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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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April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 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
two subsets with no fewer than four common elements.
micliva   39
N 2 minutes ago by de-Kirschbaum
Source: All-Russian Olympiad 1996, Grade 9, First Day, Problem 4
In the Duma there are 1600 delegates, who have formed 16000 committees of 80 persons each. Prove that one can find two committees having no fewer than four common members.

A. Skopenkov
39 replies
micliva
Apr 18, 2013
de-Kirschbaum
2 minutes ago
3 knightlike moves is enough
sarjinius   2
N 3 minutes ago by cooljoseph
Source: Philippine Mathematical Olympiad 2025 P6
An ant is on the Cartesian plane. In a single move, the ant selects a positive integer $k$, then either travels [list]
[*] $k$ units vertically (up or down) and $2k$ units horizontally (left or right); or
[*] $k$ units horizontally (left or right) and $2k$ units vertically (up or down).
[/list]
Thus, for any $k$, the ant can choose to go to one of eight possible points.
Prove that, for any integers $a$ and $b$, the ant can travel from $(0, 0)$ to $(a, b)$ using at most $3$ moves.
2 replies
sarjinius
Mar 9, 2025
cooljoseph
3 minutes ago
16th ibmo - uruguay 2001/q3.
carlosbr   21
N 13 minutes ago by de-Kirschbaum
Source: Spanish Communities
Let $S$ be a set of $n$ elements and $S_1,\ S_2,\dots,\ S_k$ are subsets of $S$ ($k\geq2$), such that every one of them has at least $r$ elements.

Show that there exists $i$ and $j$, with $1\leq{i}<j\leq{k}$, such that the number of common elements of $S_i$ and $S_j$ is greater or equal to: $r-\frac{nk}{4(k-1)}$
21 replies
carlosbr
Apr 15, 2006
de-Kirschbaum
13 minutes ago
Weird Geo
Anto0110   1
N 19 minutes ago by cooljoseph
In a trapezium $ABCD$, the sides $AB$ and $CD$ are parallel and the angles $\angle ABC$ and $\angle BAD$ are acute. Show that it is possible to divide the triangle $ABC$ into 4 disjoint triangle $X_1. . . , X_4$ and the triangle $ABD$ into 4 disjoint triangles $Y_1,. . . , Y_4$ such that the triangles $X_i$ and $Y_i$ are congruent for all $i$.
1 reply
Anto0110
5 hours ago
cooljoseph
19 minutes ago
Dear Sqing: So Many Inequalities...
hashtagmath   35
N an hour ago by ohiorizzler1434
I have noticed thousands upon thousands of inequalities that you have posted to HSO and was wondering where you get the inspiration, imagination, and even the validation that such inequalities are true? Also, what do you find particularly appealing and important about specifically inequalities rather than other branches of mathematics? Thank you :)
35 replies
1 viewing
hashtagmath
Oct 30, 2024
ohiorizzler1434
an hour ago
Hard FE R^+
DNCT1   5
N 2 hours ago by jasperE3
Find all functions $f:\mathbb{R^+}\to\mathbb{R^+}$ such that
$$f(3x+f(x)+y)=f(4x)+f(y)\quad\forall x,y\in\mathbb{R^+}$$
5 replies
DNCT1
Dec 30, 2020
jasperE3
2 hours ago
Maximum of Incenter-triangle
mpcnotnpc   4
N 2 hours ago by mpcnotnpc
Triangle $\Delta ABC$ has side lengths $a$, $b$, and $c$. Select a point $P$ inside $\Delta ABC$, and construct the incenters of $\Delta PAB$, $\Delta PBC$, and $\Delta PAC$ and denote them as $I_A$, $I_B$, $I_C$. What is the maximum area of the triangle $\Delta I_A I_B I_C$?
4 replies
mpcnotnpc
Mar 25, 2025
mpcnotnpc
2 hours ago
Something nice
KhuongTrang   26
N 2 hours ago by KhuongTrang
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
26 replies
KhuongTrang
Nov 1, 2023
KhuongTrang
2 hours ago
Tiling rectangle with smaller rectangles.
MarkBcc168   59
N 3 hours ago by Bonime
Source: IMO Shortlist 2017 C1
A rectangle $\mathcal{R}$ with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of $\mathcal{R}$ are either all odd or all even.

Proposed by Jeck Lim, Singapore
59 replies
MarkBcc168
Jul 10, 2018
Bonime
3 hours ago
Existence of AP of interesting integers
DVDthe1st   34
N 3 hours ago by DeathIsAwe
Source: 2018 China TST Day 1 Q2
A number $n$ is interesting if 2018 divides $d(n)$ (the number of positive divisors of $n$). Determine all positive integers $k$ such that there exists an infinite arithmetic progression with common difference $k$ whose terms are all interesting.
34 replies
DVDthe1st
Jan 2, 2018
DeathIsAwe
3 hours ago
Strange Geometry
Itoz   1
N 4 hours ago by hukilau17
Source: Own
Given a fixed circle $\omega$ with its center $O$. There are two fixed points $B, C$ and one moving point $A$ on $\omega$. The midpoint of the line segment $BC$ is $M$. $R$ is a fixed point on $\omega$. Line $AO$ intersects$\odot(AMR)$ at $P(\ne A)$, and line $BP$ intersects $\odot(BOC)$ at $Q(\ne B)$.

Find all the fixed points $R$ such that $\omega$ is always tangent to $\odot (OPQ)$ when $A$ varies.
Hint
1 reply
Itoz
Yesterday at 2:00 PM
hukilau17
4 hours ago
find all pairs of positive integers
Khalifakhalifa   2
N 4 hours ago by Haris1


Find all pairs of positive integers \((a, b)\) such that:
\[
a^2 + b^2 \mid a^3 + b^3
\]
2 replies
Khalifakhalifa
May 27, 2024
Haris1
4 hours ago
D860 : Flower domino and unconnected
Dattier   4
N 5 hours ago by Haris1
Source: les dattes à Dattier
Let G be a grid of size m*n.

We have 2 dominoes in flowers and not connected like here
IMAGE
Determine a necessary and sufficient condition on m and n, so that G can be covered with these 2 kinds of dominoes.

4 replies
Dattier
May 26, 2024
Haris1
5 hours ago
Equal Distances in an Isosceles Setting
mojyla222   3
N 5 hours ago by sami1618
Source: IDMC 2025 P4
Let $ABC$ be an isosceles triangle with $AB=AC$. The circle $\omega_1$, passing through $B$ and $C$, intersects segment $AB$ at $K\neq B$. The circle $\omega_2$ is tangent to $BC$ at $B$ and passes through $K$. Let $M$ and $N$ be the midpoints of segments $AB$ and $AC$, respectively. The line $MN$ intersects $\omega_1$ and $\omega_2$ at points $P$ and $Q$, respectively, where $P$ and $Q$ are the intersections closer to $M$. Prove that $MP=MQ$.

Proposed by Hooman Fattahi
3 replies
mojyla222
Yesterday at 5:05 AM
sami1618
5 hours ago
Geometry . Concurrency
IstekOlympiadTeam   6
N Aug 12, 2020 by N0_NAME
Source: Estonia IMO TST 1997 Day 1 P1
In a triangle $ABC$ points $A_1,B_1,C_1$ are the midpoints of $BC,CA,AB$ respectively,and $A_2,B_2,C_2$ are the midpoints of the altitudes from $A,B,C$ respectively. Show that the lines $A_1A_2,B_1B_2,C_1,C_2$ are concurrent.


6 replies
IstekOlympiadTeam
Nov 27, 2015
N0_NAME
Aug 12, 2020
Geometry . Concurrency
G H J
G H BBookmark kLocked kLocked NReply
Source: Estonia IMO TST 1997 Day 1 P1
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IstekOlympiadTeam
542 posts
#1 • 1 Y
Y by Adventure10
In a triangle $ABC$ points $A_1,B_1,C_1$ are the midpoints of $BC,CA,AB$ respectively,and $A_2,B_2,C_2$ are the midpoints of the altitudes from $A,B,C$ respectively. Show that the lines $A_1A_2,B_1B_2,C_1,C_2$ are concurrent.
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jayme
9777 posts
#2 • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,

you can see

http://jl.ayme.pagesperso-orange.fr/Docs/The%20cross-cevian%20point.pdf p. 20...

Sincerely
Jean-Louis
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ThE-dArK-lOrD
4071 posts
#3 • 2 Y
Y by Adventure10, Mango247
IstekOlympiadTeam wrote:
In a triangle $ABC$ points $A_1,B_1,C_1$ are the midpoints of $BC,CA,AB$ respectively,and $A_2,B_2,C_2$ are the midpoints of the altitudes from $A,B,C$ respectively. Show that the lines $A_1A_2,B_1B_2,C_1,C_2$ are concurrent.
All of them pass through Symmedian point $K$ of triangle $ABC$.
For the proof, see Theorem 11 on page 10 of this article.
This post has been edited 3 times. Last edited by ThE-dArK-lOrD, May 22, 2018, 10:24 AM
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eshan
471 posts
#4 • 8 Y
Y by purumehta, Euler149, IstekOlympiadTeam, mssmath, rkm0959, aops1221math, Adventure10, Mango247
My Solution.

Let $AD,BE,CF$ be the altitudes of $\triangle ABC.$

As $A_2$ is the mid-point of $AD,~C_1$ is the mid-point of $AB$ and $B_1$ is the mid-point of $AC,$ we have $\triangle AA_2C_1\cong ADB,~\triangle AB_1C_2\cong ADC.$

So $C_1,~A_2,~B_1$ are sollinear.

Which means, $A_2\in B_1C_1.$

Similarly, $B_2\in C_1A_1,~C_2\in A_1B_1.$

So we can say that $A_1A_2,~B_1B_2,~C_1C_2$ are $3$ cevians of $\triangle A_1B_1C_1.$

Now, $\frac{C_1A_2}{A_2B_1}=\frac{BD}{DC},$ which we get by mid-point theorem, as $BD=2C_1A_2,~DC=2A_2B_1.$

Similarly, $\frac{B_1C_2}{C_2A_1}=\frac{AF}{FB},~\frac{A_1B_2}{B_2C_1}=\frac{CE}{EA}.$

So $\frac{C_1A_2}{A_2B_1}\cdot \frac{B_1C_2}{C_2A_1}\cdot \frac{A_1B_2}{B_2C_1}=\frac{BD}{DC}\cdot \frac{AF}{FB}\cdot \frac{CE}{EA}=1,$ as altitudes of a triangle are concurrent.

Thus by converse of Ceva's theorem, $A_1A_2,~B_1B_2,~C_1C_2$ are concurrent. $\Square$
Attachments:
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suli
1498 posts
#5 • 3 Y
Y by mssmath, Adventure10, Mango247
Notice that by homothety $A_2, B_2, C_2$ are points on the medial triangle, i.e. $A_1 A_2, B_1 B_2, C_1 C_2$ are cevians of $A_1 B_1 C_1$ which are concurrent by Ceva's theorem and the homothety which states $\frac{B_1 A_2}{C_1 A_2} = \frac{BD}{CD}$ where $AD$ is altitude in $ABC$. (And yes, another application of Ceva's theorem on the orthic triangle is needed.)
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DanDumitrescu
196 posts
#6 • 2 Y
Y by Adventure10, Mango247
This problem is just a particular case of Cevian Nest Theorem.
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N0_NAME
112 posts
#7
Y by
My bary solution.
Let $K = AA_2 \cap BC$, $\alpha = \angle BAC$, $\beta = \angle CBA$, $\gamma = \angle ACB$ and $d$ is the diameter of the circumcircle of $\triangle ABC$.
$BK = AB\cdot cos \beta = d\cdot sin \gamma cos \beta$, $CK = AC\cdot cos \gamma = d\cdot sin \beta cos \gamma$.
$\implies K = [0: sin \beta cos \gamma: sin \gamma cos \beta] \implies A_2 = [sin \beta cos \gamma + sin \gamma cos \beta:sin \beta cos \gamma: sin \gamma cos \beta] = [sin \alpha: sin \beta cos \gamma: sin \gamma cos \beta]$.
$A_1 = [0:1:1]$.
$L =  [sin^2 \alpha  : sin^2 \beta : sin^2 \gamma]$.
$sin^2\alpha = sin\alpha (sin\alpha) + 0(sin^2 \beta \cdot sin^2 \gamma -  sin \beta cos \gamma \cdot  sin \gamma cos \beta)$.
$sin^2 \beta = sin^2\beta \cdot sin^2\gamma + sin^2\beta \cdot cos^2\gamma =  (sin\beta cos\gamma + sin \gamma cos \beta)(sin\beta cos\gamma) + sin^2\beta \cdot sin^2\gamma - sin \beta cos \gamma \cdot  sin \gamma cos \beta$
$ = sin\alpha (sin\beta cos\gamma)+ 1(sin^2\beta \cdot sin^2\gamma - sin \beta cos \gamma \cdot  sin \gamma cos \beta)$.
Similarly, $sin^2\gamma= sin\alpha (sin\gamma cos\beta)+ 1(sin^2\beta \cdot sin^2\gamma - sin \beta cos \gamma \cdot  sin \gamma cos \beta)$.
$\implies L \in A_1A_2$
Similarly, $L \in B_1B_2$ and $L \in C_1C_2$.
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