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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
polynomial problem from russia 2002
vineet   8
N 2 minutes ago by LawofCosine
Source: 2002 All-Russian MO, Grade 10, Problem 1
The polynomials $P$, $Q$, $R$ with real coefficients, one of which is degree $2$ and two of degree $3$, satisfy the equality $P^2+Q^2=R^2$. Prove that one of the polynomials of degree $3$ has three real roots.
8 replies
vineet
Feb 25, 2003
LawofCosine
2 minutes ago
Find the minimum
sqing   32
N 4 minutes ago by sqing
Source: Zhangyanzong
Let $a,b$ be positive real numbers such that $a^2b^2+\frac{4a}{a+b}=4.$ Find the minimum value of $a+2b.$
32 replies
+1 w
sqing
Sep 4, 2018
sqing
4 minutes ago
IMO Shortlist 2012, Geometry 2
lyukhson   89
N 8 minutes ago by ezpotd
Source: IMO Shortlist 2012, Geometry 2
Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$. The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$. Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$. Prove that $D,H,F,G$ are concyclic.
89 replies
lyukhson
Jul 29, 2013
ezpotd
8 minutes ago
2025 Xinjiang High School Mathematics Competition Q11
sqing   3
N 33 minutes ago by sqing
Source: China
Let $ a,b,c >0  . $ Prove that
$$  \left(1+\frac {a} { b}\right)\left(1+\frac {a}{ b}+\frac {b}{ c}\right) \left(1+\frac {a}{b}+\frac {b}{ c}+\frac {c}{ a}\right)  \geq 16 $$
3 replies
sqing
Saturday at 4:30 PM
sqing
33 minutes ago
IMO Shortlist 2010 - Problem G7
Amir Hossein   21
N 37 minutes ago by MathLuis
Three circular arcs $\gamma_1, \gamma_2,$ and $\gamma_3$ connect the points $A$ and $C.$ These arcs lie in the same half-plane defined by line $AC$ in such a way that arc $\gamma_2$ lies between the arcs $\gamma_1$ and $\gamma_3.$ Point $B$ lies on the segment $AC.$ Let $h_1, h_2$, and $h_3$ be three rays starting at $B,$ lying in the same half-plane, $h_2$ being between $h_1$ and $h_3.$ For $i, j = 1, 2, 3,$ denote by $V_{ij}$ the point of intersection of $h_i$ and $\gamma_j$ (see the Figure below). Denote by $\widehat{V_{ij}V_{kj}}\widehat{V_{kl}V_{il}}$ the curved quadrilateral, whose sides are the segments $V_{ij}V_{il},$ $V_{kj}V_{kl}$ and arcs $V_{ij}V_{kj}$ and $V_{il}V_{kl}.$ We say that this quadrilateral is $circumscribed$ if there exists a circle touching these two segments and two arcs. Prove that if the curved quadrilaterals $\widehat{V_{11}V_{21}}\widehat{V_{22}V_{12}}, \widehat{V_{12}V_{22}}\widehat{V_{23}V_{13}},\widehat{V_{21}V_{31}}\widehat{V_{32}V_{22}}$ are circumscribed, then the curved quadrilateral $\widehat{V_{22}V_{32}}\widehat{V_{33}V_{23}}$ is circumscribed, too.

Proposed by Géza Kós, Hungary

IMAGE
21 replies
Amir Hossein
Jul 17, 2011
MathLuis
37 minutes ago
Interesting inequality
sqing   5
N 44 minutes ago by sqing
Source: Own
Let $a,b\geq 0, 2a+2b+ab=5.$ Prove that
$$a+b^3+a^3b+\frac{101}{8}ab\leq\frac{125}{8}$$
5 replies
sqing
2 hours ago
sqing
44 minutes ago
Number of functions satisfying sum inequality
CyclicISLscelesTrapezoid   19
N an hour ago by john0512
Source: ISL 2022 C5
Let $m,n \geqslant 2$ be integers, let $X$ be a set with $n$ elements, and let $X_1,X_2,\ldots,X_m$ be pairwise distinct non-empty, not necessary disjoint subset of $X$. A function $f \colon X \to \{1,2,\ldots,n+1\}$ is called nice if there exists an index $k$ such that \[\sum_{x \in X_k} f(x)>\sum_{x \in X_i} f(x) \quad \text{for all } i \ne k.\]Prove that the number of nice functions is at least $n^n$.
19 replies
CyclicISLscelesTrapezoid
Jul 9, 2023
john0512
an hour ago
Inequality em981
oldbeginner   21
N an hour ago by sqing
Source: Own
Let $a, b, c>0, a+b+c=3$. Prove that
\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]
21 replies
oldbeginner
Sep 22, 2016
sqing
an hour ago
Interesting inequality
sqing   3
N an hour ago by sqing
Source: Own
Let $ (a+b)^2+(a-b)^2=1. $ Prove that
$$0\geq (a+b-1)(a-b+1)\geq -\frac{3}{2}-\sqrt 2$$$$ -\frac{9}{2}+2\sqrt 2\geq (a+b-2)(a-b+2)\geq -\frac{9}{2}-2\sqrt 2$$
3 replies
sqing
2 hours ago
sqing
an hour ago
Simple triangle geometry [a fixed point]
darij grinberg   50
N an hour ago by ezpotd
Source: German TST 2004, IMO ShortList 2003, geometry problem 2
Three distinct points $A$, $B$, and $C$ are fixed on a line in this order. Let $\Gamma$ be a circle passing through $A$ and $C$ whose center does not lie on the line $AC$. Denote by $P$ the intersection of the tangents to $\Gamma$ at $A$ and $C$. Suppose $\Gamma$ meets the segment $PB$ at $Q$. Prove that the intersection of the bisector of $\angle AQC$ and the line $AC$ does not depend on the choice of $\Gamma$.
50 replies
darij grinberg
May 18, 2004
ezpotd
an hour ago
Inspired by RMO 2006
sqing   6
N an hour ago by sqing
Source: Own
Let $ a,b >0  . $ Prove that
$$  \frac {a^{2}+1}{b+k}+\frac { b^{2}+1}{ka+1}+\frac {2}{a+kb}  \geq \frac {6}{k+1}  $$Where $k\geq 0.03 $
$$  \frac {a^{2}+1}{b+1}+\frac { b^{2}+1}{a+1}+\frac {2}{a+b}  \geq 3  $$
6 replies
sqing
Saturday at 3:24 PM
sqing
an hour ago
IMO 2009, Problem 2
orl   143
N an hour ago by ezpotd
Source: IMO 2009, Problem 2
Let $ ABC$ be a triangle with circumcentre $ O$. The points $ P$ and $ Q$ are interior points of the sides $ CA$ and $ AB$ respectively. Let $ K,L$ and $ M$ be the midpoints of the segments $ BP,CQ$ and $ PQ$. respectively, and let $ \Gamma$ be the circle passing through $ K,L$ and $ M$. Suppose that the line $ PQ$ is tangent to the circle $ \Gamma$. Prove that $ OP = OQ.$

Proposed by Sergei Berlov, Russia
143 replies
orl
Jul 15, 2009
ezpotd
an hour ago
A sharp one with 3 var (2)
mihaig   0
2 hours ago
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$a+b+c+\sqrt{abc}\geq4.$$
0 replies
mihaig
2 hours ago
0 replies
f(1)f(2)...f(n) has at most n prime factors
MarkBcc168   39
N 2 hours ago by cursed_tangent1434
Source: 2020 Cyberspace Mathematical Competition P2
Let $f(x) = 3x^2 + 1$. Prove that for any given positive integer $n$, the product
$$f(1)\cdot f(2)\cdot\dots\cdot f(n)$$has at most $n$ distinct prime divisors.

Proposed by Géza Kós
39 replies
MarkBcc168
Jul 15, 2020
cursed_tangent1434
2 hours ago
binomial sum ratio
thewayofthe_dragon   3
N Apr 23, 2025 by P162008
Source: YT
Someone please evaluate this ratio inside the log for any given n(I feel the sum doesn't have any nice closed form).
3 replies
thewayofthe_dragon
Jun 16, 2024
P162008
Apr 23, 2025
binomial sum ratio
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G H BBookmark kLocked kLocked NReply
Source: YT
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thewayofthe_dragon
38 posts
#1
Y by
Someone please evaluate this ratio inside the log for any given n(I feel the sum doesn't have any nice closed form).
Attachments:
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thewayofthe_dragon
38 posts
#2
Y by
Can someone solve?
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P162008
211 posts
#3
Y by
thewayofthe_dragon wrote:
Can someone solve?

The argument of the logarithm equals to 1
So, option C
This post has been edited 1 time. Last edited by P162008, Apr 22, 2025, 1:09 PM
Reason: Typo
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P162008
211 posts
#4
Y by
thewayofthe_dragon wrote:
Can someone solve?

Let $\zeta = \sum_{i=0}^{r} \binom{n}{2i}\binom{n-2i}{r-i}$

Now, define a generating function $\alpha(x)$ as $\alpha(x) = \sum_{r=0}^{\infty} \zeta x^r = \sum_{r=0}^{\infty} \sum_{i=0}^{r} \binom{n}{2i} \binom{n-2i}{r-i} x^r$

On swapping the order of summation, we get

$\alpha(x) = \sum_{i=0}^{\infty} \sum_{r=i}^{\infty} \binom{n}{2i} \binom{n-2i}{r-i} x^r = \sum_{i=0}^{\infty} \binom{n}{2i} x^i \sum_{r=i}^{\infty} \binom{n-2i}{r-i} x^{r-i}$

Let $r - i = t$ then $0 \leqslant r - i < \infty$ as $ i \leqslant r < \infty$

Now, $\alpha(x) = \sum_{i=0}^{\infty} \binom{n}{2i} x^i \sum_{t=0}^{\infty} \binom{n-2i}{t} x^t = \sum_{i=0}^{\infty} \binom{n}{2i} x^i (1 + x)^{n - 2i}
= (1 + x)^n \sum_{i=0}^{\infty} \binom{n}{2i} \left(\frac{\sqrt{x}}{1 + x}\right)^{2i}$

Then, $\alpha(x) = (1 + x)^n \left[\frac{(1 + \frac{\sqrt{x}}{1 + x})^n + (1 - \frac{\sqrt{x}}{1 + x})^n}{2}\right] = \frac{1}{2} \left[(1 + \sqrt{x} + x)^n + (1 - \sqrt{x} + x)^n\right]$

$\boxed{\therefore \zeta = \left[x^r\right] \frac{1}{2} \left[(1 + \sqrt{x} + x)^n + (1 - \sqrt{x} + x)^n\right]}$

Similarly, Let $\xi = \sum_{i=r}^{n} \binom{n}{i} \binom{2i}{2r} \left(\frac{1}{2}\right)^{2i - 2r} \left(\frac{3}{4}\right)^{n- i}$

Again, define a generating function $\beta(x)$ as $\beta(x) = \sum_{r=0}^{\infty} \xi x^r = \sum_{r=0}^{\infty} \sum_{i=r}^{n} \binom{n}{i} \binom{2i}{2r} \left(\frac{1}{2}\right)^{2i-2r} \left(\frac{3}{4}\right)^{n- i} x^r$

On swapping the order of summation, we get

$\beta(x) = \sum_{i=0}^{\infty} \sum_{r=0}^{i} \binom{n}{i} \binom{2i}{2r} \left(\frac{1}{2}\right)^{2i-2r} \left(\frac{3}{4}\right)^{n- i} x^r = \sum_{i=0}^{\infty} \binom{n}{i} \left(\frac{3}{4}\right)^{n - i} \left(\frac{1}{4}\right)^i \sum_{r=0}^{i} \binom{2i}{2r} \left(2\sqrt{x}\right)^{2r}$

$\beta(x) = \sum_{i=0}^{\infty} \binom{n}{i} \left(\frac{3}{4}\right)^{n - i} \left(\frac{1}{4}\right)^i \left[\frac{(1 + 2\sqrt{x})^{2i} + (1 - 2\sqrt{x})^{2i}}{2}\right] = \sum_{i=0}^{\infty} \binom{n}{i} \left(\frac{3}{4}\right)^{n - i} \left(\frac{1}{4}\right)^i \left[\frac{(4x + 4\sqrt{x} + 1)^i + (4x - 4\sqrt{x} + 1)^i}{2}\right]$

Then, $\beta(x) = \frac{1}{2} \left[\sum_{i=0}^{\infty} \binom{n}{i} \left(\frac{3}{4}\right)^{n-i} \left(x + \sqrt{x} + \frac{1}{4}\right)^i + \sum_{i=0}^{\infty} \binom{n}{i} \left(\frac{3}{4}\right)^{n-i} \left(x - \sqrt{x} + \frac{1}{4}\right)^i \right] = \frac{1}{2} \left[(1 + \sqrt{x} + x)^n + (1 - \sqrt{x} + x)^n\right]$

$\boxed{\therefore \xi = \left[x^r\right] \frac{1}{2} \left[(1 + \sqrt{x} + x)^n + (1 - \sqrt{x} + x)^n\right]}$

Finally, $\boxed{\log_{10} \left(\frac{\zeta}{\xi}\right) = \log_{10} 1 = \boxed{0}}$ :love:
This post has been edited 18 times. Last edited by P162008, Apr 28, 2025, 2:08 PM
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