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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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0 replies
jlacosta
Apr 2, 2025
0 replies
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Inspired by old results
sqing   3
N a minute ago by sqing
Source: Own
Let \( a, b, c \) be real numbers.Prove that
$$ \frac{(a - b + c)^2}{ (a^2+ a+1)(b^2+b+1)(c^2+ c+1)} \leq 4$$$$ \frac{(a + b + c)^2}{ (a^2+ a+1)(b^2 +b+1)(c^2+ c+1)} \leq \frac{2(69 + 11\sqrt{33})}{27}$$
3 replies
sqing
4 hours ago
sqing
a minute ago
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x^n + 1 = y^{n+1}
orl   8
N 10 minutes ago by AshAuktober
Source: IMO 1980 Finland, problem 3
Prove that the equation \[ x^n + 1 = y^{n+1}, \] where $n$ is a positive integer not smaller then 2, has no positive integer solutions in $x$ and $y$ for which $x$ and $n+1$ are relatively prime.
8 replies
orl
May 6, 2004
AshAuktober
10 minutes ago
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A cyclic inequality
KhuongTrang   13
N 15 minutes ago by KhuongTrang
Source: own-CRUX
IMAGE
Link
13 replies
2 viewing
KhuongTrang
Apr 2, 2025
KhuongTrang
15 minutes ago
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Quadric function
soryn   4
N 21 minutes ago by soryn
If f(x)=ax^2+bx+c, a,b,c integers, |a|>=3, and M îs the set of integers x for which f(x) is a prime number and f has exactly one integer solution,prove that M has at most three elements.
4 replies
soryn
Apr 18, 2025
soryn
21 minutes ago
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Complex Numbers Question
franklin2013   2
N 4 hours ago by osszhangbanvan
Hello everyone! This is one of my favorite complex numbers questions. Have fun!

$f(z)=z^{720}-z^{120}$. How many complex numbers $z$ are there such that $|z|=1$ and $f(z)$ is an integer.

Hint
2 replies
franklin2013
Yesterday at 4:08 PM
osszhangbanvan
4 hours ago
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Inequalities
sqing   25
N 5 hours ago by sqing
Let $ a,b $ be reals such that $ a^2+ab+b^2 =3$ . Prove that
$$ \frac{4}{ 3}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{11}{4 }$$$$ \frac{13}{ 4}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{2}{3 }$$$$ \frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq -\frac{17}{6 }$$$$ \frac{19}{ 6}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2}$$Let $ a,b $ be reals such that $ a^2-ab+b^2 =1 $ . Prove that
$$ \frac{3}{ 2}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }+ab \geq \frac{4}{15 }$$$$ \frac{14}{ 15}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }-ab \geq -\frac{1}{2 }$$$$ \frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq \frac{13}{42 }$$$$ \frac{41}{ 42}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2 }$$
25 replies
sqing
Apr 16, 2025
sqing
5 hours ago
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Three variables inequality
Headhunter   4
N 6 hours ago by lbh_qys
$\forall a\in R$ ,$~\forall b\in R$ ,$~\forall c \in R$
Prove that at least one of $(a-b)^{2}$, $(b-c)^{2}$, $(c-a)^{2}$ is not greater than $\frac{a^{2}+b^{2}+c^{2}}{2}$.

I assume that all are greater than it, but can't go more.
4 replies
Headhunter
Yesterday at 6:58 AM
lbh_qys
6 hours ago
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Indonesia Regional MO 2019 Part A
parmenides51   23
N Today at 2:08 AM by chinawgp
Indonesia Regional MO
Year 2019 Part A

Time: 90 minutes Rules


p1. In the bag there are $7$ red balls and $8$ white balls. Audi took two balls at once from inside the bag. The chance of taking two balls of the same color is ...


p2. Given a regular hexagon with a side length of $1$ unit. The area of the hexagon is ...


p3. It is known that $r, s$ and $1$ are the roots of the cubic equation $x^3 - 2x + c = 0$. The value of $(r-s)^2$ is ...


p4. The number of pairs of natural numbers $(m, n)$ so that $GCD(n,m) = 2$ and $LCM(m,n) = 1000$ is ...


p5. A data with four real numbers $2n-4$, $2n-6$, $n^2-8$, $3n^2-6$ has an average of $0$ and a median of $9/2$. The largest number of such data is ...


p6. Suppose $a, b, c, d$ are integers greater than $2019$ which are four consecutive quarters of an arithmetic row with $a <b <c <d$. If $a$ and $d$ are squares of two consecutive natural numbers, then the smallest value of $c-b$ is ...


p7. Given a triangle $ABC$, with $AB = 6$, $AC = 8$ and $BC = 10$. The points $D$ and $E$ lies on the line segment $BC$. with $BD = 2$ and $CE = 4$. The measure of the angle $\angle DAE$ is ...


p8. Sequqnce of real numbers $a_1,a_2,a_3,...$ meet $\frac{na_1+(n-1)a_2+...+2a_{n-1}+a_n}{n^2}=1$ for each natural number $n$. The value of $a_1a_2a_3...a_{2019}$ is ....


p9. The number of ways to select four numbers from $\{1,2,3, ..., 15\}$ provided that the difference of any two numbers at least $3$ is ...


p10. Pairs of natural numbers $(m , n)$ which satisfies $$m^2n+mn^2 +m^2+2mn = 2018m + 2019n + 2019$$are as many as ...


p11. Given a triangle $ABC$ with $\angle ABC =135^o$ and $BC> AB$. Point $D$ lies on the side $BC$ so that $AB=CD$. Suppose $F$ is a point on the side extension $AB$ so that $DF$ is perpendicular to $AB$. The point $E$ lies on the ray $DF$ such that $DE> DF$ and $\angle ACE = 45^o$. The large angle $\angle AEC$ is ...


p12. The set of $S$ consists of $n$ integers with the following properties: For every three different members of $S$ there are two of them whose sum is a member of $S$. The largest value of $n$ is ....


p13. The minimum value of $\frac{a^2+2b^2+\sqrt2}{\sqrt{ab}}$ with $a, b$ positive reals is ....


p14. The polynomial P satisfies the equation $P (x^2) = x^{2019} (x+ 1) P (x)$ with $P (1/2)= -1$ is ....


p15. Look at a chessboard measuring $19 \times 19$ square units. Two plots are said to be neighbors if they both have one side in common. Initially, there are a total of $k$ coins on the chessboard where each coin is only loaded exactly on one square and each square can contain coins or blanks. At each turn. You must select exactly one plot that holds the minimum number of coins in the number of neighbors of the plot and then you must give exactly one coin to each neighbor of the selected plot. The game ends if you are no longer able to select squares with the intended conditions. The smallest number of $k$ so that the game never ends for any initial square selection is ....
23 replies
parmenides51
Nov 11, 2021
chinawgp
Today at 2:08 AM
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VOLUNTEERING OPPORTUNITIES OPEN TO HIGH/MIDDLE SCHOOLERS
im_space_cadet   13
N Today at 12:30 AM by im_space_cadet
Hi everyone!
Do you specialize in contest math? Do you have a passion for teaching? Do you want to help leverage those college apps? Well, I have something for all of you.

I am im_space_cadet, and during the fall of last year, I opened my non-profit DeltaMathPrep which teaches students preparing for contest math the problem-solving skills they need in order to succeed at these competitions. Currently, we are very much understaffed and would greatly appreciate the help of more tutors on our platform.

Each week on Saturday and Wednesday, we meet once for each competition: Wednesday for AMC 8 and Saturday for AMC 10 and we go over a past year paper for the entire class. On both of these days, we meet at 9PM EST in the night.

This is a great opportunity for anyone who is looking to have a solid activity to add to their college resumes that requires low effort from tutors and is very flexible with regards to time.

This is the link to our non-profit for anyone who would like to view our initiative:
https://www.deltamathprep.org/

If you are interested in this opportunity, please send me a DM on AoPS or respond to this post expressing your interest. I look forward to having you all on the team!

Thanks,
im_space_cadet
13 replies
im_space_cadet
Yesterday at 2:27 PM
im_space_cadet
Today at 12:30 AM
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100th post
MathJedi108   1
N Yesterday at 11:10 PM by mdk2013
Well I guess this is my 100th post, it would be really funny if it isn't can yall share your favorite experience on AoPS here?
1 reply
MathJedi108
Yesterday at 10:59 PM
mdk2013
Yesterday at 11:10 PM
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Find all triples
pedronis   2
N Yesterday at 10:43 PM by Kempu33334
Find all triples of positive integers $(n, r, s)$ such that $n^2 + n + 1$ divides $n^r + n^s + 1$.
2 replies
pedronis
Apr 19, 2025
Kempu33334
Yesterday at 10:43 PM
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Median geometry
Sedro   4
N Yesterday at 10:01 PM by Sedro
In triangle $ABC$, points $D$, $E$, and $F$ are the midpoints of sides $BC$, $CA$, and $AB$, respectively. Prove that the area of the triangle with side lengths $AD$, $BE$, and $CF$ has area $\tfrac{3}{4}[ABC]$.
4 replies
Sedro
Yesterday at 6:03 PM
Sedro
Yesterday at 10:01 PM
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geometry
carvaan   1
N Yesterday at 6:38 PM by Lankou
The difference between two angles of a triangle is 24°. All angles are numerically double digits. Find the number of possible values of the third angle.
1 reply
carvaan
Yesterday at 5:46 PM
Lankou
Yesterday at 6:38 PM
J
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weird permutation problem
Sedro   1
N Yesterday at 6:07 PM by Sedro
Let $\sigma$ be a permutation of $1,2,3,4,5,6,7$ such that there are exactly $7$ ordered pairs of integers $(a,b)$ satisfying $1\le a < b \le 7$ and $\sigma(a) < \sigma(b)$. How many possible $\sigma$ exist?
1 reply
Sedro
Yesterday at 2:09 AM
Sedro
Yesterday at 6:07 PM
J
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J
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triangle ABC
Pirkuliyev Rovsen   1
N Aug 30, 2013 by Tsikaloudakis
Given a triangle $ABC$,let $K$ be the midpoint of $AB$ and $L$ a be point on $AC$ such that $AL=LC+CB$.Prove that ${\angle}KLB=90^\circ$ if and only if $AC=3CB$.
1 reply
Pirkuliyev Rovsen
Aug 30, 2013
Tsikaloudakis
Aug 30, 2013
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triangle ABC
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Pirkuliyev Rovsen
5047 posts
Pirkuliyev Rovsen
#1 h  Aug 30, 2013, 1:33 PM • 2 Y
Y by Adventure10, Mango247
Given a triangle $ABC$,let $K$ be the midpoint of $AB$ and $L$ a be point on $AC$ such that $AL=LC+CB$.Prove that ${\angle}KLB=90^\circ$ if and only if $AC=3CB$.
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Tsikaloudakis
979 posts
Tsikaloudakis
#2 h  Aug 30, 2013, 3:31 PM • 3 Y
Y by Pirkuliyev Rovsen, Adventure10, Mango247
We consider the midpoint M of the AL. Εxtend the AC by department CE = BC and we have:
\[\begin{array}{l} \left. \begin{array}{l} K\hat LB = {90^o}\\ AK = KB\\ AL = LE \end{array} \right\}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} L\hat BE = {90^o}\\ \\ \left. \begin{array}{l} L\hat BE = {90^o}\\ BC = CE\\ AL = LE\\ AM = ML \end{array} \right\}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} BC = CE = CL = LM = MA \end{array}\]
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