ka July Highlights and 2025 AoPS Online Class Information
jwelsh0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!
[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]
MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
and train with the best! Please note that early bird pricing ends August 19th!
Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.
Our full course list for upcoming classes is below:
All classes start 7:30pm ET/4:30pm PT unless otherwise noted.
Prealgebra 2
Friday, Jul 25 - Nov 21
Sunday, Aug 17 - Dec 14
Tuesday, Sep 9 - Jan 13
Thursday, Sep 25 - Jan 29
Sunday, Oct 19 - Feb 22
Monday, Oct 27 - Mar 2
Wednesday, Nov 12 - Mar 18
Introduction to Algebra A
Tuesday, Jul 15 - Oct 28
Sunday, Aug 17 - Dec 14
Wednesday, Aug 27 - Dec 17
Friday, Sep 5 - Jan 16
Thursday, Sep 11 - Jan 15
Sunday, Sep 28 - Feb 1
Monday, Oct 6 - Feb 9
Tuesday, Oct 21 - Feb 24
Sunday, Nov 9 - Mar 15
Friday, Dec 5 - Apr 3
Introduction to Counting & Probability
Wednesday, Jul 2 - Sep 17
Sunday, Jul 27 - Oct 19
Monday, Aug 11 - Nov 3
Wednesday, Sep 3 - Nov 19
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Friday, Oct 3 - Jan 16
Sunday, Oct 19 - Jan 25
Tuesday, Nov 4 - Feb 10
Sunday, Dec 7 - Mar 8
Introduction to Number Theory
Tuesday, Jul 15 - Sep 30
Wednesday, Aug 13 - Oct 29
Friday, Sep 12 - Dec 12
Sunday, Oct 26 - Feb 1
Monday, Dec 1 - Mar 2
Introduction to Algebra B
Friday, Jul 18 - Nov 14
Thursday, Aug 7 - Nov 20
Monday, Aug 18 - Dec 15
Sunday, Sep 7 - Jan 11
Thursday, Sep 11 - Jan 15
Wednesday, Sep 24 - Jan 28
Sunday, Oct 26 - Mar 1
Tuesday, Nov 4 - Mar 10
Monday, Dec 1 - Mar 30
Introduction to Geometry
Monday, Jul 14 - Jan 19
Wednesday, Aug 13 - Feb 11
Tuesday, Aug 26 - Feb 24
Sunday, Sep 7 - Mar 8
Thursday, Sep 11 - Mar 12
Wednesday, Sep 24 - Mar 25
Sunday, Oct 26 - Apr 26
Monday, Nov 3 - May 4
Friday, Dec 5 - May 29
Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)
Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET)
Intermediate: Grades 8-12
Intermediate Algebra
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22
Friday, Aug 8 - Feb 20
Tuesday, Aug 26 - Feb 24
Sunday, Sep 28 - Mar 29
Wednesday, Oct 8 - Mar 8
Sunday, Nov 16 - May 17
Thursday, Dec 11 - Jun 4
Precalculus
Wednesday, Aug 6 - Jan 21
Tuesday, Sep 9 - Feb 24
Sunday, Sep 21 - Mar 8
Monday, Oct 20 - Apr 6
Sunday, Dec 14 - May 31
Advanced: Grades 9-12
Calculus
Sunday, Sep 7 - Mar 15
Wednesday, Sep 24 - Apr 1
Friday, Nov 14 - May 22
Contest Preparation: Grades 6-12
MATHCOUNTS/AMC 8 Basics
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 17 - Nov 9
Wednesday, Sep 3 - Nov 19
Tuesday, Sep 16 - Dec 9
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Oct 6 - Jan 12
Thursday, Oct 16 - Jan 22
Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!)
MATHCOUNTS/AMC 8 Advanced
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 17 - Nov 9
Tuesday, Aug 26 - Nov 11
Thursday, Sep 4 - Nov 20
Friday, Sep 12 - Dec 12
Monday, Sep 15 - Dec 8
Sunday, Oct 5 - Jan 11
Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!)
Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!)
AMC 10 Problem Series
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 10 - Nov 2
Thursday, Aug 14 - Oct 30
Tuesday, Aug 19 - Nov 4
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!)
Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)
AMC 10 Final Fives
Friday, Aug 15 - Sep 12
Sunday, Sep 7 - Sep 28
Tuesday, Sep 9 - Sep 30
Monday, Sep 22 - Oct 13
Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, Oct 8 - Oct 29
Thursday, Oct 9 - Oct 30
AMC 12 Problem Series
Wednesday, Aug 6 - Oct 22
Sunday, Aug 10 - Nov 2
Monday, Aug 18 - Nov 10
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)
AMC 12 Final Fives
Thursday, Sep 4 - Sep 25
Sunday, Sep 28 - Oct 19
Tuesday, Oct 7 - Oct 28
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:
To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.
More specifically:
For new threads:
a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.
Examples: Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿) Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"
b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.
Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".
c) Good problem statement:
Some recent really bad post was:
[quote][/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.
For answers to already existing threads:
d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve , do not answer with " is a solution" only. Either you post any kind of proof or at least something unexpected (like " is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.
e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.
To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!
Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).
The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
Find all positive integers such that the sequence ,, is periodic modulo from some point onward; that is, there exists a positive integer such that the sequence , for , is periodic modulo .
Let be a positive integer. On an board, players and take turns in a game. On each turn, a player selects an edge of the board (not on the board border) and makes a cut along that edge. If after the move the board is split into more than one piece, then that player loses the game.
Player moves first. Depending on the value of , determine whether one of the players has a winning strategy.
Source: 2011 Lusophon Mathematical Olympiad - Problem 6
Let be a positive real number. The scorpion tries to catch the flea on a chessboard. The length of the side of each small square of the chessboard is . In this game, the flea and the scorpion move alternately. The flea is always on one of the vertexes of the chessboard and, in each turn, can jump from the vertex where it is to one of the adjacent vertexes. The scorpion moves on the boundary line of the chessboard, and, in each turn, it can walk along any path of length less than .
At the beginning, the flea is at the center of the chessboard and the scorpion is at a point that he chooses on the boundary line. The flea is the first one to play. The flea is said to escape if it reaches a point of the boundary line, which the scorpion can't reach in the next turn. Obviously, for big values of , the scorpion has a strategy to prevent the flea's escape. For what values of can the flea escape? Justify your answer.
Let be a positive integer and an alphabet with letters. Find the largest positive integer (as a function of ) for which there exists a sequence of letters satisfying the following conditions: a), for all . b), for all . c) There do not exist indices such that and .
In the Cartesian plane, a point is called integer if its coordinates are integers. A triangle with integer vertices has exactly 3 integer points in its interior and exactly 8 integer points on its perimeter (including the vertices). If the incenter of the triangle is an integer point and the other 2 interior integer points lie on its incircle, find the lengths of the triangle’s sides.
Source: IMO ShortList 1999, combinatorics problem 2
If a rectangle can be tiled using pieces like those shown in the diagram, prove that is even. Show that there are more than ways to file a fixed rectangle with pieces. (symmetric constructions are supposed to be different.)
Let denote the assertion. gives . gives Case 1: .
Then, for all , so for all , which is indeed a solution. Case 2: . for gives so vanishes on all nonzero reals. Now, suppose . Then, gives contradiction. So as well, which means in this case, is the zero function.
Therefore, the only solutions are and for all .
This post has been edited 1 time. Last edited by Pleaseletmewin, Mar 20, 2022, 7:39 AM
Notice that the only constant function satisfying is the zero function henceforth assume is non-constant and there exists some number such that . Let denote the given assertion
.
.
Fixing such that gives that is surjective. Claim: is injective at . Proof. Assume there exists such that then gives but since is surjective it means that is constant which violates our assumption. Now, taking to be a free variable choose such that and plugging gives so for all nonzero for all real (because is also satisfied by the relation). To conclude, the two functions satisfying the original equation are and .
This post has been edited 1 time. Last edited by MrOreoJuice, Mar 20, 2022, 11:34 AM
First, note that is a solution because for all and is a solution because for all and . Hence, we can assume from now on that there exist (not necessarily distinct) and such that and . Let .
. As the RHS can take any real value, so is surjective.
.
. Since and is surjective, can take any real value. However, this implies that for all , contradicting the fact that . Hence, the only such functions are and .
The answer is and only, which both work, so we show that no other solutions exist. Suppose , and let denote the assertion.
Pick some with . By letting vary in , it follows that is surjective.
Now suppose for some . Then gives . If , then by varying again and using surjectivity shows that , which is impossible because then it's not surjective. Thus and is injective about .
To finish, from we have , as desired.
meh.
We claim that the only solutions to the F.E. are and .
Let the assertion of the functional Equation, clearly works so here we will assume that there exists such that Case 1:
We use our in a smart way, we got so now Case 2:
Assume that there exists a real such that , then by Hence is injective at and now by Hence all the functions that work are thus we are done
Find all functions such that for all real numbers and , Proposed by novus677
Say denote the assertion. Now,note that, .This motivates us for injectivity at .
Say there exists such that .
Now, we take this oppurtunity and set, So
Now we are left with 1 case where no such exists.The function is injective at and thus, So the only set of solutions satisfying the equation is
We claim that the only solutions are and , which work.
Let denote the assertion that . By , we have , so . If there exists such that , then for nonzero , gives , and since , we have . This gives the solution . Otherwise, implies that . Then, gives , so .
We claim that the only solutions are and , which obviously work.
Let denote the assertion, or is injective
CASE : f is injective
then
then combined with injectivity gives
then
However the thing wrapped in function in RHS is clearly surjective,
The solutions to this functional equation are , which obviously works. Let be the usual assertion.
Now we separate the problem into two cases.
Case 1: There exists a not equal to such that . or .
Clearly, is impossible since Therefore in this case, the only possible solution for is
Case 2: There doesn't exist a not equal to such that .
This clearly shows that is bijective since . .
Therefore in this case, the only possible solution for is
Assume is nonzero for at least one input. Taking such that is nonzero, is bijective. Set , then , so . Then set , so , so , so is the identity as desired.