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
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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 a scalene triangle with circumcircle and incenter . Ray meets at and meets again at ; the circle with diameter cuts again at . Lines and meet at , and is the midpoint of . The circumcircles of and intersect at points and . Prove that passes through the midpoint of either or .
p1. Square has side length . Let the midpoint of be . What is the area of the overlapping region between the circle centered at with radius and the circle centered at with radius ? (You may express your answer using inverse trigonometry functions of noncommon values.)
p2. Find the number of times occurs when for the function .
p3. Stanford is building a new dorm for students, and they are looking to offer room configurations: Configuration : a one-room double, which is a square with side length of , Configuration : a two-room double, which is two connected rooms, each of them squares with a side length of .
To make things fair for everyone, Stanford wants a one-room double (rooms of configuration ) to be exactly m larger than the total area of a two-room double. Find the number of possible pairs of side lengths , where ,, such that .
p4. The island nation of Ur is comprised of islands. One day, people decide to create island-states as follows. Each island randomly chooses one of the other five islands and builds a bridge between the two islands (it is possible for two bridges to be built between islands and if each island chooses the other). Then, all islands connected by bridges together form an island-state. What is the expected number of island-states Ur is divided into?
p5. Let and be the roots of the polynomial . Compute .
p6. Carol writes a program that finds all paths on an 10 by 2 grid from cell (1, 1) to cell (10, 2) subject to the conditions that a path does not visit any cell more than once and at each step the path can go up, down, left, or right from the current cell, excluding moves that would make the path leave the grid. What is the total length of all such paths? (The length of a path is the number of cells it passes through, including the starting and ending cells.)
p7. Consider the sequence of integers an defined by , for prime and for . Find the smallest such that is a perfect power of .
p8. Let be a triangle whose -excircle, -excircle, and -excircle have radii ,, and , respectively. If and the perimeter of is , what is the area of ?
p9. Consider the set of functions satisfying:
(a)
(b) ,
(c) ,
(d) .
If can be written as where are distinct primes, compute .
p10. You are given that and that the first (leftmost) two digits of are 10. Compute the number of integers with such that starts with either the digit or (in base ).
p11. Let be the circumcenter of . Let be the midpoint of , and let and be the feet of the altitudes from and , respectively, onto the opposite sides. intersects at . The line passing through and perpendicular to intersects the circumcircle of at (on the major arc ) and , and intersects at . Point lies on the line such that is perpendicular to . Given that and , compute .
p12. Let be the isosceles triangle with side lengths . Arpit and Katherine simultaneously choose points and within this triangle, and compute , the squared distance between the two points. Suppose that Arpit chooses a random point within . Katherine plays the (possibly randomized) strategy which given Arpit’s strategy minimizes the expected value of . Compute this value.
p13. For a regular polygon with sides, let denote the regular polygon with sides such that the vertices of are the midpoints of every other side of . Let denote the polygon that results after applying f a total of k times. The area of where is a pentagon of side length , can be expressed as for some positive integers where is not divisible by the square of any prime and does not share any positive divisors with and . Find .
p14. Consider the function . This function can be expressed in the form for sequences of integers ,. Determine .
p15. In , let be the centroid and let the circumcenters of ,, and be , and , respectively. The line passing through and the midpoint of intersects at . If the radius of circle is , the radius of circle is , and , what is the length of ?
PS. You should use hide for answers. Collected here.
Consider an -by- board of unit squares for some odd positive integer . We say that a collection of identical dominoes is a maximal grid-aligned configuration on the board if consists of dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: then covers all but one square on the board. We are allowed to slide (but not rotate) a domino on the board to cover the uncovered square, resulting in a new maximal grid-aligned configuration with another square uncovered. Let be the number of distinct maximal grid-aligned configurations obtainable from by repeatedly sliding dominoes. Find the maximum value of as a function of .
I want to create a problem set of some of the hardest combi problems that are yet to appear in any contest. Can anyone help me out? Also can anyone give me some tips to create combi problems.
Let be a triangle such that . Points lie on segments , respectively, such that lines and are the angle bisectors of triangle . Find the measure of angle .
Converting from base to base 10, we get for a postive integer because its a divisor. Then you can simplify this to and since n or b cant be negative that means n has to be from 1 to 9 exclusive. Then casework from 2-8, and u get n=7 then b=21 and n=8 then b=40 so its 70
First, note that so it follows that must be a divisor of The only divisors of greater than are and so it follows that the possible values of are and yielding an answer of
9b+7 is divisible by b+7, so 9b+7 - 9(b+7) will still be divisible by b+7, so -56 is divisible by b+7. we first try -56-7=-49, so one b possibility could be 49. 49-7=42, 42/2 = 21, 21 is another possibility. 21-7=14, 14/2 = 7, which is less than 9, so we only have 21 and 49, giving us 70.
By the definition of a divisor, for an integer . Clearly does not work. So now we can just try every single value.
no gives which doesn't work gives which doesn't work gives which doesn't work gives which doesn't work gives which doesn't work gives which does work gives which does work
By the definition of a divisor, for an integer . Clearly does not work. So now we can just try every single value.
no gives which doesn't work gives which doesn't work gives which doesn't work gives which doesn't work gives which doesn't work gives which does work gives which does work
I was stuck on this problem for some reason I don't know why
The best way to tackle this problem is to convert everything to variables so first 17 base b = b + 7 and 97 base b = 9b+7 so we can just make a variable when multiplied by it it equals 97 base b
So first to do that we can simplify it like 9b+7 = x(b+7) for some value of x then when we multiply that out we get 9b+7=bx+7x. Because we want to solve for xb we have to subtract that on both sides to get 9b+7-bx= 7x then in any situation like this we have to factor out the b, but first we can subtract 7 on both sides.
When we do that we get b(9-x)= 7(x-1) which when we divide both sides we get
b= (7(x-1))/(9-x) we get this easy equation to solve because we know that x has to be a single digit number because anything greater than 9 won't work so we get that after guess and check x= 7 and 8 so when we plug that in we get 21 and 49, so when we add those together you get
This post has been edited 2 times. Last edited by sadas123, Feb 9, 2025, 4:44 PM
b+7 is divisible by 9b+7, but because b+7 is divisible by 9b+63, then b+7 is divisible by 56. And B+7 is greater than 7, so b=21 or 49[Click][sounds stupid but I got b+7 is divisible by 8b, and did a lot of stupid stuff to get -6+-5+-3+0+1+7+21+49=64 cause I forgot b is greater than 9, but luckily I realized this at the end.]
Find the sum of all integer bases for which is a divisor of
We need , so . Let , do casework and bounding.
If then so , doesn't work.
If then so , doesn't work.
If then so , which does work.
If then so , which does work.
If then so , doesn't work.
Our answer is .
(97)b =9b+7
(17)b=b+7
So b+7 divides 9b+7
gcd(9b+7,b+7)=b+7
gcd(-56,b+7)=b+7
So we get b+7 divides -56
Now , 56=2*2*2*7 and factors greater than 9+7 are 28 and 56 itself
So, b=21 or 49
Sum =70(answer)