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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.

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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
Convex Quadrilateral with Bisector Diagonal
matinyousefi   8
N 29 minutes ago by lpieleanu
Source: Germany TST 2017
In a convex quadrilateral $ABCD$, $BD$ is the angle bisector of $\angle{ABC}$. The circumcircle of $ABC$ intersects $CD,AD$ in $P,Q$ respectively and the line through $D$ parallel to $AC$ cuts $AB,AC$ in $R,S$ respectively. Prove that point $P,Q,R,S$ lie on a circle.
8 replies
matinyousefi
Apr 11, 2020
lpieleanu
29 minutes ago
Kids in clubs
atdaotlohbh   0
30 minutes ago
There are $6k-3$ kids in a class. Is it true that for all positive integers $k$ it is possible to create several clubs each with 3 kids such that any pair of kids are both present in exactly one club?
0 replies
atdaotlohbh
30 minutes ago
0 replies
Turbo's en route to visit each cell of the board
Lukaluce   22
N an hour ago by HamstPan38825
Source: EGMO 2025 P5
Let $n > 1$ be an integer. In a configuration of an $n \times n$ board, each of the $n^2$ cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate $90^{\circ}$ counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of $n$, the maximum number of good cells over all possible starting configurations.

Proposed by Melek Güngör, Turkey
22 replies
Lukaluce
Apr 14, 2025
HamstPan38825
an hour ago
n lamps
pohoatza   47
N an hour ago by yayyayyay
Source: IMO Shortlist 2006, Combinatorics 1, AIMO 2007, TST 2, P1
We have $ n \geq 2$ lamps $ L_{1}, . . . ,L_{n}$ in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows: if the lamp $ L_{i}$ and its neighbours (only one neighbour for $ i = 1$ or $ i = n$, two neighbours for other $ i$) are in the same state, then $ L_{i}$ is switched off; – otherwise, $ L_{i}$ is switched on.
Initially all the lamps are off except the leftmost one which is on.

$ (a)$ Prove that there are infinitely many integers $ n$ for which all the lamps will eventually be off.
$ (b)$ Prove that there are infinitely many integers $ n$ for which the lamps will never be all off.
47 replies
pohoatza
Jun 28, 2007
yayyayyay
an hour ago
prove that at least one of them is divisible by some other member of the set.
Martin.s   0
an hour ago
Given \( n + 1 \) integers \( a_1, a_2, \ldots, a_{n+1} \), each less than or equal to \( 2n \), prove that at least one of them is divisible by some other member of the set.
0 replies
Martin.s
an hour ago
0 replies
estimate for \( a_1 \) is the best possible
Martin.s   0
an hour ago
Let \( a_1 < a_2 < \cdots < a_n < 2n \) be positive integers such that no one of them is divisible by any other member of the sequence. Then
\[
a_1 \geq 2^k,
\]where \( k \) is defined by the inequalities
\[
3^k < 2n < 3^{k+1}.
\]This estimate for \( a_1 \) is the best possible.
0 replies
Martin.s
an hour ago
0 replies
best possible estimate.
Martin.s   0
an hour ago
Let \( a_1 < a_2 < \cdots < a_n < 2n \) be a sequence of positive integers. Then
\[
\max \left( (a_i, a_j) \right) > \frac{38n}{147} - c,
\]where \( c \) is a constant independent of \( n \), and \( (a_i, a_j) \) denotes the greatest common divisor of \( a_i \) and \( a_j \). This is the best possible estimate.
0 replies
Martin.s
an hour ago
0 replies
Polynomial
Fang-jh   6
N an hour ago by yofro
Source: Chinese TST 2007 1st quiz P3
Prove that for any positive integer $ n$, there exists only $ n$ degree polynomial $ f(x),$ satisfying $ f(0) = 1$ and $ (x + 1)[f(x)]^2 - 1$ is an odd function.
6 replies
Fang-jh
Jan 3, 2009
yofro
an hour ago
collinear wanted, toucpoints of incircle related
parmenides51   2
N an hour ago by Tamam
Source: 2018 Thailand October Camp 1.2
Let $\Omega$ be the inscribed circle of a triangle $\vartriangle ABC$. Let $D, E$ and $F$ be the tangency points of $\Omega$ and the sides $BC, CA$ and $AB$, respectively, and let $AD, BE$ and $CF$ intersect $\Omega$ at $K, L$ and $M$, respectively, such that $D, E, F, K, L$ and $M$ are all distinct. The tangent line of $\Omega$ at $K$ intersects $EF$ at $X$, the tangent line of $\Omega$ at $L$ intersects $DE$ at $Y$ , and the tangent line of $\Omega$ at M intersects $DF$ at $Z$. Prove that $X,Y$ and $Z$ are collinear.
2 replies
parmenides51
Oct 15, 2020
Tamam
an hour ago
show that there are at least eight vertices where exactly three edges meet.
Martin.s   0
an hour ago
If all the faces of a convex polyhedron have central symmetry, show that there are at least eight vertices where exactly three edges meet. (The cube has exactly eight such vertices.)
0 replies
Martin.s
an hour ago
0 replies
Nice problem
Martin.s   0
an hour ago
If \(p\) is a prime and \(n \geq p\), then
\[
n! \sum_{pi+j=n} \frac{1}{p^i i! j!} \equiv 0 \pmod{p}.
\]
0 replies
Martin.s
an hour ago
0 replies
2-var inequality
sqing   12
N an hour ago by ytChen
Source: Own
Let $ a,b> 0 ,a^3+ab+b^3=3.$ Prove that
$$ (a+b)(a+1)(b+1) \leq 8$$$$ (a^2+b^2)(a+1)(b+1) \leq 8$$Let $ a,b> 0 ,a^3+ab(a+b)+b^3=3.$ Prove that
$$ (a+b)(a+1)(b+1) \leq \frac{3}{2}+\sqrt[3]{6}+\sqrt[3]{36}$$
12 replies
1 viewing
sqing
Yesterday at 1:35 PM
ytChen
an hour ago
How many friends can sit in that circle at most?
Arytva   4
N 2 hours ago by JohannIsBach

A group of friends sits in a ring. Each friend picks a different whole number and holds a stone marked with it. Then they pass their stone one seat to the right so everyone ends up with two stones: one they made and one they received. Now they notice something odd: if your original number is $x$, your right-neighbor’s is $y$, and the next person over is $z$, then for every trio in the circle they see

$$
x + z = (2 - x)\,y.
$$
They want as many friends as possible before this breaks (since all stones must stay distinct).

How many friends can sit in that circle at most?
4 replies
Arytva
Today at 10:00 AM
JohannIsBach
2 hours ago
Bisectors, perpendicularity and circles
JuanDelPan   15
N 2 hours ago by zuat.e
Source: Pan-American Girls’ Mathematical Olympiad 2022, Problem 3
Let $ABC$ be an acute triangle with $AB< AC$. Denote by $P$ and $Q$ points on the segment $BC$ such that $\angle BAP = \angle CAQ < \frac{\angle BAC}{2}$. $B_1$ is a point on segment $AC$. $BB_1$ intersects $AP$ and $AQ$ at $P_1$ and $Q_1$, respectively. The angle bisectors of $\angle BAC$ and $\angle CBB_1$ intersect at $M$. If $PQ_1\perp AC$ and $QP_1\perp AB$, prove that $AQ_1MPB$ is cyclic.
15 replies
JuanDelPan
Oct 27, 2022
zuat.e
2 hours ago
Yet another cyclic quadrilateral
Matematika   16
N Dec 12, 2024 by lelouchvigeo
Source: European Mathematical Cup 2012, Senior Division, Problem 2
Let $ABC$ be an acute triangle with orthocenter $H$. Segments $AH$ and $CH$ intersect segments $BC$ and $AB$ in points $A_1$ and $C_1$ respectively. The segments $BH$ and $A_1C_1$ meet at point $D$. Let $P$ be the midpoint of the segment $BH$. Let $D'$ be the reflection of the point $D$ in $AC$. Prove that quadrilateral $APCD'$ is cyclic.

Proposed by Matko Ljulj.
16 replies
Matematika
Jul 27, 2013
lelouchvigeo
Dec 12, 2024
Yet another cyclic quadrilateral
G H J
Source: European Mathematical Cup 2012, Senior Division, Problem 2
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Matematika
139 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ABC$ be an acute triangle with orthocenter $H$. Segments $AH$ and $CH$ intersect segments $BC$ and $AB$ in points $A_1$ and $C_1$ respectively. The segments $BH$ and $A_1C_1$ meet at point $D$. Let $P$ be the midpoint of the segment $BH$. Let $D'$ be the reflection of the point $D$ in $AC$. Prove that quadrilateral $APCD'$ is cyclic.

Proposed by Matko Ljulj.
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pepat
20 posts
#2 • 3 Y
Y by Adventure10, Mango247, and 1 other user
I suppose segments should be lines. The following solution is probably not the easiest one but anyway:

Solution
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sunken rock
4402 posts
#3 • 2 Y
Y by Adventure10, Mango247
Let $B_1=BH\cap AC$. It is known that $BB_1$ is the bisector of $\angle A_1B_1C_1$ and $A_1B_1C_1P$ is cyclic (and all those points lie onto the 9-point circle of $\Delta ABC$), hence $B_1D\cdot B_1P=A_1B_1\cdot C_1B_1\ (\ 1\ )$. This comes easily from 2 similar triangles.
But $AC$ is diameter of $CAC_1A_1$ and $A_1B_1$ is symmetrical of $B_1C_1$ w.r.t. $BB_1$, hence $C_1B_1$ intersect circle $\odot (CAC_1A_1)$ at $C_2$ and $B_1C_2=B_1A_1$, consequently $(1)$ becomes $B_1D\cdot B_1P=B_1C_1\cdot B_1C_2$, but from power of point $B_1$ w.r.t. $(CAC_1A_2)$ we have $B_1C_1\cdot B_1C_2=B_1C\cdot B_1A$; under the circumstances $B_1C\cdot B_1A=B_1D\cdot B_1P=B_1D'\cdot B_1P$, done.

Hope I did not mess-up the indices.

Best regards,
sunken rock
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Panoz93
61 posts
#4 • 3 Y
Y by electrovector, Adventure10, Mango247
It suffices to prove that $D$ is the orthocenter of triangle $APC$. For this just notice that $AD,DC$ are the polars of $C,A$ with respect to the circumcircle of $C_{1}HA_{1}B$.
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bojler
16 posts
#5 • 2 Y
Y by Adventure10, Mango247
Let $BB_1$ be the third altitude in triangle $ABC$. Note that triangles $AB_1H$ and $BB_1C$ are similar thus
$AB_1\cdot B_1C=B_1H\cdot B_1B$ .
$H(B,H,D,B_1)$ so $PH^2=PD\cdot PB_1$ .
$B_1P\cdot B_1D'=B_1P\cdot B_1D=B_1P\cdot (B_1P-DP)=PB_1^2-PH^2=B_1H\cdot B_1B=AB_1\cdot B_1C$.
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jayme
9801 posts
#6 • 2 Y
Y by Adventure10, Jacob._07
Dear Mathlinkers,
for #, see http://www.artofproblemsolving.com/community/c6h1069884_orthocenter

Sincerely
Jean-Louis
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orestis100
1 post
#7 • 2 Y
Y by Adventure10, Mango247
Can someone please prove why H(B,H,D,B1)?
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AlastorMoody
2125 posts
#8 • 2 Y
Y by Pluto1708, Adventure10
Apply Brokard's Theorem on $\odot (BHC_1A_1) $, $D $ is the orthocenter of $\Delta PCA $, hence the conclusion
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itslumi
284 posts
#9 • 1 Y
Y by Maxito12345
AlastorMoody wrote:
Apply Brokard's Theorem on $\odot (BHC_1A_1) $, $D $ is the orthocenter of $\Delta PCA $, hence the conclusion

Same solution,Can this be done whith inverzion???
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jayme
9801 posts
#10
Y by
Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%207.pdf p. 73...

Sincerely
Jean-Louis
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Olympikus
87 posts
#11
Y by
orestis100 wrote:
Can someone please prove why H(B,H,D,B1)?

Take perspective at $C_1$ and map to $AC$. Use the Ceva-Menelaus lemma : $H(A,C;B_1,A_1C_1\cap AC)$.
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Mikeglicker
258 posts
#12 • 3 Y
Y by Mango247, Mango247, Mango247
Why showing that D is the orthocenter of APC is suffices?
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Mikeglicker
258 posts
#13
Y by
Mikeglicker wrote:
Why showing that D is the orthocenter of APC is suffices?

Anybody?
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arzhang2001
251 posts
#14
Y by
do you ignore it:
if we let $M$ Be the midpoint of $A_1C_1$ Then should $MDAC$ be Cyclic.
This post has been edited 3 times. Last edited by arzhang2001, Jun 29, 2020, 7:11 AM
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miltosk
9 posts
#15
Y by
Mikeglicker wrote:
Mikeglicker wrote:
Why showing that D is the orthocenter of APC is suffices?

Anybody?

Because then $$DAC=DPC$$as acute angles with perpendicular sides and since $$DAC=D'AC$$due to the isosceles triangle $$DAD'$$, $$DPC=D'AC$$and then its obvious.
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iman007
270 posts
#17
Y by
$\text{Brokard kills this problem}$ :D
$\text{basically we should prove that} \thickspace \angle{ADC}=180^{\circ} -\angle{APC}$
$\text{which means to prove that} \thickspace P\thickspace \text{is the orthocenter of the triangle} \thickspace ADC$
$\textbf{Brokard theorem} \thickspace \text{on}\thickspace (A_1BC_1H) \thickspace\text{finishes the problem}$
\(\boxed{\mathcal{Q.E.D}}\)
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lelouchvigeo
183 posts
#18
Y by
Nice
Since $ACD$ are autopolar by Brocard's and $P$ is the center of circle $BA_1HC_1.$ We can observe that $D$ is the orthocenter of $\triangle ACP.$
Since the reflection of the orthocenter lies on the circle. We are done
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N Quick Reply
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