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)
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Introduction to Counting & Probability
Thursday, May 15 - Jul 31
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Introduction to Number Theory
Friday, May 9 - Aug 1
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Introduction to Algebra B
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Introduction to Geometry
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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
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MATHCOUNTS/AMC 8 Basics
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Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
AMC 10 Final Fives
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AMC 12 Problem Series
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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
Let be a triangle. The incircle of touches the sides and at the points and , respectively. Let be the point where the lines and meet, and let and be points such that the two quadrilaterals and are parallelogram.
Prove that .
Chords and of circle meet at point . The line through parallel to meets again at , and meets again at . Let and let be the reflection of over . Show that passes through the midpoint of .
Consider the triangle with . The circumcircle of has radius . There is a point in the interior of the line segment such that and the length of is . The perpendicular bisector of intersects at the points and .
For context, this is a made-up functional equation mentioned at the end of section 6 of https://web.evanchen.cc/handouts/FuncEq-Intro/FuncEq-Intro.pdf, where it's shown that injective would let you finish. However, is not necessarily injective, e.g. if is constant. I make no promise that there is a nice solution to this equation.
Case 1: .
Note . . Taking .
If , is constant. Easy to find solutions.
If ,.
Original equation becomes .
Putting ,.
Hence, which is a well known FE. (Cauchy+multiplicative).
Only solution (or which dosent fit).
Case 2: . . or .
This means that if then .
If , for some . .
This means that for some . gives constant.
Else is injective and it is easy to finish.
I can't remember if this is what I thought of back in April but it should be correct.
Furthermore, it is interesting to see if anyone can remove the condition / use a weaker one. I probably tried this back then and failed.
This post has been edited 2 times. Last edited by gghx, Sep 21, 2021, 12:24 AM
Case 1: .
Note . . Taking .
If , is constant. Easy to find solutions.
If ,.
Original equation becomes .
Putting ,.
Hence, which is a well known FE. (Cauchy+multiplicative).
Only solution (or which dosent fit).
Case 2: . . or .
This means that if then .
If , for some . .
This means that for some . gives constant.
Else is injective and it is easy to finish.
I can't remember if this is what I thought of back in April but it should be correct.
Furthermore, it is interesting to see if anyone can remove the condition / use a weaker one. I probably tried this back then and failed.
Can you help me how to prove that there are only 2 cases for ?
Claim: is injective (barring constant solutions)
Suppose for some .
Comparing, we get .
Comparing, we get .
So , whether or .
Comparing, we get . If one of is zero, then the other isn't, so is constant. The only constant solutions are and , so assume from now on that , we get .
Comparing, we get . Setting such that , this becomes . Then: Let , this is just .
Comparing, we get . Setting , we get , so is a constant function whose solutions we already encountered.
Swapping we get , so using injectivity and setting , we get , testing gives .
This post has been edited 1 time. Last edited by jasperE3, Apr 27, 2025, 8:34 PM