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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
This question just asks if you can factorise 12 factorial or not
Sadigly   1
N 17 minutes ago by COCBSGGCTG3
Source: Azerbaijan Junior MO 2025 P1
A teacher creates a fraction using numbers from $1$ to $12$ (including $12$). He writes some of the numbers on the numerator, and writes $\times$ (multiplication) between each number. Then he writes the rest of the numbers in the denominator and also writes $\times$ between each number. There is at least one number both in numerator and denominator. The teacher ensures that the fraction is equal to the smallest possible integer possible.

What is this positive integer, which is also the value of the fraction?
1 reply
1 viewing
Sadigly
Friday at 7:34 AM
COCBSGGCTG3
17 minutes ago
Prime sums of pairs
Assassino9931   5
N 24 minutes ago by aidan0626
Source: Al-Khwarizmi Junior International Olympiad 2025 P5
Sevara writes in red $8$ distinct positive integers and then writes in blue the $28$ sums of each two red numbers. At most how many of the blue numbers can be prime?

Marin Hristov, Bulgaria
5 replies
Assassino9931
Yesterday at 9:35 AM
aidan0626
24 minutes ago
Inequality, inequality, inequality...
Assassino9931   11
N 31 minutes ago by Assassino9931
Source: Al-Khwarizmi Junior International Olympiad 2025 P6
Let $a,b,c$ be real numbers such that \[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]Find the smallest possible value of $a^2 + b^2 + c^2$.

Binh Luan and Nhan Xet, Vietnam
11 replies
+1 w
Assassino9931
Yesterday at 9:38 AM
Assassino9931
31 minutes ago
Anything real in this system must be integer
Assassino9931   2
N 38 minutes ago by Assassino9931
Source: Al-Khwarizmi International Junior Olympiad 2025 P1
Determine the largest integer $c$ for which the following statement holds: there exists at least one triple $(x,y,z)$ of integers such that
\begin{align*} x^2 + 4(y + z) = y^2 + 4(z + x) = z^2 + 4(x + y) = c \end{align*}and all triples $(x,y,z)$ of real numbers, satisfying the equations, are such that $x,y,z$ are integers.

Marek Maruin, Slovakia
2 replies
Assassino9931
Friday at 9:26 AM
Assassino9931
38 minutes ago
Iranian geometry configuration
Assassino9931   3
N an hour ago by Assassino9931
Source: Al-Khwarizmi Junior International Olympiad 2025 P7
Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$, such that $CD$ is not a diameter of its circumcircle. The lines $AD$ and $BC$ intersect at point $P$, so that $A$ lies between $D$ and $P$, and $B$ lies between $C$ and $P$. Suppose triangle $PCD$ is acute and let $H$ be its orthocenter. The points $E$ and $F$ on the lines $BC$ and $AD$, respectively, are such that $BD \parallel HE$ and $AC\parallel HF$. The line through $E$, perpendicular to $BC$, intersects $AD$ at $L$, and the line through $F$, perpendicular to $AD$, intersects $BC$ at $K$. Prove that the points $K$, $L$, $O$ are collinear.

Amir Parsa Hosseini Nayeri, Iran
3 replies
Assassino9931
Yesterday at 9:39 AM
Assassino9931
an hour ago
China South East Mathematical Olympiad 2014 Q3B
sqing   4
N an hour ago by AGCN
Source: China Zhejiang Fuyang , 27 Jul 2014
Let $p$ be a primes ,$x,y,z $ be positive integers such that $x<y<z<p$ and $\{\frac{x^3}{p}\}=\{\frac{y^3}{p}\}=\{\frac{z^3}{p}\}$.
Prove that $(x+y+z)|(x^5+y^5+z^5).$
4 replies
sqing
Aug 17, 2014
AGCN
an hour ago
P>2D
gwen01   5
N an hour ago by Binod98
Source: Baltic Way 1992 #18
Show that in a non-obtuse triangle the perimeter of the triangle is always greater than two times the diameter of the circumcircle.
5 replies
gwen01
Feb 18, 2009
Binod98
an hour ago
Inequality
Sadigly   3
N 3 hours ago by pooh123
Source: Azerbaijan Junior MO 2025 P5
For positive real numbers $x;y;z$ satisfying $0<x,y,z<2$, find the biggest value the following equation could acquire:


$$(2x-yz)(2y-zx)(2z-xy)$$
3 replies
Sadigly
Friday at 7:59 AM
pooh123
3 hours ago
Calculus
youochange   2
N 3 hours ago by youochange
Find the area enclosed by the curves $e^x,e^{-x},x^2+y^2=1$
2 replies
youochange
Yesterday at 2:38 PM
youochange
3 hours ago
A strong inequality problem
hn111009   0
3 hours ago
Source: Somewhere
Let $a,b,c$ be the positive number satisfied $a^2+b^2+c^2=3.$ Find the minimum of $$P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{3abc}{2(ab+bc+ca)}.$$
0 replies
hn111009
3 hours ago
0 replies
help me please,thanks
tnhan.129   0
3 hours ago
find f: R+ -> R such that:
f(x)/x + f(y)/y = (1/x + 1/y).f(sqrt(xy))
0 replies
tnhan.129
3 hours ago
0 replies
Easy divisibility
a_507_bc   2
N 3 hours ago by TUAN2k8
Source: ARO Regional stage 2023 9.4~10.4
Let $a, b, c$ be positive integers such that no number divides some other number. If $ab-b+1 \mid abc+1$, prove that $c \geq b$.
2 replies
a_507_bc
Feb 16, 2023
TUAN2k8
3 hours ago
Inspired by old results
sqing   0
3 hours ago
Source: Own
Let $a,b,c,d$ be real numbers such that $a^2+b^2+c^2 =3$. Prove that$$\frac{9}{5}>(a-b)(b-c)(2a-1)(2c-1)\geq -16$$
0 replies
sqing
3 hours ago
0 replies
integer functional equation
ABCDE   149
N 3 hours ago by ezpotd
Source: 2015 IMO Shortlist A2
Determine all functions $f:\mathbb{Z}\rightarrow\mathbb{Z}$ with the property that \[f(x-f(y))=f(f(x))-f(y)-1\]holds for all $x,y\in\mathbb{Z}$.
149 replies
ABCDE
Jul 7, 2016
ezpotd
3 hours ago
JBMO 2013 Problem 2
Igor   43
N Apr 10, 2025 by EVS383
Source: Proposed by Macedonia
Let $ABC$ be an acute-angled triangle with $AB<AC$ and let $O$ be the centre of its circumcircle $\omega$. Let $D$ be a point on the line segment $BC$ such that $\angle BAD = \angle CAO$. Let $E$ be the second point of intersection of $\omega$ and the line $AD$. If $M$, $N$ and $P$ are the midpoints of the line segments $BE$, $OD$ and $AC$, respectively, show that the points $M$, $N$ and $P$ are collinear.
43 replies
Igor
Jun 23, 2013
EVS383
Apr 10, 2025
JBMO 2013 Problem 2
G H J
Source: Proposed by Macedonia
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Igor
137 posts
#1 • 7 Y
Y by Miku_, ImSh95, Adventure10, Mango247, ItsBesi, DEKT, and 1 other user
Let $ABC$ be an acute-angled triangle with $AB<AC$ and let $O$ be the centre of its circumcircle $\omega$. Let $D$ be a point on the line segment $BC$ such that $\angle BAD = \angle CAO$. Let $E$ be the second point of intersection of $\omega$ and the line $AD$. If $M$, $N$ and $P$ are the midpoints of the line segments $BE$, $OD$ and $AC$, respectively, show that the points $M$, $N$ and $P$ are collinear.
This post has been edited 1 time. Last edited by Igor, Jun 24, 2013, 12:17 AM
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Math-lover123
304 posts
#2 • 7 Y
Y by MrOreoJuice, Miku_, ImSh95, Math_legendno12, Adventure10, ehuseyinyigit, and 1 other user
Let $H$ be the orthocenter of $ABC$.Then it is well known that $AO$ and $AH$ are isogonal WRT $\angle BAC$,so $A$,$H$,$D$ are collinear.
It is also well known that $BH=BE$,so $MD=MB=1/2BH=OP$.
Let $MD$ intersect $AC$ at $K$.
$\angle DKA=180-\angle KAD-\angle KDA=180-90+\gamma -(90-\angle MDB)=90+\gamma -(90-\angle MDB)=90+\gamma -(90-\angle CAD)=90+\gamma -\gamma =90  $.
So $OP$ and $MD$ are parallel.So triangles $MDN$ and $NOP$ are congruent and conclusion follows. :)
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BBAI
563 posts
#3 • 5 Y
Y by ImSh95, Adventure10, Mango247, Orazmuhammet, and 1 other user
Let $N =MP \cap OD$.We will prove $N$ is the midpoint of $OD$.As $AD$ is isogonal to $AO$ ,so $H$ lies on $AD$.By the well known property ,$D$ is the midpoint of $HE$ and $M$ is the midpoint of $BE$ So $MD \mid\mid BH $ Also $BH$ and $OP$ are parallel So $MD \mid\mid OP$.Now $MD=BM =\frac{1}{2}BH=\frac{1}{2}   .2OP=OP$ So $\triangle MND$ and $\triangle NOP$ are congruent $ \Longrightarrow NO=OD$.Hence done.
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MMEEvN
252 posts
#4 • 4 Y
Y by ImSh95, Adventure10, Rotten_, Orazmuhammet
Let $H$ be the orthocenter.A little angle chasing proves that $MD||OP$ Also since $\Delta EDB \sim \Delta CDA \Rightarrow \frac{MD}{DH}=\frac{MD}{DE}=\frac{DP}{DC}$ But $\Delta HCD \sim \Delta OAP \Rightarrow \frac{DP}{DC}=\frac{AP}{DC}=\frac{PO}{DH}\Rightarrow  MD=OP \Rightarrow MDPO$ a parallelogram.Done!
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seby97
344 posts
#5 • 4 Y
Y by AwesomeYRY, ImSh95, Adventure10, Mango247
Analitic solution:
Let D(0,0),A(0,a),B(b,0),C(c,0).We find $O(\dfrac{b+c}{2},\dfrac{a^2+bc}{2a})$,and $E(0,\dfrac{bc}{a})$.So,$M(\dfrac{b}{2},\dfrac{bc}{2a}),N(\dfrac{b+c}{4},\dfrac{a^2+bc}{4a}),P(\dfrac{c}{2},\dfrac{a}{2})$.Because MN,MP has the same slope(equal with $\dfrac{a^2-bc}{a(c-b)}$),the conclusion follows
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trigisfun
44 posts
#6 • 2 Y
Y by ImSh95, Adventure10
Hello,
Can someone explain why $\frac{1}{2}BH=OP$?
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Math-lover123
304 posts
#7 • 2 Y
Y by ImSh95, Adventure10
It can be easily proved using trigonometry that $ \frac{1}{2}BH=OP= \frac{1}{2}\cot \beta $
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mavropnevma
15142 posts
#8 • 4 Y
Y by ohmcfifth, ImSh95, Adventure10, and 1 other user
The following analytical solution has some unexpected consequences, detailed in the sequel.

Choose a coordinatesystem $xOy$ where $O(0,0), A(a,u), B(-b,v), C(b,v), A'(-a,-u)$.
From $\angle BAD = \angle CAO$, i.e. $\angle BAE = \angle CAA'$, follows $E(a,-u)$ (thus $AD\perp BC$), hence $D(a,v)$.
But then $M\left (\dfrac{a-b}{2},\dfrac{v-u}{2}\right )$, $N\left (\dfrac {a} {2},\dfrac{v}{2}\right )$, $P\left (\dfrac{a+b}{2},\dfrac{u+v}{2}\right )$.
The slope of the line $MN$ is $\displaystyle \dfrac {\dfrac {v} {2} - \dfrac {v-u} {2}} {\dfrac {a} {2} - \dfrac {a-b} {2}} = \dfrac {u} {b}$, while the slope of the line $NP$ is $\displaystyle \dfrac {\dfrac {u+v} {2} - \dfrac {v} {2}} {\dfrac {a+b} {2} - \dfrac {a} {2}} = \dfrac {u} {b}$; as seen, they coincide, proving the colinearity.

We can certify now a suspicion lingering from the very beginning, namely that $AB<AC$ is a useless restriction (not even the isosceles case produces any discomfort). Neither the requirement $\triangle ABC$ be acute-angled is instrumental (true, some degenerate cases may appear, but they only increase the wealth of the configurations; also, point $D$ may now be outside the side $BC$). It is strange the choice of the jury to abide by these irrelevant conditions !?!
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BBAI
563 posts
#9 • 3 Y
Y by ImSh95, Adventure10, Mango247
trigisfun wrote:
Hello,
Can someone explain why $\frac{1}{2}BH=OP$?
See this an important part of proving the existence of the euler line i.e $O,G,H$ are collinear.
if $BO \cap  \odot ABC =X$ First prove that $HAXC$ is a parallelogram and then you can see that $H ,P,X$ are collinear .Then by using similarity of $\triangle BHX$ and $OMX$ ,we get the required result.
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exmath89
2572 posts
#10 • 2 Y
Y by ImSh95, Adventure10
Solution
This post has been edited 1 time. Last edited by exmath89, Jun 24, 2013, 3:55 PM
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Igor
137 posts
#11 • 4 Y
Y by fleurdelis, ImSh95, Adventure10, Mango247
mavropnevma wrote:
We can certify now a suspicion lingering from the very beginning, namely that $AB<AC$ is a useless restriction (not even the isosceles case produces any discomfort). Neither the requirement $\triangle ABC$ be acute-angled is instrumental (true, some degenerate cases may appear, but they only increase the wealth of the configurations; also, point $D$ may now be outside the side $BC$). It is strange the choice of the jury to abide by these irrelevant conditions !?!

In my opinion, it was done on purpose so that students won't lose time checking other configurations.
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IDMasterz
1412 posts
#12 • 3 Y
Y by ImSh95, Adventure10, antimonio
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(15cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -0.2961682945153994, xmax = 25.30864012021040, ymin = -9.572516904583015, ymax = 4.131465063861762;  /* image dimensions */
pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); pen qqwuqq = rgb(0.000000000000000,0.3921568627450985,0.000000000000000); pen qqttcc = rgb(0.000000000000000,0.2000000000000002,0.8000000000000009); pen qqzzff = rgb(0.000000000000000,0.6000000000000006,1.000000000000000); pen ttzzqq = rgb(0.2000000000000002,0.6000000000000006,0.000000000000000); 

draw((11.30000000000001,2.080000000000002)--(7.460000000000008,-4.500000000000005)--(19.08000000000002,-5.160000000000006)--cycle, zzttqq); 
 /* draw figures */
draw((11.30000000000001,2.080000000000002)--(7.460000000000008,-4.500000000000005), zzttqq); 
draw((7.460000000000008,-4.500000000000005)--(19.08000000000002,-5.160000000000006), zzttqq); 
draw((19.08000000000002,-5.160000000000006)--(11.30000000000001,2.080000000000002), zzttqq); 
draw(circle((13.34420968681167,-3.523459756437201), 5.964692321123185)); 
draw((11.30000000000001,2.080000000000002)--(10.91511855898421,-4.696245976672086)); 
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[/asy]
Lemma: It is well known that the reflection of the orthocentre $H$ through the midpoint of $BC$ and $AC$ is the antipode of $A, B$ respectively.

If we let primes denote the antipodes of points, we have $BA'CH$ (Bach hehe) and $AB'CH$ are parallelograms.

Part 1: Through dilations with ratio $2$ about $E$ and $A$, we take $MD \to BH$ and $OP \to A'C$. But, $BA'CH$ is a parallogram, so $BH \parallel A'C \implies MD \parallel OP$.

Part 2: Through dilations with ratio $2$ about $E$ and $H$, we take $OM \to E'B$ and $DP \to EB'$. But $EB'BE'$ is a rectangle (points and their antipodes), so $E'B \parallel EB' \implies OM \parallel DP$.

Therefore, $OMDP$ is a parallelogram, so the $MP$ goes through the midpoint of $OD$, or $M, N, P$ are collinear, as desired $\blacksquare$.
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mathuz
1524 posts
#13 • 4 Y
Y by ImSh95, Adventure10, Mango247, and 1 other user
We have $ACEB$ is cyclic quadriteral and dioganals perpendicular - $AE\bot BC.$
So, $N$ - center of the eight points circle. :wink:
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liimr
34 posts
#14 • 2 Y
Y by ImSh95, Adventure10
useful lemma:
Given cyclic quadrilateral $ABCD$ with perpendicular diagonals. Let $AC\cap BD=E$ and foot of perpendicular from $E$ to the side $AB$ be $X_{1}$ and that perpendicular line meets $CD$ at $Y_{1}$. Define $X_{2}, Y_{2}, X_{3}, Y_{3}, X_{4}, Y_{4}$. Then these 8 points are concyclic.
This post has been edited 1 time. Last edited by liimr, Jun 26, 2013, 4:49 PM
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BBAI
563 posts
#15 • 2 Y
Y by ImSh95, Adventure10
liimr wrote:
useful lemma:
Given cyclic quadrilateral $ABCD$ with perpendicular diagonals. Let $AC\cap BD=E$ and foot of perpendicular to the side $AB$ be $X_{1}$ and meets $CD$ at $Y_{1}$. Define $X_{2}, Y_{2}, X_{3}, Y_{3}, X_{4}, Y_{4}$. Then these 8 points are concyclic.
From Where the foot of the perpendicular is drawn to $AB$?
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liimr
34 posts
#16 • 3 Y
Y by ImSh95, Adventure10, Mango247
BBAI wrote:
From Where the foot of the perpendicular is drawn to $AB$?
I edited my post I think now it is understandable :D
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Math-lover123
304 posts
#17 • 2 Y
Y by ImSh95, Adventure10
liimr can you prove the lemma?
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War-Hammer
670 posts
#18 • 3 Y
Y by liimr, ImSh95, Adventure10
It's just angle chasing , and maybe Brahmagupta's lemma can be use here.
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liimr
34 posts
#19 • 4 Y
Y by ImSh95, sayemsub15, Adventure10, Mango247
Math-lover123 wrote:
liimr can you prove the lemma?

proof is just angle chasing. Firstly prove that $X_{1}, X_{2}, X_{3}, X_{4}$ are concyclic then try to prove $X_{1}, X_{2}, X_{3}, Y_{3}$ are concyclic then others will follow directly.
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ABCDE
1963 posts
#20 • 5 Y
Y by PatrikP, raven_, ImSh95, Adventure10, Mango247
Not too hard.

Click to reveal hidden text
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mathbuzz
803 posts
#21 • 4 Y
Y by Mehmet_Koca, ImSh95, Adventure10, Mango247
it's obvious to see that , $AD$ is perpendicular to $BC$ and $E$ is the reflection of $H$ w.r.t. $BC$. now , WLOG , assume that , the circumcircle of $ABC$ is the unit circle centred at origin. then $h=a+b+c$ . also , we get , $d=(a+b+c-bc/a)/2$. so , $e=-bc/a$ . hence , $p=\frac{a+c}{2}$. $n=[a+b+c-bc/a]/4$
$m=b(a-c)/2$ . then it is easy to see that $\mu_{NP}=(b-a-c-bc/a)/(1/b-a/bc-1/a-1/c)$ and ,
$\mu_{PM}=\frac{a+c-b+bc/a}{a/bc+1/a+1/c-1/b}$ . hence $M,P,N$ are collinear. :D
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Fitim
207 posts
#22 • 3 Y
Y by ImSh95, Adventure10, Mango247
Author of this problem is Stefan Lozanovski (Стефан Лозановски). :)
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sunken rock
4394 posts
#23 • 3 Y
Y by ImSh95, Adventure10, Mango247
$\triangle BDE\sim\triangle ADC$ and being right-angled means the median $DP$ of $\triangle ADC$ is altitude in $\triangle BDE$, so $DP\parallel OM$ ($OM$ is perpendicular bisector of $BE$), and similarly the median $DM$ is altitude of $\triangle ADC$, hence $DM\parallel OP$ ( the latter being perpendicular bisector of $AC$), hence $OMDP$ is parallelogram, done.

Best regards,
sunken rock
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fclvbfm934
759 posts
#24 • 2 Y
Y by ImSh95, Adventure10
It's well known that the orthocenter is the isogonal conjugate of the circumcenter; therefore, we know that $AD \perp BC$. Then we know that $BM=MD = ME$, and $\angle BCA = \angle BEA = \angle MDE$. Extend $ME$ to intersect $AC$ at $P'$. Therefore, $\angle ADP' = \angle BCA = 90^{\circ} - \angle EAC$, so we see that $\angle DP'A = 90^{\circ}$, so $DM || OP$. Let $AO$ intersect $\omega$ at $Q$. Then, we know that $CQ = BE$, so $OP = MD$. Therefore, $MDPO$ is a parallelogram, so the diagonals bisect each other, and we are done.
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bosanac007
27 posts
#25 • 3 Y
Y by ImSh95, Adventure10, Mango247
Triangles POC and DEC are similar, using PC:CD=PO:DE and AC:BE=CD:DE. Since AC=2PC and BE=2ME=2MD we get MD=PO.
N is midpoint of DO so DN=NO. We will prove angle MND=PNO, respectively triangles MDN and NOP are congruent.
Easy to prove that angle MDN=PON. And we have MD=PO , DN=NO and angle MDN=PON, so triangles MDN and NOP are congruent.
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AlexLewandowski
157 posts
#26 • 3 Y
Y by ImSh95, Adventure10, Mango247
Using some simple trig bash, we find that OPDM is a ||gm. If PM meets OD in N', then N'D = N'O.
This implies that N and N' coincides and hence the result follows.
This post has been edited 1 time. Last edited by AlexLewandowski, Jan 18, 2017, 3:59 PM
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MathStudent2002
934 posts
#27 • 3 Y
Y by v_Enhance, ImSh95, Adventure10
Very challenging and beautiful problem! Solution with v_Enhance, msinghal, linpaws, and mathcool2009.

By $\sqrt{bc}$ inversion, $AD$, and $AO$ map to each other, so $AD$ passes through the reflection of $A$ over $BC$, hence $D$ is the foot of the $A$-altitude. Consider a homothety centered at $D$ with factor two, so that $M$ maps to $M'$, $P$ maps to $P'$, and $N$ to $O$. Also, let $Q,R$ be the reflections of $D$ over the midpoints of $AB$, $CE$, so that $QP'RM'$ is a rectangle. Since its center is the intersection of the perpendicular bisectors of $P'R$ and $RM'$, which are precisely the perpendicular bisectors of $AE$ and $BC$, which meet at $O$, $O$ is on the diagonal $P'M'$. $\blacksquare$
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WizardMath
2487 posts
#28 • 3 Y
Y by ImSh95, Adventure10, Mango247
Let $H$ be the orthocenter of $ABC$. From Brahmagupta's theorem, $MD$ is perpendicular to $AC$, which is perpendicular to $OP$.
Also we have $MD = 0.5 BE = 0.5 BH = OP$, so $MDOP$ is a parallelogram, and thus we are done.
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Drunken_Master
328 posts
#29 • 3 Y
Y by ImSh95, Adventure10, Mango247
Straight-forward with complex numbers.
Set $(ABC)$ as unit circle.

$m=\frac{b+(\frac{-bc}a)}2$
$p=\frac{a+c}2$
$d=\frac{1}2 \left(a+b+c-bc\overline{a} \right)$
Midpoint of $MP$ is $\frac{1}2 \left (\frac{b+(\frac{-bc}a)}2+\frac{a+c}2 \right)=\frac{1}2 \left(\frac{1}2(a+b+c-bc\overline{a}) \right)=\frac{o+d}2=n$

Hence, $N$ is midpoint of $MP \implies M,N,P$ are collinear, as desired. $\blacksquare$
This post has been edited 4 times. Last edited by Drunken_Master, Feb 21, 2018, 12:15 PM
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Leartia
93 posts
#30 • 3 Y
Y by Circumcircle, ImSh95, Adventure10
(A non-guessing solution using Complex numbers,isogonal conjugates)
The orthocenter $H$ is the isogonal conjugate of the circumcenter $O$, combining this with the given angle condition we have $AD \perp BC$.Working in the complex plane we set $|A|=|B|=|C|=|E|=1$.
Since $AE \perp BC$ we have $ae+bc=0$ so $e=\frac{-bc}{a}$.
$d=\frac{1}{2} (a+b+c-\frac{bc}{a})$,
$n=\frac{1}{4}(a+b+c-\frac{bc}{a})$,
$m=\frac{1}{2}(b-\frac{bc}{a})$
$p=\frac{1}{2}(a+c)$
$P,M,N$ are collinear if $\frac{p-n}{p-m}$ is a real number. $\frac{p-n}{p-m}=\frac{\frac{1}{2}(a+c)-\frac{1}{4}(a+b+c-\frac{bc}{a})}{\frac{1}{2}(a+c)-\frac{1}{2}(b-\frac{bc}{a})}=\frac{\frac{1}{4}(a+c-b-e)}{\frac{1}{2}(a+c-b-e)}=\frac{1}{2}$ So $M,N$ and $P$ are collinear.
This post has been edited 1 time. Last edited by Leartia, Aug 1, 2019, 11:07 AM
Reason: ...
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itslumi
284 posts
#32 • 2 Y
Y by ImSh95, Adventure10
this problem becom very easy in the moment when you realized a known confiraguration. That MDPO is a parallelogram.
This post has been edited 1 time. Last edited by itslumi, Jul 26, 2020, 1:46 PM
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MrOreoJuice
594 posts
#34 • 1 Y
Y by ImSh95
It is well known that the isogonal of $AC$ is the $A$-Altitude of $\triangle ABC \implies AD$ is the $A$-Altitude.
Let $(ABC)$ be the unit circle.
$$h = a + b + c$$$$d = \frac{a + b + c - \frac{bc}{a}}{2}$$Also since $E$ is the reflection of $H$ across $BC \in (ABC)$
$$\implies d = \frac{h + e}{2} \implies e = 2d - h$$$$p = \frac{a+c}{2}$$$$n = \frac{d + o}{2} = \boxed{\frac{d}{2}}$$$$m = \frac{b + e}{2} = \frac{b - h + 2d}{2} = d - \frac{a+c}{2} = d - p$$$$\implies m + p = d$$$$\implies \frac{m+p}{2} = \frac d2 = n$$Hence $N$ is the midpoint of $PM$ and we are done. :)
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sttsmet
139 posts
#35 • 1 Y
Y by ImSh95
@mathur and @liimr in posts 13 and 14 respectively, I read the 8 point theorem and I found it really impressive! But when it comes to the accual problem, I couldnt combine these two.... It is clear that the ABEC has perpendicular diameters and N is the midpoint of BE and P the midpoint of AC, but I cant keep applying this theorem. Can anybody helps me with that???
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JustKeepRunning
2958 posts
#36 • 1 Y
Y by ImSh95
It is well known that $N$ is the center of the eight point circle of quadrilateral $ABCD$ with diameter $MP,$ and the conclusion follows.

Note: This problem is also killed by the so-called parallelogram trick(which is in author's book).
This post has been edited 1 time. Last edited by JustKeepRunning, Jun 3, 2021, 8:46 PM
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Albert123
204 posts
#37 • 1 Y
Y by ImSh95
Note that: $AD$ is altitude from $A$ to $BC$
So:
$MO \perp BE$ and $DP \perp BE$ $\implies MO \parallel DP$
$OP \perp AC$ and $MD \perp AC$ $\implies OP \parallel MD$
$\implies MDOP$ is paralelogram.
Then: $M,N,P$ are collinear.$\blacksquare$
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aansc1729
111 posts
#38 • 1 Y
Y by ImSh95
Extend $ AO $ so that it meets $\omega $ at $F$. As $\angle B A E=\angle C A F$ so, $BE=CF$ $\Rightarrow$ $BCEF$ is an isoceles trapezoid with $ BC \parallel EF $. Now $\angle A E F=90^{\circ} \Rightarrow \angle A D C=90^{\circ}$.
Let $\angle B A E=\angle C A O=x$ and $\angle M E D=\angle M D E=\theta$, then by some angle chase, we get $\angle B M D=2 \theta$ $\Rightarrow$ $\angle D M O=90-2 \theta$. Similarly, we get $\angle D P O=\angle D P C-\angle O P C=180-2 \theta-90=90-2 \theta$. Also $\angle M D P=360-(\angle A D P+\angle A D B+\angle B D M)=90+2 \theta$ and $\angle M O P=360-(\angle M O E+\angle C O E+\angle P O C)=90+2 \theta$. As the opposite angles are equal, $MDOP$ is a parallelogram and $N$ is the midpoint of the diagonal $OD$ and so we get $M,N,P$ are collinear. :)
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tony88
5 posts
#39 • 2 Y
Y by ImSh95, Mango247
trigisfun wrote:
Hello,
Can someone explain why 1/2BH=OP?

see Property 10.3.2.
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john0512
4187 posts
#40
Y by
Note that by isogonality $AD\perp BC$.

Claim: $MDPO$ is a parallelogram. Extend $DM$ to meet $AC$ at $Q$. Note that since $DM$ is a median in $\triangle DBE$, $DQ$ is a symmedian in $\triangle ADC$, and is therefore also an altitude (since it is a right triangle). Therefore, $DM\perp AC$, so $DM\parallel OP$.

Let $H$ be the orthocenter. Furthermore, $BH=2OP$, so $BE=2OP$ since $E$ is the reflection of $H$ over $D$. Then $MD=ME=\frac{1}{2}BE$, so $MD=OP$ which shows the claim.

Since $MDPO$ is a parallelogram, $MP$ passes through the midpoint of $OD$, so we are done.
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Taco12
1757 posts
#41
Y by
Clearly, we have $AD \perp BC$. Now, apply complex numbers with $(ABC)$ as the unit circle. We then have $d=\frac{a+b+c-\frac{bc}{a}}{2}, e=\frac{bc}{a}$. The three midpoints are just

\begin{align*}
\frac{a+b+c-\frac{bc}{a}}{4} \\
\frac{ab-bc}{2a} \\
\frac{a+c}{2} \\
\end{align*}
These are collinear by the Collinearity Criterion. $\blacksquare$
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Spectator
657 posts
#42
Y by
Wow so nice

Consider the rectangle passing through $A, B, C$ and $E$ such that the sides are parallel to $BC$ and $AE$. Note that a homothety of factor $2$ centered at $D$ maps $M$ and $P$ to opposite vertices of the rectangle and $N$ to $O$. It suffices to prove that $O$ lies on the diagonal from the opposite vertices, which is true because $O$ lies on the perpendicular bisector of $BC$ and $AE$.
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ismayilzadei1387
219 posts
#43
Y by
from $\angle BAH$=$\angle  CAO$ lemma $AD$ is perpendicular to $BC$
from a little angle chasing you will observe that if $MDPO$ is parallelogram then we are done
we can see that $OM$ and $DP$ are perpendicular to $BE$
similarly $MD$ and $OP$ are perpendicular to $AC$
Thus $OM//DP$ and $MD//OP$ since that
$MDPO$ is parallelogram.
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Mathandski
757 posts
#44
Y by
Subjective Rating (MOHs) $       $
Attachments:
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cherry265
16 posts
#45
Y by
It is well known that $D$ is the foot of the perpendicular from $A$ to $BC$. Then $(ABC)$ unit circle and braindead complex bash easily kills.
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EVS383
21 posts
#46
Y by
Coordinates have never been easier.

First, $AD \perp BC$ as $O$ and $H$ are isogonal conjugates. Now, put everything on the unit circle using coordinates.
\begin{align*}
A &= (\cos A, \sin A) \\ 
B &=(-\cos T, \sin T) \\
C &= (\cos T, \sin T) \\
D &= (\cos A, \sin T) \\
E &= (\cos A, -\sin A) \\
\end{align*}It's clear that the midpoint of $M$ and $P$ is $N$.
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