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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:
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[*]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
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/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.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Something nice
KhuongTrang   31
N 5 minutes ago by NguyenVanHoa29
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
31 replies
KhuongTrang
Nov 1, 2023
NguyenVanHoa29
5 minutes ago
Nordic 2025 P3
anirbanbz   8
N 29 minutes ago by lksb
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
8 replies
+1 w
anirbanbz
Mar 25, 2025
lksb
29 minutes ago
another functional inequality?
Scilyse   32
N 37 minutes ago by ihategeo_1969
Source: 2023 ISL A4
Let $\mathbb R_{>0}$ be the set of positive real numbers. Determine all functions $f \colon \mathbb R_{>0} \to \mathbb R_{>0}$ such that \[x \big(f(x) + f(y)\big) \geqslant \big(f(f(x)) + y\big) f(y)\]for every $x, y \in \mathbb R_{>0}$.
32 replies
Scilyse
Jul 17, 2024
ihategeo_1969
37 minutes ago
Mount Inequality erupts in all directions!
BR1F1SZ   1
N 42 minutes ago by sami1618
Source: Austria National MO Part 1 Problem 1
Let $a$, $b$ and $c$ be pairwise distinct nonnegative real numbers. Prove that
\[
(a + b + c) \left( \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2} \right) > 4.
\](Karl Czakler)
1 reply
BR1F1SZ
an hour ago
sami1618
42 minutes ago
Division involving difference of squares
BR1F1SZ   1
N an hour ago by grupyorum
Source: Austria National MO Part 1 Problem 4
Determine all integers $n$ that can be written in the form
\[
n = \frac{a^2 - b^2}{b},
\]where $a$ and $b$ are positive integers.

(Walther Janous)
1 reply
BR1F1SZ
an hour ago
grupyorum
an hour ago
Erasing the difference of two numbers
BR1F1SZ   0
an hour ago
Source: Austria National MO Part 1 Problem 3
Consider the following game for a positive integer $n$. Initially, the numbers $1, 2, \ldots, n$ are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers $3,6,11$ and $17$ are on the board, then $3$ can be erased as $6 - 3=3$, or $6$ as $17 - 11=6$, or $11$ as $17 - 6=11$.)

For which values of $n$ is it possible to end with only one number remaining on the board?

(Michael Reitmeir)
0 replies
BR1F1SZ
an hour ago
0 replies
Centroid, altitudes and medians, and concyclic points
BR1F1SZ   0
an hour ago
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
0 replies
BR1F1SZ
an hour ago
0 replies
4 wise men and 100 hats. 3 must guess their numbers
NO_SQUARES   2
N 2 hours ago by NO_SQUARES
Source: 239 MO 2025 10-11 p5
There are four wise men in a row, each sees only those following him in the row, i.e. the $1$st sees the other three, the $2$nd sees the $3$rd and $4$th, and the $3$rd sees only the $4$th. The devil has $100$ hats, numbered from $1$ to $100$, he puts one hat on each wise man, and hides the extra $96$ hats. After that, each wise man (in turn: first the first, then the second, etc.) loudly calls a number, trying to guess the number of his hat. The numbers mentioned should not be repeated. When all the wise men have spoken, they take off their hats and check which one of them has guessed. Can the sages to act in such a way that at least three of them knowingly guessed?
2 replies
NO_SQUARES
5 hours ago
NO_SQUARES
2 hours ago
\sqrt{2-a}+\sqrt{2-b}+\sqrt{2-c}\geqslant 2+\sqrt{(2-a)(2-b)(2-c)}
NO_SQUARES   2
N 2 hours ago by ektorasmiliotis
Source: 239 MO 2025 8-9 p4
Positive numbers $a$, $b$ and $c$ are such that $a^2+b^2+c^2+abc=4$. Prove that \[\sqrt{2-a}+\sqrt{2-b}+\sqrt{2-c}\geqslant 2+\sqrt{(2-a)(2-b)(2-c)}.\]
2 replies
NO_SQUARES
6 hours ago
ektorasmiliotis
2 hours ago
BMO 2024 SL A4
MuradSafarli   2
N 2 hours ago by GreekIdiot
A4.
Let \(a \geq b \geq c \geq 0\) be real numbers such that \(ab + bc + ca = 3\).
Prove that:
\[
3 + (2 - \sqrt{3}) \cdot \frac{(b-c)^2}{b+(\sqrt{3}-1)c} \leq a+b+c
\]and determine all the cases when the equality occurs.
2 replies
MuradSafarli
Apr 27, 2025
GreekIdiot
2 hours ago
Aime type Geo
ehuseyinyigit   0
2 hours ago
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
0 replies
ehuseyinyigit
2 hours ago
0 replies
1996 St. Petersburg City Mathematical Olympiad
Sadece_Threv   2
N 2 hours ago by reni_wee
Source: 1996 St. Petersburg City Mathematical Olympiad
Find all positive integers $n$ such that $3^{n-1}+5^{n-1}$ divides $3^{n}+5^{n}$
2 replies
Sadece_Threv
Jul 29, 2024
reni_wee
2 hours ago
IMO 2010 Problem 5
mavropnevma   55
N 2 hours ago by maromex
Each of the six boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, $B_6$ initially contains one coin. The following operations are allowed

Type 1) Choose a non-empty box $B_j$, $1\leq j \leq 5$, remove one coin from $B_j$ and add two coins to $B_{j+1}$;

Type 2) Choose a non-empty box $B_k$, $1\leq k \leq 4$, remove one coin from $B_k$ and swap the contents (maybe empty) of the boxes $B_{k+1}$ and $B_{k+2}$.

Determine if there exists a finite sequence of operations of the allowed types, such that the five boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ become empty, while box $B_6$ contains exactly $2010^{2010^{2010}}$ coins.

Proposed by Hans Zantema, Netherlands
55 replies
mavropnevma
Jul 8, 2010
maromex
2 hours ago
NT ineq: sum 1/a_i < (m+n)/m , {a_1,a_2,...,a_n} subset of {1,2,...,m}
parmenides51   1
N 3 hours ago by DVDTSB
Source: 2006 MOP Homework Blue NT 6
Let $m$ and $n$ be positive integers with $m > n \ge 2$. Set $S =\{1,2,...,m\}$, and set $T = \{a_1,a_2,...,a_n\}$ is a subset of $S$ such that every element of $S$ is not divisible by any pair of distinct elements of $T$. Prove that
$$\frac{1}{a_1}+\frac{1}{a_2}+ ...+ \frac{1}{a_n} < \frac{m+n}{m}$$
1 reply
parmenides51
Apr 12, 2020
DVDTSB
3 hours ago
Grasshopper jumps in an equilateral triangle
shobber   1
N Oct 6, 2005 by yetti
Source: Pan African 2001
Let $ABC$ be an equilateral triangle and let $P_0$ be a point outside this triangle, such that $\triangle{AP_0C}$ is an isoscele triangle with a right angle at $P_0$. A grasshopper starts from $P_0$ and turns around the triangle as follows. From $P_0$ the grasshopper jumps to $P_1$, which is the symmetric point of $P_0$ with respect to $A$. From $P_1$, the grasshopper jumps to $P_2$, which is the symmetric point of $P_1$ with respect to $B$. Then the grasshopper jumps to $P_3$ which is the symmetric point of $P_2$ with respect to $C$, and so on. Compare the distance $P_0P_1$ and $P_0P_n$. $n \in N$.
1 reply
shobber
Oct 4, 2005
yetti
Oct 6, 2005
Grasshopper jumps in an equilateral triangle
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G H BBookmark kLocked kLocked NReply
Source: Pan African 2001
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shobber
3498 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ABC$ be an equilateral triangle and let $P_0$ be a point outside this triangle, such that $\triangle{AP_0C}$ is an isoscele triangle with a right angle at $P_0$. A grasshopper starts from $P_0$ and turns around the triangle as follows. From $P_0$ the grasshopper jumps to $P_1$, which is the symmetric point of $P_0$ with respect to $A$. From $P_1$, the grasshopper jumps to $P_2$, which is the symmetric point of $P_1$ with respect to $B$. Then the grasshopper jumps to $P_3$ which is the symmetric point of $P_2$ with respect to $C$, and so on. Compare the distance $P_0P_1$ and $P_0P_n$. $n \in N$.
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yetti
2643 posts
#2 • 3 Y
Y by Adventure10, Mango247, DiaaSaid
Let a = AB be the side of the equilateral triangle $\triangle ABC$. Since $AP_0 = AP_1,\ BP_1 = BP_2$, AB is a midline of the triangle $\triangle P_0P_1P_2$. Hence, the point $P_2$ is obtained from $P_0$ by a translation in the direction $\vec{AB}$ by the distance 2a. Similarly, $P_4$ is obtained from $P_2$ by a translation in the direction $\vec{CA}$ by the distance 2a and $P_6$ from $P_4$ by a translation in the direction $\vec{BC}$ by the distance 2a. It follows that the points $P_0, P_2, P_4$ form an equilateral triangle with the side lengths 2a and the sides $P_0P_2 \parallel AB,\ P_2P_4 \parallel CA, P_4P_0 \parallel BC$ and that the point $P_6 \equiv P_0$ are identical. For similar reasons, $P_3$ is obtained from $P_1$ by a translation in the direction $\vec{BC}$ by the distance 2a, $P_5$ is obtained from $P_3$ by a translation in the direction $\vec{AB}$ by the distance 2a, and $P_7$ is obtained from $P_5$ by a translation in the direction $\vec{CA}$ by the distance 2a. It follows that the points $P_1, P_3, P_5$ also form an equilateral triangle with the side lengths 2a and the sides $P_1P_3 \parallel BC,\ P_3P_5 \parallel AB, P_5P_1 \parallel CA$ and that the point $P_7 \equiv P_1$ are identical. Thus the equilateral triangles $\triangle P_0P_2P_4 \cong \triangle P_3P_5P_1$ are congruent and have parallel sides and the second is obtained from the first by a translation in the direction $\vec{P_0P_3}$ by the distance $P_0P_3$.

Up to now, we did not need the fact that the triangle $\triangle ABC$ is equilateral or that the triangle $\triangle AP_0C$ is isosceles, only the fact that $\vec{AB} + \vec{BC} + \vec{CA} = 0$, which holds for any triangle. Now we use these facts. Since $P_6 \equiv P_0$ are identical and $P_0A = P_0C, AP_0 = AP_1, CP_0 = CP_5$, the triangle $\triangle P_1P_0P_5$ is also isosceles. This means that the point $P_0$ lies on the altitude of the equilateral triangle $\triangle P_3P_5P_1$ from the vertex $P_3$. Since in addition, the isosceles triangle $\triangle P_1P_0P_5$ is right,

$P_0P_3 = P_1P_5 \dfrac{\sqrt 3}{2} - \dfrac{P_1P_5}{5} = a (\sqrt 3 - 1)$

Obviously, $P_0P_5 = P_0P_1 = 2AP_0 = 2CA \dfrac{\sqrt 2}{2} = a \sqrt 2$ and $P_0P_2 = P_0P_4 = 2a$. As a result,

$P_0P_2 = P_0P_4 = P_0P_1 \sqrt 2$

$P_0P_3 = P_0P_1 \dfrac{\sqrt 3 - 1}{\sqrt 2} = P_0P_1 \dfrac{\sqrt 6 - \sqrt 2}{2}$

$P_0P_5 = P_0P_1$

$P_0P_6 = 0$

$P_0P_n = P_0P_{(n \mod 6)}$
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