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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
Isosceles Triangle Geo
oVlad   5
N 2 minutes ago by Tamam
Source: Romania Junior TST 2025 Day 1 P2
Consider the isosceles triangle $ABC$ with $\angle A>90^\circ$ and the circle $\omega$ of radius $AC$ centered at $A.$ Let $M$ be the midpoint of $AC.$ The line $BM$ intersects $\omega$ a second time at $D.$ Let $E$ be a point on $\omega$ such that $BE\perp AC.$ Let $N$ be the intersection of $DE$ and $AC.$ Prove that $AN=2\cdot AB.$
5 replies
oVlad
Apr 12, 2025
Tamam
2 minutes ago
combinatorics problem
henderson   6
N 7 minutes ago by aokmh3n2i2rt
$65$ distinct natural numbers not exceeding $2016$ are given. Prove that among these numbers we can find four $a,b,c,d$ such that $a+b-c-d$ is divisible by $2016.$
6 replies
henderson
Jan 23, 2016
aokmh3n2i2rt
7 minutes ago
Easy Geometry
rkm0959   13
N 9 minutes ago by Rounak_iitr
Source: 2015 Korean Mathematical Olympiad P2
Let the circumcircle of $\triangle ABC$ be $\omega$. A point $D$ lies on segment $BC$, and $E$ lies on segment $AD$. Let ray $AD \cap \omega = F$. A point $M$, which lies on $\omega$, bisects $AF$ and it is on the other side of $C$ with respect to $AF$. Ray $ME \cap \omega = G$, ray $GD \cap \omega = H$, and $MH \cap AD = K$. Prove that $B, E, C, K$ are cyclic.
13 replies
rkm0959
Nov 1, 2015
Rounak_iitr
9 minutes ago
help me~~
Imshyso   0
14 minutes ago
Given triangle ABC inscribed in (O) with orthocenter H. Let K be the midpoint of AH. Take E,F on AC, AB so that BKE=CKF=90. Prove that E,O,F are collinear.
0 replies
+1 w
Imshyso
14 minutes ago
0 replies
Basic ideas in junior diophantine equations
Maths_VC   5
N 28 minutes ago by Royal_mhyasd
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
5 replies
Maths_VC
May 27, 2025
Royal_mhyasd
28 minutes ago
Find values for a, b ,c
Ferum_2710   2
N 29 minutes ago by Jupiterballs
Source: Romania JBMO tst 2023 day2 p4
Let $M \geq 1$ be a real number. Determine all natural numbers $n$ for which there exist distinct natural numbers $a$, $b$, $c > M$, such that
$n = (a,b) \cdot (b,c) + (b,c) \cdot (c,a) + (c,a) \cdot (a,b)$
(where $(x,y)$ denotes the greatest common divisor of natural numbers $x$ and $y$).
2 replies
Ferum_2710
Apr 30, 2023
Jupiterballs
29 minutes ago
Center lies on circumcircle of other
Philomath_314   41
N 41 minutes ago by Adywastaken
Source: INMO P1
In triangle $ABC$ with $CA=CB$, point $E$ lies on the circumcircle of $ABC$ such that $\angle ECB=90^{\circ}$. The line through $E$ parallel to $CB$ intersects $CA$ in $F$ and $AB$ in $G$. Prove that the center of the circumcircle of triangle $EGB$ lies on the circumcircle of triangle $ECF$.

Proposed by Prithwijit De
41 replies
Philomath_314
Jan 21, 2024
Adywastaken
41 minutes ago
find question
mathematical-forest   7
N an hour ago by whwlqkd
Are there any contest questions that seem simple but are actually difficult? :-D
7 replies
mathematical-forest
Thursday at 10:19 AM
whwlqkd
an hour ago
Inspired by a cool result
DoThinh2001   1
N an hour ago by arqady
Source: Old?
Let three real numbers $a,b,c\geq 0$, no two of which are $0$. Prove that:
$$\sqrt{\frac{a^2+bc}{b^2+c^2}}+\sqrt{\frac{b^2+ca}{c^2+a^2}}+\sqrt{\frac{c^2+ab}{a^2+b^2}}\geq 2+\sqrt{\frac{ab+bc+ca}{a^2+b^2+c^2}}.$$
Inspiration
1 reply
DoThinh2001
Today at 12:08 AM
arqady
an hour ago
Crossing ٍٍChords
matinyousefi   1
N an hour ago by Trenod
Source: Iranian Combinatorics Olympiad 2020 P3
$1399$ points and some chords between them is given.
$a)$ In every step we can take two chords $RS,PQ$ with a common point other than $P,Q,R,S$ and erase exactly one of $RS,PQ$ and draw $PS,PR,QS,QR$ let $s$ be the minimum of chords after some steps. Find the maximum of $s$ over all initial positions.
$b)$ In every step we can take two chords $RS,PQ$ with a common point other than $P,Q,R,S$ and erase both of $RS,PQ$ and draw $PS,PR,QS,QR$ let $s$ be the minimum of chords after some steps. Find the maximum of $s$ over all initial positions.

Proposed by Afrouz Jabalameli, Abolfazl Asadi
1 reply
matinyousefi
Apr 24, 2020
Trenod
an hour ago
Nice NT with powers of two
oVlad   7
N 2 hours ago by SimplisticFormulas
Source: Romania TST 2024 Day 1 P3
Let $n{}$ be a positive integer and let $a{}$ and $b{}$ be positive integers congruent to 1 modulo 4. Prove that there exists a positive integer $k{}$ such that at least one of the numbers $a^k-b$ and $b^k-a$ is divisible by $2^n.$

Cătălin Liviu Gherghe
7 replies
oVlad
Jul 31, 2024
SimplisticFormulas
2 hours ago
Inequality in triangle
Nguyenhuyen_AG   0
2 hours ago
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]
0 replies
Nguyenhuyen_AG
2 hours ago
0 replies
D,E,F are collinear.
TUAN2k8   2
N 2 hours ago by TUAN2k8
Source: Own
Help me with this:
2 replies
TUAN2k8
May 28, 2025
TUAN2k8
2 hours ago
Combinatorial identity
MehdiGolafshan   4
N 2 hours ago by watery
Let $n$ is a positive integer. Prove that
$$\sum_{k=0}^{n-1}\frac{1}{k+1}\binom{n-1}{k} = \frac{2^n-1}{n}.$$
4 replies
MehdiGolafshan
Jan 16, 2023
watery
2 hours ago
An equation from the past with different coefficients
Assassino9931   13
N Apr 27, 2025 by grupyorum
Source: Balkan MO Shortlist 2024 N2
Let $n$ be an integer. Prove that $n^4 - 12n^2 + 144$ is not a perfect cube of an integer.
13 replies
Assassino9931
Apr 27, 2025
grupyorum
Apr 27, 2025
An equation from the past with different coefficients
G H J
Source: Balkan MO Shortlist 2024 N2
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Assassino9931
1381 posts
#1 • 1 Y
Y by PikaPika999
Let $n$ be an integer. Prove that $n^4 - 12n^2 + 144$ is not a perfect cube of an integer.
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GreekIdiot
276 posts
#2 • 1 Y
Y by PikaPika999
Similar to HMO 2025 p4.
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Assassino9931
1381 posts
#3 • 1 Y
Y by PikaPika999
@above Indeed, the technique from there is useful (and popular), but the very essential similarity is with this JBMO Shortlist 2020 problem.
This post has been edited 1 time. Last edited by Assassino9931, Apr 27, 2025, 1:05 PM
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Sadigly
229 posts
#4
Y by
Also AZE BMO TST P3
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MR.1
138 posts
#5
Y by
GreekIdiot wrote:
Similar to HMO 2025 p4.

what is hmo?
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Sadigly
229 posts
#6
Y by
MR.1 wrote:
GreekIdiot wrote:
Similar to HMO 2025 p4.

what is hmo?

Hellenic Math Olympiad
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MuradSafarli
112 posts
#7
Y by
Similar to Greece Math olympiad 2025
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Tamam
22 posts
#8 • 1 Y
Y by Lahmacuncu
$n^4-12n^2+144=(n^2-6n+12)(n^2+6n+12)$ from Euclidian Algorithm we can deduce that $gcd(n^2-6n+12,n^2+6n+12)=(n,12)$
If $n$ is even let $n=2k$ we will get that $16(k^4-3k^2+9)=m^3$ from here we know $v_2$ of $LHS$ is 4 which is not divisible by 3
If $n$ is divisible by 3 let $n=3k$ we get $9(9k^4-12k^2+1)=m^3$ since m is divisivible by 3 $LHS$ should be divisible by 27 which yields a contradiction.
So both the terms in the product are perfect cubes.
Now let $a=n-3$ which is even. Let $(n-3)^2+3=n^2-6n+12=b^3$. So $a^2+4=b^3+1=(b+1)(b^2-b+1)$ Since $b\equiv 3 (mod 4)$ there exists a prime congurent to 3 modulo 4 such that $p|b^2-b+1|a^2+2^2$ but from this we get $p=2$ which is a contradiction. So no solution.
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MuradSafarli
112 posts
#9
Y by
My solution in the tst
Attachments:
BMO 2024 SL N2.pdf (76kb)
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dangerousliri
932 posts
#10 • 3 Y
Y by MuradSafarli, farhad.fritl, ehuseyinyigit
This problem was proposed by me (Dorlir Ahmeti, Albania).
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megarnie
5611 posts
#11
Y by
Suppose the expression was a cube for some $n$. Factor as $(n^2 - 6n + 12)(n^2 + 6n + 12)$.

Claim: $n^2 - 6n + 12 $ and $n^2 + 6n + 12$ are relatively prime.
Proof: Suppose otherwise. Firstly, if $p$ divides both $n^2 - 6n + 12$ and $n^2 + 6n + 12$, then $p$ divides $12n$, and since $p$ divides $12(n^2 - 6n +12)$, $p$ divides $144$, so $p$ is $2$ or $3$. But if $p = 3$, then $3 \mid n$, but this means $n^2 - 6n + 12$ and $n^2 + 6n + 12$ are both $12 \equiv 3 \pmod 9$, so their product has a $\nu_3$ of $2$, so it is not a perfect cube. Thus, $\gcd(n^2 - 6n + 12, n^2 + 6n + 12)$ is a power of $2$. Since the gcd is not $1$, it is even, so $2$ divides $n$. Since $n$ and $n - 6$ are even and different mod $4$, $8 \mid n(n-6) = n^2 - 6n$, so since $8 \mid 12n$, $8$ also divides $n^2 - 6n + (12n) = n^2 + 6n$. Thus, $n^2 - 6n + 12$ and $n^2 + 6n + 12$ are $12 \equiv 4 \pmod 8$, so their product has a $\nu_2$ of $4$, so it is not a perfect cube. $\square$

Thus, $n^2 - 6n + 12$ and $n^2 + 6n + 12$ are perfect cubes. Taking modulo $7$ gives that $n^2 - n + 5$ and $n^2 + n + 5$ are both in $\{-1,0,1\}$ modulo $7$. Multiplying by $4$ and taking mod $7$ gives $(2n-1)^2  - 2$ and $(2n+1)^2 - 2$ are in $\{-4,0,4\}\pmod 7$.

Hence $(2n-1)^2$ and $(2n+1)^2$ are in $\{-2,2,6\}\pmod 7$, and since $-2$ and $6$ are not QRs modulo $7$, they are both $2$ modulo $7$.

Thus, $(2n+1)^2 - (2n-1)^2 = 8n \equiv 0 \pmod 7$, so $7 \mid n$, but then $(2n-1)^2 \equiv 1 \pmod 7$, absurd. Therefore, no such integer $n$ exists.
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eg4334
636 posts
#12
Y by
First, factor it as $(n^2-6n+12)(n^2+6n+12)$. The only prime factors they can share are $2$ or $3$, because they must divide $12n$. If they are even, then $n=2a$ so we need $16(a^4-3a^2+9)$ to be a cube which is impossible because the parenthesses is odd. Similarly, if they are divisible by $3$ then $n=3a$ but $9(9a^4-12a^2+16)$ cannot be a cube by $v_3$ analysis. Therefore for the sake of our problem they are relatively pirme. Then let $a^3=n^2-6n+12, (a+3x)^3 = n^2+6n+12$, because their difference must be a multiple of $3$. Then $12n=9a^2x+27ax^2+27x^3 \implies n = \frac{3a^2x+9ax^2+9x^3}{4}$. Put this into the latter equation to get $16(a+3x)^3 = (3a^2x+9ax^2+9x^3)^2 + 24(3a^2x+9ax^2+9x^3)+192$. Its easy to see that when $x \geq 2$, this is impossible by size constraints because the right side is so big. When $x=1$, we have $12n = 9a^2+27a+27$ which is impossible by $\pmod{4}$. Therefore there is no such $n$,.
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ektorasmiliotis
110 posts
#13
Y by
in HMO: x^2 +108= is not a perfect cube
This post has been edited 1 time. Last edited by ektorasmiliotis, Apr 27, 2025, 7:48 PM
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grupyorum
1435 posts
#14
Y by
Set $n^4-12n^2+144=k^3$. First, I prove $k\equiv 0,3\pmod{4}$ are impossible. Suppose $4\mid k$ and $k=4u$. Notice that $v_2(LHS)=4$ for $v_2(n)\ge 2$, so $v_2(n)=1$ necessarily. Setting $n=2t$ for $t$ odd, we find $16t^4 - 48t^2 + 144 = 64u^3\Rightarrow t^4-3t^2+9 = 4u^3$ which is impossible by parity. Likewise, if $k\equiv 3\pmod{4}$, then $n^4-12n^2+144\equiv n^4\equiv 3\pmod{4}$, a contradiction.

In the remainder, we will rule out $k\in\{1,2\}\pmod{4}$. Suppose first $k\equiv 1\pmod{4}$. Note that
\[
(n^2-6)^2 + 10^2 = k^3 - 8 =(k-2)(k^2+2k+4).
\]Since $k-2\equiv -1\pmod{4}$, there is a prime $q\equiv 3\pmod{4}$ such that $q\mid \alpha^2+\beta^2$ for $\alpha=n^2-6$ and $\beta = 10$. It is well known that $q\mid\alpha,\beta$ in this case, but $q\mid 10$ is clearly impossible.

Lastly, suppose $k\equiv 2\pmod{4}$. We have
\[
(n^2-6)^2 + 9^2 = k^3 - 27 = (k-3)(k^2+3k+9).
\]Once again, $k-3\equiv -1\pmod{4}$, so there is a prime $p\equiv 3\pmod{4}$ dividing both $n^2-6$ and $9$. At the same time, $3\mid n$ is impossible: otherwise $v_3(n^4)\ge 4$, $v_3(12n^2)\ge 3$ whereas $v_3(144)=2$ gives $v_3(LHS)=2$, not possible for a perfect cube.
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