Happy Thanksgiving! Please note that there are no classes November 25th-December 1st.

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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
a+b+c+abc=4 with two equality cases
KhuongTrang   48
N 4 minutes ago by KhuongTrang
Source: own
Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c+abc=4.$ Prove that
$$\color{blue}{\sqrt{14a+b+c} +\sqrt{14b+a+c} +\sqrt{14c+b+a}\le 2+2\sqrt{5}\cdot\sqrt{a+b+c+2}. }$$Equality holds iff $a=b=c=1$ or $a=b=2,c=0$ and any permutations.
48 replies
KhuongTrang
Mar 11, 2024
KhuongTrang
4 minutes ago
a^2+ b^2+ c^2
sqing   3
N 24 minutes ago by sqing
Source: Own
Let $ a,b, c$ be reals such that $a +b =2 , b^2+2c^2= 9 $ and $ 3c^2+4a^2 =48.$ Prove that$$a^2+ b^2+ c^2=14$$Let $ a,b ,c$ be reals such that $a +b =1 , b^2+2c^2= 9 $ and $ 3c^2+4a^2 =48.$ Prove that$$a^2+ b^2+ c^2=\frac{7(101-6\sqrt{41})}{25}$$
3 replies
1 viewing
sqing
Yesterday at 1:48 AM
sqing
24 minutes ago
Needed help on inequalities
fAaAtDoOoG   10
N 34 minutes ago by arqady
Hello guys, this is my first post. I've encountered an inequality and struggled to solve it. If anyone can solve it, that would be awesome. Any help would be greatly appreciated!!!

$$\frac{b+c}{\sqrt{a^{2}+bc}} + \frac{c+a}{\sqrt{b^{2}+ac}} + \frac{a+b}{\sqrt{c^{2}+ab}} > 4, a,b,c \in \mathbb{R}^{+}$$
10 replies
fAaAtDoOoG
Yesterday at 1:28 AM
arqady
34 minutes ago
Olympiad with Beginner Friendly Theory
WheatNeat   10
N 37 minutes ago by WheatNeat
I want to practice on olympiad level problems, but I still don't have all the theory I need to do most national and such olympiads/TSTs (currently working through vol 2 and only a third of the way through). Are there any good olympiads that I should practice on that are relatively beginner friendly (theory-wise)?
10 replies
WheatNeat
Nov 19, 2024
WheatNeat
37 minutes ago
No more topics!
A bunch of similar triangles all around the circumcircle
Tintarn   2
N Nov 17, 2024 by Alex-Five
Source: Baltic Way 2024, Problem 12
Let $ABC$ be an acute triangle with circumcircle $\omega$ such that $AB<AC$. Let $M$ be the midpoint of the arc $BC$ of~$\omega$ containing the point~$A$, and let $X\neq M$ be the other point on $\omega$ such that $AX=AM$. Points $E$ and $F$ are chosen on sides $AC$ and $AB$ of the triangle $ABC$ such that $EX=EC$ and $FX=FB$. Prove that $AE=AF$.
2 replies
Tintarn
Nov 16, 2024
Alex-Five
Nov 17, 2024
A bunch of similar triangles all around the circumcircle
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Source: Baltic Way 2024, Problem 12
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Tintarn
8982 posts
#1
Y by
Let $ABC$ be an acute triangle with circumcircle $\omega$ such that $AB<AC$. Let $M$ be the midpoint of the arc $BC$ of~$\omega$ containing the point~$A$, and let $X\neq M$ be the other point on $\omega$ such that $AX=AM$. Points $E$ and $F$ are chosen on sides $AC$ and $AB$ of the triangle $ABC$ such that $EX=EC$ and $FX=FB$. Prove that $AE=AF$.
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SimplisticFormulas
23 posts
#2
Y by
Neat!

Claim 1: $XAFE$ is cyclic
Proof: Indeed, note that simce $AX=AM$, $\angle FBX=\angle ABX=\angle ACX=\angle ECX$ which gives us $\angle AEX=\angle 2 \cdot \angle ECX=2 \cdot \angle FBX=\angle AFX$, proving the claim.

Let $XE \cap (ABC)=Y \neq X$. Observe that since $\angle YXC=\angle MCA$, we get $YC=AM$, implying that $MYCA$ is an isosceles trapezium.
$\implies \angle AFE=\angle AXE=\angle AXY=\angle ACY=\angle MAC=\angle MBC=90-\frac{A}{2}$
$\implies \angle AEF=90-\frac{A}{2}=\angle AFE$
$\implies AE=AF$, as required.$\blacksquare$
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Alex-Five
4 posts
#3
Y by
Firstly, $\triangle XBF$ is similar to $\triangle EXC$ by spiral similarity.
Therefore $AXFE$ is cyclic.
We now finish by angle chase:
$\angle AEF=\angle AEX+\angle XEF=\angle AFX +\angle FAX \\ =2\angle ABX+
\angle BAX = 2\angle ACX + \angle XCB = \angle MCB $.

Therefore:
$\angle AFE =180- \angle FAE - \angle FEA =180- \angle BMC - \angle MCB = \angle MBC = \angle MCB $.
Hence $\angle AEF = \angle AFE $ and the assertion follows.
This post has been edited 3 times. Last edited by Alex-Five, Nov 17, 2024, 12:49 PM
Reason: Spelling mistakes (again...)
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