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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!

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0 replies
jlacosta
Nov 1, 2024
0 replies
Problem 1 (First Day)
Valentin Vornicu   131
N 12 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
1 viewing
Valentin Vornicu
Jul 12, 2004
lnzhonglp
12 minutes ago
Maximization
Butterfly   1
N 23 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
Butterfly
an hour ago
lbh_qys
23 minutes ago
Strange Conditional Sequence
MarkBcc168   20
N 24 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
24 minutes ago
Looking for a nice proof
KhuongTrang   60
N 27 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
27 minutes ago
No more topics!
New beautiful concurrency circle related lemma?
Circumcircle   7
N Nov 21, 2024 by ohiorizzler1434
Source: Kosovo Math Olympiad 2025, Grade 11, Problem 4
Let $ABC$ be a given triangle. Let $A_1$ and $A_2$ be points on the side $BC$. Let $B_1$ and $B_2$ be points on the side $CA$. Let $C_1$ and $C_2$ be points on the side $AB$. Suppose that the points $A_1,A_2,B_1,B_2,C_1$ and $C_2$ lie on a circle. Prove that the lines $AA_1, BB_1$ and $CC_1$ are concurrent if and only if $AA_2, BB_2$ and $CC_2$ are concurrent.
7 replies
Circumcircle
Nov 16, 2024
ohiorizzler1434
Nov 21, 2024
New beautiful concurrency circle related lemma?
G H J
Source: Kosovo Math Olympiad 2025, Grade 11, Problem 4
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Circumcircle
64 posts
#1 • 1 Y
Y by ItsBesi
Let $ABC$ be a given triangle. Let $A_1$ and $A_2$ be points on the side $BC$. Let $B_1$ and $B_2$ be points on the side $CA$. Let $C_1$ and $C_2$ be points on the side $AB$. Suppose that the points $A_1,A_2,B_1,B_2,C_1$ and $C_2$ lie on a circle. Prove that the lines $AA_1, BB_1$ and $CC_1$ are concurrent if and only if $AA_2, BB_2$ and $CC_2$ are concurrent.
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Haris1
23 posts
#2
Y by
Ceva and POP , a similar definiton like this problem was in the plane geometry book which they considerd it to be cyclocevian conjugate.
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dangerousliri
924 posts
#3 • 1 Y
Y by ehuseyinyigit
This problem was proposed by me. One can prove similar results if the points are in any Conic Sections. I never seen this problem before and hopefully I discovered this first.
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tobiSALT
50 posts
#4
Y by
By Ceva's theorem, $AA_1, BB_1, CC_1$ are concurrent if and only if
$$ \frac{BA_1}{A_1C} \cdot \frac{CB_1}{B_1A} \cdot \frac{AC_1}{C_1B} = 1 $$Similarly, $AA_2, BB_2, CC_2$ are concurrent if and only if
$$ \frac{BA_2}{A_2C} \cdot \frac{CB_2}{B_2A} \cdot \frac{AC_2}{C_2B} = 1 $$Since $A_1, A_2, B_1, B_2, C_1, C_2$ lie on a circle, by the intersecting chords theorem, we have
$$ BA_1 \cdot BA_2 = BC_1 \cdot BC_2 $$$$ CA_1 \cdot CA_2 = CB_1 \cdot CB_2 $$$$ AB_1 \cdot AB_2 = AC_1 \cdot AC_2 $$Dividing the first equation by the second, we have
$$ \frac{BA_1 \cdot BA_2}{CA_1 \cdot CA_2} = \frac{BC_1 \cdot BC_2}{CB_1 \cdot CB_2} \implies \frac{BA_1}{CA_1} \cdot \frac{BA_2}{CA_2} = \frac{BC_1}{CB_1} \cdot \frac{BC_2}{CB_2} $$Dividing the third equation by the first, we have
$$ \frac{AB_1 \cdot AB_2}{BA_1 \cdot BA_2} = \frac{AC_1 \cdot AC_2}{BC_1 \cdot BC_2} \implies \frac{AB_1}{BA_1} \cdot \frac{AB_2}{BA_2} = \frac{AC_1}{BC_1} \cdot \frac{AC_2}{BC_2} $$Dividing the second equation by the third, we have
$$ \frac{CA_1 \cdot CA_2}{AB_1 \cdot AB_2} = \frac{CB_1 \cdot CB_2}{AC_1 \cdot AC_2} \implies \frac{CA_1}{AB_1} \cdot \frac{CA_2}{AB_2} = \frac{CB_1}{AC_1} \cdot \frac{CB_2}{AC_2} $$Assume that $AA_1, BB_1, CC_1$ are concurrent. Then, we have
$$ \frac{BA_1}{A_1C} \cdot \frac{CB_1}{B_1A} \cdot \frac{AC_1}{C_1B} = 1 $$We want to show that $AA_2, BB_2, CC_2$ are concurrent, which is equivalent to
$$ \frac{BA_2}{A_2C} \cdot \frac{CB_2}{B_2A} \cdot \frac{AC_2}{C_2B} = 1 $$From $\frac{BA_1}{CA_1} \cdot \frac{BA_2}{CA_2} = \frac{BC_1}{CB_1} \cdot \frac{BC_2}{CB_2}$, we have
$$ \frac{BA_2}{CA_2} = \frac{CA_1}{BA_1} \cdot \frac{BC_1}{CB_1} \cdot \frac{BC_2}{CB_2} $$Similarly, from $\frac{AB_1}{BA_1} \cdot \frac{AB_2}{BA_2} = \frac{AC_1}{BC_1} \cdot \frac{AC_2}{BC_2}$,
$$ \frac{AB_2}{BA_2} = \frac{BA_1}{AB_1} \cdot \frac{AC_1}{BC_1} \cdot \frac{AC_2}{BC_2} $$From $\frac{CA_1}{AB_1} \cdot \frac{CA_2}{AB_2} = \frac{CB_1}{AC_1} \cdot \frac{CB_2}{AC_2}$,
$$ \frac{CA_2}{AB_2} = \frac{AB_1}{CA_1} \cdot \frac{CB_1}{AC_1} \cdot \frac{CB_2}{AC_2} $$Then,
$$ \frac{BA_2}{A_2C} \cdot \frac{CB_2}{B_2A} \cdot \frac{AC_2}{C_2B} = \left( \frac{CA_1}{BA_1} \cdot \frac{BC_1}{CB_1} \cdot \frac{BC_2}{CB_2} \right) \left( \frac{AB_1}{CB_1} \cdot \frac{BA_2}{AB_2} \right) \left( \frac{AC_1}{CB_2} \cdot \frac{AB_2}{CA_2} \right) $$$$ = \left( \frac{CA_1}{BA_1} \cdot \frac{BC_1}{CB_1} \cdot \frac{BC_2}{CB_2} \right) \left( \frac{AB_1}{CB_1} \cdot \frac{BA_1}{AB_1} \cdot \frac{AC_1}{BC_1} \cdot \frac{AC_2}{BC_2} \right) \left( \frac{AC_1}{CB_2} \cdot \frac{AB_1}{CA_1} \cdot \frac{AC_1}{CB_1} \cdot \frac{AC_2}{CB_2} \right) $$Therefore,
$$ \frac{BA_2}{A_2C} \cdot \frac{CB_2}{B_2A} \cdot \frac{AC_2}{C_2B} = \frac{CA_1}{BA_1} \cdot \frac{AB_1}{CB_1} \cdot \frac{AC_1}{CB_2} \cdot \frac{AC_1}{BC_1} \cdot \frac{BC_2}{CB_2} \cdot \frac{BC_1}{CB_1} \cdot \frac{AC_2}{BC_2} $$The same argument holds for the converse.

So $AA_1, BB_1, CC_1 \text{ are concurrent} \iff AA_2, BB_2, CC_2 \text{ are concurrent}$
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ItsBesi
90 posts
#5
Y by
I found the exact same solution as #2 and #4 during the contest.
Very nice problem btw, really enjoyed solving it :)
Here is how I wrote it in the contest:

Let $\Omega$ be the circle that points $A_1,A_2,B_1,B_2,C_1,C_2$ pass through

From Power of the Point Theorem (POP) we get:

$AC_1 \cdot AC_2=Pow(A,\Omega)=AB_2 \cdot AB_1 \implies AC_1 \cdot AC_2= AB_2 \cdot AB_1 \implies \boxed{\frac{AC_1}{AB_1}=\frac{AB_2}{AC_2}}$ $...(1)$

Simmilarly we find:

$BA_2 \cdot BA_1=Pow(B, \Omega)=BC_2 \cdot BC_1 \implies BA_2 \cdot BA_1=BC_2 \cdot BC_1 \implies \boxed{\frac{BA_1}{BC_1}=\frac{BC_2}{BA_2} }$ $...(2)$

Finally we get:
$CB_1 \cdot CB_2 = Pow(C,\Omega)=CA_1 \cdot CA_2 \implies CB_1 \cdot CB_2=CA_1 \cdot CA_2 \implies \boxed{\frac{CB_1}{CA_1}=\frac{CA_2}{CB_2} }$ $...(3)$

First part:


Second part:
This post has been edited 3 times. Last edited by ItsBesi, Nov 16, 2024, 11:11 PM
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AshAuktober
615 posts
#6
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Ceva's theorem with power of a point works.
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AshAuktober
615 posts
#7
Y by
Okay, apparently this problem exists from before. (I found it in my notes for 2024 indian national olympiad camp.)

Nonetheless, all the problems were quite nice from kosovo mo, and organising for 6 grades is a daunting task. So great job to the proposers and organisers.
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ohiorizzler1434
399 posts
#8
Y by
What the sigma? Blud just stated the cyclocevian conjugate. Now that's rizz! https://mathworld.wolfram.com/CyclocevianConjugate.html
This post has been edited 1 time. Last edited by ohiorizzler1434, Nov 21, 2024, 2:44 AM
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