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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
inequality
xytunghoanh   3
N 7 minutes ago by sqing
For $a,b,c\ge 0$. Let $a+b+c=3$.
Prove or disprove
\[\sum ab +\sum ab^2 \le 6\]
3 replies
xytunghoanh
2 hours ago
sqing
7 minutes ago
f(f(n))=2n+2
Jackson0423   1
N 17 minutes ago by jasperE3
Source: 2013 KMO Second Round

Let \( f : \mathbb{N} \to \mathbb{N} \) be a function satisfying the following conditions for all \( n \in \mathbb{N} \):
\[
\begin{cases}
f(n+1) > f(n) \\
f(f(n)) = 2n + 2
\end{cases}
\]Find the value of \( f(2013) \).
1 reply
Jackson0423
Yesterday at 4:07 PM
jasperE3
17 minutes ago
Proving ZA=ZB
nAalniaOMliO   8
N an hour ago by Mathgloggers
Source: Belarusian National Olympiad 2025
Point $H$ is the foot of the altitude from $A$ of triangle $ABC$. On the lines $AB$ and $AC$ points $X$ and $Y$ are marked such that the circumcircles of triangles $BXH$ and $CYH$ are tangent, call this circles $w_B$ and $w_C$ respectively. Tangent lines to circles $w_B$ and $w_C$ at $X$ and $Y$ intersect at $Z$.
Prove that $ZA=ZH$.
Vadzim Kamianetski
8 replies
nAalniaOMliO
Mar 28, 2025
Mathgloggers
an hour ago
Hard geometry
Lukariman   1
N an hour ago by Lukariman
Given circle (O) and chord AB with different diameters. The tangents of circle (O) at A and B intersect at point P. On the small arc AB, take point C so that triangle CAB is not isosceles. The lines CA and BP intersect at D, BC and AP intersect at E. Prove that the centers of the circles circumscribing triangles ACE, BCD and OPC are collinear.
1 reply
Lukariman
an hour ago
Lukariman
an hour ago
Incircle triangles inequality
MathMystic33   1
N an hour ago by Quantum-Phantom
Source: 2025 Macedonian Team Selection Test P5
Let $\triangle ABC$ be a triangle with side‐lengths $a,b,c$, incenter $I$, and circumradius $R$. Denote by $P$ the area of $\triangle ABC$, and let $P_1,\;P_2,\;P_3$ be the areas of triangles $\triangle ABI$, $\triangle BCI$, and $\triangle CAI$, respectively. Prove that
\[
\frac{abc}{12R}
\;\le\;
\frac{P_1^2 + P_2^2 + P_3^2}{P}
\;\le\;
\frac{3R^3}{4\sqrt[3]{abc}}.
\]
1 reply
MathMystic33
Yesterday at 6:06 PM
Quantum-Phantom
an hour ago
Concurrency of tangent touchpoint lines on thales circles
MathMystic33   2
N an hour ago by Diamond-jumper76
Source: 2024 Macedonian Team Selection Test P4
Let $\triangle ABC$ be an acute scalene triangle. Denote by $k_A$ the circle with diameter $BC$, and let $B_A,C_A$ be the contact points of the tangents from $A$ to $k_A$, chosen so that $B$ and $B_A$ lie on opposite sides of $AC$ and $C$ and $C_A$ lie on opposite sides of $AB$. Similarly, let $k_B$ be the circle with diameter $CA$, with tangents from $B$ touching at $C_B,A_B$, and $k_C$ the circle with diameter $AB$, with tangents from $C$ touching at $A_C,B_C$.
Prove that the lines $B_AC_A, C_BA_B, A_CB_C$ are concurrent.
2 replies
MathMystic33
Yesterday at 7:41 PM
Diamond-jumper76
an hour ago
Find all possible values of q-p
yunxiu   18
N 2 hours ago by Jupiterballs
Source: 2012 European Girls’ Mathematical Olympiad P5
The numbers $p$ and $q$ are prime and satisfy
\[\frac{p}{{p + 1}} + \frac{{q + 1}}{q} = \frac{{2n}}{{n + 2}}\]
for some positive integer $n$. Find all possible values of $q-p$.

Luxembourg (Pierre Haas)
18 replies
yunxiu
Apr 13, 2012
Jupiterballs
2 hours ago
Inspired by nhathhuyyp5c
sqing   3
N 2 hours ago by sqing
Source: Own
Let $ x,y>0, x+2y- 3xy\leq 4. $Prove that
$$ \frac{1}{x^2} + 2y^2 + 3x + \frac{4}{y} \geq 3\left(2+\sqrt[3]{\frac{9}{4} }\right)$$
3 replies
sqing
3 hours ago
sqing
2 hours ago
Easy but Nice 12
TelvCohl   1
N 2 hours ago by Luis González
Source: Own
Given a $ \triangle ABC $ with orthocenter $ H $ and a point $ P $ lying on the Euler line of $ \triangle ABC. $ Prove that the midpoint of $ PH $ lies on the Thomson cubic of the pedal triangle of $ P $ WRT $ \triangle ABC. $
1 reply
TelvCohl
Mar 8, 2025
Luis González
2 hours ago
Similar Problems
Saucepan_man02   2
N 2 hours ago by quasar_lord
Could anyone post some problems which are similar to the below problem:

Find the real solution of: $$x^9+9/8 x^6+27/64 x^3-x+219/512.$$
Sol(outline)
2 replies
Saucepan_man02
May 12, 2025
quasar_lord
2 hours ago
Inspired by old results
sqing   0
2 hours ago
Source: Own
Let $ a,b,c>0 $ . Prove that
$$\frac{a+kb}{b+c}+\frac{b+kc}{c+a}+\frac{c+ka}{a+b}\geq \frac{3(k+1)}{2}$$W here $-1 \leq k \leq  \frac{537}{90}.$
0 replies
sqing
2 hours ago
0 replies
orthocenter on sus circle
DVDTSB   2
N 2 hours ago by Diamond-jumper76
Source: Romania TST 2025 Day 2 P1
Let \( ABC \) be an acute triangle with \( AB < AC \), and let \( O \) be the center of its circumcircle. Let \( A' \) be the reflection of \( A \) with respect to \( BC \). The line through \( O \) parallel to \( BC \) intersects \( AC \) at \( F \), and the tangent at \( F \) to the circle \( \odot(BFC) \) intersects the line through \( A' \) parallel to \( BC \) at point \( M \). Let \( K \) be a point on the ray \( AB \), starting at \( A \), such that \( AK = 4AB \).
Show that the orthocenter of triangle \( ABC \) lies on the circle with diameter \( KM \).

Proposed by Radu Lecoiu

2 replies
DVDTSB
Yesterday at 12:18 PM
Diamond-jumper76
2 hours ago
problem 5
termas   74
N 2 hours ago by maromex
Source: IMO 2016
The equation
$$(x-1)(x-2)\cdots(x-2016)=(x-1)(x-2)\cdots (x-2016)$$is written on the board, with $2016$ linear factors on each side. What is the least possible value of $k$ for which it is possible to erase exactly $k$ of these $4032$ linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?
74 replies
termas
Jul 12, 2016
maromex
2 hours ago
I think I know why this problem was rejected by IMO PSC several times...
mshtand1   1
N 3 hours ago by sarjinius
Source: Ukrainian Mathematical Olympiad 2025. Day 2, Problem 11.8
Exactly $102$ country leaders arrived at the IMO. At the final session, the IMO chairperson wants to introduce some changes to the regulations, which the leaders must approve. To pass the changes, the chairperson must gather at least \(\frac{2}{3}\) of the votes "FOR" out of the total number of leaders. Some leaders do not attend such meetings, and it is known that there will be exactly $81$ leaders present. The chairperson must seat them in a square-shaped conference hall of size \(9 \times 9\), where each leader will be seated in a designated \(1 \times 1\) cell. It is known that exactly $28$ of these $81$ leaders will surely support the chairperson, i.e., they will always vote "FOR." All others will vote as follows: At the last second of voting, they will look at how their neighbors voted up to that moment — neighbors are defined as leaders seated in adjacent cells \(1 \times 1\) (sharing a side). If the majority of neighbors voted "FOR," they will also vote "FOR." If there is no such majority, they will vote "AGAINST." For example, a leader seated in a corner of the hall has exactly $2$ neighbors and will vote "FOR" only if both of their neighbors voted "FOR."

(a) Can the IMO chairperson arrange their $28$ supporters so that they vote "FOR" in the first second of voting and thereby secure a "FOR" vote from at least \(\frac{2}{3}\) of all $102$ leaders?

(b) What is the maximum number of "FOR" votes the chairperson can obtain by seating their 28 supporters appropriately?

Proposed by Bogdan Rublov
1 reply
mshtand1
Mar 14, 2025
sarjinius
3 hours ago
Find an angle
Rushil   15
N Oct 30, 2024 by namanrobin08
Source: Indian RMO 1998 Problem 1
Let $ABCD$ be a convex quadrilateral in which $\angle BAC = 50^{\circ}, \angle CAD = 60^{\circ}$and $\angle BDC = 25^{\circ}$. If $E$ is the point of intersection of $AC$ and $BD$, find $\angle AEB$.
15 replies
Rushil
Oct 26, 2005
namanrobin08
Oct 30, 2024
Find an angle
G H J
G H BBookmark kLocked kLocked NReply
Source: Indian RMO 1998 Problem 1
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Rushil
1592 posts
#1 • 3 Y
Y by Biographicsmean, Adventure10, Mango247
Let $ABCD$ be a convex quadrilateral in which $\angle BAC = 50^{\circ}, \angle CAD = 60^{\circ}$and $\angle BDC = 25^{\circ}$. If $E$ is the point of intersection of $AC$ and $BD$, find $\angle AEB$.
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frt
1294 posts
#2 • 6 Y
Y by raltz, mihirb, opptoinfinity, Rushil_india, Adventure10, Mango247
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srulikbd
400 posts
#3 • 1 Y
Y by Adventure10
could you explain me please why does Since $\angle BDC=\frac{1}{2}\angle BAC$, $A$ is the circumcenter of $\triangle BCD$?
thanks :)
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D.P.L
223 posts
#4 • 2 Y
Y by Adventure10, Mango247
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aznluster
570 posts
#5 • 2 Y
Y by Adventure10, Mango247
frt wrote:
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Can you direct me to the proof of your first statement?
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geoistrivial
41 posts
#6 • 2 Y
Y by Excursion, Adventure10
aznluster wrote:
frt wrote:
Click to reveal hidden text

Can you direct me to the proof of your first statement?

It is actually a false statement. It is true that if a central angle and inscribed angle of a circle both cut the same arc, the central angle has twice the value of the inscribed angle. However, his statement is false because there are many other points that satisfy such constraints. Indeed, if you draw a diagram, A is clearly not the circumcenter.
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sunken rock
4394 posts
#7 • 1 Y
Y by Adventure10
Draw the circle $\odot (ABC)$, take $O$ the midpoint of its arc $BC$ containing $A$, draw the circle $(O, OB)$.
From $A$, draw a halfline $(Ax)$, directed to the halfplane not containing $B$, such as $\angle CAx=60^\circ$, it will intersect the second circle at $D$; see that for many $A$ on the first circle will find a $D$ on the second one, and the required angle is not constant.

Best regards,
sunken rock
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geoistrivial
41 posts
#8 • 2 Y
Y by Adventure10, Mango247
I believe this problem does not give enough information. Perhaps this is the real one: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=150&t=451779
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10000th User
3049 posts
#9 • 1 Y
Y by Adventure10
Surprisingly, it turns out that $A$ IS indeed the circumcenter of $\triangle BCD$ even with the additional information! :what?:

EDIT: Yes, the additional information is necessary because I was able to construct two different quadrilaterals with Rushil's conditions.
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Sayan
2130 posts
#10 • 1 Y
Y by Adventure10
No extra condition is necessary. A trigonometric approach shows that $\angle AEB = 95^\circ$
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coolcheetah157
139 posts
#11 • 1 Y
Y by Adventure10
@Sayan - what trig approach did you use?
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tst
118 posts
#12 • 2 Y
Y by Adventure10, Mango247
A would be the circumcentre because DC chord and BC chord subtend double angle at A....
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amar_04
1915 posts
#14 • 7 Y
Y by AlastorMoody, Path_to_Almighty, stranger_02, srijonrick, Bumblebee60, Adventure10, Mango247
Not a proper solution has been posted here, so I'm posting, sorry for bumping this thread.

Extend $CA$ to a point $K$ such that $2\angle BKD=\angle BAD=55^\circ$ and as $\angle BCD=125^\circ$. So, $BCDK$ is a cyclic quadrilateral. So, $\angle BKC=25^\circ\implies \angle KBA=25^\circ$. So, $KA=AB$. Similarly $KA=AD$. So, $AB=AD$. So,$\angle ADB=35^\circ\implies \angle AEB=95^\circ$.
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srijonrick
168 posts
#15
Y by
Sorry for bumping this thread again, but in the book An Excursion in Mathematics an extra condition is given: $\angle CBD=30^{\circ}$, which is not there in the question, mods kindly add it.
P.S:- The source (the book) is really authentic. :)
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Hexagon_6-
23 posts
#16
Y by
This is tough for early student , but legend first notice that angle BDC=1/2CAB >> A is centre now you feel easy
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namanrobin08
99 posts
#17
Y by
Please note that if the angle subtended by two points $P, Q$ at $X$ is half of that subtended by them at some other point $Y$, it does NOT imply that $X$ is the circumcentre of $ABY$.

But in the present question, it happens to be the circumcentre. This is NOT because that half angle thing, but becaue that half angle thing happens twice at the same point.
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