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 Number Theory
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Introduction to Algebra B
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Introduction to Geometry
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Paradoxes and Infinity
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Intermediate: Grades 8-12
Intermediate Algebra
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MATHCOUNTS/AMC 8 Advanced
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AMC 10 Problem Series
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AMC 10 Final Fives
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Introduction to Programming with Python
Thursday, May 22 - Aug 7
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Tuesday, Jun 17 - Sep 2
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Some users don't want to learn, some other simply ignore advises.
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To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.
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- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
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[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".
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[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.
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Let be the unit circle in the complex plane. Let be the map given by . We define and for . The smallest positive integer such that is called the period of . Determine the total number of points in of period .
(Hint : )
Let be a triangle and let be its circumcenter and its incenter.
Let be the radical center of its three mixtilinears and let be the isogonal conjugate of .
Let be the Gergonne point of the triangle .
Prove that line is parallel with line .
For the set of positive integers , no matter how its elements are partitioned into two subsets, at least one of the subsets must contain three numbers ( is allowed) such that . Find the minimal .
Points Lying on its Cevian Triangle's Thomson Cubic
Feuerbach-Gergonne1
N2 hours ago
by golue3120
Source: Own
Given and a point , let be the cevian triangle of with respect to . Let be the orthocenter of , and denote the isotomic conjugate of with respect to by , respectively. Let the centroid of be , and denote the isogonal conjugate of with respect to by . Prove that or in brief
Let be the circumcenter and the orthocenter of an acute triangle . Prove that the area of one of the triangles , and is equal to the sum of the areas of the other two.
There are line segments in the plane such that every two segments cross and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it facing the other endpoint. Then he will clap his hands times. Every time he claps,each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will ever occupy the same intersection point at the same time.
(a) Prove that Geoff can always fulfill his wish if is odd.
(b) Prove that Geoff can never fulfill his wish if is even.
Let be the assertion .
Now, let satisfy and . We have Note that is a solution. Assume there exists an , such that . We get
Thus, is injective.
From here we easily get the other two solutions.
For , we get that has a zero point. Let . Plugging in , we get .
Now for , for all we have , and by changing by we get that is odd.
1) There exists t different from s. t. .
For , for all , we have . So for all , but can get every real value while varying, so is a zero function.
2) :
Take : , so , so for all is either or .
We can easily check that implies .
So is zero, fixed or for all . These three functions satisfy the condition.
This post has been edited 3 times. Last edited by MilosMilicev, May 15, 2018, 1:12 PM
Denote as the assertion and assume that such function exists. is a solution, and it is the only constant solution. Let's look at nonconstant solutions now.
Since is nonconstant, we may take such that is nonzero.
We will use this to show that is surjective. For any , take .
Next, we will show that is injective. First, we will show that if if and only if .
First, take to have for all .
If , by the fact that is surjective we get for all .
This is an obvious contradiction, so . Now if and , by we have , so again, contradiction to surjective condition. We now have .
To improve this to injectivity, assume and . Then by and , it is easy to note that is a periodic function with period . Since is nonzero, this cannot hold. Therefore, is injective.
Now gives and gives .
So by injectivity we have , and .
This gives us that is equal to either or . Now we take care of the "point-wise trap".
Assume nonzero reals exist such that and . gives the desired contradiction.
So in the end we have three solutions, ,, and . These clearly work, done.
This post has been edited 3 times. Last edited by rkm0959, Feb 19, 2018, 1:15 PM
Determine all functions satisfying for all real numbers and .
Assume that for any constant. Now take that is not constant.
Let the assertion of the given FE. Now assume that exists more than 1 cero no the function.
Let the set of the ceros of and take then . Then is periodic at .
Note that is also a cero by .
Then: Now take then
In fact if we skip the case when then Assume that exists such that then i will prove that . Then is injective.
That means exists an only and hence (contradiction!!) (We assumed that exists more than 1 cero, thats why contradiction)
Hence has an unique
The same proof for get injective.
Then: Then the solutions are:
Thus we are done
This post has been edited 1 time. Last edited by MathLuis, May 4, 2021, 1:01 PM
Note that is the only constant solution. Now we look for non-constant solutions.
Let be the assertion.
If there exist a such that , then and so is constant, contradiction. so . is odd. so for all .
So we have and as solutions.
Let for some we have and . , contradiction.
So we have our three solutions.
I have a much simpler solution but I don't know that it is correct or not, kindly check.
Substituting in the original equation gives can be any real number (except if either or ) and it is not possible that (R represents every real number/) and so the above equation is true if and only if either or .
Case 1 - , it is indeed a solution.
Case 2 - which gives
Combining, we get and
This post has been edited 1 time. Last edited by logrange, May 9, 2021, 7:03 PM
I claim all the solutions are , which all satisfy the given relation. Disregard the trivial solution and let . Then for all gives that is surjective. In particular putting in we get . Assume there exists some other with image . Then gives and , so periodic is modulo . Subtracting leaves a contradiction, so is injective at .
Now for some let be a solution to , which must exist. Putting in we get , or .
Putting in once again gives and multiplying gives , so for any particular positive . By , this applies to any real .
Let be some non-zero reals with and , then putting in gives: Here, we only have two cases, the first one is that and have the same sign, this gives , or . The second one is , or . Either way, we get a contradiction and only the linear solutions work.
I denote by plugging in some and into the given equation.
Firstly, suppose that there exists a real number such that , otherwise for all and thus f is bijective.
since f is bijective and therefore injective, we know:
Case 1:
then we have but that does not hold true for all so for some
Plugging back in the original equation, we get or So... ,, are all solutions.
Consider and , this gives , thus . Then gives , and . Now consider with . gives . The left hand side evaluates to . Thus we are either forced or is all zero. Assume the former. Now we show injective, consider with ,, then forces , so either or (this second case doesn't matter because by definition). Thus we can conclude for all . Then consider , this gives , which is a contradiction. Thus is injective. Then take to get , so is linear going thru the origin, testing functions we get , along with our original solution of
First assume that there exists such that . implies .So either 1. for all or 2..So we proved that and is injective at (since and at we can say this). implies so is odd,and gives or .Plug in and this will cause a contradiction so ;or ; or for all .
This post has been edited 2 times. Last edited by rafigamath, Feb 26, 2024, 1:00 PM Reason: typo
We claim the only solutions are , and . Clearly all of these functions work. Now we prove these are the only such functions.
Let the above assertion be represented by .
Clearly the only constant solution is . Now assume to be nonconstant. Claim 1:
Proof: Note that gives us .
Now gives us . Claim 2:
Proof: Observe that gives us Claim 3:
Proof: Assume FtSoC that such exists. Then gives us . However we have assumed to be nonconstant. Contradiction. Claim 4:
Proof: yields Claim 5:
Proof: Assume FtSoC that for some nonzero , we have .
Then . If , both of which lead to a contradiction.
Therefore we are done.
Q. E. D.
Let denote the given assertion. is a solution, we want to find the other solutions, so we can assume there exists such that so is surjective. . If , then by surjectivity there exists such that so , contradiction. Therefore, . . .
Suppose . Then . Now if then is constant, but is surjective, contradiction. As a result, .
Now,
Now, as we don't fall into the pointwise trap , we assume there exists such that . .
We write all the four possibilities:
- , contradition.
- , contradition.
- , contradition.
- , contradition.
So, no such exists. This means or , which we can check and see that both are indeed solutions.
To sum up, we proved the only solutions are ,, and .