ka May Highlights and 2025 AoPS Online Class Information
jlacosta0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.
Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.
Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!
Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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.
Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28
Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19
Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30
Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14
Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19
Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)
Intermediate: Grades 8-12
Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22
MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21
AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22
Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that = 2∠AMP.
Suppose we have distinct positive integers , and an odd prime not dividing any of them, and an integer such that if one considers the infinite sequence and looks at the highest power of that divides each of them, these powers are not all zero, and are all at most . Prove that there exists some (which may depend on ) such that whenever divides an element of this sequence, the maximum power of that divides that element is exactly .
Expected Intersections from Random Pairing on a Circle
tom-nowy2
Nan hour ago
by lele0305
Let be a positive integer. Consider points on the circumference of a circle.
These points are randomly divided into pairs, and line segments are drawn connecting the points in each pair.
Find the expected number of intersection points formed by these segments, assuming no three segments intersect at a single point.
Find the maximum number of Permutation of set {} such that for every 2 different number and in this set at last in one of the permutation comes exactly after
Let and then we have .obviously the set of prime divisors of and are the same(note that ).(the case is trivial) we'll have two cases:
1-there exist an odd prime such that then prove that (prove this using ).
the rest is solved using inequality.But the other case remaining is:
2-there doesn't exist such :it means that must be .(prove this using again )
I will instead characterize every solution given that and —this is the version of the problem I received.
Clearly works, and by size we require hence no other solution exists for , so assume . Then as a consequence of Zsigmondy there must exist some prime dividing that doesn't divide and thus , unless:
for some , and hence . Then either which is impossible or , which yields and its permutation .
for some , so from which we find that must be a power of , say . Then from which we extract the very ugly answer for all such that .
Since everything is reversible, it follows that the solutions are , and for all such that . (clearly, this implies the original problem)
Assume Then, we have So, we have the solutions Now, assume one of is 1. WLOG let it be Therefore, we have For some prime suppose that such that Then, So, since we have Note that and since So, meaning If then and so Solving, we have If then and meaning Solving, we find no solutions. Now, assume that . Therefore, there exists some prime such that WLOG, let Since we have meaning Since, we must have In conclusion, the solutions are
Observe that and this clearly yields a solution. So assume that Observe that by size
Now, if we get which indeed is a solution as long as
WLOG assume that Then let We get By Zsigmondy's Theorem there is prime dividing but not which is a contradiction, unless we have in which This would give and testing possibilities yields so
To summarize, the solutions are for that yields integer values for the last one.