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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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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
Centroid, altitudes and medians, and concyclic points
BR1F1SZ   3
N 22 minutes ago by EeEeRUT
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
3 replies
BR1F1SZ
Monday at 9:45 PM
EeEeRUT
22 minutes ago
IMO Genre Predictions
ohiorizzler1434   60
N 26 minutes ago by Yiyj
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
60 replies
ohiorizzler1434
May 3, 2025
Yiyj
26 minutes ago
Combinatorics
P162008   1
N 27 minutes ago by jasperE3
Let $m,n \in \mathbb{N}.$ Let $[n]$ denote the set of natural numbers less than or equal to $n.$

Let $f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$

Compute the sum of the digits of $f(4,4).$
1 reply
P162008
32 minutes ago
jasperE3
27 minutes ago
square root problem
kjhgyuio   5
N 30 minutes ago by Solar Plexsus
........
5 replies
kjhgyuio
May 3, 2025
Solar Plexsus
30 minutes ago
Diodes and usamons
v_Enhance   47
N an hour ago by EeEeRUT
Source: USA December TST for the 56th IMO, by Linus Hamilton
A physicist encounters $2015$ atoms called usamons. Each usamon either has one electron or zero electrons, and the physicist can't tell the difference. The physicist's only tool is a diode. The physicist may connect the diode from any usamon $A$ to any other usamon $B$. (This connection is directed.) When she does so, if usamon $A$ has an electron and usamon $B$ does not, then the electron jumps from $A$ to $B$. In any other case, nothing happens. In addition, the physicist cannot tell whether an electron jumps during any given step. The physicist's goal is to isolate two usamons that she is sure are currently in the same state. Is there any series of diode usage that makes this possible?

Proposed by Linus Hamilton
47 replies
v_Enhance
Dec 17, 2014
EeEeRUT
an hour ago
3-var inequality
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b\geq  0 ,a^3-ab+b^3=1  $. Prove that
$$  \frac{1}{2}\geq     \frac{a}{a^2+3 }+ \frac{b}{b^2+3}   \geq  \frac{1}{4}$$$$  \frac{1}{2}\geq     \frac{a}{a^3+3 }+ \frac{b}{b^3+3}   \geq  \frac{1}{4}$$$$  \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2}  \geq  \frac{1}{3}$$$$  \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2}  \geq  \frac{1}{3}$$Let $ a,b\geq  0 ,a^3+ab+b^3=3  $. Prove that
$$  \frac{1}{2}\geq     \frac{a}{a^2+3 }+ \frac{b}{b^2+3}   \geq  \frac{1}{4}(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{3}-1)$$$$  \frac{1}{2}\geq     \frac{a}{a^3+3 }+ \frac{b}{b^3+3}   \geq  \frac{1}{2\sqrt[3]{9}}$$$$  \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2}  \geq  \frac{4\sqrt[3]{3}+3\sqrt[3]{9}-6}{17}$$$$  \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2}  \geq  \frac{\sqrt[3]{3}}{5}$$
1 reply
sqing
2 hours ago
sqing
an hour ago
IMO ShortList 2001, combinatorics problem 3
orl   37
N 2 hours ago by deduck
Source: IMO ShortList 2001, combinatorics problem 3, HK 2009 TST 2 Q.2
Define a $ k$-clique to be a set of $ k$ people such that every pair of them are acquainted with each other. At a certain party, every pair of 3-cliques has at least one person in common, and there are no 5-cliques. Prove that there are two or fewer people at the party whose departure leaves no 3-clique remaining.
37 replies
orl
Sep 30, 2004
deduck
2 hours ago
area of O_1O_2O_3O_4 <=1, incenters of right triangles outside a square
parmenides51   2
N 2 hours ago by Solilin
Source: Thailand Mathematical Olympiad 2012 p4
Let $ABCD$ be a unit square. Points $E, F, G, H$ are chosen outside $ABCD$ so that $\angle AEB =\angle BF C = \angle CGD = \angle DHA = 90^o$ . Let $O_1, O_2, O_3, O_4$, respectively, be the incenters of $\vartriangle ABE, \vartriangle BCF, \vartriangle CDG, \vartriangle DAH$. Show that the area of $O_1O_2O_3O_4$ is at most $1$.
2 replies
parmenides51
Aug 17, 2020
Solilin
2 hours ago
Geo metry
TUAN2k8   3
N 2 hours 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.
3 replies
TUAN2k8
Yesterday at 10:33 AM
TUAN2k8
2 hours ago
Functional equation of nonzero reals
proglote   8
N 2 hours ago by jasperE3
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 $x+y \neq 0$.
8 replies
proglote
Oct 24, 2013
jasperE3
2 hours ago
Interesting inequalities
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b\geq  0 ,a^2-ab+b^2+a+b=3  $. Prove that
$$  \frac{39+\sqrt{13}}{78}\geq  \frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \geq  \frac{1}{2}$$Let $ a,b\geq  0 ,a^2+ab+b^2+a+b=3  $. Prove that
$$  \frac{19+\sqrt{10}}{39}\geq  \frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \geq    \frac{39+\sqrt{13}}{78}$$Let $ a,b\geq  0 ,a^2+ab+b^2+a+b=5  $. Prove that
$$  \frac{3}{5}> \frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \geq     \frac{185+3\sqrt{21}}{402}$$
1 reply
sqing
3 hours ago
sqing
2 hours ago
4-var inequality
sqing   2
N 3 hours ago by sqing
Source: SXTB (4)2025 Q2837
Let $ a,b,c,d> 0  $. Prove that
$$   \frac{1}{(3a+1)^4}+ \frac{1}{(3b+1)^4}+\frac{1}{(3c+1)^4}+\frac{1}{(3d+1)^4} \geq \frac{1}{16(3abcd+1)}$$
2 replies
sqing
Yesterday at 2:59 PM
sqing
3 hours ago
Inspired by Bet667
sqing   2
N 3 hours ago by sqing
Source: Own
Let $ a,b $ be a real numbers such that $a^2+kab+b^2\ge a^3+b^3.$Prove that$$a+b\leq k+2$$Where $ k\geq 0. $
2 replies
sqing
Yesterday at 2:46 PM
sqing
3 hours ago
find positive n so that exists prime p with p^n-(p-1)^n$ a power of 3
parmenides51   14
N 3 hours ago by MathLuis
Source: JBMO Shortlist 2017 NT5
Find all positive integers $n$ such that there exists a prime number $p$, such that $p^n-(p-1)^n$ is a power of $3$.

Note. A power of $3$ is a number of the form $3^a$ where $a$ is a positive integer.
14 replies
parmenides51
Jul 25, 2018
MathLuis
3 hours ago
Two circles, a tangent line and a parallel
Valentin Vornicu   104
N May 2, 2025 by cubres
Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2
Two circles $ G_1$ and $ G_2$ intersect at two points $ M$ and $ N$. Let $ AB$ be the line tangent to these circles at $ A$ and $ B$, respectively, so that $ M$ lies closer to $ AB$ than $ N$. Let $ CD$ be the line parallel to $ AB$ and passing through the point $ M$, with $ C$ on $ G_1$ and $ D$ on $ G_2$. Lines $ AC$ and $ BD$ meet at $ E$; lines $ AN$ and $ CD$ meet at $ P$; lines $ BN$ and $ CD$ meet at $ Q$. Show that $ EP = EQ$.
104 replies
Valentin Vornicu
Oct 24, 2005
cubres
May 2, 2025
Two circles, a tangent line and a parallel
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2
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Valentin Vornicu
7301 posts
#1 • 19 Y
Y by Davi-8191, OlympusHero, FaThEr-SqUiRrEl, lc426, centslordm, Adventure10, Designerd, jhu08, megarnie, Numbertheorydog, HWenslawski, Kanimet0, ImSh95, Zhaom, Mango247, Rounak_iitr, Math_.only., ItsBesi, cubres
Two circles $ G_1$ and $ G_2$ intersect at two points $ M$ and $ N$. Let $ AB$ be the line tangent to these circles at $ A$ and $ B$, respectively, so that $ M$ lies closer to $ AB$ than $ N$. Let $ CD$ be the line parallel to $ AB$ and passing through the point $ M$, with $ C$ on $ G_1$ and $ D$ on $ G_2$. Lines $ AC$ and $ BD$ meet at $ E$; lines $ AN$ and $ CD$ meet at $ P$; lines $ BN$ and $ CD$ meet at $ Q$. Show that $ EP = EQ$.
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arqady
30237 posts
#2 • 29 Y
Y by jam10307, shiningsunnyday, claserken, shawnee03, futurestar, Richangles, myh2910, OlympusHero, User582032, starchan, sotpidot, FaThEr-SqUiRrEl, centslordm, Adventure10, megarnie, asdf334, Numbertheorydog, HWenslawski, rayfish, Kanimet0, ImSh95, Mango247, CoC_Ali, Rounak_iitr, Math_.only., ehuseyinyigit, cubres, and 2 other users
let MN and AB intesect at K. Then $KA^2=KM*KN=KB^2.$ Hence KA=KB.
CD||AB. Hence PM=QM.
$\angle AEB=\angle ECD=\angle MAB,$ $\angle EBA=\angle EDC=\angle MBA.$
hence$\triangle AMB$ and$\triangle AEB$ congruity.
Hence $EM\perp AB.$ Hence $EM\perp PQ.$
Hence $EP=EQ.$
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yetti
2643 posts
#3 • 10 Y
Y by JasperL, FaThEr-SqUiRrEl, centslordm, Adventure10, Designerd, jhu08, ImSh95, Rounak_iitr, cubres, and 1 other user
The radical axis MN of the circles $(G_1), (G_2)$ cuts the tangent length segment AB at its midpoint C. Since $PQ \parallel AB$, the triangles $\triangle ABN \sim \triangle PQN$ are centrally similar with similarity center N. Hence, M is the midpoint of the segment PQ. The radii $G_1A, G_2B$ of the circles $(G_1), (G_2)$ are perpendicular to the tangent AB, i.e, also to the chords AM, DM. Hence they cut these chords at their midpoints, which means that the triangles $\triangle ACM, \triangle BDM$ are isosceles and the angles $\angle ACM = \angle AMC, \angle BDM = \angle BMD$ are equal. Since the lines $AB \parallel CD$ are parallel and $M \in CD$, the angles $\angle MAB = \angle AMC = \angle ACM = \angle EAB$ are equal and similarly, the angles $\angle MBA = \angle BMD = \angle BDM = \angle EBA$ are also equal. Thus the diagonal AB of the quadrilateral AMBE bisects its opposite angles at the vertices A, B, which implies that this quadrilateral is a kite with AE = AM, BE = BM and and its diagonals $AB \perp EM$ are perpendicular to each other. Consequently, the lines $EM \perp CD \equiv PQ$ are also perpendicular to each other. Since M is the midpoint of the segment PQ, the line EM is the perpendicular bisector of this segment. It follows that the triangle $\triangle EPQ$ is isosceles with EP = EQ.
Attachments:
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Arne
3660 posts
#4 • 8 Y
Y by mihajlon, OlympusHero, FaThEr-SqUiRrEl, Adventure10, ImSh95, cubres, and 2 other users
An "extra" question:

Prove that $EN$ bisects $\angle{CND}$.
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Zhero
2043 posts
#5 • 5 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres
Since $ AB$ is the common tangent of $ G_1$ and $ G_2$ and $ MN$ is the radical axis of $ G_1$ and $ G_2$, $ MN$ bisects $ AB$. A homothety centered at $ N$ that maps $ A$ to $ P$ and $ B$ to $ Q$ therefore maps the midpoint of $ AB$ to $ PQ$, so we have that $ M$ is the midpoint of $ PQ$.

It is sufficient to show that $ EM \perp CD$. Let $ R$ and $ S$ be the projections of $ A$ and $ B$ onto $ CD$, respectively; since $ AB || CD$, $ AR = BS$. We first note that $ AC = AM$, since $ m \angle AMC = m \angle EAB = m \angle ACM$ (as $ AB$ is tangent to $ G_1$). We can similarly deduce that $ BM = BD$.

It also follows from this that $ CM = 2CR$ and $ DM = 2DS$. Let $ E'$ be the point such that $ E'M \perp CD$ and $ E'M = 2AR = 2BS$. A homothety centered at $ C$ that maps $ R$ to $ M$ must therefore map $ A$ to $ E'$, and a homothety centered at $ D$ must therefore map $ B$ to $ E'$. But this means that $ E'$ lies on both $ AC$ and $ BD$, implying that $ E' = E$. It follows that $ EM \perp CD$, so we are done.
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Pikachu!!!
58 posts
#6 • 6 Y
Y by S117, FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres
Arne wrote:
An "extra" question:

Prove that $ EN$ bisects $ \angle{CND}$.

This extra question is also beautiful one.

First, from what we know Let $ \angle{ANM}=\angle{BAM}=\angle{BAE}=x$ and $ \angle{BNM}=\angle{ABM}=\angle{ABE}=y$

Therefore, $ \angle{AEB}+\angle{ANB}= (\pi-x-y)+(x+y)=\pi$ $ \longrightarrow$ $ A,N,B,E$ is concyclic.

It follows that $ \angle{BNE}=\angle{BAE}=x$ and it is known that $ \angle{BND}=\angle{BNM}=y$
We get $ \angle{DNE}=x+y$

Similarly, we also have $ \angle{CNE}=x+y$ :lol: :lol: :lol:
This is what we want to prove.
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arshakus
769 posts
#7 • 5 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres
$K$ is the midpint of $AB$. $NM$ is radical axis for $G_1,G_2$ circles. also we know that radical axis passes through the midpoints of comment tangents. Thus $N,M,K$ lies on the same line. $PQ||AB=>M$ is the midpoint of $PQ$.
Also it is easy to note this number of equations.
$\angle{PCA}=\angle{ANM}=\angle{BAE}=\angle{MAB}=\angle{PMA}=\angle{CNA}=\angle{ENB}$[*] the last equation follows from the fact that $AENB$ is cyclic.
Similarly we can get in the opposite side. Thus $\triangle {AMB}=\triangle {AEB}=>EM\perp PQ=>EP=EQ$
@Arne
from [*] it is easy to note it.
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arshakus
769 posts
#8 • 5 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres
One more "extra" question!)
Prove that $2AB=CD$
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siavosh
29 posts
#9 • 5 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres
arshakus wrote:
One more "extra" question!)
Prove that $2AB=CD$
In your solution we prove that $ \triangle{AMB}=\triangle{AEB}=> EB=BM=BD$
and $ CD\parallel AB =>  \frac{EB}{ED}=\frac{AB}{CD}=2 $
:)
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JSGandora
4216 posts
#10 • 6 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, Rounak_iitr, cubres
Let $\angle ACM=\alpha$, then since $AB||CD$, we have $\angle EAB=\angle ACM=\alpha$ and since $AB$ is tangent to circle $G_1$, we have $\angle BAM=\angle ACM=\alpha$. Thus $\angle EAB=\angle BAM=\alpha$. Similarly, if we let $\angle BDM=\beta$, we find $\angle EBA=\angle ABM=\beta$.

Now let the intersection of $EM$ and $AB$ be $F$, then $\angle AFE=90^\circ$ and $\angle CME=\angle AFE=90^\circ$ and therefore proving $EP=EQ$ is equivalent to proving $PM=QM$ since $EM\perp PQ$.

Let the intersection of ray $NM$ and $AB$ be point $G$, then $GA=GB$ because $G$ is on the radical axis of the two circles and thus $GA^2=GB^2\implies GA=GB$. Additionally, since $AB||CD$, we have $\triangle AGN~\triangle PMN$ and $\triangle BGN~\triangle QMN$ and therefore $PM=PQ$ as desired. $\blacksquare$
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BBAI
563 posts
#11 • 4 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, cubres
As the tangent is parallel to the chords So it follows that $ \triangle ACM$ and $ \triangle BDM$ are isosceles and $EA=AM=AC$ so $EM$ is perpendicular to $PQ$. Define $Z=MN \cap AB$.$Z$ lies on radical axis ,so $Z$ is the midpoint of $AB$. And hence $M$ is the midpoint of $PQ$. So done.
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subham1729
1479 posts
#12 • 4 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, cubres
OMG!! got a proof without bashing :O - if $MN$ meets $AB$ at $K$ then $AK^2=BK^2\implies AK=KB\implies PM=MQ\implies \angle{CMN}=\frac {\pi}{2}\implies EP=EQ$
This post has been edited 1 time. Last edited by subham1729, May 3, 2013, 12:04 PM
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BBAI
563 posts
#13 • 4 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, cubres
I think $E,M,N$ are not at all collinear.
I have drawn a ggb diagram. and checked it.
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sunken rock
4392 posts
#14 • 6 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres, and 1 other user
Yet additional requirement:

the quadrilateral $ANBE$ is harmonic!


Best regards,
sunken rock
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exmath89
2572 posts
#15 • 7 Y
Y by Math-Star., FaThEr-SqUiRrEl, ImSh95, Adventure10, Rounak_iitr, cubres, and 1 other user
Solution
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G
H
=
a