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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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0 replies
jlacosta
Apr 2, 2025
0 replies
2025 USA IMO
john0512   80
N 21 minutes ago by Martin.s
Congratulations to all of you!!!!!!!

Alexander Wang
Hannah Fox
Karn Chutinan
Andrew Lin
Calvin Wang
Tiger Zhang

Good luck in Australia!
80 replies
john0512
Apr 19, 2025
Martin.s
21 minutes ago
2025 PROMYS Results
Danielzh   28
N 37 minutes ago by ConfidentKoala4
Discuss your results here!
28 replies
Danielzh
Apr 18, 2025
ConfidentKoala4
37 minutes ago
{MathILy, MathILy-Er} 2025 Decisions Thread
mysterynotfound   5
N 38 minutes ago by ConfidentKoala4
Discuss your decisions here!
also share any relevant details about your decisions if you want
5 replies
mysterynotfound
Today at 3:35 AM
ConfidentKoala4
38 minutes ago
MOP Emails
hellohannah   16
N 40 minutes ago by Alex-131
So mop emails are probably coming tomorrow, feel free to discuss here. I'll probably post when I hear that they're out unless I'm asleep
16 replies
+5 w
hellohannah
Today at 4:59 AM
Alex-131
40 minutes ago
Inequalities
sqing   4
N 2 hours ago by sqing
Let $x,y\ge 0$ such that $ 13(x^3+y^3) \leq 125(1+xy)$. Prove that
$$  k(x+y)-xy\leq  5(2k-5)$$Where $k\geq 5.6797. $
$$  6(x+y)-xy\leq 35$$
4 replies
sqing
Yesterday at 1:04 PM
sqing
2 hours ago
Inscribed Semi-Circle!!!
ehz2701   2
N 5 hours ago by mathafou
A right triangle $ABC$ with legs $AB = a$ and $BC = b$ is drawn with a semicircle inscribed into the triangle. What is the smallest possible radius of the semi-circle and the largest possible radius?

2 replies
ehz2701
Sep 11, 2022
mathafou
5 hours ago
geometry
carvaan   1
N 5 hours ago by vanstraelen
OABC is a trapezium with OC // AB and ∠AOB = 37°. Furthermore, A, B, C all lie on the circumference of a circle centred at O. The perpendicular bisector of OC meets AC at D. If ∠ABD = x°, find last 2 digit of 100x.
1 reply
carvaan
Yesterday at 5:48 PM
vanstraelen
5 hours ago
Inequalities
nhathhuyyp5c   1
N Today at 9:09 AM by Mathzeus1024
Let $a, b, c$ be non-negative real numbers such that $a^2 + b^2 + c^2 = 3$. Find the maximum and minimum values of the expression
\[
P = \frac{a}{a^2 + 2} + \frac{b}{b^2 + 2} + \frac{c}{c^2 + 2}.
\]
1 reply
nhathhuyyp5c
Yesterday at 6:35 AM
Mathzeus1024
Today at 9:09 AM
weird permutation problem
Sedro   2
N Today at 8:56 AM by alexheinis
Let $\sigma$ be a permutation of $1,2,3,4,5,6,7$ such that there are exactly $7$ ordered pairs of integers $(a,b)$ satisfying $1\le a < b \le 7$ and $\sigma(a) < \sigma(b)$. How many possible $\sigma$ exist?
2 replies
Sedro
Yesterday at 2:09 AM
alexheinis
Today at 8:56 AM
In a school of $800$ students, $224$ students play cricket, $240$ students play
Vulch   2
N Today at 8:12 AM by MathBot101101
Hello everyone,
In a school of $800$ students, $224$ students play cricket, $240$ students play hockey and $336$ students play basketball. $64$ students play both basketball and hockey, $80$ students play both cricket and basketball, $40$ students play both cricket and hockey, and $24$ students play all three: basketball, hockey, and cricket. Find the number of students who do not play any game.

Edit:
In the above problem,I just want to know that why the number of students who don't play any game shouldn't be 0, because,if we add 224,240 and 336 it comes out to be 800.I have solution,but I just want to know how to explain it without theoretically.Thank you!
2 replies
Vulch
Yesterday at 11:41 PM
MathBot101101
Today at 8:12 AM
Combinatorial proof
MathBot101101   8
N Today at 5:55 AM by MathBot101101
Is there a way to prove
\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}=1-\frac{1}{{n+1)!}
without induction and using only combinatorial arguments?

Induction proof wasn't quite as pleasing for me.
8 replies
MathBot101101
Yesterday at 7:37 AM
MathBot101101
Today at 5:55 AM
Inequalities
sqing   25
N Today at 3:58 AM by sqing
Let $   a,b    $ be reals such that $  a^2+ab+b^2 =3$ . Prove that
$$ \frac{4}{ 3}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{11}{4 }$$$$ \frac{13}{ 4}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{2}{3 }$$$$ \frac{3}{ 2}\geq  \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq -\frac{17}{6 }$$$$ \frac{19}{ 6}\geq  \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2}$$Let $   a,b    $ be reals such that $  a^2-ab+b^2 =1 $ . Prove that
$$ \frac{3}{ 2}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }+ab \geq \frac{4}{15 }$$$$ \frac{14}{ 15}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }-ab \geq -\frac{1}{2 }$$$$ \frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq \frac{13}{42 }$$$$ \frac{41}{ 42}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2 }$$
25 replies
sqing
Apr 16, 2025
sqing
Today at 3:58 AM
Three variables inequality
Headhunter   4
N Today at 3:18 AM by lbh_qys
$\forall a\in R$ ,$~\forall b\in R$ ,$~\forall c \in R$
Prove that at least one of $(a-b)^{2}$, $(b-c)^{2}$, $(c-a)^{2}$ is not greater than $\frac{a^{2}+b^{2}+c^{2}}{2}$.

I assume that all are greater than it, but can't go more.
4 replies
Headhunter
Yesterday at 6:58 AM
lbh_qys
Today at 3:18 AM
Indonesia Regional MO 2019 Part A
parmenides51   23
N Today at 2:08 AM by chinawgp
Indonesia Regional MO
Year 2019 Part A

Time: 90 minutes Rules


p1. In the bag there are $7$ red balls and $8$ white balls. Audi took two balls at once from inside the bag. The chance of taking two balls of the same color is ...


p2. Given a regular hexagon with a side length of $1$ unit. The area of the hexagon is ...


p3. It is known that $r, s$ and $1$ are the roots of the cubic equation $x^3 - 2x + c = 0$. The value of $(r-s)^2$ is ...


p4. The number of pairs of natural numbers $(m, n)$ so that $GCD(n,m) = 2$ and $LCM(m,n) = 1000$ is ...


p5. A data with four real numbers $2n-4$, $2n-6$, $n^2-8$, $3n^2-6$ has an average of $0$ and a median of $9/2$. The largest number of such data is ...


p6. Suppose $a, b, c, d$ are integers greater than $2019$ which are four consecutive quarters of an arithmetic row with $a <b <c <d$. If $a$ and $d$ are squares of two consecutive natural numbers, then the smallest value of $c-b$ is ...


p7. Given a triangle $ABC$, with $AB = 6$, $AC = 8$ and $BC = 10$. The points $D$ and $E$ lies on the line segment $BC$. with $BD = 2$ and $CE = 4$. The measure of the angle $\angle DAE$ is ...


p8. Sequqnce of real numbers $a_1,a_2,a_3,...$ meet $\frac{na_1+(n-1)a_2+...+2a_{n-1}+a_n}{n^2}=1$ for each natural number $n$. The value of $a_1a_2a_3...a_{2019}$ is ....


p9. The number of ways to select four numbers from $\{1,2,3, ..., 15\}$ provided that the difference of any two numbers at least $3$ is ...


p10. Pairs of natural numbers $(m , n)$ which satisfies $$m^2n+mn^2 +m^2+2mn = 2018m + 2019n + 2019$$are as many as ...


p11. Given a triangle $ABC$ with $\angle ABC =135^o$ and $BC> AB$. Point $D$ lies on the side $BC$ so that $AB=CD$. Suppose $F$ is a point on the side extension $AB$ so that $DF$ is perpendicular to $AB$. The point $E$ lies on the ray $DF$ such that $DE> DF$ and $\angle ACE = 45^o$. The large angle $\angle AEC$ is ...


p12. The set of $S$ consists of $n$ integers with the following properties: For every three different members of $S$ there are two of them whose sum is a member of $S$. The largest value of $n$ is ....


p13. The minimum value of $\frac{a^2+2b^2+\sqrt2}{\sqrt{ab}}$ with $a, b$ positive reals is ....


p14. The polynomial P satisfies the equation $P (x^2) = x^{2019} (x+ 1) P (x)$ with $P (1/2)= -1$ is ....


p15. Look at a chessboard measuring $19 \times 19$ square units. Two plots are said to be neighbors if they both have one side in common. Initially, there are a total of $k$ coins on the chessboard where each coin is only loaded exactly on one square and each square can contain coins or blanks. At each turn. You must select exactly one plot that holds the minimum number of coins in the number of neighbors of the plot and then you must give exactly one coin to each neighbor of the selected plot. The game ends if you are no longer able to select squares with the intended conditions. The smallest number of $k$ so that the game never ends for any initial square selection is ....
23 replies
parmenides51
Nov 11, 2021
chinawgp
Today at 2:08 AM
AMC Rant
yebbisdubbis   160
N Feb 14, 2018 by GameMaster402
Source: Myself
I am going to rant about how the AMC is very stupid. I welcome your comments and let me know if you agree with me.

I love doing math. I love thinking about olympiad type problems over several days; I often look forward to weekends when I have time to work on IMO Shortlists, Russian and Romanian problems from the Contests page, and USAMTS problems. These Olympiad type problems are sometimes stupid/contrived/there's really no use in doing them except to practice for high school math competitions; especially when it comes to inequalities and advanced euclidean geometry. I don't waste my time with these, because I don't do Olympiads to practice for math competitons. I do Olympiad problems to improve my problem solving skills, for fun, and because they are challenging yet the mathematics involved is elementary. I tend to work on combinatorics and number theory problems which require problem solving only, and not some ridiculous "trick" or pointless bashes. And I don't care about speed either - I take my time, I like to ponder.

And I can say I have greatly improved my problem solving skills in the past year by doing these kinds of problems. And so I look forward to taking the USAMO, if I qualify for it, as I believe I do stand a chance. The problem for me, is qualifying. We all know that to qualify for the USAMO, you have to have a superb combined performance on the AIME and the AMC. Now, I have never really enjoyed computational math - its boring, its easy to make mistakes, and it often tests concepts which have been recycled from one contest to another, many times over - you might call these concepts "tricks". And so every year I dread the AMC, and the AIME - I don't do well with the time constraints, I always make mistakes, and I end up not making the USA(J)MO.

So I would like to rant about the stupidity of the AMC, both as a test of skill in mathematics, and its relation to the culture of math competitions in the U.S. In many other countries with national math competitions, there is no computational exam - only Olympiad. Computational contests make little sense in the first place, as they do not produce "thinkers", "mathematicians", or even "problem-solvers". From my experience, at least with the AMC, they instead produce "m(athletes)", "expert test-takers", and "top-scorers". This is NOT what real math is about, and so it is my belief that these tests do not produce the kind of thinkers that other countries, especially those in Eastern Europe, do so well. The majority of scores on the USAMO are close to 0 because the AMC selects speedy, "tricksters", trained to solve the same kinds of problems very quickly, and not ponderers, creative thinkers, and problem-solvers. One can look at where the Eastern European olympiad veterans end up - as emmenint mathematicians, many with Fields medals; and where AMC top scorers end up - on Wall Street.

The AMC is also just a ridiculous method of selecting people for the USAMO. For me it is quite difficult to make the USAMO through the AMC 12. I don't believe I can get more than 18 questions right on the AMC 12, due to time constraints, and I'm sick of practicing for this boring competition. I am afraid I will be dissapointed again this year, after I fail to make USAMO, as it will have been another oppurtunity lost to practice my Olympiad problem solving skills. I don't want to leave high school never having been able to show my skills in a contest setting.

Please comment.
160 replies
yebbisdubbis
Nov 28, 2014
GameMaster402
Feb 14, 2018
AMC Rant
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Source: Myself
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Bob_Smith
513 posts
#147 • 2 Y
Y by Adventure10 and 1 other user
How could you????

Personally, I really like AMC's. They really test your problem solving skills and application of neat tricks.
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viperstrike
1198 posts
#148 • 6 Y
Y by ThePathOfWar, DrMath, bearytasty, 62861, Adventure10, Mango247
"really test your problem solving skills" - good one!
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alderan
19 posts
#150 • 2 Y
Y by Adventure10, Mango247
I am wondering if there is a thread/post that considers if the AIME is open to anyone who takes the AMC 10/12. This would be sort of analogous to the UK system, where for their AIME equivalent, non-qualifiers can pay to take the test. After all, both AIME and AMC 10/12 grading involves running the sheets through a scantron. Will significantly "de-stress" the lives of high-schoolers...If this has been discussed before, can someone please point me to the thread?

Thanks.
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green_dog_7983
3713 posts
#151 • 2 Y
Y by Adventure10, Mango247
I mean you can just "take" the AIME after they post on AoPS.
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alderan
19 posts
#153 • 2 Y
Y by Adventure10, Mango247
green_dog_7983 wrote:
I mean you can just "take" the AIME after they post on AoPS.

:). The intent of my question was to take the test officially and have it graded, so that they are in contention for JMO. I am seeing posts where folks say that they guessed right on the last 10 (string of D's or C's or whatever) and they make AIME. Then they end up struggling on the AIME. Allowing every AMC10/12 test taker to the AIME, the stress due to the AMC10/12 is much lower...
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azagorod101
194 posts
#154 • 2 Y
Y by Adventure10, Mango247
To be fair all people in the honors lane have to take the AMC at my school. Too bad nobody except a few take them seriously.
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wu2481632
4239 posts
#156 • 2 Y
Y by rzlng, Adventure10
Quote:
I really doubt potential IMO team candidates get cut by the AMC/AIME section of the process.

This is very possible. One year, an IMO gold medalist got a 9 on the AIME before the IMO. Just this past year two USAJMO HMs missed qualifying again.
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RegularHexagon
441 posts
#157 • 2 Y
Y by Adventure10, Mango247
Just noticed that this thread was revived from 2014. I didn't intend to revive an argument, so I deleted my post.
wu2481632 wrote:
Quote:
I really doubt potential IMO team candidates get cut by the AMC/AIME section of the process.

This is very possible. One year, an IMO gold medalist got a 9 on the AIME before the IMO. Just this past year two USAJMO HMs missed qualifying again.

I understand and I know personally one of the USAJMO HMs who didn't requalify. The problem was that no alternative to the AMC/AIME was proposed. If there was a perfect way to pick out worthy olympiad contestants, we would agree that that system should be in place.
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mathchampion1
3895 posts
#158 • 2 Y
Y by Adventure10, Mango247
I think the 2018 10A was worse tbh. I got 123, but the floor will go pretty low this time.
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mathdragon2000
2458 posts
#159 • 2 Y
Y by Adventure10, Mango247
RegularHexagon wrote:
Just noticed that this thread was revived from 2014. I didn't intend to revive an argument, so I deleted my post.
wu2481632 wrote:
Quote:
I really doubt potential IMO team candidates get cut by the AMC/AIME section of the process.

This is very possible. One year, an IMO gold medalist got a 9 on the AIME before the IMO. Just this past year two USAJMO HMs missed qualifying again.

I understand and I know personally one of the USAJMO HMs who didn't requalify. The problem was that no alternative to the AMC/AIME was proposed. If there was a perfect way to pick out worthy olympiad contestants, we would agree that that system should be in place.
USAMTS would be a nice alternative, except the index calculation is fundamentally unfair. A perfect 75 on the USAMTS (which is magnitudes of order harder than a 100 on the AMC 12) still only gets you a 100 on the 12, and you'll need like a 13/14 on the AIME to make USA(J)MO :mad:

I really hate that system, in case you couldn't tell :P
This post has been edited 1 time. Last edited by mathdragon2000, Feb 13, 2018, 11:24 PM
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programjames1
3046 posts
#160 • 1 Y
Y by Adventure10
I agree. You should be able to get a higher index. However, USAMTS is a month long contest, so maybe if it were made harder.

And one more thing, you can use other resources such as computer programs in USAMTS, much different than IMO, IMO.
This post has been edited 1 time. Last edited by programjames1, Feb 13, 2018, 11:59 PM
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mathdragon2000
2458 posts
#161 • 1 Y
Y by Adventure10
@above
programjames1 wrote:
I agree. You should be able to get a higher index. However, USAMTS is a month long contest, so maybe if it were made harder.

And one more thing, you can use other resources such as computer programs in USAMTS, much different than IMO, IMO.
I see what you did there :P

I guess I can kind of see your point - maybe USAMTS could ban anything other than your brain, and make it an on-site test. (e.g. come back on three separate days, like the USA(J)MO does)
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Archimedes15
1491 posts
#162 • 1 Y
Y by Adventure10
mathdragon2000 wrote:
@above
programjames1 wrote:
I agree. You should be able to get a higher index. However, USAMTS is a month long contest, so maybe if it were made harder.

And one more thing, you can use other resources such as computer programs in USAMTS, much different than IMO, IMO.
I see what you did there :P

I guess I can kind of see your point - maybe USAMTS could ban anything other than your brain, and make it an on-site test. (e.g. come back on three separate days, like the USA(J)MO does)

This suggestion goes back to a post made by v_Enhance a long time ago in the thread. It would be impractical to offer such a large proof contest over a short period of time and would put a lot of pressure on the graders. It is better to just allow people to take the test on their own time so that the graders could grade as the submissions come in and not all at once.

The other reason I don't like that solution is because it would make it so I would have to clear out my schedule for three days. I'm not saying I wouldn't (USAMTS is pretty fun), but most wouldn't create three days worth of time to take a proof competition if they know that it might amount to nothing. USA(J)MO is different because if you have qualified for USA(J)MO, being able to say the USA(J)MO is a huge accomplishment in itself. However, taking the USAMTS isn't something most people would put on their resume unless they were the winners or something. This would result in less participation in the USAMTS competitions.
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statman
543 posts
#163 • 3 Y
Y by TheBigOne, Adventure10, Mango247
The original post seems to suggest that ending up on Wall Street is a failure. Why is this? Isn't applied math also math?
This post has been edited 1 time. Last edited by statman, Feb 14, 2018, 12:33 AM
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GameMaster402
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#166 • 5 Y
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Disclaimer: These are simply my beliefs and opinions. Also, I did not read every single post in this thread so some of my ideas may have already been said.

Hm it seems like a coincidence that I was talking to a friend about this exact thing today (he didn't know that this thread existed) and he said the exact same points as OP.

This friend qualified for JMO in 7th grade and 8th grade. He spent about 5 hours every day in 7th grade preparing for AIME and made JMO because he only focused on computational math. However, he did pretty bad on it. In 8th grade, he only focused on proof math...and his AIME score went down by 3 points, while his JMO score went up by 13. The next year he did the exact same thing, only focusing on proof math. However, his AIME score remained unchanged, his index remainded unchanged, but he did not qualify for JMO. However, his proof writing skills had increased by a significant amount he and thoroughly solved 4 JMO problems while solving at home.

I believe that the system works well in introducing students for math via the AMCs (as said by many others), but does not help olympiad qualifiers with their olympiad scores (after all, they spend most of their time trying to get into olympiad and only have a month to actually prep for olympiad). The difference between the AIME and Olympiad is quite significant, and although some AIME #13-15s might be harder than some JMO 1/4s, the overlap basically ends there. Something that my friend brought up was the creation of a new contest in between AIME/JMO difficulty, kind of like BAMO or something. However, I don't know how this would be implemented considering how our current system has remained unchanged for ~ 20 years, and it would be a lot more work for MAA.

I think I would agree with the OP on most of these points - there is obviously a correlation with Olympiad Participants and their AMC/AIME scores, but does it work the other way around? That is, is there a correlation between non-olympiad qualifiers and their proof writing skill?

Maybe a possible way to fix the system is to space out the AMC contests, such that AMC 10/12 might be in mid December, AIME would be in mid-late February, and USA(J)MO in mid-late April as usual. This would give contestants at least two months to prepare for the next level of competition, and might have a better turnout of scores. However, one major flaw is how we would do this because we've been doing the same sort of thing for almost 2 decades.
This post has been edited 1 time. Last edited by GameMaster402, Feb 14, 2018, 1:15 AM
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