The time is now - Spring classes are filling up!

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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!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile!

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!

Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
Nov 1, 2024
0 replies
Inspired by Ruji2018252
sqing   2
N 6 minutes ago by sqing
Source: Own
Let $ a,b,c\in [0,1] $ and $ a+b+c=1 .$ Prove that
$$\dfrac{4}{3}\leq\dfrac{1-a^2}{a+2}+\dfrac{2-b^2}{b+2}+\dfrac{1-c^2}{c+2} \leq 3-\dfrac{4\sqrt 6}{7}$$$$\dfrac{11}{6}\leq\dfrac{2-a^2}{a+2}+\dfrac{1-b^2}{b+2}+\dfrac{2-c^2}{c+2}  \leq  \dfrac{4}{7}(6-\sqrt 6)$$$$\dfrac{10}{3}\leq\dfrac{2-a^2}{a+1}+\dfrac{1-b^2}{b+1}+\dfrac{2-c^2}{c+1}  \leq 4 $$$$\dfrac{5}{2}\leq\dfrac{1-a^2}{a+1}+\dfrac{2-b^2}{b+1}+\dfrac{1-c^2}{c+1} \leq 3$$
2 replies
1 viewing
sqing
2 hours ago
sqing
6 minutes ago
Nice Quadrilateral Geo
amuthup   48
N 26 minutes ago by L13832
Source: 2021 ISL G4
Let $ABCD$ be a quadrilateral inscribed in a circle $\Omega.$ Let the tangent to $\Omega$ at $D$ meet rays $BA$ and $BC$ at $E$ and $F,$ respectively. A point $T$ is chosen inside $\triangle ABC$ so that $\overline{TE}\parallel\overline{CD}$ and $\overline{TF}\parallel\overline{AD}.$ Let $K\ne D$ be a point on segment $DF$ satisfying $TD=TK.$ Prove that lines $AC,DT,$ and $BK$ are concurrent.
48 replies
amuthup
Jul 12, 2022
L13832
26 minutes ago
thanks u!
Ruji2018252   4
N an hour ago by Victoria_Discalceata1
$a,b,c\in [0;2]$ and $a+b+c=3$ find min and max:
$$P=\dfrac{2-a^2}{a+1}+\dfrac{2-b^2}{b+1}+\dfrac{2-c^2}{c+1}$$
4 replies
2 viewing
Ruji2018252
3 hours ago
Victoria_Discalceata1
an hour ago
ISL functional equation problem
Mhremath   3
N an hour ago by pco
2. (ISL 01/A4) Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
\[
f(xy)f(x)-f(y) = (x-y)f(x)f(y)
\]for all $x,y$.
3 replies
Mhremath
Yesterday at 5:59 PM
pco
an hour ago
No more topics!
Inequality
TRcrescent27   10
N Jul 23, 2016 by leonardg
Source: Similar 2013 Turkey JMO
Let $a,b,c$ be real numbers such that $a+b+c=0$ and $a^2+b^2+c^2=2$. Determine the maximum value of
$$|a^2b^2(a-b)+b^2c^2(b-c)+c^2a^2(c-a)|$$
10 replies
TRcrescent27
Jul 19, 2016
leonardg
Jul 23, 2016
Inequality
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G H BBookmark kLocked kLocked NReply
Source: Similar 2013 Turkey JMO
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TRcrescent27
32 posts
#1 • 1 Y
Y by Adventure10
Let $a,b,c$ be real numbers such that $a+b+c=0$ and $a^2+b^2+c^2=2$. Determine the maximum value of
$$|a^2b^2(a-b)+b^2c^2(b-c)+c^2a^2(c-a)|$$
This post has been edited 1 time. Last edited by TRcrescent27, Jul 19, 2016, 5:56 PM
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leonardg
1554 posts
#2 • 1 Y
Y by Adventure10
2,of course .
This post has been edited 2 times. Last edited by leonardg, Jul 19, 2016, 8:35 PM
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happykiwi
33 posts
#3 • 2 Y
Y by Adventure10, Mango247
leonardg wrote:
2,of course .

Do you mind explaining? :P
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leonardg
1554 posts
#4 • 2 Y
Y by Adventure10, Mango247
happykiwi wrote:
Do you mind explaining? :P

2=two=zwei=dva=doi=dos .
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oty
2290 posts
#5 • 2 Y
Y by Adventure10, Mango247
I think it to obivious for leonardg to write a full solution , but in my opinion this forum is also about helping . here is my solution , a lot of calcul :blush:
TRcrescent27 wrote:
Let $a,b,c$ be real numbers such that $a+b+c=0$ and $a^2+b^2+c^2=2$. Determine the maximum value of
$$|a^2b^2(a-b)+b^2c^2(b-c)+c^2a^2(c-a)|$$
$LHS \leq \sum  |ab| |ab(a-b)| \leq \sqrt{( \sum |ab|^{2} ) (\sum |ab(a-b)|^{2})}$ now
$\sum |ab(a-b)|^2=\sum a^{2}b^{2}(a^{2}+b^{2})-2(\sum a^{3}b^{3})$
$=2(\sum a^{2}b^{2})-3a^{2}b^2c^2-2(\sum a^{3}b^{3})$
now from $(\sum a^{3}b^{3})=(\sum ab)^{3}-3abc(\sum a^{2}b+ab^{2})-6a^{2}b^{2}c^{2}$ which is
$(\sum a^{3}b^{3})=(\sum ab)^{3}-3abc(\sum  ab(a+b))-6a^{2}b^{2}c^{2}=(\sum ab)^{3}+3a^{2}b^{2}c^{2}$
hence $\sum |ab(a-b)|^{2}=2-3a^{2}b^{2}c^{2}-2(-1+3a^{2}b^{2}c^{2})=4-9a^{2}b^{2}c^{2}$
on the other hand we have
$(\sum a^{2}b^{2})=(\sum ab)^{2}-2abc(a+b+c) =(\sum ab )^{2}$
and for $\sum a^{2} =(\sum a)^{2}-2(\sum ab)=2$ we get sum $\sum ab=-1$ all this give us
$LHS \leq \sqrt{4-9a^{2}b^{2}c^{2}} \leq \sqrt{4}=2$ equality happen for $c=0$ $a=-b=1$
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Scorpion.k48
195 posts
#6 • 3 Y
Y by baladin, Adventure10, Mango247
We have : $P=|a^2b^2(a-b)+b^2c^2(b-c)+c^2a^2(c-a)|=|(a-b)(b-c)(c-a)(ab+bc+ca)|=|(a-b)(b-c)(c-a)|$.
WLOG $a\ge b\ge c$, hence $a-c\le \sqrt{2(a^2+b^2+c^2)}=2$,

$ |(a-b)(b-c)(c-a) |=(a-b)(b-c)(a-c)$

$\le (\frac{a-b+b-c}{2})^2(a-c)=\frac{1}{4}(a-c)^3 \le 2 $.

when $a=1,b=0,c=-1$,the maximum of $P $ is $2 $
This post has been edited 1 time. Last edited by Scorpion.k48, Jul 20, 2016, 1:52 AM
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leonardg
1554 posts
#7 • 4 Y
Y by 62861, baladin, Adventure10, Mango247
oty wrote:
I think it to obivious for leonardg to write a full solution , but in my opinion this forum is also about helping . here is my solution , a lot of calcul :blush:
No , it's not obvious at all . But before writing always comes the mistery .
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Nguyenhuyen_AG
3264 posts
#8 • 1 Y
Y by Adventure10
If $a,\,b,\,c$ are real numbers and $t \geqslant 0$ such that $a+b+c=0,$ $a^2+b^2+c^2 = 6t^2.$ Prove that
\[\left|(a-b)(b-c)(c-a)\right| \leqslant 6\sqrt{3}t^3.\]
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leonardg
1554 posts
#9 • 1 Y
Y by Adventure10
Nguyenhuyen_AG wrote:
If $a,\,b,\,c$ are real numbers and $t \geqslant 0$ such that $a+b+c=0,$ $a^2+b^2+c^2 = 6t^2.$ Prove that
\[\left|(a-b)(b-c)(c-a)\right| \leqslant 6\sqrt{3}t^3.\]

See above . Is the same thing . And see below too :
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oty
2290 posts
#10 • 2 Y
Y by Adventure10, Mango247
the identity $|\sum a^{2}b^{2} (a-b)|=|\sum ab| |(a-b)(b-c)(c-a)|$ in the condition $a+b+c=0$ is pretty interesting , beautiffull Dear leonardg
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leonardg
1554 posts
#11 • 2 Y
Y by Adventure10, Mango247
Thank you ,oty .
Actually , it is based on the Vandermonde determinants , easy deductible .
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