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
Perfect powers of functions on a finite set
Tintarn   1
N 5 minutes ago by pieater314159
Source: Baltic Way 2024, Problem 9
Let $S$ be a finite set. For a positive integer $n$, we say that a function $f\colon S\to S$ is an $n$-th power if there exists some function $g\colon S\to S$ such that
\[
    f(x) = \underbrace{g(g(\ldots g(x)\ldots))}_{\mbox{\scriptsize $g$ applied $n$ times}}
\]for each $x\in S$.

Suppose that a function $f\colon S\to S$ is an $n$-th power for each positive integer $n$. Is it necessarily true that $f(f(x)) = f(x)$ for each $x\in S$?
1 reply
Tintarn
3 hours ago
pieater314159
5 minutes ago
Prime divisors of n+2^floor(sqrt(n))
topologicalsort   0
11 minutes ago
Source: Bulgarian Autumn Tournament 2024, 10.3
Find all polynomials $P$ with integer coefficients, for which there exists a number $N$, such that for every natural number $n \geq N$, all prime divisors of $n+2^{\lfloor \sqrt{n} \rfloor}$ are also divisors of $P(n)$.
0 replies
1 viewing
topologicalsort
11 minutes ago
0 replies
APQ tangent to BC
topologicalsort   0
23 minutes ago
Source: Bulgarian Autumn Tournament 2024, 10.2
Let $ABC$ be a scalene acute triangle, where $AL$ $(L \in BC)$ is the internal bisector of $\angle BAC$ and $M$ is the midpoint of $BC$. Let the internal bisectors of $\angle AMB$ and $\angle CMA$ intersect $AB$ and $AC$ in $P$ and $Q$, respectively. Prove that the circumcircle of $APQ$ is tangent to $BC$ if and only if $L$ belongs to it.
0 replies
topologicalsort
23 minutes ago
0 replies
Simple system of equations
topologicalsort   0
35 minutes ago
Source: Bulgarian Autumn Tournament 2024, 10.1
Find all real solutions to the system of equations: $$\begin{cases} (x^2+xy+y^2)\sqrt{x^2+y^2} = 88 \\ (x^2-xy+y^2)\sqrt{x^2+y^2} = 40  \end{cases}$$
0 replies
topologicalsort
35 minutes ago
0 replies
No more topics!
Maximize (a-b)(b-c)(c-a)/(a+b+c)^3
kiyoras_2001   9
N Oct 13, 2021 by NTstrucker
Source: Own?
For $a, b, c\ge 0$ prove that
$$(a-b)(b-c)(c-a)\le \frac{(a+b+c)^3}{6\sqrt{3}}.$$
9 replies
kiyoras_2001
Aug 25, 2020
NTstrucker
Oct 13, 2021
Maximize (a-b)(b-c)(c-a)/(a+b+c)^3
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kiyoras_2001
569 posts
#1
Y by
For $a, b, c\ge 0$ prove that
$$(a-b)(b-c)(c-a)\le \frac{(a+b+c)^3}{6\sqrt{3}}.$$
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Victoria_Discalceata1
711 posts
#2 • 2 Y
Y by kiyoras_2001, llplp
I think it is Potla's old problem, I am not sure though.
It suffices to consider $c>b>a$. Then the LHS is decreasing in $a$ while the RHS is increasing in $a$ thus it suffices to consider $a=0$. Left to prove is $(b+c)^3\ge 6\sqrt{3}bc(c-b)$. It holds trivially when $b=0$ and otherwise taking $c=tb$, where $t>1$, we obtain $(t+7-4\sqrt{3})(t-2-\sqrt{3})^2\ge 0$ which is obvious.
This post has been edited 1 time. Last edited by Victoria_Discalceata1, Aug 25, 2020, 6:12 PM
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EagleEye
497 posts
#3 • 2 Y
Y by kiyoras_2001, Inconsistent
sniped :oops:
WLOG $a$ be the largest among $a,b,c$. If $a\geq b\geq c$, then $\text{LHS}\leq 0$. So, nothing to prove.
Suppose $a\geq c\geq b$. Let $x=a-c$ and $y=c-b$. Then, $x,y\geq 0$ and $a+b+c = 3a + x + 2y \geq x+2y$.
\begin{align*}
(a-b)(b-c)(c-a) &= xy(x+y)\\
&= \frac{1}{2(1+\sqrt{3})^3} \cdot (1+\sqrt{3})x \cdot (1+\sqrt{3})^2y\cdot 2(x+y)\\
&\underset{\text{AM-GM}}{\leq} \frac{1}{2(1+\sqrt{3})^3} \left(\frac{(1+\sqrt{3})x + (1+\sqrt{3})^2y + 2(x+y)}{3} \right)^3 \\
&=\frac{1}{2(1+\sqrt{3})^3} \left(\frac{1+\sqrt{3}}{\sqrt{3}}(x+2y) \right)^3 = \frac{(x+2y)^3}{6\sqrt{3}} \leq \frac{(a+b+c)^3}{6\sqrt{3}}
\end{align*}
This post has been edited 2 times. Last edited by EagleEye, Aug 25, 2020, 6:20 PM
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kiyoras_2001
569 posts
#4
Y by
Thank you Victoria_Discalceata1 and EagleEye.
Victoria_Discalceata1 wrote:
I think it is Potla's old problem, I am not sure though.
Victoria_Discalceata1, I guess you mean this but of course my problem is different.
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Victoria_Discalceata1
711 posts
#5 • 2 Y
Y by kiyoras_2001, Inconsistent
kiyoras_2001 wrote:
Victoria_Discalceata1, I guess you mean this but of course my problem is different.
Among others. There was a short paper written by him on the topic linked to one of his posts but I may not remember correctly. I will PM you if I find it.
Of course your problem can be also bashed with pqr/uvw which is not very complicated but a bit longer.
When the LHS is nonpositive the equality obviously holds. When the LHS is positive we square both sides and we are left with $$6^29^2r^2+6^3p\left(2p^2-9q\right)r+\left(p^2+12q\right)\left(p^2-6q\right)^2\ge 0.$$We are done if $t=\frac{p^2}{q}\ge\frac{9}{2}$ and otherwise $$\Delta_r=6^43^3q^3\left(3(t-2)(t-3)(2t-9)-(t-3)^3- 9\right)<0$$as desired. Equality occurs when $r=0$ and $p^2=6q$.
This post has been edited 4 times. Last edited by Victoria_Discalceata1, Aug 27, 2020, 3:17 AM
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KaiRain
878 posts
#6 • 1 Y
Y by kiyoras_2001
kiyoras_2001 wrote:
For $a, b, c\ge 0$ prove that
$$(a-b)(b-c)(c-a)\le \frac{(a+b+c)^3}{6\sqrt{3}}.$$
There is a generalization for this inequality, you see, https://artofproblemsolving.com/community/q1h1790922p11848720.
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kiyoras_2001
569 posts
#7 • 2 Y
Y by Mango247, Mango247
Thanks.
Thanks.
Thanks.
This post has been edited 1 time. Last edited by kiyoras_2001, Aug 26, 2020, 11:15 AM
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VMF-er
512 posts
#8 • 1 Y
Y by kiyoras_2001
kiyoras_2001 wrote:
For $a, b, c\ge 0$ prove that
$$(a-b)(b-c)(c-a)\le \frac{(a+b+c)^3}{6\sqrt{3}}.$$

We just need to show that $(a+b+c)^{6}\geq 108(b-c)^{2}(c-a)^{2}(a-b)^{2}$
WLOG, assume that $c=min\lbrace a,b,c\rbrace$.
$\Rightarrow RHS\leq a^{2}b^{2}(a-b)^{2}=\frac{1}{4}.(2ab).(2ab).(a-b)^{2}\leq \frac{1}{4}.[\frac{2ab+2ab+(a-b)^{2}}{3}]^{3}=\frac{(a+b)^{6}}{108}\leq LHS$
Equality holds at $(a,b,c)=(2+\sqrt{3},1,0)$ or any cyclic permutations. $\square$
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sqing
38870 posts
#9
Y by
For $a, b, c\ge 0 .$ Prove that
$$(a-b)(b-2c)(c-a)\le \frac{2(10+7\sqrt{7})(a+b+c)^3}{243}$$$$(a-b)(b-2c)(c-2a)\le \frac{4(10+7\sqrt{7})(a+b+c)^3}{243}$$$$(a-b)(b-2c)(c-3a)\le \frac{(35+13\sqrt{13})(a+b+c)^3}{108}$$
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NTstrucker
156 posts
#10
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A special case of 2011 Brazil MO P6.
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