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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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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
trigonometric inequality
MATH1945   11
N 2 minutes ago by sqing
Source: ?
In triangle $ABC$, prove that $$sin^2(A)+sin^2(B)+sin^2(C) \leq \frac{9}{4}$$
11 replies
1 viewing
MATH1945
May 26, 2016
sqing
2 minutes ago
Centrally symmetric polyhedron
genius_007   0
4 minutes ago
Source: unknown
Does there exist a convex polyhedron with an odd number of sides, where each side is centrally symmetric?
0 replies
genius_007
4 minutes ago
0 replies
Combinatorial Game
Cats_on_a_computer   2
N 14 minutes ago by Cats_on_a_computer

Let n>1 be odd. A row of n spaces is initially empty. Alice and Bob alternate moves (Alice first); on each turn a player may either
1. Place a stone in any empty space, or
2. Remove a stone from a non-empty space S, then (if they exist) place stones in the nearest empty spaces immediately to the left and to the right of S.

Furthermore, no move may produce a position that has appeared earlier. The player loses when they cannot make a legal move.
Assuming optimal play, which move(s) can Alice make on her first turn?
2 replies
Cats_on_a_computer
28 minutes ago
Cats_on_a_computer
14 minutes ago
2-var inequality
sqing   4
N 17 minutes ago by sqing
Source: Own
Let $ a,b> 0 $ and $2a+2b+ab=5. $ Prove that
$$ \frac{a^2}{b^2}+\frac{1}{a^2}-a^2\geq  1$$$$ \frac{a^3}{b^3}+\frac{1}{a^3}-a^3\geq  1$$
4 replies
sqing
an hour ago
sqing
17 minutes ago
A game of digits and seventh powers
v_Enhance   28
N 24 minutes ago by blueprimes
Source: Taiwan 2014 TST3 Quiz 1, P2
Alice and Bob play the following game. They alternate selecting distinct nonzero digits (from $1$ to $9$) until they have chosen seven such digits, and then consider the resulting seven-digit number by concatenating the digits in the order selected, with the seventh digit appearing last (i.e. $\overline{A_1B_2A_3B_4A_6B_6A_7}$). Alice wins if and only if the resulting number is the last seven decimal digits of some perfect seventh power. Please determine which player has the winning strategy.
28 replies
v_Enhance
Jul 18, 2014
blueprimes
24 minutes ago
n containers distribute gas to a car for a single loop along the track
parmenides51   16
N 33 minutes ago by v_Enhance
Source: Spanish Mathematical Olympiad 1997 P6
The exact quantity of gas needed for a car to complete a single loop around a track is distributed among $n$ containers placed along the track. Prove that there exists a position starting at which the car, beginning with an empty tank of gas, can complete a loop around the track without running out of gas. The tank of gas is assumed to be large enough.
16 replies
parmenides51
Jul 31, 2018
v_Enhance
33 minutes ago
Factorial: n!|a^n+1
Nima Ahmadi Pour   68
N 39 minutes ago by blueprimes
Source: IMO Shortlist 2005 N4, Iran preparation exam
Find all positive integers $ n$ such that there exists a unique integer $ a$ such that $ 0\leq a < n!$ with the following property:
\[ n!\mid a^n + 1
\]

Proposed by Carlos Caicedo, Colombia
68 replies
Nima Ahmadi Pour
Apr 24, 2006
blueprimes
39 minutes ago
Geometry problem
Whatisthepurposeoflife   1
N an hour ago by Royal_mhyasd
Source: Derived from MEMO 2024 I3
Triangle ∆ABC is scalene the circle w that goes through the points A and B intersects AC at E BC at D let the Lines BE and AD intersect at point F. And let the tangents A and B of circle w Intersect at point G.
Prove that C F and G are collinear
1 reply
Whatisthepurposeoflife
an hour ago
Royal_mhyasd
an hour ago
Inspired by SunnyEvan
sqing   2
N an hour ago by SunnyEvan
Source: Own
Let $ a,b,c   $ be reals such that $ a^2+b^2+c^2=2(ab+bc+ca). $ Prove that$$ \frac{1}{12} \leq \frac{a^2b+b^2c+c^2a}{(a+b+c)^3} \leq \frac{5}{36} $$Let $ a,b,c   $ be reals such that $ a^2+b^2+c^2=5(ab+bc+ca). $ Prove that$$ -\frac{1}{25} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{197}{675} $$
2 replies
sqing
May 17, 2025
SunnyEvan
an hour ago
\frac{x^2}{x+y}+\frac{y^2}{1-x}+\frac{(1-x-y)^2}{1-y}\geq\frac{1}{2} if 0<x,y<1
parmenides51   6
N an hour ago by JH_K2IMO
Source: KJMO 2009 p3
For two arbitrary reals $x, y$ which are larger than $0$ and less than $1.$ Prove that$$\frac{x^2}{x+y}+\frac{y^2}{1-x}+\frac{(1-x-y)^2}{1-y}\geq\frac{1}{2}.$$
6 replies
parmenides51
May 2, 2019
JH_K2IMO
an hour ago
3 var inequality
SunnyEvan   9
N an hour ago by SunnyEvan
Let $ a,b,c \in R $ ,such that $ a^2+b^2+c^2=4(ab+bc+ca)$Prove that :$$ \frac{7-2\sqrt{14}}{48} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{7+2\sqrt{14}}{48} $$
9 replies
SunnyEvan
May 17, 2025
SunnyEvan
an hour ago
2-var inequality
sqing   6
N an hour ago by sqing
Source: Own
Let $ a,b>0 , a^2+b^2-ab\leq 1 . $ Prove that
$$a^3+b^3 -\frac{a^4}{b+1}  -\frac{b^4}{a+1} \leq 1 $$
6 replies
sqing
Yesterday at 2:15 AM
sqing
an hour ago
A sharp one with 3 var (3)
mihaig   2
N 2 hours ago by mihaig
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$a^2+b^2+c^2+5abc\geq8.$$
2 replies
mihaig
Yesterday at 5:17 PM
mihaig
2 hours ago
an equation from the a contest
alpha31415   2
N 2 hours ago by User141208
Find all (complex) roots of the equation:
(z^2-z)(1-z+z^2)^2=-1/7
2 replies
alpha31415
May 21, 2025
User141208
2 hours ago
Indian Mathematical Olympiad 1986 - Problem 9
Leon   7
N Apr 14, 2021 by denery
Show that among all quadrilaterals of a given perimeter the square has the largest area.
7 replies
Leon
Aug 13, 2009
denery
Apr 14, 2021
Indian Mathematical Olympiad 1986 - Problem 9
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G H BBookmark kLocked kLocked NReply
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Leon
256 posts
#1 • 3 Y
Y by samrocksnature, Adventure10, Mango247
Show that among all quadrilaterals of a given perimeter the square has the largest area.
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Kunihiko_Chikaya
14514 posts
#2 • 3 Y
Y by vp2016, Adventure10, Mango247
Let the quadrilateral has side length $ a,\ b,\ c,\ d$ with $ p=\frac{a+b+c+d}{2}$,

we have $ [ABCD]=\sqrt{(p-a)(p-b)(p-c)(p-d)}=(\sqrt[4]{(p-a)(p-b)(p-c)(p-d)})^2$

$ \leq \left\{\frac{(p-a)+(p-b)+(p-c)+(p-d)}{4}\right\}^2=\frac{p^2}{4}$,

the equality holds when $ a=b=c=d$.
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Bugi
1857 posts
#3 • 4 Y
Y by Adventure10, Mango247, Mango247, Mango247
I think that formula holds only if the quadrilateral is cyclic.

But we can do like this:

Solution
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Kunihiko_Chikaya
14514 posts
#4 • 1 Y
Y by Adventure10
Bugi wrote:
I think that formula holds only if the quadrilateral is cyclic.

You are right.

Related problem

1969 Czechoslovakia Mathematical Olympiad

A convex quadrilateral $ ABCD$ with sides $ AB = a,\ BC = b,\ CD = c,\ DA = d$ and angles $ \alpha = \angle{DAB},\ \beta = \angle{ABC},\ \gamma = \angle{BCD},\ \delta = \angle{CDA}$
is given. Let $ \displaystyle s = \frac {a + b + c + d}{2}$ and $ P$ be the area of the quadrilateral. Prove that
\[ P^2 = (s - a)(s - b)(s - c)(s - d) - abcd\cos ^ 2 \displaystyle \frac {\alpha + \gamma}{2}.\]
Now the name of the nation changed,\ didn't that?

It occured to me that Ellipse has something to do with the solution.
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Mathias_DK
1312 posts
#5 • 3 Y
Y by Titttu, Adventure10, Mango247
Leon wrote:
Show that among all quadrilaterals of a given perimeter the square has the largest area.
Lemma 1:
Given a triangle $ \triangle ABC$, with $ BC$ and $ AB+AC$ fixed, then the area of $ \triangle ABC$ is largest when $ AB=AC$.
Proof:
The locus of all points $ P$ such that $ PB+PC = AB+AC$ is an ellipse. Since the area of $ \triangle PBC$ is the distance from the line $ BC$ to $ P$ times $ BC$, we see that $ \triangle PBC$ has the maximal area when $ P$ are furthest away from $ BC$. Because of symmetry this must be when $ P$ lies on the bisector of $ BC$, and hence $ AB=AC$.

Let $ \square ABCD$ be the quadrilateral with the largest area. Using lemma 1 on $ \triangle ABC$ we can wlog assume that $ AB=BC$. And using on $ \triangle ADC$: wlog $ AD=DC$. Hence $ \triangle BCD$ is concurrent to $ \triangle DAB$. So it is enough to find the maximal area of $ \triangle BCD$ when $ BC+CD$ is fixed. Using Lemma 1, we see that we can wlog assume that $ BC=CD$, so $ \square ABCD$ must be a rhombus. Let $ E$ be the projection of $ C$ on $ BD$. Then $ \triangle BEC$ is a rightangled triangle, with $ BC$ fixed. Now $ BE^2+EC^2 = BC^2$ is also fixed, and the area of $ \triangle BEC$ is $ \frac{1}{2}BE \times EC$. Since $ BE^2 + EC^2$ is fixed we see that $ BE \times EC$ is largest whenever $ BE=EC$.(Wellknown) Doing this for $ \triangle DEC$ also we see that $ \angle BCD = 90^\circ$, and since we could use this symmetric argument for all the angles in the quadrilateral, $ \square ABCD$ must be a square.
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Maharjun
109 posts
#6 • 3 Y
Y by lckihi, Adventure10, Mango247
I have made use of the result that if the sum of two, or more positive quantities is constant. then their product is maximum when they are all equal. For two quantities, it can be proven using A.M G.M as follows-

Click here if interested in proof of above inequality for two quantities

Step 1:(prove that the quadrilateral is a rhombus,using hero's formula for area of triangle)
solution

step 2 -(prove that the rhombus is a square, trivial)
solution
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guptaamitu1
658 posts
#7
Y by
Maharjun wrote:
I have made use of the result that if the sum of two, or more positive quantities is constant. then their product is maximum when they are all equal. For two quantities, it can be proven using A.M G.M as follows-

Click here if interested in proof of above inequality for two quantities

Step 1:(prove that the quadrilateral is a rhombus,using hero's formula for area of triangle)
solution

step 2 -(prove that the rhombus is a square, trivial)
solution

I think that this solution is wrong (because according to me, step 1 of your solution is not correct). Basically, you are there trying to show that if we replace $(a,b)$ by $({a + b\over 2},{a + b\over 2})$ then the area of the quadrilateral will increase. But that is not always true. Since if you do this replacement, then the value of $e$ also gets effected. So you cannot guarantee that the area of the quadrilateral will increase.
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denery
180 posts
#8
Y by
imagine a person using jensen" s inequality to prove it area of quadrilateral is a convex function :D
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