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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 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
postaffteff
JetFire008   9
N 3 minutes ago by drago.7437
Source: Internet
Let $P$ be the Fermat point of a $\triangle ABC$. Prove that the Euler line of the triangles $PAB$, $PBC$, $PCA$ are concurrent and the point of concurrence is $G$, the centroid of $\triangle ABC$.
9 replies
JetFire008
Mar 15, 2025
drago.7437
3 minutes ago
A scary fish and a fiend
nukelauncher   96
N 10 minutes ago by Mathandski
Source: USA November TST for IMO 2021 and TST for EGMO 2021, Problem 2, by Zack Chroman and Daniel Liu
Let $ABC$ be a scalene triangle with incenter $I$. The incircle of $ABC$ touches $\overline{BC},\overline{CA},\overline{AB}$ at points $D,E,F$, respectively. Let $P$ be the foot of the altitude from $D$ to $\overline{EF}$, and let $M$ be the midpoint of $\overline{BC}$. The rays $AP$ and $IP$ intersect the circumcircle of triangle $ABC$ again at points $G$ and $Q$, respectively. Show that the incenter of triangle $GQM$ coincides with $D$.

Zack Chroman and Daniel Liu
96 replies
+1 w
nukelauncher
Nov 16, 2020
Mathandski
10 minutes ago
Iterated FE on positive integers
MarkBcc168   61
N 15 minutes ago by pi271828
Source: ELMO 2020 P1
Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f : \mathbb{N} \to \mathbb{N}$ such that $$f^{f^{f(x)}(y)}(z)=x+y+z+1$$for all $x,y,z \in \mathbb{N}$.

Proposed by William Wang.
61 replies
MarkBcc168
Jul 28, 2020
pi271828
15 minutes ago
IMO 2023 P2
799786   89
N an hour ago by kaede_Arcadia
Source: IMO 2023 P2
Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.
89 replies
799786
Jul 8, 2023
kaede_Arcadia
an hour ago
combinatorıc
o.k.oo   0
an hour ago
A total of 3300 handshakes were made at a party attended by 600 people. It was observed
that the total number of handshakes among any 300 people at the party is at least N. Find
the largest possible value for N.
0 replies
o.k.oo
an hour ago
0 replies
Problem 2, Olympic Revenge 2013
hvaz   66
N an hour ago by MonkeyLuffy
Source: XII Olympic Revenge - 2013
Let $ABC$ to be an acute triangle. Also, let $K$ and $L$ to be the two intersections of the perpendicular from $B$ with respect to side $AC$ with the circle of diameter $AC$, with $K$ closer to $B$ than $L$. Analogously, $X$ and $Y$ are the two intersections of the perpendicular from $C$ with respect to side $AB$ with the circle of diamter $AB$, with $X$ closer to $C$ than $Y$. Prove that the intersection of $XL$ and $KY$ lies on $BC$.
66 replies
hvaz
Jan 26, 2013
MonkeyLuffy
an hour ago
Functional equation wrapped in f's
62861   35
N 2 hours ago by ihatemath123
Source: RMM 2019 Problem 5
Determine all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
\[f(x + yf(x)) + f(xy) = f(x) + f(2019y),\]for all real numbers $x$ and $y$.
35 replies
62861
Feb 24, 2019
ihatemath123
2 hours ago
JBMO Shortlist 2023 C1
Orestis_Lignos   6
N 2 hours ago by zhenghua
Source: JBMO Shortlist 2023, C1
Given is a square board with dimensions $2023 \times 2023$, in which each unit cell is colored blue or red. There are exactly $1012$ rows in which the majority of cells are blue, and exactly $1012$ columns in which the majority of cells are red.

What is the maximal possible side length of the largest monochromatic square?
6 replies
Orestis_Lignos
Jun 28, 2024
zhenghua
2 hours ago
Number Theory
MuradSafarli   6
N 2 hours ago by krish6_9
Find all natural numbers \( a, b, c \) such that

\[
2^a \cdot 3^b + 1 = 5^c.
\]
6 replies
1 viewing
MuradSafarli
6 hours ago
krish6_9
2 hours ago
Equilateral triangle geo
MathSaiyan   1
N 2 hours ago by ricarlos
Source: PErA 2025/3
Let \( ABC \) be an equilateral triangle with circumcenter \( O \). Let \( X \) and \( Y \) be two points on segments \( AB \) and \( AC \), respectively, such that \( \angle XOY = 60^\circ \). If \( T \) is the reflection of \( O \) with respect to line \( XY \), prove that lines \( BT \) and \( OY \) are parallel.
1 reply
MathSaiyan
Yesterday at 1:47 PM
ricarlos
2 hours ago
IMO 2009, Problem 5
orl   86
N 3 hours ago by Ilikeminecraft
Source: IMO 2009, Problem 5
Determine all functions $ f$ from the set of positive integers to the set of positive integers such that, for all positive integers $ a$ and $ b$, there exists a non-degenerate triangle with sides of lengths
\[ a, f(b) \text{ and } f(b + f(a) - 1).\]
(A triangle is non-degenerate if its vertices are not collinear.)

Proposed by Bruno Le Floch, France
86 replies
orl
Jul 16, 2009
Ilikeminecraft
3 hours ago
Diagonals BD,CE concurrent with diameter AO in cyclic ABCDE
WakeUp   10
N 3 hours ago by zhenghua
Source: Romanian TST 2002
Let $ABCDE$ be a cyclic pentagon inscribed in a circle of centre $O$ which has angles $\angle B=120^{\circ},\angle C=120^{\circ},$ $\angle D=130^{\circ},\angle E=100^{\circ}$. Show that the diagonals $BD$ and $CE$ meet at a point belonging to the diameter $AO$.

Dinu Șerbănescu
10 replies
WakeUp
Feb 5, 2011
zhenghua
3 hours ago
Parallel lines in two-circle configuration
Tintarn   3
N 3 hours ago by zhenghua
Source: Francophone 2024, Senior P3
Let $ABC$ be an acute triangle, $\omega$ its circumcircle and $O$ its circumcenter. The altitude from $A$ intersects $\omega$ in a point $D \ne A$ and the segment $AC$ intersects the circumcircle of $OCD$ in a point $E \ne C$. Finally, let $M$ be the midpoint of $BE$. Show that $DE$ is parallel to $OM$.
3 replies
Tintarn
Apr 4, 2024
zhenghua
3 hours ago
IMO Shortlist 2013, Algebra #5
lyukhson   33
N 3 hours ago by HamstPan38825
Source: IMO Shortlist 2013, Algebra #5
Let $\mathbb{Z}_{\ge 0}$ be the set of all nonnegative integers. Find all the functions $f: \mathbb{Z}_{\ge 0} \rightarrow \mathbb{Z}_{\ge 0} $ satisfying the relation
\[ f(f(f(n))) = f(n+1 ) +1 \]
for all $ n\in \mathbb{Z}_{\ge 0}$.
33 replies
lyukhson
Jul 9, 2014
HamstPan38825
3 hours ago
Variable point on the median
MarkBcc168   47
N Yesterday at 2:36 AM by HamstPan38825
Source: APMO 2019 P3
Let $ABC$ be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $BC$. A variable point $P$ is selected in the line segment $AM$. The circumcircles of triangles $BPM$ and $CPM$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $DP$ and $EP$ intersect (a second time) the circumcircles to triangles $CPM$ and $BPM$ at $X$ and $Y$, respectively. Prove that as $P$ varies, the circumcircle of $\triangle AXY$ passes through a fixed point $T$ distinct from $A$.
47 replies
MarkBcc168
Jun 11, 2019
HamstPan38825
Yesterday at 2:36 AM
Variable point on the median
G H J
G H BBookmark kLocked kLocked NReply
Source: APMO 2019 P3
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MarkBcc168
1593 posts
#1 • 3 Y
Y by ike.chen, Adventure10, Rounak_iitr
Let $ABC$ be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $BC$. A variable point $P$ is selected in the line segment $AM$. The circumcircles of triangles $BPM$ and $CPM$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $DP$ and $EP$ intersect (a second time) the circumcircles to triangles $CPM$ and $BPM$ at $X$ and $Y$, respectively. Prove that as $P$ varies, the circumcircle of $\triangle AXY$ passes through a fixed point $T$ distinct from $A$.
Z K Y
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Anaskudsi
112 posts
#2 • 6 Y
Y by sub_math, Ali3085, JustinLee2017, Nahul2005, Muaaz.SY, Adventure10
-We claim that the fixed point $T$ is the intersection of the circumcircle of the triangle $(ABC) $ and the circumcircle of the triangle $(AXY) $.
$\textcolor{red}{\textbf{\underline{Claim.1:}}} BD \parallel CX$ and $CE \parallel BY$.

$\textcolor{blue}{\textbf{The proof:}} $
We have:
$$\angle{BDX} =\angle{BDP} =\angle{BMP} =\angle{CXP} = \angle{CXD} $$So $BD \parallel CX$. In the same way we can prove that $CE \parallel BY$

$\textcolor{red}{\textbf{\underline{Claim.2:}}}  BCXY$ is a cyclic quadrilateral.

$\textcolor{blue}{\textbf{The proof:}}  $
Let $BD \cap CE=Z$ by radical axis theorem on the quadrilaterals $(BDMP), (ECMP)$ and $(BDCE)$ we can prove that $P, M $ and $Z$ are collinear.
Let $BY \cap CX =V $. From Claim.1 we know that $BVCZ$ is a parallelogram, so $Z, M$ and $V$ are collinear, so $P, M$ and $V$ are collinear.
We also know that :
$$VY \cdot VB= VP \cdot VM= VX \cdot VC$$So $BCXY$ is a cyclic quadrilateral.

We will prove that the fixed point $T$ is the intersection of the circumcircle of the triangle $(ABC) $ and the circumcircle of the triangle $(AXY) $.

By using radical axis theorem in the quadrilaterals $(BCXY)$, $(ACBT)$ and $(AXYT)$ we conclude that $BC, XY$ and $AT$ are concurrent. So it is enough to prove that as $P$ varies, the line $XY$ passes through a fixed point $K$ on $BC$.


$\textcolor{red}{\textbf{\underline{Claim.3:}}}  XY$ passes through a fixed point $K$ on $BC$.

$\textcolor{blue}{\textbf{The proof:}}  $
We will use barycentric coordinates with respect to the triangle $ABC$.
We know that $M=(0:1:1)$. Let $P=(m:1:1)$.
Let us find the equation of $(PMB)$ it is easy to prove that $v=0$ and that $w=\frac{a^2}{2} $ (since $B, M$ lies on that circle). Now since $P$ lies on that circle so we have :
$$(um+\frac{a^2}{2} )(m+2)=a^2+b^2m+c^2 \Longrightarrow u=\frac{2b^2+2c^2-a^2}{2(m+2)}=L$$So the equation of $(PMB) $ is
$$ -\sum a^2yz + (\sum x)(Lx+\frac{a^2}{2}z)=0$$Now we will find the coordinates of $D=(x:y:z)$. Since $D \in (ABC) $ so :
$$\sum a^2yz=0$$Since $D \in (PMB) $ so :
$$- \sum a^2yz + (\sum x)(Lx+\frac{a^2}{2}z)=0 \Longrightarrow (\sum x)(Lx+\frac{a^2}{2}z)=0$$$$\Longrightarrow - 2Lx = a^2z \Longrightarrow D=(a^2:t:-2L)$$Now since $D \in (ABC) $ so :
$$-2Lta^2-2La^2b^2+a^2c^2t=0 \Longrightarrow t=\frac{2Lb^2}{c^2-2L}$$$$\Longrightarrow D=(a^2(c^2-2L):2Lb^2:-2L(c^2-2L))$$
Let $P_\infty$ be a point at infinity along the line $BD$, it is easy to prove that $P_\infty=(a^2(c^2-2L):(c^2-2L)(2L-a^2):-2L(c^2-2L))$.
Let $X=(x:y:z) $, Since $BD \parallel CX$ so $C, X$ and $P_\infty$ are collinear, we have :
\[\begin{vmatrix}
x & y & z \\
a^2(c^2-2L) & (c^2-2L)(2L-a^2) & -2L(c^2-2L) \\
0 & 0 & 1
\end{vmatrix}=0.\]$$\Longrightarrow (2L-a^2)x=a^2y$$So $X=(a^2:2L-a^2:t_{1})$ for some real number $t_{1}$.
Similarly $Y=(a^2:t_{2}:2L-a^2)$.
Let $XY \cap BC =K=(0:y:z) $, we need to prove that $\frac{z}{y}$ does not depend on $m$.
Since $X, Y$ and $K$ are collinear, we have :
\[\begin{vmatrix}
a^2 & 2L-a^2 & t_{1} \\
a^2 & t_{2} & 2L-a^2 \\
0 & y & z
\end{vmatrix}=0.\]
\[\Longleftrightarrow 
\begin{vmatrix}
1 & 2L-a^2 & t_{1} \\
1 & t_{2} & 2L-a^2 \\
0 & y & z
\end{vmatrix}=0.\]$$\Longleftrightarrow (2L-a^2-t_{1})y=-(2L-a^2-t_{2})z \Longleftrightarrow \frac{z}{y} =\frac{2L-a^2-t_{1}}{-(2L-a^2-t_{2})}$$So we need to prove that $\frac{2L-a^2-t_{1}}{2L-a^2-t_{2}}$ does not depend on $m$.
Since $X, P$ and $D$ are collinear, we have:
\[\begin{vmatrix}
-a^2 & -(2L-a^2) & -t_{1} \\
a^2(c^2-2L) & 2Lb^2 & -2L(c^2-2L) \\
m & 1 & 1
\end{vmatrix}=0.\]
Now we will add $(2L-a^2)$ times
the third row to the first row. This gives:

\[\begin{vmatrix}
-a^2+2Lm-a^2m & 0 & 2L-a^2-t_{1} \\
a^2(c^2-2L) & 2Lb^2 & -2L(c^2-2L) \\
m & 1 & 1
\end{vmatrix}=0.\]
$$\Longrightarrow (2L-a^2-t_{1})(a^2(c^2-2L) - 2Lb^2m) +(2Lm-a^2-a^2m)(2Lb^2-2L(2L-c^2))=0$$$$-(2L-a^2-t_{1})=\frac{(a^2+a^2m-2Lm)(2Lb^2-2L(2L-c^2))}{a^2(2L-c^2)+ 2Lb^2m} $$
Let $d=a^2+a^2m-2Lm$, it is easy to see that $d$ is symmetric with respect to $b$ and $c$. Now we have (remember that $L=\frac{2b^2+2c^2-a^2}{2(m+2)}$) :
$$-(2L-a^2-t_{1})=\frac{d(2Lb^2-2L(2L-c^2))}{a^2(2L-c^2)+ 2Lb^2m} $$$$=\frac{d(\frac{2b^4+2b^2c^2-a^2b^2}{m+2} - (\frac{2b^2-a^2-c^2m}{m+2})(\frac{2b^2+2c^2-a^2}{m+2}))} {\frac{2a^2b^2-a^4-a^2c^2m+2b^4m+2b^2c^2m-a^2b^2m}{m+2}}$$
$$=\frac{d(m(2b^4+2c^4+4b^2c^2-a^2b^2-a^2c^2)-(2b^2+2c^2-a^2)(2b^2-a^2)+2b^2(2b^2+2c^2-a^2))}{(m+2)(m(2b^2-a^2)(b^2+c^2)+a^2(2b^2-a^2))}$$$$=\frac{d(m(b^2+c^2)(2b^2+2c^2-a^2)+a^2(2b^2+2c^2-a^2)) }{(m+2)(2b^2-a^2)(m(b^2+c^2)+a^2)}$$$$=\frac{d(2b^2+2c^2-a^2)}{(m+2)(2b^2-a^2)}$$$$\Longrightarrow -(2L-a^2-t_{1})=\frac{d(2b^2+2c^2-a^2)}{(m+2)(2b^2-a^2)}$$Similarly we can prove that :
$$-(2L-a^2-t_{2})=\frac{d(2b^2+2c^2-a^2)}{(m+2)(2c^2-a^2)}$$So we have :
$$\frac{-(2L-a^2-t_{1})}{-(2L-a^2-t_{2})} =\frac{\frac{d(2b^2+2c^2-a^2)}{(m+2)(2b^2-a^2)}}{\frac{d(2b^2+2c^2-a^2)}{(m+2)(2c^2-a^2)}}=\frac{2c^2-a^2}{2b^2-a^2}$$So $\frac{z}{y} =\frac{2c^2-a^2}{a^2-2b^2}$ and it does not depend on $m$.
As desired.
$\blacksquare$
This post has been edited 1 time. Last edited by Anaskudsi, Jun 11, 2019, 5:18 PM
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MarkBcc168
1593 posts
#3 • 25 Y
Y by Supercali, p_square, huricane, MoriartyBS, brokendiamond, Jerry37284, Reef334, rashah76, Godfather2043, Pluto04, aops29, myh2910, AllanTian, Wizard0001, Pitagar, ChanandlerBong, guptaamitu1, sabkx, Adventure10, Mango247, ZVFrozel, khina, Rounak_iitr, Funcshun840, GeoKing
[asy]
size(10cm,0);
defaultpen(fontsize(10pt));
pair A = (1,4);
pair B = (0,0);
pair C = (4,0);
pair M = (B+C)/2;
pair O = circumcenter(A,B,C);
pair P = 0.85*M + 0.15*A;
pair Ob = circumcenter(B,P,M);
pair Oc = circumcenter(C,P,M);
pair E = 2*foot(B,O,Ob) - B;
pair F = 2*foot(C,O,Oc) - C;
pair X = 2*foot(Oc,P,E) - P;
pair Y = 2*foot(Ob,P,F) - P;
pair D = 2*foot(O,A,M) - A;
pair C1 = 2*foot(O,P,E) - E;
pair B1 = 2*foot(O,P,F) - F;
pair K = extension(B,E,C,F);
pair L = 2*M-K;
pair T = extension(X,Y,B,C);
pair Q = 2*foot(O,A,T)-A;

draw(A--B--C--cycle,linewidth(1.5));
draw(C1--E--F--B1);
draw(B--B1,blue);
draw(C--C1,blue);
draw(A--K,blue);
draw(Y--B--K--C--X--L--cycle);
draw(X--Y);
draw(circumcircle(A,B,C));
draw(circumcircle(B,P,M),red);
draw(circumcircle(C,P,M),red);
draw(B--T--Y);
draw(T--C1,dashed);
draw(A--T,dashed);
draw(circumcircle(A,X,Y),dashed);

dot("$A$",A,N);
dot("$B$",B,SW);
dot("$C$",C,SE);
dot("$D$",E,S);
dot("$E$",F,S);
dot("$M$",M,1.5*dir(55));
dot("$P$",P,dir(200));
dot("$B_1$",B1,NW);
dot("$C_1$",C1,N);
dot("$X$",X,2*dir(100));
dot("$Y$",Y,1.5*N);
dot("$K$",K,S);
dot("$L$",L,dir(150));
dot("$Q$",T,W);
dot("$T$",Q,NW);
[/asy]
The proof proceed in five steps.

Step 1 : Introducing $K,L$.

First, by Radical Axis theorem on $\omega_B, \omega_C$ and $\Omega$, lines $BD, CE, AM$ are concurrent at $K$. Moreover, by Reim's theorem on $\overline{XPD}$ and $\overline{CMB}$ gives $CX\parallel BD$. Similarly $BY\parallel CE$. Thus $BY, CX$ meet at point $L$ such that $BLCK$ is parallelogram.

Step 2 : $B,C,X,Y$ are concyclic.

Clearly $L$ is reflection of $K$ across $M$ so $L\in AM$. Thus $LB\cdot LY = LP\cdot LM = LC\cdot LX$ or $B,C,X,Y$ are concyclic.

Step 3 : $XY\parallel DE$

Notice that reflections $X_1, Y_1$ of $X,Y$ across $M$ lies on $BD, CE$ respectively. By reflection, $B,C,X_1, Y_1$ are concyclic hence Reim's theorem on $\odot(BCX_1Y_1)$ and $\odot(BCDE)$ gives $X_1Y_1\parallel DE$. Hence $XY\parallel DE$ as claimed.

Step 4 : Key construction.

Now let $B_1, C_1$ be points on $\Omega$ such that $BB_1\parallel CC_1\parallel AM$. By Reim's theorem on $PDB_1$ and $BMC$ gives $P, E, B_1$ are colinear. Similarly $P, D, C_1$ are colinear. As $XY\parallel DE$, applying Reim's theorem again gives $B_1, C_1, X, Y$ are concyclic.

Step 5 : Conclusion.

Now let $Q=B_1C_1\cap BC$, which is a fixed point. By Radical Axis on $\odot(BCXY)$, $\odot(B_1C_1XY)$, $\Omega$, we get $Q\in XY$. By Radical Axis on $\odot(AXY)$, $\odot(ABC)$, $\odot(BCXY)$, we get $\odot(AXY)$ pass through $AQ\cap\Omega$, which is fixed point so we are done.
This post has been edited 3 times. Last edited by MarkBcc168, Jun 12, 2020, 4:05 AM
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juckter
322 posts
#4 • 36 Y
Y by 62861, TLP.39, p_square, Cycle, e_plus_pi, AlastorMoody, GorgonMathDota, Kayak, Pathological, anantmudgal09, Pluto1708, lminsl, RAMUGAUSS, khina, SpecialBeing2017, Pure_IQ, amar_04, rashah76, Kagebaka, FAA2533, aops29, Aryan-23, tapir1729, opptoinfinity, myh2910, rcorreaa, tigerzhang, Nuterrow, Quidditch, CyclicISLscelesTrapezoid, jrsbr, sabkx, kamatadu, Adventure10, Funcshun840, giangtruong13
I neglected to mention this on the results thread out of fear that the AoPS community would hate me, but I'm the author of this problem :)
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MarkBcc168
1593 posts
#5 • 6 Y
Y by amar_04, aops29, DapperPeppermint, k12byda5h, Adventure10, Mango247
juckter wrote:
I neglected to mention this on the results thread out of fear that the AoPS community would hate me, but I'm the author of this problem :)

Unfortunately, I did not solve this in the exam but I found the solution quickly afterward with the aid of Geogebra. In my opinion, this problem is a reasonable APMO P3 only if it's given in the statement that the fixed point lies on $\odot(ABC)$ as it's nearly impossible to try degenerate cases or claim the point using compass (Diagram has too many steps to construct).

Apart from the difficulty to guess the point, the configuration in this problem is rich and nice. Congrats on finding it.
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khina
993 posts
#6 • 8 Y
Y by danepale, Leartia, amar_04, aops29, Wizard0001, tigerzhang, Fakesolver19, Adventure10
this problem gives me crippling depression
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IndoMathXdZ
691 posts
#7 • 5 Y
Y by GorgonMathDota, huricane, Pure_IQ, Adventure10, Mango247
Quite nice but unfortunately didn't realize this at the exam due to difficulty of constructing the diagram :(
Claim 01. $BY, PM, CX$ concur.
Proof. By radical axis on $BDPM$, $PMEC$ and $BDCE$ gives us that $BD, PM, \text{and} EC$ concur.
Reim Theorem gives us that $BD \parallel XC$, and $BY \parallel EC$. So, we must have $BY, PM, XC$ concur.
Let this three lines concur at point $G$, now notice that this gives us that $BYXC$ is cyclic.

Radical axis again on $AZBC$, $BYXC$, $AZYX$ gives us that $AZ, XY, BC$ concur.

Therefore, to prove that $Z$ is fixed as $P$ varies, we just need to prove that $XY \cap BC$ is fixed as $P$ varies.
By Menelaos, it suffices to prove that $\frac{CX}{XG} \cdot \frac{GY}{BY}$ is constant.

Claim 02. $\frac{CX}{XG} \cdot \frac{GY}{BY} $ is constant.
Proof.
\[ \frac{CX}{XG} \cdot \frac{GY}{BY} = \frac{CX}{BY} \cdot \frac{GY}{GX} = \frac{CX}{BY} \cdot \frac{GC}{GB} = \frac{CX}{BY} \cdot \frac{BE}{CD} \]Let $R_1$ and $R_2$ be radii of circumcircle $CPM$ and $BPM$ respectively.
But, we know that
\[ \frac{CX}{BY}  = \frac{R_1 \cdot \sin \angle XPC}{R_2 \cdot \sin \angle YPB} = \frac{PC \sin \angle DPC}{PB \sin \angle BPE} \]So,
\[ \frac{CX}{BY} \cdot \frac{BE}{CD} = \frac{PC \cdot BE \cdot \sin DPC}{PB \cdot CD \cdot \sin BPE} = \frac{\sin (\angle AMC - \angle BAC)}{\sin (180^{\circ} - \angle BAC - \angle AMC)}\]
This post has been edited 1 time. Last edited by IndoMathXdZ, Jun 11, 2019, 1:01 AM
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MessingWithMath
340 posts
#8 • 3 Y
Y by Adventure10, Mango247, Mango247
Rip tried inversion on this for way to long but any inversion doesn't give anything nice. Does anyone have a inversion solution somehow?
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jbaca
225 posts
#10 • 2 Y
Y by Adventure10, Mango247
Solution. Basically, the official solution, but anyway...
Let $T$ be the second intersection point of $(AXY)$ and $\Gamma$. We show that $T$ is the fixed point we need.
By the radical axis theorem, $CE$, $MP$ and $DB$ concur at a point, say $Q$. By Reim's theorem, $BD\parallel CY$ and $CE\parallel BX$. Let $R=\overline{BX}\cap \overline{CY}$. The previous parallelisms imply that $QCRB$ is a paralelogram, therefore, $R$ lies on $MP$. Since $BD\parallel YR$, we obtain that $PXRY$ is cyclic; then, the radical axes of $(PXRY)$, $(AXTY)$ and $(ABC)$ are concurrent, i.e. $AT,\ XY$ and $BC$ meet at a common point $S$; hence, it suffices to show that $S$ is a fixed point. Notice that
$$SB\cdot SC=SA\cdot ST=XS\cdot SY$$i.e. $BXCY$ is cyclic. Let $L=\overline{EX}\cap \Gamma,\ L\neq E$ and $K=\overline{PY}\cap \Gamma,\ K\neq D$. Using directed angles, we get
$$\measuredangle LPM=\measuredangle ECM=\measuredangle ECB=\measuredangle PLB$$in other words, $BL\parallel PM$. Similarly, we prove that $PM\parallel KC$. Moreover, note that
\begin{eqnarray*}
\measuredangle KLP&=&\measuredangle KLB-\measuredangle PLB=\measuredangle KCB-\measuredangle LPM=\measuredangle AMB-\measuredangle XBC\\
&=&\measuredangle PYC-\measuredangle XYC= \measuredangle PYX
	\end{eqnarray*}which gives that $KYLX$ is cyclic; therefore, the radical axis theorem applied to $(KYLX),\ (AXTY)$ and $(AKTL)$ leads to infer that $XY$, $AT$ y $KL$ are concurrent, i.e. $S$ coincides with the intersection point of $KL$ and $BC$. Since $K$ and $L$ are fixed, we conclude that $S$ is fixed, as required. $\blacksquare$
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rocketscience
466 posts
#11 • 2 Y
Y by Adventure10, Mango247
Define $U = DP \cap \Gamma$ and $V = EP \cap \Gamma$. Observe that
\[\measuredangle BMP = \measuredangle BDP = \measuredangle BCU\]so $AM \parallel UC$. Similarly, $AM \parallel VB$ and it follows that $U, V$ are fixed points.

By Reim's, $BD \parallel CX$ and $BY \parallel CE$, and by radical axis, we have a point $R = BD \cap PM \cap CE$. Define $R' = BY \cap CX$. Since $BRCR'$ is a parallelogram, we have $R' \in RM$. Hence $R'$ is on the radical axis of $\odot(BMP), \odot(CMP)$ and thus $BCXY$ is cyclic.

Now, as $BC, XY$ antiparallel in $\angle BR'C$ and $BC, DE$ antiparallel in $\angle BRC$, we have $DE \parallel XY$. Then since $UV, DE$ antiparallel in $\angle DPE$ we have $UVYX$ cyclic. Hence by radical axis we have a point $Z = UV \cap XY \cap BC$, which is fixed. Finally, if we define the fixed point $T = AZ \cap \Gamma$, then $ZT \cdot ZA = ZX \cdot ZY$ so $T \in \odot(AXY)$ is the desired fixed point. $\Box$
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p_square
442 posts
#12 • 3 Y
Y by guptaamitu1, Adventure10, Mango247
By proof is almost the same as MarkBcc168's so I won't write down the whole proof.

Alternative method to his Step 3 and 4
This post has been edited 2 times. Last edited by p_square, Jun 11, 2019, 4:57 PM
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AlastorMoody
2125 posts
#13 • 5 Y
Y by Pluto1708, Aryan-23, DPS, aops29, Adventure10
The fixed point part was tough!
APMO 2019 P3 wrote:
Let $ABC$ be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $BC$. A variable point $P$ is selected in the line segment $AM$. The circumcircles of triangles $BPM$ and $CPM$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $DP$ and $EP$ intersect (a second time) the circumcircles to triangles $CPM$ and $BPM$ at $X$ and $Y$, respectively. Prove that as $P$ varies, the circumcircle of $\triangle AXY$ passes through a fixed point $T$ distinct from $A$.
Solution: Let $G$ be the radical center WRT $\odot (ABC),$ $\odot (BMP)$ and $\odot (CMP)$. Apply Reim's Theorem on $\odot (BMP)$ and $\odot (CMP)$ $\implies$ $BY||CE$ and $XC||BD$. If $G'$ is the reflection of $G$ over $M$, then $BG'CG$ forms a parallelogram $\implies$ $BY$ $\cap$ $CX$ $\cap$ $AM$ $=$ $G'$
$$G'Y \cdot G'B = G'P \cdot G'M = G'X \cdot G'C \implies BYCX \text{ is cyclic}$$$\angle YPD=180^{\circ}-\angle YBD=\angle DGE$ $\implies$ $PDGE$ is cyclic and, $\angle BYP$ $=$ $\angle PMC$ $=$ $\angle G'XP$ $\implies$ $G'XYP$ is cyclic $\implies$ $\angle PED$ $=$ $\angle BGG'$ $=$ $\angle PG'X$ $=$ $\angle PYX$ $\implies$ $XY||DE$. Let $DP, EP$ $\cap$ $\odot (ABC)$ $=$ $K, N$. Apply Reim's Theorem on $\odot (BMP)$, $\odot (CMP)$ and $\odot (ABC)$ $\implies$ $CK|| AM || BN$ $\implies$ $N, K$ are fixed points. Since, $XY$ $,$ $KN$ are antiparallel WRT $\angle NPK$ $\implies$ $NYXK$ is cyclic. Let $\odot (AXY)$ $\cap$ $\odot (ABC)$ $=$ $T$. Apply Radical Axes Theorem on $\odot (ABC)$, $\odot (BYXC)$, $\odot (AXYT)$ and $\odot (NYXK)$, $\odot (AXYT)$, $\odot (ABC)$ $\implies$ $AT$ $\cap$ $NK$ $\cap$ $XY$ $\cap$ $BC$ $=$ $F$ is a fixed point $\implies$ $T$ is the desired fixed point
This post has been edited 1 time. Last edited by AlastorMoody, Jun 12, 2019, 8:44 AM
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zuss77
520 posts
#14 • 1 Y
Y by Adventure10
MarkBcc168 wrote:
In my opinion, this problem is a reasonable APMO P3 only if it's given in the statement that the fixed point lies on $\odot(ABC)$ as it's nearly impossible to try degenerate cases


It somewhat natural to consider the case when $D$, $P$, $E$ are collinear. In this case $Y=D$, $X=E$, ($AXY$) coincide with circumcircle.
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v_Enhance
6858 posts
#15 • 10 Y
Y by e_plus_pi, AlastorMoody, Imayormaynotknowcalculus, Tuleuchina, v4913, CrazyMathMan, Pitagar, Chokechoke, Adventure10, Rounak_iitr
This solution is divided into three main parts.

Part I: the parallelogram. First, we let $K$ denote the concurrence point of lines $BD$, $PM$, $CE$ (by radical axis). Note also that $\measuredangle BYP = \measuredangle BMP = \measuredangle PMC = \measuredangle PEC$, which implies $\overline{BY} \parallel \overline{CE}$. Similarly $\overline{CX} \parallel \overline{BD}$. Thus we may construct parallelogram $BLCK$ with $L$ the concurrence point of lines $BY$, $AM$, $CX$.

Note now that:
  • $BCDE$ is cyclic.
  • $BYXC$ is cyclic, by power of a point from $L$ ($LY \cdot LB = LP \cdot LM = LX \cdot LC$).
  • Thus $\overline{XY} \parallel \overline{DE}$; say bay $\measuredangle PYX = \measuredangle PLX = \measuredangle PKD = \measuredangle PED$.



[asy] size(11cm); pair A = dir(119); pair B = dir(210); pair C = dir(330); pair M = (B+C)/2; draw(A--B--C--cycle, deepcyan); filldraw(unitcircle, invisible, deepcyan);

pair O = origin; pair S = extension(O, foot(O, A, M), B, C); pair T = -A+2*foot(O, A, S);

pair K = 1.75*M-0.75*A; pair L = B+C-K; pair D = -B+2*foot(O, B, K); pair E = -C+2*foot(O, C, K);

filldraw(B--L--C--K--cycle, invisible, orange); draw(circumcircle(B, D, M), dashed+deepgreen); draw(circumcircle(C, E, M), dashed+deepgreen); pair P = -M+2*foot(circumcenter(B, M, D), K, M); pair X = extension(D, P, C, L); pair Y = extension(E, P, B, L);

draw(A--K, red); draw(A--S, lightblue); draw(B--S, deepcyan); pair Bp = extension(E, Y, B, B+A-M); pair Cp = extension(D, X, C, C+A-M); draw(D--Cp, orange); draw(E--Bp, orange); draw(C--E, orange); draw(B--Bp, red); draw(C--Cp, red);

draw(S--Cp, blue); draw(S--X, blue);

dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$O$", O, dir(135)); dot("$S$", S, dir(S)); dot("$T$", T, dir(T)); dot("$K$", K, dir(K)); dot("$L$", L, dir(45)); dot("$D$", D, dir(D)); dot("$E$", E, dir(300)); dot("$P$", P, dir(P)); dot("$X$", X, dir(80)); dot("$Y$", Y, dir(100)); dot("$B'$", Bp, dir(Bp)); dot("$C'$", Cp, dir(Cp));

/* TSQ Source:

!size(11cm); A = dir 119 B = dir 210 C = dir 330 M = (B+C)/2 A--B--C--cycle deepcyan unitcircle 0.1 lightcyan / deepcyan

O = origin R135 S = extension O foot O A M B C T = -A+2*foot O A S

K = 1.75*M-0.75*A L = B+C-K R45 D = -B+2*foot O B K E = -C+2*foot O C K R300

B--L--C--K--cycle 0.1 yellow / orange circumcircle B D M dashed deepgreen circumcircle C E M dashed deepgreen P = -M+2*foot circumcenter B M D K M X = extension D P C L R80 Y = extension E P B L R100

A--K red A--S lightblue B--S deepcyan B' = extension E Y B B+A-M C' = extension D X C C+A-M D--Cp orange E--Bp orange C--E orange B--Bp red C--Cp red

S--Cp blue S--X blue

*/ [/asy]



Part II: the isosceles trapezoid. Let line $EP$ meet $\Gamma$ again at $B'$. Define $C'$ similarly. We apply Reim's theorem three times:
  • Since $\measuredangle BB'E = \measuredangle ECB = \measuredangle MPE$, we have $\overline{BB'} \parallel \overline{PM} \equiv \overline{AM}$.
  • Similarly, $\overline{CC'} \parallel \overline{AM}$.
  • Since $\measuredangle XYP = \measuredangle PED = \measuredangle BCX$ (using $\overline{XY} \parallel \overline{DE}$) we get $B'C'XY$ cyclic as well.

Part III: finishing. We now describe the fixed point. By radical axis now, lines $B'C'$, $YX$, $BC$ meet at $S$.

Since $BB'C'C$ was an isosceles trapezoid with legs parallel to $\overline{AM}$, we could also have described $S$ as the point on $BC$ such that $\overline{SO} \perp \overline{AM}$, where $O$ is the center of $\Gamma$. This is independent of $P$.

Consequently, if $\overline{AS}$ meets $\Gamma$ again at $T$, then $ATXY$ is cyclic and moreover $T$ does not depend on $P$.

Remark: In my opinion, Part II is the toughest part, as it requires the introduction of the magical points $B'$ and $C'$. Part I is routine: working nearly only with points already given or which are natural to add. Meanwhile Part III is mostly the identification of $S$, which can eventually be done with a good figure. Parts I and III also motivate each other.
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rocketscience
466 posts
#16 • 3 Y
Y by Blast_S1, Adventure10, Mango247
MarkBcc168 wrote:
In my opinion, this problem is a reasonable APMO P3 only if it's given in the statement that the fixed point lies on $\odot(ABC)$ as it's nearly impossible to try degenerate cases or claim the point using compass (Diagram has too many steps to construct).

Apart from the difficulty to guess the point, the configuration in this problem is rich and nice. Congrats on finding it.

My thoughts on finding the point. It's reasonable to guess that the fixed point is the second intersection of $(AXY)$ with the circumcircle by considering the isosceles case, where the diagram remains intact in all regards except that the circle has no fixed point other than $A$.

To find the exact point, it's helpful to consider $P=M$, whence $(BMP)$ and $(CMP)$ are tangent to the median, $XYDE$ is a parallelogram, and $X, Y \in (BHC)$. Then, constructing the parallels to $AM$ through $B, C$ (the important step in the general solution) is natural if one attempts the inversion $\odot(M, MB)$.
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