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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 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
one cyclic formed by two cyclic
CrazyInMath   20
N 29 minutes ago by pingupignu
Source: EGMO 2025/3
Let $ABC$ be an acute triangle. Points $B, D, E$, and $C$ lie on a line in this order and satisfy $BD = DE = EC$. Let $M$ and $N$ be the midpoints of $AD$ and $AE$, respectively. Suppose triangle $ADE$ is acute, and let $H$ be its orthocentre. Points $P$ and $Q$ lie on lines $BM$ and $CN$, respectively, such that $D, H, M,$ and $P$ are concyclic and pairwise different, and $E, H, N,$ and $Q$ are concyclic and pairwise different. Prove that $P, Q, N,$ and $M$ are concyclic.
20 replies
CrazyInMath
Yesterday at 12:38 PM
pingupignu
29 minutes ago
Rioplatense 2022 - Level 3 - Problem 3
407420   3
N an hour ago by alfonsoramires
Let $n$ be a positive integer. Given a sequence of nonnegative real numbers $x_1,\ldots ,x_n$ we define the transformed sequence $y_1,\ldots ,y_n$ as follows: the number $y_i$ is the greatest possible value of the average of consecutive terms of the sequence that contain $x_i$. For example, the transformed sequence of $2,4,1,4,1$ is $3,4,3,4,5/2$.
Prove that
a) For every positive real number $t$, the number of $y_i$ such that $y_i>t$ is less than or equal to $\frac{2}{t}(x_1+\cdots +x_n)$.
b) The inequality $\frac{y_1+\cdots +y_n}{32n}\leq \sqrt{\frac{x_1^2+\cdots +x_n^2}{32n}}$ holds.
3 replies
407420
Dec 6, 2022
alfonsoramires
an hour ago
Inequality with a,b,c
GeoMorocco   1
N an hour ago by Sedro
Source: Morocco Training
Let $a,b,c$ be positive real numbers. Prove that:
$$\sqrt[3]{a^3+b^3}+\sqrt[3]{b^3+c^3}+\sqrt[3]{c^3+a^3}\geq \sqrt[3]{2}(a+b+c)$$
1 reply
GeoMorocco
4 hours ago
Sedro
an hour ago
pairwise coprime sum gcd
InterLoop   23
N an hour ago by TestX01
Source: EGMO 2025/1
For a positive integer $N$, let $c_1 < c_2 < \dots < c_m$ be all the positive integers smaller than $N$ that are coprime to $N$. Find all $N \ge 3$ such that
$$\gcd(N, c_i + c_{i+1}) \neq 1$$for all $1 \le i \le m - 1$.
23 replies
InterLoop
Yesterday at 12:34 PM
TestX01
an hour ago
problem 5
termas   73
N an hour ago by zuat.e
Source: IMO 2016
The equation
$$(x-1)(x-2)\cdots(x-2016)=(x-1)(x-2)\cdots (x-2016)$$is written on the board, with $2016$ linear factors on each side. What is the least possible value of $k$ for which it is possible to erase exactly $k$ of these $4032$ linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?
73 replies
termas
Jul 12, 2016
zuat.e
an hour ago
sequence infinitely similar to central sequence
InterLoop   13
N 2 hours ago by TestX01
Source: EGMO 2025/2
An infinite increasing sequence $a_1 < a_2 < a_3 < \dots$ of positive integers is called central if for every positive integer $n$, the arithmetic mean of the first $a_n$ terms of the sequence is equal to $a_n$.

Show that there exists an infinite sequence $b_1$, $b_2$, $b_3$, $\dots$ of positive integers such that for every central sequence $a_1$, $a_2$, $a_3$, $\dots$, there are infinitely many positive integers $n$ with $a_n = b_n$.
13 replies
InterLoop
Yesterday at 12:38 PM
TestX01
2 hours ago
2ab+1 | a^2 + b^2 + 1
goldeneagle   23
N 2 hours ago by Pseudo_Matter
Source: Iran 3rd round 2013 - Number Theory Exam - Problem 2
Suppose that $a,b$ are two odd positive integers such that $2ab+1 \mid a^2 + b^2 + 1$. Prove that $a=b$.
(15 points)
23 replies
goldeneagle
Sep 11, 2013
Pseudo_Matter
2 hours ago
Hard excircle geo
a_507_bc   9
N 2 hours ago by lksb
Source: MEMO 2023 T6
Let $ABC$ be an acute triangle with $AB < AC$. Let $J$ be the center of the $A$-excircle of $ABC$. Let $D$ be the projection of $J$ on line $BC$. The internal bisectors of angles $BDJ$ and $JDC$ intersectlines $BJ$ and $JC$ at $X$ and $Y$, respectively. Segments $XY$ and $JD$ intersect at $P$. Let $Q$ be the projection of $A$ on line $BC$. Prove that the internal angle bisector of $QAP$ is perpendicular to line $XY$.

Proposed by Dominik Burek, Poland
9 replies
a_507_bc
Aug 25, 2023
lksb
2 hours ago
Easy functional equation
Sadigly   1
N 2 hours ago by jasperE3
Let $\alpha\neq0$ be a real number. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$f(f(x+y))=f(x+y)+f(x)f(y)+\alpha xy$$for all $x;y\in\mathbb{R}$
1 reply
Sadigly
Yesterday at 7:03 PM
jasperE3
2 hours ago
2013 China Second Round Olympiad (C) Test 2 Q3
sqing   2
N 2 hours ago by Jjesus
Source: 13 Oct 2013
The integers $n>1$ is given . The positive integer $a_1,a_2,\cdots,a_n$ satisfing condition :
(1) $a_1<a_2<\cdots<a_n$;
(2) $\frac{a^2_1+a^2_2}{2},\frac{a^2_2+a^2_3}{2},\cdots,\frac{a^2_{n-1}+a^2_n}{2}$ are all perfect squares .
Prove that :$a_n\ge 2n^2-1.$
2 replies
sqing
Oct 15, 2013
Jjesus
2 hours ago
10 but not 11 Consecutive Divisors
codyj   2
N 3 hours ago by Pippex23
Source: OMM 2007 1
Find all integers $N$ with the following property: for $10$ but not $11$ consecutive positive integers, each one is a divisor of $N$.
2 replies
codyj
Jul 19, 2014
Pippex23
3 hours ago
i love mordell
MR.1   7
N 3 hours ago by iniffur
Source: own
find all pairs of $(m,n)$ such that $n^2-79=m^3$
7 replies
MR.1
Apr 10, 2025
iniffur
3 hours ago
Right angles
USJL   10
N 4 hours ago by bin_sherlo
Source: 2018 Taiwan TST Round 3
Let $I$ be the incenter of triangle $ABC$, and $\ell$ be the perpendicular bisector of $AI$. Suppose that $P$ is on the circumcircle of triangle $ABC$, and line $AP$ and $\ell$ intersect at point $Q$. Point $R$ is on $\ell$ such that $\angle IPR = 90^{\circ}$.Suppose that line $IQ$ and the midsegment of $ABC$ that is parallel to $BC$ intersect at $M$. Show that $\angle AMR = 90^{\circ}$

(Note: In a triangle, a line connecting two midpoints is called a midsegment.)
10 replies
USJL
Apr 2, 2020
bin_sherlo
4 hours ago
sequence positive
malinger   35
N 4 hours ago by Bonime
Source: ISL 2006, A2, VAIMO 2007, P4, Poland 2007
The sequence of real numbers $a_0,a_1,a_2,\ldots$ is defined recursively by \[a_0=-1,\qquad\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}=0\quad\text{for}\quad n\geq 1.\]Show that $ a_{n} > 0$ for all $ n\geq 1$.

Proposed by Mariusz Skalba, Poland
35 replies
malinger
Apr 22, 2007
Bonime
4 hours ago
incircle excenter midpoints
danepale   8
N Jan 10, 2025 by Om245
Source: Middle European Mathematical Olympiad T-6
Let the incircle $k$ of the triangle $ABC$ touch its side $BC$ at $D$. Let the line $AD$ intersect $k$ at $L \neq D$ and denote the excentre of $ABC$ opposite to $A$ by $K$. Let $M$ and $N$ be the midpoints of $BC$ and $KM$ respectively.

Prove that the points $B, C, N,$ and $L$ are concyclic.
8 replies
danepale
Sep 21, 2014
Om245
Jan 10, 2025
incircle excenter midpoints
G H J
Source: Middle European Mathematical Olympiad T-6
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danepale
99 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let the incircle $k$ of the triangle $ABC$ touch its side $BC$ at $D$. Let the line $AD$ intersect $k$ at $L \neq D$ and denote the excentre of $ABC$ opposite to $A$ by $K$. Let $M$ and $N$ be the midpoints of $BC$ and $KM$ respectively.

Prove that the points $B, C, N,$ and $L$ are concyclic.
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wiseman
216 posts
#2 • 3 Y
Y by lazizbek42, Adventure10, Mango247
$\rightarrow I_a =$The $A-$excenter of $\triangle{ABC}$.
$\rightarrow \ell =$The line tangent to $k(I,r)$ at $L$.
$\rightarrow (AC,k(I,r))=E$ , $(AB,k(I,r))=F$.
$\rightarrow (BC,EF,\ell)=H$.
$\rightarrow (I_aM,LH)=K$.
$\rightarrow IQ \perp AD$ , $Q \in AD$.
$\rightarrow I_aD' \perp BC$ , $D' \in BC$.
$\rightarrow$ The facts that $BC,EF,\ell$ are concurrent and $I_aM \parallel AD$ and $HB.HC=HD.HM$are well-known, so I don't post a proof for them unless needed.
$\rightarrow HD=HL$ $,$ $DL \parallel MK \Rightarrow HL.HK=HD.HM=HB.HC \Rightarrow BLKC$ is cyclic.
Now the only thing we have to prove is that $KM.MN=BM.CM$.
$\rightarrow MI_a \parallel AD \Rightarrow \triangle{IQD} \sim \triangle{MD'I_a} \Rightarrow \frac{I_aM}{r}=\frac{2r_a}{DL}=\frac{2r_a.HM}{MK.HD}$

$\Rightarrow \frac{MK.I_aM}{2}= \frac{r.r_a.HM}{HD} = \frac{BD.CD.HM}{HD} = \frac{BD.BM.HC}{HD} = BM.CM \blacksquare$(Note that in the two last conclusions, we used the fact that $HB.HC=HD.HM$.)
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TelvCohl
2312 posts
#3 • 4 Y
Y by DanDumitrescu, Adventure10, Mango247, and 1 other user
My solution:

Let $ D', D'' $ be the reflection of $ D, D' $ with respect to $ M, K $.
Let $ X $ be the intersection of the polar of $ A $ with respect to $ (I) $ and $ BC $ .
Let $ N' $ be the reflection of $ N $ with respect to the perpendicular bisector of $ BC $ .

By homothety with center $ A $ we get $ A, L, D, D'' $ are collinear.

Since it's well known that $ KD'\perp BC $ ,
so we get $ N' $ lie on $ AD'' $ .
Since $ AD $ is the polar of $ X $ with respect to $ (I) $ ,
so we get $ XL $ is tangent to $ (I) $ and $ (X,D;B,C)=-1 $ .
Since $ XL=XD $ and $ N'M=N'D $ ,
so $ \angle XLD =\angle LDX =\angle N'DM =\angle DMN' $ .
i.e. $ L,X,M,N' $ are concyclic
Since $ DL \cdot DN'=DX \cdot DM=DB \cdot DC $ ,
so we get $ B, C, N, N', L $ are concyclic.

Q.E.D
This post has been edited 1 time. Last edited by TelvCohl, May 6, 2015, 8:40 PM
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Pinionrzek
54 posts
#4 • 3 Y
Y by atharvatekawade, ILOVEMYFAMILY, Adventure10
Let $E$, $F$ be points of tangency of the incircle with $AC$ and $AB$. It is easy to show that $MK \parallel AD$. Besides $FD \parallel BK$. Hence we get $\angle KBM= \angle BDF = \angle FLD$ and $ \angle FDL = \angle BKM$, thus $\Delta FDL \sim \Delta BKM$. Notice that $BL$ is a symmedian of the triangle $FDL$ so $\angle BLF= \angle GLD= \angle KBN$, where $G$ is a midpoint of $FD$. From that we get that $\angle NBM= \angle BLD$. Analogously $\angle DLC=\angle NCM$, thus $B, N, C, L$ are concyclic.
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PatrikP
18 posts
#5 • 2 Y
Y by Adventure10, Mango247
Pinionrzek wrote:
It is easy to show that $MK \parallel AD$.

Can someone show this?
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JackXD
151 posts
#6 • 3 Y
Y by PatrikP, Adventure10, Mango247
PatrikP wrote:
Pinionrzek wrote:
It is easy to show that $MK \parallel AD$.

Can someone show this?

Note that the homothety mapping the incircle to the excircle maps $D$ to the antipode of $E$.This with $DM=EM$ makes it clear.
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anantmudgal09
1979 posts
#7 • 2 Y
Y by Adventure10, Mango247
Let $\ell$ be the perpendicular bisector of $BC$. Define points $K'$ and $N'$ as the images of $K$ and $N$, respectively in the symmetry about $\ell$. Let $T$ be a point on line $BC$ such that $(B, C; D, T)=-1$ and note that $TL$ is tangent to $\omega$.

It is well-known that $B, I, C, K$ are concyclic, so $B, I, C, K'$ are also concyclic. By power of point we get $$DI\cdot DK'=DB\cdot DC=DM\cdot DT,$$so $T, I, M, K'$ are concyclic. Thus, $$\angle (AD, BC)=\angle TDL=\angle TID=\angle TMK'=\angle N'DM,$$so $A, D, N'$ are collinear.

Finally, observe that $\angle TLN'=\angle TMN'$, so $T, L, M, N'$ are concyclic. By power of point, $$DB\cdot DC=DT\cdot DM=DL\cdot DN',$$so $B, L, C, N'$ are concyclic. The result follows as $BCNN'$ is an isosceles trapezoid. $\square$
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lazizbek42
548 posts
#8
Y by
$$AD||MK$$triangle $FQD$ similar triangle $BMK$.
$$ \angle BQD=\angle NBM $$Similarity:$$\angle CQD=\angle NCM$$$B,Q,C,N$ concyclic.
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Om245
163 posts
#9 • 1 Y
Y by Captainscrubz
Same solution as #1 but bit more motivated

Claim 1: $MK \parallel AD$
Let $A$-excircle touch $BC$ at $D'$ and $R$ be it's antipode in excircle. Then by homothety at $A$ sending incircle to excircle sends $D$ to $R$ hence $\overline{A-D-R}$. $M$ is midpoint of $DD'$ thus by MPT in $\triangle DD'R$ we get $MK \parallel AD$.

Let $S$ be reflection of $M$ above $TI$
Claim 2: $S \in (LBC)$
Let $E,F$ be other touch point of incircle with $AC,AB$ then $T =EF \cap BC$. $A$ lie on polar of $T$, therefore $TL$ tangent to incircle.
Note we have $MK \perp TI$. Perpendicular bisector of $MS$ and $DL$ is $TI$ only. This give us $L,S,M,D$ cyclic.
As $(T,D;B,C)=-1$ by using power of point we get $$TB \cdot TC =TM \cdot TD = TL \cdot TS$$Which give us $S \in (LBC)$.


Now the motivation for introducing $S$ come in picture. We claim $N$ lie on $(SLBC)$.
Note $\angle IXK = 90$ hence $X \in (IBKC)$. We use power of point at $M$
$$MB\cdot MC = MX \cdot MK = 2\cdot MX \cdot \frac{MK}{2}= MS \cdot MN$$which yields $N \in (SBC)$. Hence $L,B,C,N$ cyclic.
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