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]
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All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.
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Intermediate: Grades 8-12
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Source: Al-Khwarizmi International Junior Olympiad 2025 P1
Determine the largest integer for which the following statement holds: there exists at least one triple of integers such that and all triples of real numbers, satisfying the equations, are such that are integers.
Four balls of radius 1 are mutually tangent, three resting on the floor and the fourth resting on the others. A tetrahedron, each of whose edges have length , is circumscribed around the balls. Then equals
Given a pyramid where is a parallelogram.
The intersection of the diagonals of the base is point .
Point is connected to the midpoint of , point to the midpoint of ,
point to the midpoint of and point to the midpoint of .
a) Prove: the four lines are concurrent in a point .
b) Calulate .
Given a regular tetrahedron with edges .
Construct at the apex three perpendiculars to the three lateral faces.
Take a point on each perpendicular at a distance from the apex such that these three points lie above the apex.
Calculate the lenghts of the sides of the triangle.
a) Given is a right cylinder of height and radius .
The sun shines on this solid at an angle of .
What is the area of the shadow that this solid casts on the plane of the botom base?
b) Given is a right cone of height and radius .
The sun shines on this solid at an angle of .
What is the area of the shadow that this solid casts on the plane of the base?
Given a tetrahedron and a plane , parallel with the edges and . .
a) Prove: the intersection of the tetrahedron with the plane is a parallelogram.
b) If and ,
calculates the lenghts of the sides of this parallelogram.
Given the points
a) Determine the point , above the xy-plane, such that the pyramid is a regular tetrahedron.
b) Calculate the volume.
c) Calculate the radius of the inscribed sphere and the radius of the circumscribed sphere.
volume 9f a pentagonal base pyramid circumscribed around a right circular cone
FOL1
NMay 6, 2025
by Mathzeus1024
A pentagonal base pyramid is circumscribed around a right circular cone, whose height is equal to the radius of the base. The total surface area of the pyramid is d times greater than that of the cone. Find the volume of the pyramid if the lateral surface area of the cone is equal to .
This is the solution for all integer numbers .
The equation is equivalent to Case 1. If then or .
If then .
If , we also have then . We get . Case 2. If . It follows that or . Since so , a contradiction. Case 3. If . With then . We obtain .
With , from the equation we get . The equation become
Equation has roots when . It is easy to see that if then . Therefore .
If then . Hence or , a contradiction since .
If then . Hence or , a contradiction since .
If then . Hence or . Since and so . We find . Hence .
If then . It follows that or . We see that . If then , a contradiction.
If then . It follows that or . Since if then then . We find . Ir follows that .
Another simple solution: from the condition, number m^6-m is of the form n^3-5n^2. For m>=4 we have that (m^2+1)^3-5(m^2+1)<m^6-m<(m^2+2)^3-5(m^2+2), so no solutions because n^3-5n^2 is strictly increasing in naturals which are >=5. After checking other cases we get that (m,n)={(1,5),(3,11)}.
This post has been edited 1 time. Last edited by MilosMilicev, Jan 13, 2017, 6:01 PM
We are going to solve this by bounding Motivation for the bounding
The motivation:
I got the motivation to solve it in this way because of a similiar problem from 2016 jbmo tst turkey problem 3,In the first thought the problems dont look quite similiar but if we look closely the bounding is the same.The main hint that gave me for this bound of perfect cubes is: Since they can be represented as:
The 2 very powerful lemmas
We will have to prove to lemmas: Lemma 1:Proof:2) Now let so our inequality becomes:
Expanding the left-hand side: so the inequality becomes:
Subtracting from both sides:
Multiplying both sides by (and flipping the inequality):
We now show this holds for all . This can be proven either by induction or by observing that the function is strictly increasing for , and .
Hence, the inequality holds for all , and we have:
Therefore, lies strictly between two perfect cubes. Lemma 2:
Now we will prove that lies strictly between two consecutive cubes.
**Proof:** First, we show that
Expanding the left-hand side: so the inequality becomes:
Subtracting from both sides:
Adding to both sides:
This inequality clearly holds for all , since even at , we get:
Now we show that:
Expanding the left-hand side: so the inequality becomes:
Subtracting from both sides:
Adding to both sides:
Thus, the inequality becomes:
This inequality holds for all , since the left-hand side is linear and the right-hand side is quadratic. For example, at :
Therefore, for all , we have:
Hence, lies strictly between two consecutive cubes.
Finishing the problem
Now, by **Lemma 1** and **Lemma 2**, and the identity we conclude that:
We substitute this into the original equation:
Let us expand both sides.
**Left-hand side (LHS):**
**Right-hand side (RHS):**
Now subtract both sides:
Simplifying:
So we have:
Now, observe that since , all terms on the left-hand side are positive, while the right-hand side is zero, which is a contradiction.
Alternatively, notice that this simplifies to:
Now we consider modulo divisibility: since divides the left-hand side, it must divide the right-hand side:
Thus, the only possible values of are the positive divisors of 4. Since , the only candidate is .
We test :
Left-hand side:
Right-hand side:
So , and therefore is **not** a solution.
Hence, there is no solution for any , and since , this implies .
**Conclusion:** There is no solution to the equation for any and .
A very great problemmm
This post has been edited 1 time. Last edited by TopGbulliedU, Apr 9, 2025, 2:50 PM