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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
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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0 replies
jlacosta
May 1, 2025
0 replies
Primes and sets
mathisreaI   40
N 19 minutes ago by Tinoba-is-emotional
Source: IMO 2022 Problem 3
Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of $S$ around the circle such that the product of any two neighbors is of the form $x^2+x+k$ for some positive integer $x$.
40 replies
mathisreaI
Jul 13, 2022
Tinoba-is-emotional
19 minutes ago
integer functional equation
ABCDE   155
N an hour ago by heheman
Source: 2015 IMO Shortlist A2
Determine all functions $f:\mathbb{Z}\rightarrow\mathbb{Z}$ with the property that \[f(x-f(y))=f(f(x))-f(y)-1\]holds for all $x,y\in\mathbb{Z}$.
155 replies
ABCDE
Jul 7, 2016
heheman
an hour ago
Three numbers cannot be squares simultaneously
WakeUp   37
N an hour ago by LeYohan
Source: APMO 2011
Let $a,b,c$ be positive integers. Prove that it is impossible to have all of the three numbers $a^2+b+c,b^2+c+a,c^2+a+b$ to be perfect squares.
37 replies
WakeUp
May 18, 2011
LeYohan
an hour ago
Problem G5 - IMO Shortlist 2007
April   30
N an hour ago by Double07
Source: ISL 2007, G5, AIMO 2008, TST 3, P2
Let $ ABC$ be a fixed triangle, and let $ A_1$, $ B_1$, $ C_1$ be the midpoints of sides $ BC$, $ CA$, $ AB$, respectively. Let $ P$ be a variable point on the circumcircle. Let lines $ PA_1$, $ PB_1$, $ PC_1$ meet the circumcircle again at $ A'$, $ B'$, $ C'$, respectively. Assume that the points $ A$, $ B$, $ C$, $ A'$, $ B'$, $ C'$ are distinct, and lines $ AA'$, $ BB'$, $ CC'$ form a triangle. Prove that the area of this triangle does not depend on $ P$.

Author: Christopher Bradley, United Kingdom
30 replies
1 viewing
April
Jul 13, 2008
Double07
an hour ago
Inequalities
sqing   3
N 2 hours ago by DAVROS
Let $ a,b,c\geq 0 ,a+b+c\leq 3. $ Prove that
$$a^2+b^2+c^2+ab +2ca+2bc +  abc \leq \frac{251}{27}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{2}{5}abc  \leq \frac{4861}{540}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{7}{20}abc  \leq \frac{2381411}{26460}$$
3 replies
sqing
Today at 12:47 PM
DAVROS
2 hours ago
n is divisible by 5
spiralman   0
3 hours ago
n is an integer. There are n integers such that they are larger or equal to 1, and less or equal to 6. Sum of them is larger or equal to 4n, while sum of their square is less or equal to 22n. Prove n is divisible by 5.
0 replies
spiralman
3 hours ago
0 replies
It is given that $M=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{23}=\frac{n}{23!},
Vulch   2
N 3 hours ago by P162008
It is given that $M=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{23}=\frac{n}{23!},$ where $n$ is a natural number.What is the remainder when $n$ is divided by $13?$
2 replies
Vulch
Apr 9, 2025
P162008
3 hours ago
Symmetric System's Cyclic Sum
worthawholebean   6
N Today at 4:09 PM by kilobyte144
If $ a+1=b+2=c+3=d+4=a+b+c+d+5$, then $ a+b+c+d$ is

$ \text{(A)}\ -5 \qquad
\text{(B)}\ -10/3 \qquad
\text{(C)}\ -7/3 \qquad
\text{(D)}\ 5/3 \qquad
\text{(E)}\ 5$
6 replies
worthawholebean
Feb 11, 2008
kilobyte144
Today at 4:09 PM
Inequalities
sqing   5
N Today at 2:45 PM by sqing
Let $ a,b,c>0. $ Prove that$$a^2+b^2+c^2+abc-k(a+b+c)\geq 3k+2-2(k+1)\sqrt{k+1}$$Where $7\geq k \in N^+.$
$$a^2+b^2+c^2+abc-3(a+b+c)\geq-5$$
5 replies
sqing
Yesterday at 2:23 PM
sqing
Today at 2:45 PM
Find the value of k/m
zolfmark   4
N Today at 2:31 PM by Shan3t
If we have
K=1/1+1/3+1/4+......1/2019

m=2018+2017/2+2016/3+......2/2017+1/2018
4 replies
zolfmark
Mar 1, 2024
Shan3t
Today at 2:31 PM
geometry
luckvoltia.112   2
N Today at 1:35 PM by luckvoltia.112
ChGiven an acute triangle ABC inscribed in circle $(O)$ The altitudes $BE, CF$ , intersect
each other at $H$. The tangents at $B$ and $C $of $(O)$ intersect at $S$. Let $M $be the midpoint of $BC$. $EM$ intersects $SC$
at $I$, $FM$ intersects $SB$ at $J.$
a) Prove that the points $I, S, M, J$ lie on the same circle.
b) The circle with diameter $AH$ intersects the circle $(O)$ at the second point $T.$ The line $AH$ intersects
$(O)$ at the second point $K$. Prove that $S,K,T$ are collinear.
2 replies
luckvoltia.112
May 18, 2025
luckvoltia.112
Today at 1:35 PM
positive integer solutions
zolfmark   1
N Today at 1:07 PM by Mathzeus1024
positive integer solutions
x^2+y^2+xy=283
1 reply
zolfmark
Jan 5, 2018
Mathzeus1024
Today at 1:07 PM
Minimize z.
Entrepreneur   1
N Today at 12:18 PM by Lankou
Minimize $z = 6x + 3y.$ Subject to the constraints:
$$\begin{cases}
4x+y\ge80,\\
x+5y \ge115,\\
3x+2y\le150,\\
x,y\ge0.
\end{cases}$$
1 reply
Entrepreneur
Today at 11:30 AM
Lankou
Today at 12:18 PM
by vectors
zolfmark   1
N Today at 11:54 AM by Mathzeus1024
by vectors shoe that
in ABC .
: cosA + cosB + cosC ≤ 3/2
1 reply
zolfmark
Oct 26, 2023
Mathzeus1024
Today at 11:54 AM
Erasing a and b and replacing them with a - b + 1
jl_   1
N Apr 23, 2025 by maromex
Source: Malaysia IMONST 2 2023 (Primary) P5
Ruby writes the numbers $1, 2, 3, . . . , 10$ on the whiteboard. In each move, she selects two distinct numbers, $a$ and $b$, erases them, and replaces them with $a+b-1$. She repeats this process until only one number, $x$, remains. What are all the possible values of $x$?
1 reply
jl_
Apr 23, 2025
maromex
Apr 23, 2025
Erasing a and b and replacing them with a - b + 1
G H J
G H BBookmark kLocked kLocked NReply
Source: Malaysia IMONST 2 2023 (Primary) P5
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jl_
9 posts
#1
Y by
Ruby writes the numbers $1, 2, 3, . . . , 10$ on the whiteboard. In each move, she selects two distinct numbers, $a$ and $b$, erases them, and replaces them with $a+b-1$. She repeats this process until only one number, $x$, remains. What are all the possible values of $x$?
Z K Y
The post below has been deleted. Click to close.
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maromex
202 posts
#2
Y by
The sum of all numbers on the board minus how many numbers there are on the board is invariant; after each move, the amount of numbers decreases by $1$, and the sum of all numbers decreases by $a + b - (a + b - 1) = 1$ as well. This value is equal to $1 + 2 + \ldots + 10 - 10 = 45$, and it will always be equal to $45$. We add $1$ to get the sum of all numbers when there is $1$ number, and the only possible value of $x$ is $46$.
This post has been edited 1 time. Last edited by maromex, Apr 23, 2025, 11:06 AM
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