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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.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. 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)!

Be sure to mark your calendars for the following events:
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April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]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
Coincide
giangtruong13   1
N 14 minutes ago by giangtruong13
Source: Hanoi Specialized School's Math Test (Round 2 - Phase 1)
Let $ABCD$ be a trapezoid inscribed in circle $(O)$, $AD||BC, AD < BC$. Let $P$ is the symmetric point of $A$ across $BC$, $AP$ intersects $BC$ at $K$. Let $M$ is midpoint of $BC$ and $H$ is orthocenter of triangle $ABC$. On $BD$ take a point $F$ so that $AF||HM$. Prove that: $ FK,AC,PD$ coincide
1 reply
giangtruong13
Yesterday at 4:05 PM
giangtruong13
14 minutes ago
Website to learn math
hawa   54
N 22 minutes ago by Samujjal101
Hi, I'm kinda curious what website do yall use to learn math, like i dont find any website thats fun to learn math
54 replies
hawa
Apr 9, 2025
Samujjal101
22 minutes ago
About my new website
Samujjal101   0
24 minutes ago
Hi everybody!
I'm registering some of the finest minds in math into my website.. it's not completely developed.. but still if you want we would be very grateful to have you!
Text to display
Maths-matchmaker is a website for connecting math minds together with a mission to unite together. It uses a matching algorithm to match 1:1 with like minded peers based on their interests or topics in math
0 replies
Samujjal101
24 minutes ago
0 replies
JBMO Shortlist 2021 G4
Lukaluce   3
N 28 minutes ago by S11_DooronbekovNurakhmed
Source: JBMO Shortlist 2021
Let $ABCD$ be a convex quadrilateral with $\angle B = \angle D = 90^{\circ}$. Let $E$ be the point of intersection of $BC$ with $AD$ and let $M$ be the midpoint of $AE$. On the extension of $CD$, beyond the point $D$, we pick a point $Z$ such that $MZ = \frac{AE}{2}$. Let $U$ and $V$ be the projections of $A$ and $E$ respectively on $BZ$. The circumcircle of the triangle $DUV$ meets again $AE$ at the point $L$. If $I$ is the point of intersection of $BZ$ with $AE$, prove that the lines $BL$ and $CI$ intersect on the line $AZ$.
3 replies
Lukaluce
Jul 2, 2022
S11_DooronbekovNurakhmed
28 minutes ago
About my new website
Samujjal101   0
29 minutes ago
Hi everybody!
I'm registering some of the finest minds in math into my website.. it's not completely developed.. but still if you want we would be very grateful to have you!
Text to display
Maths-matchmaker is a website for connecting math minds together with a mission to unite together. It uses a matching algorithm to match 1:1 with like minded peers based on their interests or topics in math
0 replies
1 viewing
Samujjal101
29 minutes ago
0 replies
Determine all the sets of six consecutive positive integers
sqing   7
N 40 minutes ago by Adventure1000
Source: JBMO 2017, Q1
Determine all the sets of six consecutive positive integers such that the product of some two of them . added to the product of some other two of them is equal to the product of the remaining two numbers.
7 replies
sqing
Jun 26, 2017
Adventure1000
40 minutes ago
Polygon with minimum internal angle 120^\circ
Kunihiko_Chikaya   1
N 43 minutes ago by Mathzeus1024
How many sides can have the polygon with minimum internal angle of $ 120^\circ$ by adding every $ 5^\circ$?
Sorry for my poor English.
1 reply
Kunihiko_Chikaya
Aug 9, 2007
Mathzeus1024
43 minutes ago
One more problem defined only with lines
Assassino9931   2
N an hour ago by sami1618
Source: Balkan MO 2024 Shortlist G6
Let $ABC$ be a triangle and the points $K$ and $L$ on $AB$, $M$ and $N$ on $BC$, and $P$ and $Q$ on $AC$ be such that $AK = LB < \frac{1}{2}AB, BM = NC < \frac{1}{2}BC$ and $CP = QA < \frac{1}{2}AC$. The intersections of $KN$ with $MQ$ and $LP$ are $R$ and $T$ respectively, and the intersections of $NP$ with $LM$ and $KQ$ are $D$ and $E$, respectively. Prove that the lines $DR, BE$ and $CT$ are concurrent.
2 replies
1 viewing
Assassino9931
Yesterday at 10:31 PM
sami1618
an hour ago
Two parallels
jayme   3
N an hour ago by ND_
Source: own?
Dear Mathlinkers,

1. ABCD a square
2. (A) the circle with center at A passing through B
3. P the points of intersection of the segment AC and (A)
4. I the midpoint of AB
5. Q the point of intersection of the segment IC and (A)
6. X the point of intersection of the parallel to AB through Q and BC
7. Y the point of intersection of the segment AX and (A)

Prove : CY is parallel to IP.

Jean-Louis
3 replies
jayme
Today at 8:25 AM
ND_
an hour ago
Interesting inequality
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b,c>0  . $ Prove that
$$\frac{a}{b}+ \frac{kb^2}{c^2} + \frac{c}{a}\geq 5\sqrt[5]{\frac{k}{16}}$$Where $ k >0. $
$$\frac{a}{b}+ \frac{16b^2}{c^2} + \frac{c}{a}\geq 5$$$$\frac{a}{b}+ \frac{ b^2}{2c^2} + \frac{c}{a}\geq \frac{5}{2} $$
1 reply
sqing
an hour ago
sqing
an hour ago
Junior Balkan Mathematical Olympiad 2021- P3
Lukaluce   35
N an hour ago by Rayvhs
Source: JBMO 2021
Let $ABC$ be an acute scalene triangle with circumcenter $O$. Let $D$ be the foot of the altitude from $A$ to the side $BC$. The lines $BC$ and $AO$ intersect at $E$. Let $s$ be the line through $E$ perpendicular to $AO$. The line $s$ intersects $AB$ and $AC$ at $K$ and $L$, respectively. Denote by $\omega$ the circumcircle of triangle $AKL$. Line $AD$ intersects $\omega$ again at $X$.
Prove that $\omega$ and the circumcircles of triangles $ABC$ and $DEX$ have a common point.
35 replies
Lukaluce
Jul 1, 2021
Rayvhs
an hour ago
amazing balkan combi
egxa   3
N an hour ago by CatinoBarbaraCombinatoric
Source: BMO 2025 P4
There are $n$ cities in a country, where $n \geq 100$ is an integer. Some pairs of cities are connected by direct (two-way) flights. For two cities $A$ and $B$ we define:

$(i)$ A $\emph{path}$ between $A$ and $B$ as a sequence of distinct cities $A = C_0, C_1, \dots, C_k, C_{k+1} = B$, $k \geq 0$, such that there are direct flights between $C_i$ and $C_{i+1}$ for every $0 \leq i \leq k$;
$(ii)$ A $\emph{long path}$ between $A$ and $B$ as a path between $A$ and $B$ such that no other path between $A$ and $B$ has more cities;
$(iii)$ A $\emph{short path}$ between $A$ and $B$ as a path between $A$ and $B$ such that no other path between $A$ and $B$ has fewer cities.
Assume that for any pair of cities $A$ and $B$ in the country, there exist a long path and a short path between them that have no cities in common (except $A$ and $B$). Let $F$ be the total number of pairs of cities in the country that are connected by direct flights. In terms of $n$, find all possible values $F$

Proposed by David-Andrei Anghel, Romania.
3 replies
+1 w
egxa
Yesterday at 1:57 PM
CatinoBarbaraCombinatoric
an hour ago
The daily problem!
Leeoz   141
N 3 hours ago by steve4916
Every day, I will try to post a new problem for you all to solve! If you want to post a daily problem, you can! :)

Please hide solutions and answers, hints are fine though! :)

Problems usually get harder throughout the week, so Sunday is the easiest and Saturday is the hardest!

Past Problems!
141 replies
Leeoz
Mar 21, 2025
steve4916
3 hours ago
random achievements
Bummer12345   22
N 3 hours ago by steve4916
What are some random math achievements that you have accomplished but possess no real meaning?

For example, I solved #10 on the 2024 national mathcounts team round, though my team got a 5 Click to reveal hidden text and ended up getting 30-somethingth place
22 replies
Bummer12345
Mar 25, 2025
steve4916
3 hours ago
Something Horrible-a Challenge
Xueshuxue   24
N Apr 7, 2025 by Solocraftsolo
Hello, I was wondering if it's possible to make 8 with the numbers 5, 3, 5, and 7 under the following rules:
-You can only use 5 twice, 3 once, and 7 once.
-You must use all the numbers.
You can stack numbers to form larger numbers (example: I could take 3 and 5 and turn it into 35 or 53, or use 7, 3, and 5 to make 375.)
-You are allowed to use parentheses.
(Also, I already found out that no 3 digital numbers will work for the solution.)
24 replies
Xueshuxue
Mar 28, 2025
Solocraftsolo
Apr 7, 2025
Something Horrible-a Challenge
G H J
G H BBookmark kLocked kLocked NReply
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Xueshuxue
45 posts
#1
Y by
Hello, I was wondering if it's possible to make 8 with the numbers 5, 3, 5, and 7 under the following rules:
-You can only use 5 twice, 3 once, and 7 once.
-You must use all the numbers.
You can stack numbers to form larger numbers (example: I could take 3 and 5 and turn it into 35 or 53, or use 7, 3, and 5 to make 375.)
-You are allowed to use parentheses.
(Also, I already found out that no 3 digital numbers will work for the solution.)
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Inaaya
293 posts
#2
Y by
(5x3)-7 !!!
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sanaops9
823 posts
#3
Y by
you need to use both fives rbo

you didn't mention what functions to use so $\lfloor{\dfrac{55+3}{7}\rfloor}$
This post has been edited 1 time. Last edited by sanaops9, Mar 28, 2025, 7:28 PM
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derekwang2048
1223 posts
#4
Y by
if you're allowed to use more than just the basic operations, this is simple
This post has been edited 2 times. Last edited by derekwang2048, Mar 28, 2025, 7:29 PM
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sanaops9
823 posts
#5
Y by
@above, you can use \lfloor and \rfloor for left floor and right floor, respectively.

EDIT: ig u figured it out
This post has been edited 2 times. Last edited by sanaops9, Mar 28, 2025, 7:31 PM
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vincentwant
1353 posts
#6 • 1 Y
Y by ARWonder
Click to reveal hidden text
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Solocraftsolo
2554 posts
#7
Y by
the answer is obviously

\[|P\left(\left\lfloor \sqrt{\sqrt{\left\lceil \sqrt{\left\lfloor \sqrt{\left\lceil \sqrt{\sqrt{\left\lfloor \sqrt{\sqrt{\left\lceil \sqrt{\sqrt{\sqrt{5537}}}\right\rceil !!}}\right\rfloor !}}\right\rceil !}\right\rfloor !}\right\rceil !}}\right\rfloor\right)|\]

where the P function is the power set and the | | means cardinality
This post has been edited 1 time. Last edited by Solocraftsolo, Mar 28, 2025, 8:51 PM
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Leeoz
178 posts
#8 • 2 Y
Y by Exponent11, TTNT
chat gpt cooked
chat gpt wrote:
To make 8 using only the numbers **3**, **5**, **5**, and **7**, one possible solution is:

\[
(5 + 5) - (7 - 3) = 8
\]
Explanation:
- First, add the two 5s: \(5 + 5 = 10\).
- Then, subtract the difference between 7 and 3: \(7 - 3 = 4\).
- Finally, subtract 4 from 10: \(10 - 4 = 8\).

This gives you 8!
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ChickensEatGrass
44 posts
#9
Y by
Leeoz wrote:
chat gpt cooked
chat gpt wrote:
To make 8 using only the numbers **3**, **5**, **5**, and **7**, one possible solution is:

\[
(5 + 5) - (7 - 3) = 8
\]
Explanation:
- First, add the two 5s: \(5 + 5 = 10\).
- Then, subtract the difference between 7 and 3: \(7 - 3 = 4\).
- Finally, subtract 4 from 10: \(10 - 4 = 8\).

This gives you 8!

what the heck
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huajun78
66 posts
#11
Y by
Leeoz wrote:
chat gpt cooked
chat gpt wrote:
To make 8 using only the numbers **3**, **5**, **5**, and **7**, one possible solution is:

\[
(5 + 5) - (7 - 3) = 8
\]
Explanation:
- First, add the two 5s: \(5 + 5 = 10\).
- Then, subtract the difference between 7 and 3: \(7 - 3 = 4\).
- Finally, subtract 4 from 10: \(10 - 4 = 8\).

This gives you 8!

$6=8$, proof by ChatGPT
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SirAppel
877 posts
#12
Y by
$(5+3)^{\lfloor \frac{7}{5} \rfloor}$
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cheltstudent
618 posts
#13
Y by
$lcm(7,5) - 3^{3}=8$
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Soupboy0
343 posts
#14 • 1 Y
Y by PuppyPenguinDolphin
$\frac{5!}{3!}-5-7$
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Owen314159
12 posts
#15
Y by
Soupboy0 wrote:
$\frac{5!}{3!}-5-7$

I think this is the most normal sol
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RandomMathGuy500
59 posts
#16
Y by
chat gpt wrote:
You can make 8 using the numbers 5, 5, 3, and 7 with the following equation:
$(7+3)-(5+5)=8$
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Total_Awesomeness
307 posts
#17
Y by
derekwang2048 wrote:
if you're allowed to use more than just the basic operations, this is simple

ur not allowed to use 8
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happymoose666
307 posts
#18
Y by
he is not using 8, it's an equal sign
This post has been edited 1 time. Last edited by happymoose666, Mar 29, 2025, 10:37 PM
Reason: yes
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Total_Awesomeness
307 posts
#19
Y by
no at the end
he says that it equals the digit 8, which is not allowed
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sanaops9
823 posts
#20
Y by
Owen314159 wrote:
Soupboy0 wrote:
$\frac{5!}{3!}-5-7$

I think this is the most normal sol
vincentwant wrote:
Click to reveal hidden text

No vincent's was prolly the best and intended sol
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pieMax2713
4187 posts
#21
Y by
$|\{5, 5, 7\}|^3$ where | | denotes cardinality
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K1mchi_
93 posts
#22 • 1 Y
Y by ChickensEatGrass
(7!/(5!+3!))/5
I just used lots of factorial
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WiseHawkCuteFriendly
581 posts
#23
Y by
chat gpt wrote:
$\frac{7+5}{3}+5=8$,
reasoning: $\frac{12}{3}+5=8$
$4+5=8$
This post has been edited 2 times. Last edited by WiseHawkCuteFriendly, Mar 31, 2025, 1:46 AM
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happymoose666
307 posts
#24
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Total_Awesomeness wrote:
no at the end
he says that it equals the digit 8, which is not allowed

But he meant $\frac{55+\lfloor{\sqrt3}\rfloor}{7}$ is equal to 8, so he solved the problem, he is not using eight
This post has been edited 2 times. Last edited by happymoose666, Mar 31, 2025, 1:21 PM
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Demetri
1314 posts
#25
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Soupboy0 wrote:
$\frac{5!}{3!}-5-7$

I think this is the most normal sol
This doesn't use factorials
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Solocraftsolo
2554 posts
#26 • 1 Y
Y by Exponent11
huajun78 wrote:

$6=8$, proof by ChatGPT

proof by exhaustion
proof by contradiction
proof by chatGPT
This post has been edited 1 time. Last edited by Solocraftsolo, Apr 7, 2025, 4:10 PM
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