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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
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
Function
Musashi123   0
3 minutes ago
f:R\{0} ->R\{0}
f(x/y+y/x)=f(x)/f(y)+f(y)/f(x)
f(xy)=f(x).f(y)
0 replies
Musashi123
3 minutes ago
0 replies
hard problem
Cobedangiu   1
N 4 minutes ago by m4thbl3nd3r
Let $a,b,c>0$ and $a+b+c=3$. Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$
1 reply
Cobedangiu
33 minutes ago
m4thbl3nd3r
4 minutes ago
real+ FE
pomodor_ap   1
N 14 minutes ago by Parsia--
Source: Own, PDC001-P7
Let $f : \mathbb{R}^+ \to \mathbb{R}^+$ be a function such that
$$f(x)f(x^2 + y f(y)) = f(x)f(y^2) + x^3$$for all $x, y \in \mathbb{R}^+$. Determine all such functions $f$.
1 reply
+1 w
pomodor_ap
3 hours ago
Parsia--
14 minutes ago
Similar triangles formed by angular condition
Mahdi_Mashayekhi   5
N 23 minutes ago by sami1618
Source: Iran 2025 second round P3
Point $P$ lies inside of scalene triangle $ABC$ with incenter $I$ such that $:$
$$ 2\angle ABP = \angle BCA , 2\angle ACP = \angle CBA $$Lines $PB$ and $PC$ intersect line $AI$ respectively at $B'$ and $C'$. Line through $B'$ parallel to $AB$ intersects $BI$ at $X$ and line through $C'$ parallel to $AC$ intersects $CI$ at $Y$. Prove that triangles $PXY$ and $ABC$ are similar.
5 replies
Mahdi_Mashayekhi
Apr 19, 2025
sami1618
23 minutes ago
Tango course
oVlad   0
34 minutes ago
Source: Romania EGMO TST 2019 Day 1 P4
Six boys and six girls are participating at a tango course. They meet every evening for three weeks (a total of 21 times). Each evening, at least one boy-girl pair is selected to dance in front of the others. At the end of the three weeks, every boy-girl pair has been selected at least once. Prove that there exists a person who has been selected on at least 5 distinct evenings.

Note: a person can be selected twice on the same evening.
0 replies
oVlad
34 minutes ago
0 replies
Inequality with three conditions
oVlad   0
36 minutes ago
Source: Romania EGMO TST 2019 Day 1 P3
Let $a,b,c$ be non-negative real numbers such that \[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]Prove that $a^2+b^2+c^2\leqslant 2abc+1.$
0 replies
oVlad
36 minutes ago
0 replies
NT with repeating decimal digits
oVlad   0
38 minutes ago
Source: Romania EGMO TST 2019 Day 1 P2
Determine the digits $0\leqslant c\leqslant 9$ such that for any positive integer $k{}$ there exists a positive integer $n$ such that the last $k{}$ digits of $n^9$ are equal to $c{}.$
0 replies
oVlad
38 minutes ago
0 replies
Easy geo
oVlad   0
40 minutes ago
Source: Romania EGMO TST 2019 Day 1 P1
A line through the vertex $A{}$ of the triangle $ABC{}$ which doesn't coincide with $AB{}$ or $AC{}$ intersectes the altitudes from $B{}$ and $C{}$ at $D{}$ and $E{}$ respectively. Let $F{}$ be the reflection of $D{}$ in $AB{}$ and $G{}$ be the reflection of $E{}$ in $AC{}.$ Prove that the circles $ABF{}$ and $ACG{}$ are tangent.
0 replies
oVlad
40 minutes ago
0 replies
Combo with cyclic sums
oVlad   0
43 minutes ago
Source: Romania EGMO TST 2017 Day 1 P4
In $p{}$ of the vertices of the regular polygon $A_0A_1\ldots A_{2016}$ we write the number $1{}$ and in the remaining ones we write the number $-1.{}$ Let $x_i{}$ be the number written on the vertex $A_i{}.$ A vertex is good if \[x_i+x_{i+1}+\cdots+x_j>0\quad\text{and}\quad x_i+x_{i-1}+\cdots+x_k>0,\]for any integers $j{}$ and $k{}$ such that $k\leqslant i\leqslant j.$ Note that the indices are taken modulo $2017.$ Determine the greatest possible value of $p{}$ such that, regardless of numbering, there always exists a good vertex.
0 replies
oVlad
43 minutes ago
0 replies
Collect ...
luutrongphuc   1
N an hour ago by GreekIdiot
Find all functions $f: \mathbb{R^+} \rightarrow \mathbb{R^+}$ such that:
$$f\left(f(xy)+1\right)=xf\left(x+f(y)\right)$$
1 reply
luutrongphuc
an hour ago
GreekIdiot
an hour ago
Easy Geometry
TheOverlord   33
N an hour ago by math.mh
Source: Iran TST 2015, exam 1, day 1 problem 2
$I_b$ is the $B$-excenter of the triangle $ABC$ and $\omega$ is the circumcircle of this triangle. $M$ is the middle of arc $BC$ of $\omega$ which doesn't contain $A$. $MI_b$ meets $\omega$ at $T\not =M$. Prove that
$$ TB\cdot TC=TI_b^2.$$
33 replies
TheOverlord
May 10, 2015
math.mh
an hour ago
Existence of a circle tangent to four lines
egxa   3
N an hour ago by mathuz
Source: All Russian 2025 10.2
Inside triangle \(ABC\), point \(P\) is marked. Point \(Q\) is on segment \(AB\), and point \(R\) is on segment \(AC\) such that the circumcircles of triangles \(BPQ\) and \(CPR\) are tangent to line \(AP\). Lines are drawn through points \(B\) and \(C\) passing through the center of the circumcircle of triangle \(BPC\), and through points \(Q\) and \(R\) passing through the center of the circumcircle of triangle \(PQR\). Prove that there exists a circle tangent to all four drawn lines.
3 replies
egxa
Apr 18, 2025
mathuz
an hour ago
An easy FE
oVlad   0
an hour ago
Source: Romania EGMO TST 2017 Day 1 P3
Determine all functions $f:\mathbb R\to\mathbb R$ such that \[f(xy-1)+f(x)f(y)=2xy-1,\]for any real numbers $x{}$ and $y{}.$
0 replies
oVlad
an hour ago
0 replies
GCD of a sequence
oVlad   0
an hour ago
Source: Romania EGMO TST 2017 Day 1 P2
Determine all pairs $(a,b)$ of positive integers with the following property: all of the terms of the sequence $(a^n+b^n+1)_{n\geqslant 1}$ have a greatest common divisor $d>1.$
0 replies
oVlad
an hour ago
0 replies
AA_2, BB_2, CC_1 are concurrent, internal bisectors, midpoints
parmenides51   4
N Mar 22, 2022 by L567
Source: 2015 British FST2 p1
An acute-angled triangle $\vartriangle ABC$ is given, and $A_1,B_1, C_1$ are the midpoints of sides $BC,CA, AB$ respectively. The internal angle bisector of $\angle AC_1C$ meets $AC$ at $L$, and the internal angle bisector of $\angle CC_1B$ meets $BC$ at $K$. The line $LK$ intersects $B_1C_1$ at $A_2$, and $A_1C_1$ at $B_2$. Prove that the lines $AA_2, BB_2, CC_1$ are concurrent.
4 replies
parmenides51
Jan 14, 2021
L567
Mar 22, 2022
AA_2, BB_2, CC_1 are concurrent, internal bisectors, midpoints
G H J
Source: 2015 British FST2 p1
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parmenides51
30630 posts
#1 • 4 Y
Y by HWenslawski, Mango247, Mango247, Mango247
An acute-angled triangle $\vartriangle ABC$ is given, and $A_1,B_1, C_1$ are the midpoints of sides $BC,CA, AB$ respectively. The internal angle bisector of $\angle AC_1C$ meets $AC$ at $L$, and the internal angle bisector of $\angle CC_1B$ meets $BC$ at $K$. The line $LK$ intersects $B_1C_1$ at $A_2$, and $A_1C_1$ at $B_2$. Prove that the lines $AA_2, BB_2, CC_1$ are concurrent.
Z K Y
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L567
1184 posts
#2 • 1 Y
Y by HWenslawski
Let $AA_2 \cap BC = X$, $BB_2 \cap AC = Y$. Let $a, b, c$ be the lengths of the three sides and $m_c$ be the length of $CC_1$. Using angle bisector theorem, we get that $AL = \frac{cb}{c + 2m_c}, LC = \frac{2m_cb}{c+2m_c}$. Since $LB_1 = LC - B_1C$, $LB_1 = b \frac{2m_c - c}{2m_c + c}$. Also, observe that $LK || AB$. So,

$\frac{BX}{XC} = \frac{C_1A_2}{A_2B_1} = \frac{AL}{LB_1} = \frac{c}{2m_c - c}$

Similarly, we get that $\frac{CY}{YA} = \frac{2m_c - c}{c}$. Since $\frac{AC_1}{C_1B} = 1$, by ceva's theorem, the three lines concur

@below Thanks for pointing out the mistakes!, ive fixed them now
This post has been edited 1 time. Last edited by L567, Jan 15, 2021, 8:31 AM
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electrovector
479 posts
#3
Y by
Dear L567, thanks for your nice solution. However there are a few mistakes.
L567 wrote:
Let $AA_1 \cap BC = X$, $BB_1 \cap AC = Y$
Let $AA_2 \cap BC = X$, $BB_2 \cap AC = Y$
L567 wrote:
Since $LB = LC - BC$, $LB = b \frac{2m_c - c}{2m_c + c}$
Since $LB_1 = LC - B_1C$, $LB_1 = b \frac{2m_c - c}{2m_c + c}$
L567 wrote:
$\frac{BX}{XC} = \frac{C_1A_2}{A_2B} = \frac{AL}{LB}$
$\frac{BX}{XC} = \frac{C_1A_2}{A_2B_1} = \frac{AL}{LB_1}$
Z K Y
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oolite
343 posts
#4
Y by
L567 wrote:
Also, observe that $LK || AB$
I'm impressed that this is so easy that you can just "observe" it! :)

It took me a while. Here's what I ended up doing:
  • Say that the line through $L$ parallel to $AB$ crosses $C_1C$ at $X$ and crosses $C_1K$ at $K^\prime$. We need to show that $K^\prime=K$. Note that $\angle C_1LX=\angle LC_1A=\angle XC_1L$ then angle-chase further in $\triangle K^\prime C_1L$ to find $LX=K^\prime X$. But since $C_1C$ is a median, this means $K^\prime$ lies on $BC$, i.e. $K^\prime = BC\cap C_1K$ which equals $K$ by definition.
Is there a neater way to prove that $KL\parallel AB$?
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L567
1184 posts
#5 • 1 Y
Y by oolite
Oops, its not "obvious", I should have elaborated, $$\frac{CK}{KB} = \frac{CC_1}{C_1B} = \frac{CC_1}{C_1A} = \frac{BL}{LA}$$so $KL || AB$
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