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
Let be an acute triangle with incenter , circumcenter , and circumcircle . Let be the midpoint of . Ray meets at . Denote by and the circumcircles of and , respectively. Line meets at and , while line meets at and . Assume that lies inside and .
Consider the tangents to at and and the tangents to at and . Given that , prove that these four lines are concurrent on .
Given an acute non-isosceles triangle with circumcircle . is the midpoint of segment and is the midpoint of of (the one that doesn't contain ). and are points on such that . Assume there exists point on segment such that circumcircle of triangle is tangent to . Let be the circumcircle of triangle . Line meets for the second time at . Let be a point on such that be a circle thaat passes through , and tangents to and be a circle that passes through and tangents to . Prove that circle with center and radias is tangent to circles and .
Clearly works, so let's now assume that is non-constant.
so otherwise must be constant, becomes .
Observe that since gives and is not constant.
Now lets define the set
If for some then from we get that and thus
Also so ( note that since otherwise but and hence a contradiction.)
Now so as well and hence for every we have that for all , i.e . .
Now pick any , then and thus .
So all the solution are:
and they clearly work.
This post has been edited 1 time. Last edited by Prod55, May 9, 2023, 8:05 PM
Clearly works, so let's now assume that is non-constant.
so otherwise must be constant, becomes .
Observe that since gives and is not constant.
Now lets define the set
If for some then from we get that and thus
Also so ( note that since otherwise but and hence a contradiction.)
Now so as well and hence for every we have that for all , i.e . .
Now pick any , then and thus .
So all the solution are:
and they clearly work.
l hope that my solution is true
Let be the given assertion
1)
From we get and from and we get f(x)=f(1)=c for all
2)
From we get then
From and we get from , and we get from here we get
Then for all
This post has been edited 1 time. Last edited by Mathlover_1, May 9, 2023, 10:48 PM
Neat F.E., let the assertion as usual, assume non-constant as the other case works. Main Weapon: If , from we get for all call this
Now gives so from we have so if then is constant so we get that , now clearly is injective at or else for such that which again gives constant so nope.
Now gives and gives so by we get so and now by we get that so by we get that , now pick and let and for any reals and we get so constant for all values but .
Hence the set of solutions is for all reals and for all and , thus we are done .
Let P(x,y) be the assertion: f(x)+f(y)=f(xy)+f(f(x)+f(y))
assume f(a)=f(b). If we substitute y=a or b, the only term that can change is f(ax). Therfore f(ax)=f(bx) for all x. P(0,0): f(2f(0))=f(0) => f(2f(0)x)=f(0). So either f is constant (which works) or f(0)=0.
P(x,0): f(x)=f(f(x)) => f(2x)=f(2f(x))
(for x,y≠0) f(P(x,y)): f(f(x)+f(y))=f(f(xy)+f(f(x)+f(y)))=
=f(f(x)+f(y))+f(xy)-f(xy(f(x)+f(y))) => f(f(x)+f(y))=f(1).
For x=y≠0 we have f(1)=f(2f(x))=f(2x).
So f is constant except 0 which it is 0 at (it works).
But what if f was from R+ to R+, you might ask.
Well: the equation f(f(x)+f(y))=f(1) still holds, so if we can rewrite the equation as f(x)+f(y)=f(xy)+f(1).
Denote m=Inf(Im(f)). Suppose f(x)<m+ε for arbitrarily small ε>0.
P(x,x): f(1)=2f(x)-f(x^2)<m+2ε. so f(1)=m.
P(x,1/x): f(x)+f(1/x)=2m => f(x)=m for all x. So the only solution is a positive constant function.
Clearly works, so let's now assume that is non-constant.
so otherwise must be constant, becomes .
Observe that since gives and is not constant.
Now lets define the set
If for some then from we get that and thus
Also so ( note that since otherwise but and hence a contradiction.)
Now so as well and hence for every we have that for all , i.e . .
Now pick any , then and thus .
So all the solution are:
and they clearly work.
l hope that my solution is true
Let be the given assertion
1)
From we get and from and we get f(x)=f(1)=c for all
2)
From we get then
From and we get from , and we get from here we get
Then for all
the last step is wrong, works if is continuous at 0 but it's not always true.
Solved with hints. Really nice and instructive problem. I think I learnt a lot from this one. The answers are for all for a fixed constant and for a fixed constant . It’s easy to see that these functions satisfy the given conditions. We now show these are the only solutions. Let denote the assertion that for real numbers and . We first start off with fixing certain values of .
Claim : or is constant. Proof : Say there exists some such that . Then, comparing and yields, Thus, letting implies that for all real numbers . Thus, if is non-constant there does not exist such such that .
Now, implies which according to our previous observation requires, so we have indeed as claimed.
In what proceed, we assume that is not constant. First of all, note that implies, and implies for all real numbers . This will come in useful later. Note that gives, and also gives Thus, setting these two expressions for equal we have and further, when in the above result, Thus, . Now, returning to we have, which implies that for all real numbers . Then, applying this result to gives so, for all reals . Now, from this based on our previous observation, it follows that for all real numbers and . Considering and for any two non-zero implies that is indeed constant over the non-zero reals. Thus, all solutions are indeed as claimed.
The only solutions are for some constant , and also and . These clearly work. Now we assume that is not constant. It suffices to show that there is a constant for which .
Let be the given assertion.
.
.
So .
.
Thus, . Since is nonconstant, .
now gives .
with gives , so also .
We have .
, so . In this way, we get as well, so gives that . Now comparing and gives . For , setting gives , as desired.
Find all functions such that for all the following holds:
Let be the assertion .
Note that all functions of the form , work, otherwise suppose is nonconstant.
Comparing the last two we get since is nonconstant, so . Repeat the same process with instead:
And now we can compare:
If then is constant, so and we get .
So setting , we get where (since we had here), which satisfies the equation (by casework on whether or is ).