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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
and train with the best! Please note that early bird pricing ends August 19th!
Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

Our full course list for upcoming classes is below:
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0 replies
jwelsh
Jul 1, 2025
0 replies
abc=1
hctb00   11
N 4 minutes ago by SunnyEvan
Source: unknown
$a,b,c>0,abc=1$,prove:\[\frac{1}{a^2+b+1}+\frac{1}{b^2+c+1}+\frac{1}{c^2+a+1}\le1\]
11 replies
hctb00
Aug 16, 2014
SunnyEvan
4 minutes ago
imo 2025 problems in SL
Darghy   10
N 8 minutes ago by BR1F1SZ
C1 G4 N7

N3 A3 C8
10 replies
Darghy
3 hours ago
BR1F1SZ
8 minutes ago
ez combi
m4thbl3nd3r   2
N 8 minutes ago by m4thbl3nd3r
Find the smallest positve integer $k$ so that one can split the set $T=\{1,2,\dots,k\}$ in to two nonempty disjoint sets $A,B$ such that every elements in $T$ belongs to either $A$ or $B$ and there are two elements in the same set whose sum is a cube.
2 replies
m4thbl3nd3r
37 minutes ago
m4thbl3nd3r
8 minutes ago
IMO 2025 P2
sarjinius   55
N 13 minutes ago by iStud
Source: 2025 IMO P2
Let $\Omega$ and $\Gamma$ be circles with centres $M$ and $N$, respectively, such that the radius of $\Omega$ is less than the radius of $\Gamma$. Suppose $\Omega$ and $\Gamma$ intersect at two distinct points $A$ and $B$. Line $MN$ intersects $\Omega$ at $C$ and $\Gamma$ at $D$, so that $C, M, N, D$ lie on $MN$ in that order. Let $P$ be the circumcentre of triangle $ACD$. Line $AP$ meets $\Omega$ again at $E\neq A$ and meets $\Gamma$ again at $F\neq A$. Let $H$ be the orthocentre of triangle $PMN$.

Prove that the line through $H$ parallel to $AP$ is tangent to the circumcircle of triangle $BEF$.
55 replies
sarjinius
Yesterday at 3:38 AM
iStud
13 minutes ago
No more topics!
A second final attempt to make a combinatorics problem
JARP091   2
N May 29, 2025 by JARP091
Source: At the time of writing this problem I do not know the source if any
Arthur Morgan is playing a game.

He has $n$ eggs, each with a hardness value $k_1, k_2, \dots, k_n$, where $\{k_1, k_2, \dots, k_n\}$ is a permutation of the set $\{1, 2, \dots, n\}$. He is throwing the eggs from an $m$-floor building.

When the $i$-th egg is dropped from the $j$-th floor, its new hardness becomes
\[
\left\lfloor \frac{k_i}{j+1} \right\rfloor.
\]If $\left\lfloor \frac{k_i}{j+1} \right\rfloor = 0$, then the egg breaks and cannot be used again.

Arthur can drop each egg from a particular floor at most once.
For which values of $n$ and $m$ can Arthur always determine the correct ordering of the eggs according to their initial hardness values?
Note: The problem might be wrong or too easy
2 replies
JARP091
May 25, 2025
JARP091
May 29, 2025
A second final attempt to make a combinatorics problem
G H J
Source: At the time of writing this problem I do not know the source if any
The post below has been deleted. Click to close.
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JARP091
124 posts
#1
Y by
Arthur Morgan is playing a game.

He has $n$ eggs, each with a hardness value $k_1, k_2, \dots, k_n$, where $\{k_1, k_2, \dots, k_n\}$ is a permutation of the set $\{1, 2, \dots, n\}$. He is throwing the eggs from an $m$-floor building.

When the $i$-th egg is dropped from the $j$-th floor, its new hardness becomes
\[
\left\lfloor \frac{k_i}{j+1} \right\rfloor.
\]If $\left\lfloor \frac{k_i}{j+1} \right\rfloor = 0$, then the egg breaks and cannot be used again.

Arthur can drop each egg from a particular floor at most once.
For which values of $n$ and $m$ can Arthur always determine the correct ordering of the eggs according to their initial hardness values?
Note: The problem might be wrong or too easy
This post has been edited 1 time. Last edited by JARP091, May 25, 2025, 2:46 PM
Reason: Wrongly LaTeXted
Z K Y
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JARP091
124 posts
#2
Y by
Bump for this problem
Z K Y
The post below has been deleted. Click to close.
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JARP091
124 posts
#3
Y by
Drop something from 1, if it breaks then its 0

if it doesnt break, it is currently 1, we have one more one, and we have a number we are trying to find out

we cant drop the number we are interested in from floor 1, otherwise we lose information

hence drop another egg from floor 2, if it doesnt break, it was 3, we started with a 2, and ther last egg is 1, if it breaks, then we either dropped a 1, or we dropped a 2, and so the possible outputs are (1,0,2) (1,0,3) (1,0,1), now we cant figure out what state the last egg is in, so it is impossible.

For n $>$ 3, n = 3 is a subproblem that cannot be solved.

Hence only possible solutions are:

i) n = 1, m $\geq$ 1

ii) n = 2, m $\geq$ 2
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