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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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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
Iran 3rd Round Geo
M11100111001Y1R   12
N 2 minutes ago by Mathgloggers
Source: Iran 2024 3rd Round Test 1 P3
Consider an acute scalene triangle $\triangle{ABC}$. The interior bisector of $A$ intersects $BC$ at $E$ and the minor arc of $\overarc {BC}$ in circumcircle of $\triangle{ABC}$ at $M$. Suppose that $D$ is a point on the minor arc of $\overarc {BC}$ such that $ED=EM$. $P$ is a point on the line segment of $AD$ such that $\angle ABP=\angle ACP \not= 0$. $O$ is the circumcenter of $\triangle{ABC}$. Prove that $OP \perp AM$.
12 replies
+2 w
M11100111001Y1R
Aug 23, 2024
Mathgloggers
2 minutes ago
find question
mathematical-forest   0
8 minutes ago
Are there any contest questions that seem simple but are actually difficult? :-D
0 replies
mathematical-forest
8 minutes ago
0 replies
Inspired by qrxz17
sqing   2
N 33 minutes ago by MathsII-enjoy
Source: Own
Let $a, b,c>0 ,(a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 27 $. Prove that $$a+b+c\geq 3\sqrt {3}$$
2 replies
sqing
2 hours ago
MathsII-enjoy
33 minutes ago
D,E,F are collinear.
TUAN2k8   1
N 34 minutes ago by Beelzebub
Source: Own
Help me with this:
1 reply
TUAN2k8
Yesterday at 1:07 AM
Beelzebub
34 minutes ago
Inspired by qrxz17
sqing   2
N an hour ago by MathsII-enjoy
Source: Own
Let $ a,b,c $ be reals such that $ (a^2+b^2)^2 + (b^2+c^2)^2 +(c^2+a^2)^2 = 28 $ and $  (a^2+b^2+c^2)^2 =16. $ Find the value of $ a^2(a^2-1) + b^2(b^2-1)+c^2(c^2-1).$
2 replies
1 viewing
sqing
2 hours ago
MathsII-enjoy
an hour ago
Prove DK and BC are perpendicular.
yunxiu   63
N an hour ago by sknsdkvnkdvf
Source: 2012 European Girls’ Mathematical Olympiad P1
Let $ABC$ be a triangle with circumcentre $O$. The points $D,E,F$ lie in the interiors of the sides $BC,CA,AB$ respectively, such that $DE$ is perpendicular to $CO$ and $DF$ is perpendicular to $BO$. (By interior we mean, for example, that the point $D$ lies on the line $BC$ and $D$ is between $B$ and $C$ on that line.)
Let $K$ be the circumcentre of triangle $AFE$. Prove that the lines $DK$ and $BC$ are perpendicular.

Netherlands (Merlijn Staps)
63 replies
yunxiu
Apr 13, 2012
sknsdkvnkdvf
an hour ago
Geometry with fix circle
falantrng   34
N an hour ago by Aiden-1089
Source: RMM 2018 Problem 6
Fix a circle $\Gamma$, a line $\ell$ to tangent $\Gamma$, and another circle $\Omega$ disjoint from $\ell$ such that $\Gamma$ and $\Omega$ lie on opposite sides of $\ell$. The tangents to $\Gamma$ from a variable point $X$ on $\Omega$ meet $\ell$ at $Y$ and $Z$. Prove that, as $X$ varies over $\Omega$, the circumcircle of $XYZ$ is tangent to two fixed circles.
34 replies
falantrng
Feb 25, 2018
Aiden-1089
an hour ago
Set of perfect powers is irreducible
Assassino9931   2
N an hour ago by navid
Source: Al-Khwarizmi International Junior Olympiad 2025 P4
For two sets of integers $X$ and $Y$ we define $X\cdot Y$ as the set of all products of an element of $X$ and an element of $Y$. For example, if $X=\{1, 2, 4\}$ and $Y=\{3, 4, 6\}$ then $X\cdot Y=\{3, 4, 6, 8, 12, 16, 24\}.$ We call a set $S$ of positive integers good if there do not exist sets $A,B$ of positive integers, each with at least two elements and such that the sets $A\cdot B$ and $S$ are the same. Prove that the set of perfect powers greater than or equal to $2025$ is good.

(In any of the sets $A$, $B$, $A\cdot B$ no two elements are equal, but any two or three of these sets may have common elements. A perfect power is an integer of the form $n^k$, where $n>1$ and $k > 1$ are integers.)

Lajos Hajdu and Andras Sarkozy, Hungary
2 replies
Assassino9931
May 9, 2025
navid
an hour ago
Stop Projecting your insecurities
naman12   53
N an hour ago by EeEeRUT
Source: 2022 USA TST #2
Let $ABC$ be an acute triangle. Let $M$ be the midpoint of side $BC$, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$, respectively. Suppose that the common external tangents to the circumcircles of triangles $BME$ and $CMF$ intersect at a point $K$, and that $K$ lies on the circumcircle of $ABC$. Prove that line $AK$ is perpendicular to line $BC$.

Kevin Cong
53 replies
naman12
Dec 12, 2022
EeEeRUT
an hour ago
Roots of unity
Henryfamz   1
N an hour ago by Mathzeus1024
Compute $$\sec^4\frac\pi7+\sec^4\frac{2\pi}7+\sec^4\frac{3\pi}7$$
1 reply
Henryfamz
May 13, 2025
Mathzeus1024
an hour ago
Shortest number theory you might've seen in your life
AlperenINAN   11
N an hour ago by Assassino9931
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)+1$ is a perfect square, then $pq + 1$ is also a perfect square.
11 replies
AlperenINAN
May 11, 2025
Assassino9931
an hour ago
AZE JBMO TST
IstekOlympiadTeam   10
N an hour ago by Assassino9931
Source: AZE JBMO TST
Prove that there are not intgers $a$ and $b$ with conditions,
i) $16a-9b$ is a prime number.
ii) $ab$ is a perfect square.
iii) $a+b$ is also perfect square.
10 replies
IstekOlympiadTeam
May 2, 2015
Assassino9931
an hour ago
Iran TST Starter
M11100111001Y1R   3
N an hour ago by dgrozev
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[
a_n < \frac{1}{1404}
\]
3 replies
M11100111001Y1R
May 27, 2025
dgrozev
an hour ago
An interesting functional equation
giannis2006   3
N an hour ago by GreekIdiot
Source: Own
Find all functions $f:R^+->R^+$ such that:
$f(xf(y))=xy-xf(x)+f(x)^2$ for all $x,y>0$

The most difficult version of this problem is the following:
Find all functions $f:R^+->R^+$ such that:
$f(xf(y+f(x)))=xy+f(x)^2$ for all $x,y>0$
3 replies
giannis2006
Jun 8, 2023
GreekIdiot
an hour ago
external bisector in 2 angle
crocodilepradita   6
N Apr 27, 2025 by deduck
Given a $\triangle ABC$ with incenter $I$. Line $BI$ and $CI$ intersects $CA$ and $AB$ at $E$ and $F$, respectively. Let $M$ and $N$ be the midpoints of $BI$ and $CI$. Line $FM$ meets the external bisector of angle $B$ at $K$ and line $EN$ meets the external bisector of angle $C$ at $L$. Prove that $K, B, L, C$ are concylic.
6 replies
crocodilepradita
Aug 22, 2024
deduck
Apr 27, 2025
external bisector in 2 angle
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crocodilepradita
145 posts
#1 • 1 Y
Y by Rounak_iitr
Given a $\triangle ABC$ with incenter $I$. Line $BI$ and $CI$ intersects $CA$ and $AB$ at $E$ and $F$, respectively. Let $M$ and $N$ be the midpoints of $BI$ and $CI$. Line $FM$ meets the external bisector of angle $B$ at $K$ and line $EN$ meets the external bisector of angle $C$ at $L$. Prove that $K, B, L, C$ are concylic.
This post has been edited 1 time. Last edited by crocodilepradita, Aug 22, 2024, 5:51 AM
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SBYT
196 posts
#2 • 1 Y
Y by ehuseyinyigit
We will prove that $CK\perp BJ$ at $K$(Then similarly $BL\perp CJ$ at $L$,we'v done)

Denote the $B-escenter$ by $T$ and the midpoint of $\stackrel\frown{AB},\stackrel\frown{AC}$ are $H,G$,respectively.
Then $DH,DG$ are perpendicular bisector of $BI,CI$,respectively.

We can know that $FM,AD,CG$ are concurrent by Pascal on $DHCGBA$.Denote the point by $S$.

Due to $\frac{BM}{BI}=\frac{1}{2}=\frac{TG}{TI}$,we can get $\frac{SK}{KM}=\frac{SC}{CG}$.
It means that $CK\parallel BI \Rightarrow CK\perp BJ$.$\Box$
Attachments:
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Khoi333
1 post
#3
Y by
BC cut FK at O, we have C(KMOF)=-1
So C(KMBI)=-1, but IM=MB
So IB || KC, so that BKC=90°
Like that, BLC=90°, end!
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crocodilepradita
145 posts
#4
Y by
Khoi333 wrote:
BC cut FK at O, we have C(KMOF)=-1
So C(KMBI)=-1, but IM=MB
So IB || KC, so that BKC=90°
Like that, BLC=90°, end!

After C(K,M;B,I)=-1, since IM=MB, how do u prove BKC=90
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cursed_tangent1434
649 posts
#5
Y by
Let $I_a,I_b,I_c$ denote the $A-$ , $B-$ and $C-$excenters respectively. Denote by $M_a$ , $M_b$ and $M_c$ the midpoints of minor arcs $BC$ , $AC$ and $AB$ respectively. By the Incenter-Excenter Lemma $M_aI=M_aB$ and $M_cI=M_cB$ implying that $M_aM_c$ is the perpendicular bisector of segment $BI$. Thus, $M_aM_c$ passes through $M$ and is perpendicular to $BI$. Similarly, $M_aM_b$ passes through $N$ and is perpendicular to $CI$. We can now show the key claim.

Claim : Lines $\overline{KL}$ and $\overline{I_bI_c}$ are parallel.

Proof : This is a simple calculation. Note,
\[\frac{KI_c}{M_cM} = \frac{FI_c}{FM_c} \text{ and } \frac{LI_b}{M_bN} = \frac{EI_b}{EM_b}\]from which we have
\[\frac{KI_c}{LI_b} = \frac{FI_c}{FM_c} \cdot \frac{EM_b}{EI_b} \cdot \frac{MM_c}{NM_b}\]Now, since $I_cBIA$ is cyclic we have that
\[FI \cdot FI_c = FA \cdot FB = FM_c \cdot FC\]from whence we have
\[\frac{FI_c}{FM_c} = \frac{FC}{FI}\]and similarly,
\[\frac{EM_b}{EI_c} = \frac{EI}{EB}\]Thus,
\[\frac{KI_c}{LI_b} = \frac{FC}{FI} \cdot \frac{EI}{EB} \cdot \frac{MM_c}{NM_b} \]Now, by the Law of Sines we have
\[\frac{FC}{FB} = \frac{\sin B}{ \sin \frac{C}{2}} \text{ and } \frac{FB}{FI} = \frac{\sin (90 - \frac{A}{2})}{\sin \frac{B}{2}} \]from which it follows that
\[\frac{FC}{FI} = \frac{\sin B \sin (90 - \frac{A}{2})}{ \sin \frac{B}{2} \sin \frac{C}{2}}\]and similarly,
\[\frac{EB}{EI} = \frac{\sin C \sin (90 - \frac{A}{2})}{ \sin \frac{C}{2} \sin \frac{B}{2}}\]Thus,
\[\frac{KI_c}{LI_b} = \frac{\sin B \sin (90 - \frac{A}{2})}{ \sin \frac{B}{2} \sin \frac{C}{2}}\cdot  \frac{ \sin \frac{C}{2} \sin \frac{B}{2}}{\sin C \sin (90 - \frac{A}{2})}\cdot \frac{\sin \frac{C}{2}}{ \sin \frac{B}{2}}  = \frac{\sin B \sin \frac{C}{2}}{\sin C \sin \frac{B}{2}} = \frac{\cos \frac{B}{2}}{\cos \frac{C}{2}}\]Further, since $I_bI_cBC$ is cyclic,
\[\frac{I_aI_c}{I_aI_b} = \frac{I_aC}{I_aB} = \frac{\sin \angle CBI_a}{ \sin \angle BCI_a} = \frac{\sin (90 - \frac{B}{2})}{\sin (90 - \frac{C}{2})} = \frac{\cos \frac{B}{2}}{\cos \frac{C}{2}}\]Thus,
\[\frac{KI_c}{LI_b} = \frac{I_aI_c}{I_aI_b}\]which implies that $KL \parallel I_bI_c$ as desired.

Now the finish is easy. Note that since $I_bI_c \perp AI$, $KL \perp AI$ as well. Hence,
\[\measuredangle LKI_a = 90 + \measuredangle II_aB = 90 + \measuredangle ICB = \measuredangle BCL\]which implies that $BCLK$ is indeed cyclic, as claimed.
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ihategeo_1969
242 posts
#6
Y by
Nice and easy. Define $\triangle I_AI_BI_C$ to be the ex-central triangle. See that $\overline{MI} \parallel \overline{KC}$ and similarly $\overline{NI} \parallel \overline{LB}$ because $(B,I;M,\overline{MI} \cap \overline{KC}) \overset K= (I_C,I;F,C)=-1$, and similar for other. Now it is just angle chase \begin{align*}
\measuredangle BKC &= \measuredangle BKM+\measuredangle MKC 
=\measuredangle BKM+\measuredangle FMI 
= 90^\circ+\measuredangle BMK+\measuredangle FMI 
= 90^\circ+\measuredangle IMF+\measuredangle FMI=90^\circ
\end{align*}Similarly $\measuredangle BLC=90^{\circ}$ and done. (200th post on HSO lessgo)
This post has been edited 1 time. Last edited by ihategeo_1969, Apr 27, 2025, 5:55 AM
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deduck
237 posts
#7
Y by
yesterday i somehow found this in my tabs without remembering ever seeing the problem or opening it how did that happen lol!

solution:
Let $FK \cap BC = X$. Note that by apolonius circle lemma for first equal sign and project through $C$ for second equal sign, $-1 = (FX;MK)= (IB; M(CK \cap BE))$. But since $M$ is midpoint of $IB$ then $CK \cap BE = \infty$ so they're parallel, so $CK \perp BK$. Similarly $BL \perp CL$ so it's cyclic.

motivation:
The first few things that I noticed was that the angles at $B$ and $C$ were sus and looked like apolonius circle lemma angles so that is why i defined $X$ as intersection of $FK$ and $BC$. Also on my paper it looks like $CK \perp BK$ and $BL \perp CL$ because I drew with ruler, and since that proves concyclic, it would be nice to show that's true, and it's connected to $BE || CK$, which is nice because harmonic sometimes gives parallelisms easily. After using apolonius circle lemma to get $(FX;MK) = -1$ i got stuck for a while and did not understand how to move the points to $(IB;M(CK \cap BE))$ but then i saw that u can project through $C$. so that finishes the problem because the intersection of the 2 lines are at infinity!!!
This post has been edited 1 time. Last edited by deduck, Apr 27, 2025, 6:57 PM
Reason: mixed up \cap and \perp
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