 13, 2024 by Mathandski
Source: USA TST for EGMO 2020, Problem 2, by Andrew Gu
Let $ABC$ be a triangle and let $P$ be a point not lying on any of the three lines $AB$, $BC$, or $CA$. Distinct points $D$, $E$, and $F$ lie on lines $BC$, $AC$, and $AB$, respectively, such that $\overline{DE}\parallel \overline{CP}$ and $\overline{DF}\parallel \overline{BP}$. Show that there exists a point $Q$ on the circumcircle of $\triangle AEF$ such that $\triangle BAQ$ is similar to $\triangle PAC$.

Andrew Gu
29 replies
a1267ab
Dec 16, 2019
Mathandski
Dec 13, 2024
Parallel lines lead to similar triangles
G H J
Source: USA TST for EGMO 2020, Problem 2, by Andrew Gu
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a1267ab
223 posts
#1 • 5 Y
Y by NJOY, kat1012, Adventure10, Mango247, Rounak_iitr
Let $ABC$ be a triangle and let $P$ be a point not lying on any of the three lines $AB$, $BC$, or $CA$. Distinct points $D$, $E$, and $F$ lie on lines $BC$, $AC$, and $AB$, respectively, such that $\overline{DE}\parallel \overline{CP}$ and $\overline{DF}\parallel \overline{BP}$. Show that there exists a point $Q$ on the circumcircle of $\triangle AEF$ such that $\triangle BAQ$ is similar to $\triangle PAC$.

Andrew Gu
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a1267ab
223 posts
#2 • 11 Y
Y by NJOY, Feridimo, kat1012, Kagebaka, AllanTian, centslordm, ike.chen, Newmaths, aidan0626, Adventure10, Funcshun840
My problem. Not terribly original, though: you may notice that it is just a generalization of ELMO SL 2013 G3, USA TST 2008/7, and USA December TST 2012/1.

Solution 1

Solution 2

Solution 3
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anantmudgal09
1980 posts
#3 • 6 Y
Y by math_pi_rate, mira74, Pluto1708, Adventure10, Mango247, NCbutAN
a1267ab wrote:
Let $ABC$ be a triangle and let $P$ be a point not lying on any of the three lines $AB$, $BC$, or $CA$. Distinct points $D$, $E$, and $F$ lie on lines $BC$, $AC$, and $AB$, respectively, such that $\overline{DE}\parallel \overline{CP}$ and $\overline{DF}\parallel \overline{BP}$. Show that there exists a point $Q$ on the circumcircle of $\triangle AEF$ such that $\triangle BAQ$ is similar to $\triangle PAC$.

Andrew Gu

Fix $ABCP$ and animate $D$ on $BC$ with constant velocity. Since $D \mapsto E$ and $D \mapsto F$ are linear, $\odot(AEF)$ passes through a fixed point. So, we show for $D=B$ and $D=C$ that $Q$ lies on this circle. Both cases are analogous, so effectively, we just check $D=B$. Then $E=B$ and say $F=F'$ with $\overline{BF'} \parallel \overline{CP}$. Then $\measuredangle BQA=\measuredangle PCA=\measuredangle BF'A$ so $Q$ lies on $\odot(ABF')$. The result follows. $\blacksquare$
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alifenix-
1547 posts
#4 • 3 Y
Y by v4913, Adventure10, Mango247
anantmudgal09 wrote:
Fix $ABCP$ and animate $D$ on $BC$ with constant velocity. Since $D \mapsto E$ and $D \mapsto F$ are linear, $\odot(AEF)$ passes through a fixed point.

You can also use barycentric coordinates to show this, which is what I did.
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mira74
1010 posts
#5 • 1 Y
Y by Adventure10
length bash
Attachments:
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Al3jandro0000
804 posts
#6 • 2 Y
Y by Adventure10, Mango247
a1267ab wrote:
My problem. Not terribly original, though: you may notice that it is just a generalization of ELMO SL 2013 G3, USA TST 2008/7, and USA December TST 2012/1.

Solution 1

Solution 2

Solution 3

In Solution 1. I think the correct is $\angle QFA=\angle QEC $ following the spiral similarity.
Z K Y
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kapilpavase
595 posts
#7 • 3 Y
Y by Kagebaka, Adventure10, Mango247
Perform $\sqrt{bc}$ inversion in $ABC$. We have that $Q$ is mapped to $P$. Hence if $E\rightarrow E', F\rightarrow F'$ then we need to show $E',P,F'$ collinear. Observe that $D\rightarrow E, D\rightarrow F$ are perspectivities while $E\rightarrow E', F\rightarrow F'$ are projectivities. Hence $E'\rightarrow F'$ is a projectivity. When $D$ is to infinity along $BC$, $E,F$ are also at infinity along respective lines and hence $E',F'$ are at $A$. Hence the projectivity fixes a point, hence it is a perspectivity. When $D=B$ line $E'F'$ is line $CP$ and when $D=C$, line $E'F'$ is line $BP$. Hence $P$ is the centre of perspectivity. We are done.
This post has been edited 3 times. Last edited by kapilpavase, Dec 18, 2019, 10:26 AM
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MarkBcc168
1595 posts
#9 • 14 Y
Y by kapilpavase, AlastorMoody, Pluto1708, amar_04, SomeUser221104, riadok, Soundricio, euclides05, IAmTheHazard, Adventure10, Mango247, Aryan-23, math_comb01, Ritwin
Here is an amusing solution using Generalized Reim's theorem.

Lemma (Generalized Reim's): Let $\mathcal{C}$ be a conic and $A,B$ be points on it. Let $\omega_1$ and $\omega_2$ be circles through $A,B$ which intersect $\mathcal{C}$ again at $\{C,D\}$ and $\{E,F\}$ respectively. Then $CD\parallel EF$.

Proof: Let $I,J$ be circle points. Consider a projective transformation which takes $A,B$ to $I,J$ respectively. Denote the image with primes. Then $\mathcal{C}', \omega_1', \omega_2'$ are all circles. Moreover, since $\omega_1, \omega_2$ intersect at $A,B,I,J$, we get the other two intersection of $\omega_1', \omega_2'$ are $I',J'$. Thus by radical axis theorem, $C'D', E'F', I'J'$ are concurrent. Taking the projective transformation back, we deduce the lemma. $\blacksquare$

Back to the main problem. Let $Q$ be the point such that $\triangle BAQ\stackrel{+}{\sim}\triangle PAC$. Clearly $X=CQ\cap DF$ lie on $\odot(AQF)$ and $Y=BQ\cap DE$ lie on $\odot(AQE)$. Moreover, by converse of Pascal's theorem on $AFXQYE$, these six points lie on a conic.

Assume for the contradiction that this conic is not a circle. Then by Generalized Reim's theorem, $FX\parallel EY$ which is a contradiction as two lines intersect at $D$!
This post has been edited 1 time. Last edited by MarkBcc168, Feb 26, 2020, 12:55 PM
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math_pi_rate
1218 posts
#10 • 2 Y
Y by anantmudgal09, Adventure10
10 min solve using moving points (Although I didn't realize the much easier solution given by Anant above :P). Anyway, here goes nothing: Fix $A,B,C,D,F$ and animate $P$ on the line $\ell$ through $B$ parallel to $DF$. Then $$P \mapsto \infty_{CP} \mapsto D\infty_{CP} \mapsto E$$is a projective map. Also, it's easy to see that $P \mapsto Q$ is also a projective map (Try thinking about spiral similarities). In fact, $\angle AQC$ is fixed (since $\angle ABP$ is also fixed), amd so $Q$ moves on a fixed circle through $A$ and $C$. Thus, it suffices to prove $A,E,F,Q$ concyclic for 3 positions of $P$ (Note that 3 is sufficient, say by inversion at $A$). Taking $P=B$ gives $E=Q=C$, taking $P=C\infty_{AD} \cap \ell$ gives $E=A$, and finally $P=\infty_{\ell}$ makes $Q=A$. Thus, all three cases degenerate to given result. Hence, done. $\blacksquare$
This post has been edited 2 times. Last edited by math_pi_rate, Dec 18, 2019, 3:51 PM
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Al3jandro0000
804 posts
#11 • 1 Y
Y by Adventure10
anantmudgal09 wrote:
a1267ab wrote:
Let $ABC$ be a triangle and let $P$ be a point not lying on any of the three lines $AB$, $BC$, or $CA$. Distinct points $D$, $E$, and $F$ lie on lines $BC$, $AC$, and $AB$, respectively, such that $\overline{DE}\parallel \overline{CP}$ and $\overline{DF}\parallel \overline{BP}$. Show that there exists a point $Q$ on the circumcircle of $\triangle AEF$ such that $\triangle BAQ$ is similar to $\triangle PAC$.

Andrew Gu

Fix $ABCP$ and animate $D$ on $BC$ with constant velocity. Since $D \mapsto E$ and $D \mapsto F$ are linear, $\odot(AEF)$ passes through a fixed point. So, we show for $D=B$ and $D=C$ that $Q$ lies on this circle. Both cases are analogous, so effectively, we just check $D=B$. Then $E=B$ and say $F=F'$ with $\overline{BF'} \parallel \overline{CP}$. Then $\measuredangle BQA=\measuredangle PCA=\measuredangle BF'A$ so $Q$ lies on $\odot(ABF')$. The result follows. $\blacksquare$

How do you localized the point $Q $ in this step:
$$\measuredangle BQA=\measuredangle PCA=\measuredangle BF'A$$.

Also you have a typo using $F $ like $E $.
This post has been edited 1 time. Last edited by Al3jandro0000, Dec 18, 2019, 5:17 PM
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AlastorMoody
2125 posts
#12 • 2 Y
Y by SomeUser221104, Adventure10
MarkBcc168 wrote:
Lemma (Generalized Reim's): Let $\mathcal{C}$ be a conic and $A,B$ be points on it. Let $\omega_1$ and $\omega_2$ be circles through $A,B$ which intersect $\mathcal{C}$ again at $\{C,D\}$ and $\{E,F\}$ respectively. Then $CD\parallel EF$.
Does the converse hold? I mean the following:
Quote:
Let $\mathcal{C}$ be a conic and $A$ $\in$ $\mathcal{C}$. Let $C,D,E,F$ $\in$ $\mathcal{C}$, such that $CD||EF$. Then, $\odot (ACD)$ $\cap$ $\odot (AEF)$ $\in$ $\mathcal{C}$
If this holds, then we can use it here
This post has been edited 3 times. Last edited by AlastorMoody, Dec 27, 2019, 8:31 PM
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Pluto1708
1107 posts
#13 • 3 Y
Y by AlastorMoody, lachimolala, Adventure10
A possible proof
Quote:
Let $\mathcal{C}$ be a conic and $A$ $\in$ $\mathcal{C}$. Let $C,D,E,F$ $\in$ $\mathcal{C}$, such that $CD||EF$. Then, $\odot (ACD)$ $\cap$ $\odot (AEF)$ $\in$ $\mathcal{C}$
Let $T=\odot{AEF}\cap \odot{ACD}$.It suffices to show \[(TC,TD,TE,TF)=(AC,AD,AE,AF)\]to show that $T\in \mathcal{C}$.Now \begin{align*}M=AD\cap\odot{AEF} \\ N=AC\cap \odot{AEF} \\ K=TC\cap \odot{AEF} \\ L=TD\cap \odot{AEF} \end{align*}.Now by Reim's Theorm we have $NL\parallel CD\parallel EF\parallel KM$.Thus we have $NMEF$ and $LKFE$ are just reflections of each other in the perpendicular bisector of $EF$.In particular we have $(N,M,E,F)=(K,L,E,F)$.Thus we have \[(AC,AD,AE,AF)\overset{A}{=}(N,M,E,F)=(K,L,E,F)\overset{T}{=}(TC,TD,TE,TF)\]so we conlcude.$ T\in \mathcal{C} \blacksquare$

[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm); 
real labelscalefactor = 0.25; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -10.15829267367829, xmax = 5.7088690931314545, ymin = -10.187082943603759, ymax = 3.1611399809851473;  /* image dimensions */
pen ttttff = rgb(0.2,0.2,1); pen qqzzqq = rgb(0,0.6,0); pen ffvvqq = rgb(1,0.3333333333333333,0); pen ccqqqq = rgb(0.8,0,0); 

draw(circle((-2.7649821719223664,-0.7014638853057545), 2.721795572670584), linewidth(0.7) + ttttff); 
draw(circle((-1.9932436628718546,-4.1746288993265255), 6.1092791987257336), linewidth(0.7) + qqzzqq); 
 /* draw figures */
draw((-5.141944381050513,-2.0274434461534434)--(-0.10406669660888554,-1.2739147326685858), linewidth(0.8)); 
draw((-7.92556230736818,-5.634384984480556)--(-2.084770743483372,1.933964647595522), linewidth(0.7) + blue); 
draw((-2.084770743483372,1.933964647595522)--(-6.3517682426391415,-8.455581575489546), linewidth(0.8) + ffvvqq); 
draw((-2.084770743483372,1.933964647595522)--(3.4271343960244307,-6.992925197911663), linewidth(0.7) + blue); 
draw((-2.084770743483372,1.933964647595522)--(4.106571525019676,-3.8347068471575865), linewidth(0.8) + ffvvqq); 
draw((-4.497267715956468,1.3979069569562004)--(-6.3517682426391415,-8.455581575489546), linewidth(0.8) + ccqqqq); 
draw((-4.497267715956468,1.3979069569562004)--(4.106571525019676,-3.8347068471575865), linewidth(0.8) + ccqqqq); 
draw((-8.069941667734536,-4.804734990140489)--(4.001877157867984,-2.9991210632341323), linewidth(0.8)); 
draw((-8.069941667734536,-4.804734990140489)--(-7.92556230736818,-5.634384984480556), linewidth(0.7) + linetype("4 4")); 
draw((3.4271343960244307,-6.992925197911663)--(-7.92556230736818,-5.634384984480556), linewidth(0.7) + linetype("4 4")); 
draw((3.4271343960244307,-6.992925197911663)--(4.001877157867984,-2.9991210632341323), linewidth(0.7) + linetype("4 4")); 
draw((4.001877157867984,-2.9991210632341323)--(4.106571525019676,-3.8347068471575865), linewidth(0.7) + linetype("4 4")); 
draw((-6.3517682426391415,-8.455581575489546)--(4.106571525019676,-3.8347068471575865), linewidth(0.7) + linetype("4 4")); 
draw((-8.069941667734536,-4.804734990140489)--(-6.3517682426391415,-8.455581575489546), linewidth(0.7) + linetype("4 4")); 
draw((-7.92556230736818,-5.634384984480556)--(-6.3517682426391415,-8.455581575489546), linewidth(0.7) + linetype("4 4")); 
 /* dots and labels */
dot((-2.084770743483372,1.933964647595522),dotstyle); 
label("$A$", (-1.9986531762279593,2.1492585657340526), NE * labelscalefactor); 
dot((-0.10406669660888554,-1.2739147326685858),dotstyle); 
label("$D$", (-0.017949129353473085,-1.0586208145300553), NE * labelscalefactor); 
dot((-5.141944381050513,-2.0274434461534434),dotstyle); 
label("$C$", (-5.055826813795101,-1.812149528014913), NE * labelscalefactor); 
dot((-8.069941667734536,-4.804734990140489),dotstyle); 
label("$E$", (-7.983824100479125,-4.58944107200196), NE * labelscalefactor); 
dot((-4.497267715956468,1.3979069569562004),dotstyle); 
label("$T$", (-4.409945059379508,1.6110237703877257), NE * labelscalefactor); 
dot((4.001877157867984,-2.9991210632341323),linewidth(4pt) + dotstyle); 
label("$F$", (4.094164707092471,-2.8240309432660076), NE * labelscalefactor); 
dot((-7.92556230736818,-5.634384984480556),linewidth(4pt) + dotstyle); 
label("$N$", (-7.833118357782153,-5.472146136369935), NE * labelscalefactor); 
dot((3.4271343960244307,-6.992925197911663),linewidth(4pt) + dotstyle); 
label("$M$", (3.5128711281184373,-6.828497820642679), NE * labelscalefactor); 
dot((4.106571525019676,-3.8347068471575865),linewidth(4pt) + dotstyle); 
label("$K$", (4.201811666161737,-3.6636772240062774), NE * labelscalefactor); 
dot((-6.3517682426391415,-8.455581575489546),linewidth(4pt) + dotstyle); 
label("$L$", (-6.2614727553708756,-8.292496463984689), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */[/asy]
This post has been edited 3 times. Last edited by Pluto1708, Dec 28, 2019, 6:41 AM
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Physicsknight
640 posts
#15 • 1 Y
Y by Adventure10
This geo is giving away a big hint. $D$ could be any point on $BC$, hence we know that $\odot(AEF)$ has to pass through a fixed point. The problem is much easier when $D$ is either $B$ or $C$, so we wish to prove that $\odot(AEF)$ passes through a fixed point. Let $O$ be the circumcenter of $\triangle AEF$. $\triangle OEF$ is always similar to a fixed triangle. It's isosceles, and $\angle EOF=2\angle A$. Notice that, as $D$ moves on $BC$ with velocity $v$, $E$ and $F$ both move with constant velocity as well, and using a little complex numbers it's easy to see that $O$ also moves with constant velocity $\Longrightarrow$ $\odot(OEF)$s are coaxial. We're done.
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amar_04
1915 posts
#18 • 5 Y
Y by mueller.25, GeoMetrix, Hexagrammum16, Bumblebee60, Adventure10
Different sketch from all others, this solution mainly focusses on Isogonal Lines. :)
USATST for EGMO 2020 P2 wrote:
Let $ABC$ be a triangle and let $P$ be a point not lying on any of the three lines $AB$, $BC$, or $CA$. Distinct points $D$, $E$, and $F$ lie on lines $BC$, $AC$, and $AB$, respectively, such that $\overline{DE}\parallel \overline{CP}$ and $\overline{DF}\parallel \overline{BP}$. Show that there exists a point $Q$ on the circumcircle of $\triangle AEF$ such that $\triangle BAQ$ is similar to $\triangle PAC$.

Andrew Gu

Let $DE\cap \odot(ABC)=X$ and $DF\cap\odot(ABC)=Y$. We claim that the point $Q=XB\cap YC\in\odot(ABC)$. $XB\cap YC\in\odot(ABC)$ by Pascal's Theorem on $AFYQXE$. Now $\angle ACP=\angle AEX=\angle AQB$. So we just need to prove that $\{AP,AQ\}$ are isogonal WRT $\triangle AEF$. Let $XQ\cap PC=T$ and $PB\cap CQ=J$.

Claim:- $APBT$ and $APCJ$ are cyclic quadrilaterals.
Let $TC\cap YK=K$. So, $\angle ACT=\angle AEX=\angle AQT\implies ACQT$ is a cyclic quad. So, $\angle ATC=\angle AQY=\angle AFY$ and as $PB\cap FY$. So, by Reim's Theorem we get that $APBT$ is a cyclic quadrilateral. Similarly we get that $APCJ$ is a cyclic quadrilateral.

So, $\angle TAB=\angle TPB=\angle JPC=\angle JAC\implies \{AT,AJ\}$ are isogonal WRT $\triangle AFE$ so as $\{AB,AC\}$. So, by Isogonal Line Lemma we get that $\{AP,AQ\}$ are also Isogonal WRT $\triangle AEF$. Hence, $\triangle BAQ\sim\triangle CAP$. So, $Q=CY\cap BC$ is the desired point. $\blacksquare$
This post has been edited 3 times. Last edited by amar_04, Feb 22, 2020, 3:31 PM
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Kagebaka
3001 posts
#19
Y by
alifenix- wrote:
anantmudgal09 wrote:
Fix $ABCP$ and animate $D$ on $BC$ with constant velocity. Since $D \mapsto E$ and $D \mapsto F$ are linear, $\odot(AEF)$ passes through a fixed point.

You can also use barycentric coordinates to show this, which is what I did.

Claim: Let the reflection of $AP$ over the angle bisector of $\angle BAC$ meet $(AEF)$ at a point $Q'.$ If we fix $P$ and vary $D$ along $BC,$ then $Q'$ is fixed too.

Proof: We use barycentric coordinates with reference triangle $\triangle ABC.$ Let $P=(t_1, t_2, t_3)$ and $D=(0,d,1-d).$ Note that the point at infinity along $PC$ must have coordinates $(t_1:t_2:-t_2-t_1),$ so $E=(x_1,0,1-x_1)$ must satisfy
\begin{align*}
0&=\begin{vmatrix}
0 & d & 1-d \\
x_1 & 0 & 1-x_1 \\
t_1 & t_2 & -t_2 - t_1 \\
\end{vmatrix} \\
0&=d((1-x_1)t_1+x_1(t_2+t_1))+(1-d)(x_1t_2) \\
0&=dt_1+x_1t_2 \\
x_1&=-\frac{dt_1}{t_2}, \\
\end{align*}so $E=(-dt_1:0:t_2+dt_1).$ Similarly, $F=(x_2,1-x_2,0)$ satisfies
\begin{align*}
0&=\begin{vmatrix}
0 & d & 1-d \\
x_2 & 1-x_2 & 0 \\
t_1 & -t_1 - t_3 & t_3 \\
\end{vmatrix} \\
0&=dx_2t_3+(1-d)(x_2(t_1+t_3)+t_1(1-x_2))\\
0&=x_2t_3+(1-d)t_1 \\
x_2&=-\frac{(1-d)t_1}{t_3}, \\
\end{align*}so $F=(-(1-d)t_1:t_3+(1-d)t_1:0).$ $(AEF)$ is parametrized by $a^2yz+b^2zx+c^2xy=(x+y+z)(ux+vy+wz);$ plugging in $A$ gives us $u=0,$ and plugging in $E$ and $F,$ we get
\begin{align*}
-b^2dt_1(t_2+dt_1)&=t_2w(t_2+dt_1) \\
w&=-\frac{b^2dt_1}{t_2} \\
\end{align*}and
\begin{align*}
-c^2(1-d)t_1(t_3+(1-d)t_1)&=t_3v(t_3+(1-d)t_1) \\
v&=-\frac{c^2(1-d)t_1}{t_3}, \\
\end{align*}respectively. Finally, plugging in $Q'=(t:\tfrac{b^2}{t_2}:\tfrac{c^2}{t_3})$ gives us
\[b^2c^2\left(\frac{a^2}{t_2t_3}+\frac{t}{t_3}+\frac{t}{t_2}\right)=-\frac{b^2c^2}{t_2t_3}\left(t+\frac{b^2}{t_2}+\frac{c^2}{t_3}\right)((1-d)t_1+dt_1),\]which in particular implies that $t$ is fixed because the terms with $d$ cancel out. $\Box$
Back to the main problem: Let $BP\cap CQ' = G$ and $BQ'\cap CP = H.$ We could probably do this next step with some easy angle chasing, but instead we'll use DDIT because it's a crime not to use it when you can. By Dual of Desargues' Involution about $A$ with respect to $BQ'CP,$ we find that $\{AB,AC\},$ $\{AP,AQ'\},$ and $\{AG,AH\}$ are swapped under some involution about the pencil of lines through $A,$ which clearly must be reflection across the angle bisector of $\angle BAC.$ This implies that $\measuredangle GAB=\measuredangle GQ'B=\measuredangle CQ'H=\measuredangle CAH,$ so $A$ is the Miquel Point of $BQ'CP.$ Thus, $\measuredangle Q'GA=\measuredangle HBA=\measuredangle HPA,$ whence $\triangle BAQ'\stackrel{+}{\sim}\triangle PAC$ by AA, so $Q'$ is the point we want. $\blacksquare$
This post has been edited 2 times. Last edited by Kagebaka, May 9, 2020, 5:06 AM
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NMN12
26 posts
#20
Y by
If we construct the point $Q$ one can notice that $\angle AQB = \angle ACP$ by definition and $\angle AQC = \angle ABP$ by $\triangle ABP\sim \triangle AQC$ which follows from the spiral similarity with center $A$. Denote these two angle relations by $(1)$.

Let $\Gamma  = (AEF)$ , $R = \Gamma \cap ED$, $S = \Gamma \cap FD$, $Q' = RB \cap SC$.
We have $B = RQ' \cap FA$, $D = RE\cap FS$, $C = SQ'\cap AE$ are collinear.
From Pascal's theorem $\implies$ $A,R,F,Q',E,S$ lie on a conic but 5 of them lie on $\Gamma \implies Q'\in \Gamma$.

Then $\angle AQ'B = \angle AQ'R = \angle AER = \angle AED = \angle ACP$, $\angle AQ'C = \angle AQ'S = \angle AFS = \angle AFD = \angle ABP$ which means that $Q'$ satisfies $(1)$. But there is only one such point (It is intersection of two circles through $AB$ and $AC$ and it is $\neq A$) $\implies Q\equiv Q' \implies Q\in \Gamma$
This post has been edited 2 times. Last edited by NMN12, May 25, 2020, 8:59 AM
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amuthup
779 posts
#21
Y by
Basically the first solution in post #2.

Assume all angles are directed.

Let $G$ be the point on line $\overline{AC}$ such that $\overline{BG}\parallel\overline{CP},$ let $H$ be the point on line $\overline{AB}$ such that $\overline{CH}\parallel\overline{BP},$ and redefine $Q$ as the point such that $\triangle ABQ\sim\triangle APC.$

$\textbf{Claim: }$ $(ABG)$ and $(ACH)$ intersect at $Q.$

$\emph{Proof: }$ Let $X$ be the second intersection of $(ABG)$ and $(ACH).$ Note that $$\angle BXA=\angle BGA=\angle BGC=\angle PCA.$$Similarly, $$\angle CXA=\angle CHA=\angle CHB=\angle PBA.$$Therefore, $A$ is the center of the spiral similarity sending $\overline{XB}$ to $\overline{CP},$ so $X=Q,$ as desired. $\blacksquare$

Therefore, $Q$ is the center of the spiral similarity sending $\overline{GB}$ to $\overline{CH}.$ Hence, to show that $Q$ lies on $(AEF),$ it suffices to show that $\frac{BF}{FH}=\frac{GA}{AC}.$

Since both $\overline{DF}$ and $\overline{CH}$ are parallel to $\overline{BP},$ we have $\overline{DF}\parallel\overline{CH},$ so $\frac{BF}{FH}=\frac{BD}{DC}.$ Similarly, we can show that $\frac{GA}{AC}=\frac{BD}{DC},$ so we are done.
This post has been edited 1 time. Last edited by amuthup, Sep 19, 2020, 4:14 PM
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GeronimoStilton
1521 posts
#22
Y by
Amazing problem.

Let $M$ denote the second intersection of $PB$ and $(PAC)$, define $N$ similarly. Let $BN\cap CM=Q$. Note that $A$ is the Miquel point of $PBQCNM$, so there is a spiral similarity about $A$ taking $BQ$ to $PC$. It remains to demonstrate that $Q$ lies on $(AEF)$.

Take $D=C$, then $C=E$ and $\measuredangle DQA=\measuredangle MQA=\measuredangle MBA=\measuredangle DFA$, so $Q$ lies on $(AEF)$. Similarly, $Q$ lies on $(AEF)$ when $D=B$. Let the value of $F$ in the case $D=C$ be $F^*$ and the value of $E$ in the case $D=B$ be $E^*$. Now, consider some intermediate value of $D$, and note that by the Mean Geometry Theorem (as $E$ and $F$ move linearly with respect to $D$), $\triangle QEF\sim\triangle QCF^*\sim\triangle QE^*B$, hence $Q$ is the center of spiral similarity between $EF$ and $CF^*$, and so lies on $(AEF)$, as desired.
This post has been edited 1 time. Last edited by GeronimoStilton, Oct 18, 2020, 8:44 PM
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kevinmathz
4680 posts
#23
Y by
Let the lines parallel to $BP$ and $CP$ through $C$ and $B$ hit at $C'$ and $B'$. Let $Q$ be the intersections of the circumcircles of $BAB'$ and $CAC'$. Note that since $AQ$ is a radical axis of the two such circles, then $\triangle QB'C \sim \triangle QBC'$. Now, since $\frac{C'F}{FB} = \frac{CD}{DB} = \frac{CE}{EB}$, then $\triangle QB'E \sim \triangle QBF$ and $\triangle QCE \sim \triangle QC'F$ and $\angle EQF = \angle CQC' = 180^{\circ} - \angle A$ so $AEQF$ is cyclic. Letting $B''$ and $C''$ be where $CP$ hits $BQ$ and where $BP$ hits $CQ$ respectively, we see $\angle QB''C = \angle QBB' = \angle QAB' = \angle QAC$ so $QB''AC$ is cyclic and so is $QC''AB$. Now, $\angle B''AB = \angle B''AC' = \angle B''CC' = \angle B''PB$ so $B''BPA$ is cyclic. Finally, $$\angle BAP = \angle BB''P = \angle BB''C = \angle QB''C = \angle QAC$$which means since $\angle ABP = \angle AC'C = \angle AQC$, then $\triangle BAP \sim \triangle QAC$ and due to similarity, $\triangle BAQ = \triangle QAC$ and we are done.
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MP8148
888 posts
#24 • 2 Y
Y by Mango247, Mango247
[asy]
size(6cm);
defaultpen(fontsize(10pt)+linewidth(0.4));
dotfactor *= 1.5;

pair A = dir(110), B = dir(210), C = dir(330), D = B+dir(B--C)*abs(B-C)*0.4, P = B+dir(B--C)*abs(B-C)*0.55+dir(90)*0.4, P1 = A+dir(A--P)*abs(A-B)*abs(A-C)/abs(A-P), Q = 2*foot(P1,A,dir(270))-P1, E = extension(D,D+dir(C--P),A,C), F = extension(D,D+dir(B--P),A,B), X = extension(B,Q,D,E);

draw(D--F--B--C--E--X--B--P, heavyblue);
draw(A--B--Q--A--P--C--A, heavyred);
draw(circumcircle(A,E,F), heavygreen+dashed);

dot("$A$", A, dir(120));
dot("$B$", B, dir(180));
dot("$C$", C, dir(30));
dot("$D$", D, dir(270));
dot("$E$", E, dir(315));
dot("$F$", F, dir(200));
dot("$X$", X, dir(150));
dot("$P$", P, dir(270));
dot("$Q$", Q, dir(225));
[/asy]
Let $Q$ be the point such that $\triangle ABQ \sim \triangle APC$ and $X = \overline{BQ} \cap \overline{DE}$. Since $$\measuredangle XQA = \measuredangle BQA = \measuredangle PCA = \measuredangle XEA,$$$AXQE$ is cyclic, so it suffices to show that $AXFQ$ is cyclic, which can be done by length chasing: \begin{align*} \frac{BF}{BX} &= \frac{\sin \angle BDF}{\sin \angle BDX} \cdot \frac{\sin \angle BXD}{\sin \angle BFD} \quad (\text{law of sines}) \\ &= \frac{\sin \angle PBC}{\sin \angle PCB} \cdot \frac{\sin \angle PAB}{\sin \angle PBA} \quad (\text{angle chase}) \\ &= \frac{PC}{PA} \quad (\text{trig ceva}) \\ &= \frac{BA}{BQ} \quad (\text{similarity})\end{align*}$\blacksquare$
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SnowPanda
186 posts
#25 • 1 Y
Y by Pranav1056
bary bash
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MathForesterCycle1
79 posts
#26
Y by
dame dame
This post has been edited 2 times. Last edited by MathForesterCycle1, Oct 17, 2021, 5:16 PM
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rafaello
1079 posts
#29
Y by
Let $Q$ be the point such that $\triangle BAQ\sim\triangle PAC$. This also means that $\triangle QAC\sim\triangle BAP$. Now, let $G=DE\cap BQ$, then we have $\measuredangle AEG=\measuredangle AED=\measuredangle ACP=\measuredangle AQB=\measuredangle AQG$, which means that $AEQG$ is cyclic. Redefine $F,H$ so that $F=AB\cap (AEQG)$ and $H=QC\cap (AEQG)$. By Pascal's theorem on hexagon $FAEGQH$, we get that $D$ lies on $FH$. Therefore, $\measuredangle PBA=\measuredangle CQA=\measuredangle HQA=\measuredangle HFA=\measuredangle DFA$, thus $DF\parallel PB$. We are done.

[asy]import olympiad;
size(10cm);defaultpen(fontsize(10pt));

pair O,A,B,C,P,Q,D,E,F,G,H;
O=(0,0);A=dir(115);B=dir(195);C=dir(345);P=dir(255)*0.74;Q=-(C-P)/(P-A)*(A-B)+B;D=0.6B+0.4C;F=extension(A,B,D,B+D-P);E=extension(A,C,D,C+D-P);G=extension(E,D,B,Q);H=extension(F,D,Q,C);


draw(A--B--C--cycle,red);draw(circumcircle(A,E,F),royalblue);draw(B--Q--C,orange);draw(F--H,orange);draw(G--E,orange);draw(B--P--C,red);


dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$P$",P,dir(P));
dot("$Q$",Q,dir(Q));
dot("$F$",F,dir(F));
dot("$E$",E,dir(E));
dot("$D$",D,dir(D));
dot("$G$",G,dir(G));
dot("$H$",H,dir(H));
[/asy]
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HamstPan38825
8859 posts
#30
Y by
The key (and difficulty) of this seems to be messing around with the symmetry and getting it to work.

Denote $X \neq D = (DEF) \cap \overline{BC}$. We claim that the Miquel Point of $EFX$ with respect to $ABC$ is our desired point $Q$. Observe that
\begin{align*}
\measuredangle BQA &= \measuredangle BQF + \measuredangle FQA \\
&= \measuredangle BXF + \measuredangle FEA \\
&= \measuredangle DEF+ \measuredangle FEA \\
&= \measuredangle DEA \\
&= \measuredangle PCA,
\end{align*}and for similar reasons $\measuredangle CQA = \measuredangle PBA$. However, the point $Q'$ such that $\triangle BAQ' \sim \triangle PAC$ also satisfies $\triangle CAQ \sim \triangle PAB$ by spiral similarity lemma, and thus fulfills the above angle conditions. In fact, $$Q \neq A = (AQ'C) \cap (AQ'B)$$by our angle conditions, so $Q = Q'$ must hold. The conclusion follows naturally because $Q$ lies on $(AEF)$ by definition.
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IAmTheHazard
5001 posts
#31 • 1 Y
Y by centslordm
bary/mmp solution by someone who does not actually know bary nor mmp

Fix $P$ and note that the choice of $Q$ is uniquely determined. Now vary $D:=(0,t,1-t)$ linearly along $\overline{BC}$, so the (normalized) components of the coordinates of $E$ and $F$ vary linearly as well because $\overline{DE},\overline{DF}$ are parallel to fixed lines. Now consider the "normalized" formula for the circle $(AEF)$, which is of the form $-a^2yz-b^2zx-c^2xy+(ux+vy+wz)=0$. Since $A$ always lies on this circle, we have $u=0$, and then by plugging in $E$ and $F$ we find that both $v$ and $w$ vary linearly in terms of $t$.

We wish to prove that $Q$ always lies on this circle. Since it is fixed, this is equivalent to some linear function in $t$ being identically zero, so we only have to check it at two points. At $D=B$, we have $E=B$, and because $\measuredangle AQB=\measuredangle ACP=\measuredangle AFB$, $ABQF=AEFQ$ is cyclic. Similarly, when $D=C$, $ACQE=AEFQ$ is cyclic, so we are done. $\blacksquare$
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MathLuis
1523 posts
#32
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Ok @v_Enhance forgive me for what i'm about to show u.
Let $Q$ the inverse of $P$ down $\sqrt{bc}$ inversion, now let $CQ \cap FD=K$ and $BQ \cap DE=L$. (Note that $\triangle APB \sim \triangle ACQ$ and $\triangle APC \sim \triangle ABQ$)
Claim 1: $ALQE, AFQK$ are cyclic.
Proof: By simple angle u just need:
$$\angle AFD=\angle ABP=\angle AQK \; \text{and} \; \angle AED=\angle ACP=\angle AQL$$Claim 2: $ALFQEK$ lies in a conic $\mathcal C$
Proof: Follows from the converse of Pascal's theorem.
Finishing: Assume FTSOC that $\mathcal C$ isn't a circle, then by Degenerate 3 conics theorem over $\mathcal C, (ALQE), (AFQK)$ we get that $LE \parallel FK$, contradiction!. Hence $\mathcal C$ is a circle, so $Q$ lies in $(AEF)$ as desired, thus we are done :D.
This post has been edited 3 times. Last edited by MathLuis, Oct 7, 2023, 4:30 PM
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shendrew7
795 posts
#33
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Construct $X$ on $AB$ with $BX \parallel DE$, and $Y$ analogously.

Claim 1: $(ABC)$, $(AEF)$, and $(AXY)$ are coaxial.

Simply note there exists a spiral similarity sending $BX \mapsto FE \mapsto YC$, as
\[\frac{BF}{FY} = \frac{BD}{DC} = \frac{XE}{EC}. \quad {\color{blue} \Box}\]
Claim 2: The second concurrency point is the desired point $Q$.

Let $Q'$ be the desired point such that $\triangle BAQ' \sim \triangle PAC$. Then
\[\measuredangle BQA = \measuredangle PCA = \measuredangle BXA, \quad \measuredangle AQC = \measuredangle ABP = \measuredangle AYC,\]
so $Q' \ne A$ lies on $(ABX)$ and $(AYC)$, and thus $Q' = Q$. $\blacksquare$
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AN1729
17 posts
#34
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Solved with Janhavi Kulkarni

Let $F' \in \overline{AB}$ be such that $CF' \parallel \overline{BP}$
Define $E'$ analogously.
Thus, $\frac{BF}{FF'} = \frac{BD}{DC} = \frac{E'E}{EC}$
Consider $Q$ such that $A$ is spiral center taking $\overline{BQ}$ to $\overline{PC}$
$\implies \triangle AQB \sim \triangle APC$
$\implies \angle AQB = \angle ACP = \angle AE'B \implies (AQE'B)$ and similarly $(AQF'C)$
Now, consider spiral similarity centered at $Q$ taking $\overline{BF'}$ to $\overline{E'C}$
Note that the same spiral similarity takes $F$ to $E$
$\implies \measuredangle QFA = \measuredangle QFF' = \measuredangle QEC = \measuredangle QEA \implies (AQEF)$
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cj13609517288
1906 posts
#35
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So basically, we want $Q$, the image of $P$ under $\sqrt{bc}$ inversion, to be on $(AEF)$.

First, check this for $D=B$. Then call this $E$ $E_1$. Then $DE_1\parallel PC$, so the image of $(ADE_1)$ will be the line through $C$ and $CP\cap AB$, that is, line $CP$. Similarly, this also works for $D=C$.

Now note that $E$ and $F$ are moving in the way of the ratio thing of the Miquel point config. So basically, indeed $Q$ will be the Miquel point of self-intersecting quadrilateral $BF_1CE_1$, and the ratio thing gives that it lies on all such circles $(AEF)$. $\blacksquare$
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Mathandski
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#36
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