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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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Rational numbers
steven_zhang123   1
N an hour ago by kokcio
Source: G635
Find all positive real numbers \( \alpha \) such that there exist infinitely many rational numbers \( \frac{p}{q} (p, q \in \mathbb{Z}, p > 0, \gcd(p, q) = 1 ) \) satisfying

\[
\left| \frac{q}{p} - \frac{\sqrt{5} - 1}{2} \right| < \frac{\alpha}{p^2}.
\]
1 reply
steven_zhang123
Today at 1:24 PM
kokcio
an hour ago
Find Triples of Integers
termas   39
N an hour ago by VideoCake
Source: IMO 2015 problem 2
Find all positive integers $(a,b,c)$ such that
$$ab-c,\quad bc-a,\quad ca-b$$are all powers of $2$.

Proposed by Serbia
39 replies
1 viewing
termas
Jul 10, 2015
VideoCake
an hour ago
Convex quadrilateral and midpoints [RMO2-2011, India]
Potla   14
N an hour ago by mqoi_KOLA
Let $ABCD$ be a convex quadrilateral. Let $E,F,G,H$ be the midpoints of $AB,BC,CD,DA$ respectively. If $AC,BD,EG,FH$ concur at a point $O,$ prove that $ABCD$ is a parallelogram.
14 replies
Potla
Dec 31, 2011
mqoi_KOLA
an hour ago
inequality marathon
EthanWYX2009   191
N an hour ago by Martin.s
There is an inequality marathon now, but the problem is too hard for me to solve, let's start a new one here, please post problems that is not too difficult.
------
P1.
Find the maximum value of ${M}$, such that for $\forall a,b,c\in\mathbb R_+,$
$$a^3+b^3+c^3-3abc\geqslant M(a^2b+b^2c+c^2a-3abc).$$
191 replies
EthanWYX2009
May 21, 2023
Martin.s
an hour ago
2025 Caucasus MO Juniors P2
BR1F1SZ   1
N an hour ago by GreekIdiot
Source: Caucasus MO
There are $30$ children standing in a circle. For each girl, it turns out that among the five people following her clockwise, there are more boys than girls. Find the greatest number of girls that can stand in a circle.
1 reply
BR1F1SZ
Yesterday at 12:55 AM
GreekIdiot
an hour ago
Unique solution
USJL   0
2 hours ago
Source: 2025 Taiwan TST Round 2 Independent Study 1-N
Find all $g:\mathbb{R}\to\mathbb{R}$ so that there exists a unique $f:\mathbb{R}\to\mathbb{R}$ satisfying $f(0)=g(0)$ and
\[f(x+g(y))+f(-x-g(-y))=g(x+f(y))+g(-x-f(-y))\]for all $x,y\in\mathbb{R}$.

Proposed by usjl
0 replies
USJL
2 hours ago
0 replies
AIME Study Forum
duckShroom   196
N 2 hours ago by Aopsauser9999
Hello, MathCosine and I have created a study forum mainly for AIME! Respond with 'join' if you want to join and I'll add you. I'll also add this post in other topics so it reaches more people.
196 replies
duckShroom
Jan 8, 2025
Aopsauser9999
2 hours ago
AMC 10 PREP Bay Area recommendation?
HCM2001   3
N 2 hours ago by HCM2001
Hi

For those from Bay Area, which school/center do you recommend for AMC 10 prep (goal is AIME qual)? so far i've look at ThinkAcademy, Alphastar, Random Math.. if you have taken classes there can you share your experiences? Thanks.
3 replies
1 viewing
HCM2001
2 hours ago
HCM2001
2 hours ago
Show that XD and AM meet on Gamma
MathStudent2002   90
N 2 hours ago by ErTeeEs06
Source: IMO Shortlist 2016, Geometry 2
Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$. The points $D$, $E$, $F$ are selected on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$, $\overline{IE}\perp \overline{AI}$, and $\overline{IF}\perp \overline{AI}$. Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\Gamma$.

Proposed by Evan Chen, Taiwan
90 replies
MathStudent2002
Jul 19, 2017
ErTeeEs06
2 hours ago
A number theory about divisors which no one fully solved at the contest
nAalniaOMliO   7
N 2 hours ago by CHESSR1DER
Source: Belarusian national olympiad 2024
Let's call a pair of positive integers $(k,n)$ interesting if $n$ is composite and for every divisor $d<n$ of $n$ at least one of $d-k$ and $d+k$ is also a divisor of $n$
Find the number of interesting pairs $(k,n)$ with $k \leq 100$
M. Karpuk
7 replies
nAalniaOMliO
Jul 24, 2024
CHESSR1DER
2 hours ago
Modified Fermat's last theorem
Euler8038   0
3 hours ago
Source: Own
Prove that, for any n, there is an infinite number of sequences composed by n pairwise coprime positive integers such that the sum of the n-th powers of the term in the sequence gives you an n-th power.

To be clear, if n=2 the conjecture is just about Pythagorean triples.

If n=3, you have to show that there exist an infinite number of triplets such that a³+b³+c³ is a cube, with a, b, c pairwise coprime positive integers.
0 replies
Euler8038
3 hours ago
0 replies
Infinite cube triplets
Euler8038   0
3 hours ago
Let a, b, x be positive coprime integers. Prove that there exist an infinite number of triplets (a, b, x) such that x³=3ab(a+b), or disprove the conjecture.
0 replies
Euler8038
3 hours ago
0 replies
k 2016 AMC International Participation Update
AMCDirector   10
N Feb 10, 2017 by whiteawesomesun
The MAA AMC program very much appreciates the increasing interest in international participation in our competitions.

Unfortunately, the rapid growth in interest and participation has made it unfeasible for our internal resources to manage the program as it currently operates. We are reevaluating the international registration and competition administration model during the 2016-2017 cycle.

For the 2016-2017 competition cycle, the MAA AMC exams will only be offered to schools with a USA, Canada, APO/FPO/DPO mailing addresses and schools registered with one of the five AMC international partners. If your school is in one of the areas represented by an AMC international partner, please contact the group manager directly for information on participating in the AMC program this year.

Thank you for your patience as we assess our options to successfully offer the AMC exams to more international audiences in the future. For more information: http://www.maa.org/math-competitions/international-registrations
10 replies
AMCDirector
Sep 6, 2016
whiteawesomesun
Feb 10, 2017
2016 AMC International Participation Update
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AMCDirector
851 posts
#1 • 11 Y
Y by Generic_Username, DrMath, mathwiz_aku, whiteawesomesun, trumpeter, Mudkipswims42, AstroCarp, hodori01, ilike3.1415926, Adventure10, Mango247
The MAA AMC program very much appreciates the increasing interest in international participation in our competitions.

Unfortunately, the rapid growth in interest and participation has made it unfeasible for our internal resources to manage the program as it currently operates. We are reevaluating the international registration and competition administration model during the 2016-2017 cycle.

For the 2016-2017 competition cycle, the MAA AMC exams will only be offered to schools with a USA, Canada, APO/FPO/DPO mailing addresses and schools registered with one of the five AMC international partners. If your school is in one of the areas represented by an AMC international partner, please contact the group manager directly for information on participating in the AMC program this year.

Thank you for your patience as we assess our options to successfully offer the AMC exams to more international audiences in the future. For more information: http://www.maa.org/math-competitions/international-registrations
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william122
1576 posts
#3 • 2 Y
Y by Adventure10, Mango247
Does this mean the foreign nations, such as China, can no longer take the AMCs? :o
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GeneralCobra19
1535 posts
#6 • 1 Y
Y by Adventure10
william122 wrote:
Does this mean the foreign nations, such as China, can no longer take the AMCs? :o
I think that is what he means.
But:
MAA wrote:
Manager Country Email
Jianping Wu China wjianp@yeah.net
Yusuf Zeybek Iraq yusuf.zeybek@fezalar.org
Adewale Solarin Nigeria asolarin2002@yahoo.com
Penny Chow Singapore penny@mathsoasis.com
Peter Shiue Taiwan shiue@unlv.nevada.edu
These countries will be able to take it.

EDIT: Skipiano corrected me. This is false.
This post has been edited 2 times. Last edited by GeneralCobra19, Sep 7, 2016, 1:57 AM
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hwl0304
1840 posts
#7 • 5 Y
Y by FlyingCucumber, EpicSkills32, Emathmaster, Adventure10, Mango247
Guys, realize that the restriction of international contestants is actually a good thing for the majority of people. Remember back last year, when AMC 10/12 B discussion had to be postponed because of international test takers? The two posts that DPatrick made as announcement got a total of nearly 250 downvotes when they still existed. (here). People are realizing now that those international students should have the right to take the tests and are getting less self-centered due to this topic surfacing, but understand that many American and Canadian people actually are benefited from this restriction, including many of the people protesting, believe it or not (I'm not saying that the others don't matter; Internationals: I'm sorry you are not allowed to take the test, but also consider the fact that you are removing a significant inconvenience for the majority of test takers and are also helping the math community in general).

(PS: this isn't meant to come out as harsh, just saying to not hate on this decision so much)
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First
2352 posts
#8 • 2 Y
Y by Adventure10, Mango247
Unless you are like a permanent resident you could try to represent your country for the IMO and take their version of the AMC's
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KenV
1198 posts
#10 • 1 Y
Y by Adventure10
hwl0304 wrote:
Guys, realize that the restriction of international contestants is actually a good thing for the majority of people. Remember back last year, when AMC 10/12 B discussion had to be postponed because of international test takers? The two posts that DPatrick made as announcement got a total of nearly 250 downvotes when they still existed. (here). People are realizing now that those international students should have the right to take the tests and are getting less self-centered due to this topic surfacing, but understand that many American and Canadian people actually are benefited from this restriction, including many of the people protesting, believe it or not (I'm not saying that the others don't matter; Internationals: I'm sorry you are not allowed to take the test, but also consider the fact that you are removing a significant inconvenience for the majority of test takers and are also helping the math community in general).

(PS: this isn't meant to come out as harsh, just saying to not hate on this decision so much)

I agree. You always have to consider the other side- in fact, it wouldn't make sense for MAA to make the decision if they did not believe it would improve the overall quality of the test.
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fofx
231 posts
#11 • 4 Y
Y by budu, soojoong, Adventure10, Mango247
This is extremely unfair to students living outside the US with American citizenship that have studied hard but it doesn't seem like anything will change this.

I'm not familiar with whether my school has an APO/FPO/DPO. It's the AES in New Delhi and it's associated with the American Embassy. Would this count as one of the three addresses?
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LenNon
16 posts
#12 • 1 Y
Y by Adventure10
It used to be that the international students outside America can only take AMC12 A test. The policy says that the students who are qualified for AIME should get higher than 100 or the top 5 percent of the test takers. Due to the huge number of students outside America in AMC12A, it is kind of easier to get into AIME through AMC12.( I may be wrong)
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jmerry
12096 posts
#13 • 3 Y
Y by fofx, Adventure10, Mango247
fofx wrote:
I'm not familiar with whether my school has an APO/FPO/DPO. It's the AES in New Delhi and it's associated with the American Embassy. Would this count as one of the three addresses?
Yes, although which one it is I couldn't say. APO (Army/Air Force) and FPO (Fleet - Navy/Marines) are military while DPO (Diplomatic) is specifically for embassies, but DPO is a recent addition to cover embassies that weren't already being handled by one of the others. You should ask someone at the school for details.
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HKIS200543
380 posts
#15 • 1 Y
Y by Adventure10
If one took the AMC 10 through a foreign partner, are they still eligible for AIME qualification?
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whiteawesomesun
1207 posts
#16 • 2 Y
Y by Adventure10, Mango247
yes yes///
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