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k a AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
There is still time to train for the November 6th and November 12th AMC 10A/12A and AMC 10B/12B, respectively! Enroll in our weekend seminars to be held on November 2nd and 3rd (listed below) and you will learn problem strategies, test taking techniques, and be able to take a full practice test! Note that the “B” seminars will have different material from the “A” seminars which were held in October.

[list][*]Special AMC 10 Problem Seminar B
[*]Special AMC 12 Problem Seminar B[/list]
For those who want to take a free practice test before the AMC 10/12 competitions, you can simulate a real competition experience by following this link. As you assess your performance on these exams, be sure to gather data!

[list][*]Which problems did you get right?
[list][*]Was the topic a strength (e.g. number theory, geometry, counting/probability, algebra)?
[*]How did you prepare?
[*]What was your confidence level?[/list]
[*]Which problems did you get wrong?
[list][list][*]Did you make an arithmetic error?
[*]Did you misread the problem?
[*]Did you have the foundational knowledge for the problem?
[*]Which topics require more fluency through practice (e.g. number theory, geometry, counting/probability, algebra)?
[*]Did you run out of time?[/list][/list]
Once you have analyzed the results with the above questions, you will have a plan of attack for future contests! BEST OF LUCK to all competitors at this year’s AMC 10 and AMC 12!

Did you know that the day after both the AMC 10A/12A and AMC 10B/12B you can join a free math jam where our AoPS team will go over the most interesting problems? Find the schedule below under “Mark your calendars”.

Mark your calendars for these upcoming free math jams!
[list][*]November 20th: Amherst College Info Session, 7:30 pm ET: Matt McGann, Dean of Admission and Financial Aid at Amherst College, and Nathan Pflueger, math professor at Amherst College, will host an info session exploring both Amherst College specifically and liberal arts colleges generally. Topics include opportunities in math, the admission process, and financial aid for both US and international students.
[*]November 7th: 2024 AMC 10/12 A Discussion, Thursday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 A, administered November 6. We will discuss some of the most interesting problems from each test!
[*]November 13th: 2024 AMC 10/12 B Discussion, Wednesday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 B, administered November 12. We will discuss some of the most interesting problems from each test![/list]
AoPS Spring classes are open for enrollment. Get a jump on the New Year and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

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Whether you've taken our classes at AoPS Online or AoPS Academies or reached the top echelons of our competition training with our Worldwide Online Olympiad Training (WOOT) program, you can now add your AoPS experience to the education section on your LinkedIn profile.

Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

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0 replies
jlacosta
Nov 1, 2024
0 replies
Problem 1 (First Day)
Valentin Vornicu   131
N 10 minutes ago by lnzhonglp
1. Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$.
131 replies
Valentin Vornicu
Jul 12, 2004
lnzhonglp
10 minutes ago
Maximization
Butterfly   1
N 21 minutes ago by lbh_qys

Assume $a,b>0$ satisfy $a^2+b^2+\frac{1}{a}+\frac{1}{b}=\frac{27}{4}.$ Prove $2a+\frac{a}{4b}\le 5.$
1 reply
+1 w
Butterfly
an hour ago
lbh_qys
21 minutes ago
Strange Conditional Sequence
MarkBcc168   20
N 22 minutes ago by cushie27
Source: APMO 2019 P2
Let $m$ be a fixed positive integer. The infinite sequence $\{a_n\}_{n\geq 1}$ is defined in the following way: $a_1$ is a positive integer, and for every integer $n\geq 1$ we have
$$a_{n+1} = \begin{cases}a_n^2+2^m & \text{if } a_n< 2^m \\ a_n/2 &\text{if } a_n\geq 2^m\end{cases}$$For each $m$, determine all possible values of $a_1$ such that every term in the sequence is an integer.
20 replies
MarkBcc168
Jun 11, 2019
cushie27
22 minutes ago
Looking for a nice proof
KhuongTrang   60
N 25 minutes ago by KhuongTrang
Source: own
Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that$$\color{blue}{a^2+b^2+c^2+3\ge 2\sqrt{2(a^3b+b^3c+c^3a)+3abc}.}$$
60 replies
KhuongTrang
Jul 5, 2024
KhuongTrang
25 minutes ago
No more topics!
Hard inequality
proglote   11
N Oct 21, 2011 by SCP
Source: Brazil MO 2011, problem 6
Let $a_{1}, a_{2}, a_{3}, ... a_{2011}$ be nonnegative reals with sum $\frac{2011}{2}$, prove :

$|\prod_{cyc} (a_{n} - a_{n+1})| = |(a_{1} - a_{2})(a_{2} - a_{3})...(a_{2011}-a_{1})| \le \frac{3 \sqrt3}{16}.$
11 replies
proglote
Oct 16, 2011
SCP
Oct 21, 2011
Hard inequality
G H J
Source: Brazil MO 2011, problem 6
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proglote
958 posts
#1 • 3 Y
Y by Davi-8191, Adventure10, Mango247
Let $a_{1}, a_{2}, a_{3}, ... a_{2011}$ be nonnegative reals with sum $\frac{2011}{2}$, prove :

$|\prod_{cyc} (a_{n} - a_{n+1})| = |(a_{1} - a_{2})(a_{2} - a_{3})...(a_{2011}-a_{1})| \le \frac{3 \sqrt3}{16}.$
This post has been edited 1 time. Last edited by proglote, Oct 16, 2011, 10:30 PM
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dinoboy
2903 posts
#2 • 3 Y
Y by proglote, Adventure10, Mango247
What? How do $2011$ non-negative integers sum to $\frac{2011}{2}$?
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proglote
958 posts
#3 • 2 Y
Y by Adventure10, Mango247
dinoboy wrote:
What? How do $2011$ non-negative integers sum to $\frac{2011}{2}$?

Sorry I meant reals :blush: edited
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hvaz
148 posts
#4 • 2 Y
Y by Adventure10, Mango247
Dear proglote,

according to the Brazilian Math Olympiad rules, you are not allowed to post this problem until tuesday, once there are a few people who didn't took the test yet.

Sincerely,

hvaz
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Amir Hossein
5451 posts
#5 • 3 Y
Y by proglote, Adventure10, Mango247
Unlocked as requested.
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proglote
958 posts
#6 • 2 Y
Y by Adventure10, Mango247
Thank you Amir.

I think it holds for any $n \ge 3$ variables :

Let $a_{1}, a_{2}, a_{3}, ... a_{n}$ be nonnegative reals with sum $\frac{n}{2}$, prove :

$|\prod_{cyc} (a_{n} - a_{n+1})| = |(a_{1} - a_{2})(a_{2} - a_{3})...(a_{n}-a_{1})| \le \frac{3 \sqrt3}{16}.$
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SCP
1502 posts
#7 • 2 Y
Y by Adventure10, Mango247
If $n$ is even:
Take $a_{2k+1}=0$ and $a_{2k}=1$
then we get $1>\frac{3 \sqrt3}{16}.$
So your generaziation can't be always true, but I think it's true for odd ones.
****
Do you post all problems of Brazilian MO?
***
after the edit: don't understand Portuguese and couldn't find there the questions even
This post has been edited 1 time. Last edited by SCP, Oct 19, 2011, 7:38 PM
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jlucascsa
10 posts
#8 • 2 Y
Y by Adventure10, Mango247
$ a_{1} - a_{2011} =0 $ in your example.
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proglote
958 posts
#9 • 1 Y
Y by Adventure10
SCP wrote:
Where do I miss in understanding the question?

In your case $a_{2011} - a_{1} = 0$ so it's not a valid counterexample.
SCP wrote:
Do you post all problems of Brazilian MO?

I only post some of them.. They are all available at http://www.obm.org.br/opencms/, I can translate them if you are interested.
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SCP
1502 posts
#10 • 2 Y
Y by Adventure10, Mango247
So I missed first the parity.
EDIT:
I have whool proof and it holds for all odd numbers and there can be always equality.
Will post proof later.
It goes if the min. is $a_i$ we can better let it $0$ and add $a_i$ to the maximum.
Then we have some $a_i,a_j,a_k,0$ it will be best that $a_j$ is zero and we add that to other...
The we get $0,a,b,0,x,0,x,0,x\cdots$ and solve it with AM-GM.
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math154
4302 posts
#11 • 3 Y
Y by MathPanda1, Adventure10, Mango247
First we perform the following algorithm (while loop):
    1. If
$a_{i-1}\ge a_i\le a_{i+1}$ for any $i$ (indices taken $\pmod{2011}$), then WLOG $a_j\to a_j+a_i$ and $a_i\to0$, where $a_j$ is maximal.
2. If $a_{i+1}\le a_{i+2}\le a_{i+3}\le a_{i+4}$ or $a_{i+1}\ge a_{i+2}\ge a_{i+3}\ge a_{i+4}$ for any $i$, then WLOG swap $a_{i+2},a_{i+3}$.
3. If $0=a_{i-2}\le a_{i-1}\le a_i\ge a_{i+1}\ge a_{i+2}=0$ for any $i$, WLOG $a_{i-1}\le a_{i+1}$, then WLOG $a_{i-1}\to a_{i-1}+a_i$ and $a_i\to 0$ (since $(a_{i-1}+a_i)^2 a_{i+1}^2 \ge (a_i-a_{i-1})(a_i-a_{i+1})a_{i-1}a_{i+1} \ge 0$).
4. If nothing happened in steps 1, 2, and 3, stop. Otherwise, repeat the algorithm.

Clearly we end up with a bunch of zeros with either $1$ or $2$ positive numbers in between consecutive pairs, where we assume that no two consecutive numbers are equal. Suppose $p$ pairs have $1$ in between and $q$ pairs have $2$ in between; then $2p+3q=2011$ (so for some integer $r\ge0$, $(p,q)=(1004-3r,1+2r)$), so let $s_1,\ldots,s_p$ ($s_1+\cdots+s_p=S$) denote the sums of the numbers in between the $p$ pairs of consecutive $0$'s with $1$ numbers in between and $t_1,\ldots,t_q$ ($t_1+\cdots+t_q=T=2011/2-S$) the sums of the numbers in between the $q$ consecutive $0$'s with $2$ numbers in between. It's not hard to check that in $[0,1]$, $|x(1-2x)(1-x)|\le 1/(6\sqrt{3})$ and so
\begin{align*}
\prod_{n=1}^{2011}|a_n-a_{n+1}|\le s_1^2\cdots s_p^2\frac{t_1^3\cdots t_q^3}{(6\sqrt{3})^q}
&\le (S/p)^{2p}\frac{(T/q)^{3q}}{(6\sqrt{3})^q} \\
&= (S/2p)^{2p}(T/3q)^{3q}\frac{2^{2p}3^{3q}}{(6\sqrt{3})^q} \\
&\le \left(\frac{2011/2}{2p+3q}\right)^{2p+3q}\frac{2^{2p}3^{3q}}{(6\sqrt{3})^q} = \left(\frac{3\sqrt{3}}{16}\right)^{1+2r}\le \frac{3\sqrt{3}}{16},
\end{align*}as desired.
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SCP
1502 posts
#12 • 2 Y
Y by Adventure10, Mango247
math154 wrote:
First we perform the following algorithm (while loop):
    1. If
$a_{i-1}\ge a_i\le a_{i+1}$ for any $i$ (indices taken $\pmod{2011}$), then WLOG $a_j\to a_j+a_i$ and $a_i\to0$, where $a_j$ is maximal.
2. If $a_{i+1}\le a_{i+2}\le a_{i+3}\le a_{i+4}$ or $a_{i+1}\ge a_{i+2}\ge a_{i+3}\ge a_{i+4}$ for any $i$, then WLOG swap $a_{i+2},a_{i+3}$.
3. If $0=a_{i-2}\le a_{i-1}\le a_i\ge a_{i+1}\ge a_{i+2}=0$ for any $i$, WLOG $a_{i-1}\le a_{i+1}$, then WLOG $a_{i-1}\to a_{i-1}+a_i$ and $a_i\to 0$ (since $(a_{i-1}+a_i)^2 a_{i+1}^2 \ge (a_i-a_{i-1})(a_i-a_{i+1})a_{i-1}a_{i+1} \ge 0$).
4. If nothing happened in steps 1, 2, and 3, stop. Otherwise, repeat the algorithm.

If we have wtih this algoritm $2$ positive integers between two zeros (of course we will have, cause $2011$ is odd), the other ones are of form $0x0x0x0x0x0x0x$ by $Am_GM$
looking to $0a_1a_2\cdots a_n0$ the prodcut of differences is maximum $\frac{2(a_1+\cdots+a_n)}{n+1})^{n+1}$ with that equality.

Writing $t=\frac{n}{2}-\frac{(n-3)x}{2}$ we get $ab(a-b)x^{n-3}\le a(t-a)(t-2a)x^{n-3}\le t^3 \frac{\sqrt{3}}{18}t^3x^{n-3}$

Where $t^3x^{n-3}\le \frac{t^3(1.5x)^{n-3}}{1.5^{n-3}}\le 1.5^3$ with AM-GM and the proof is clear.
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