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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)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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
Trivial Factoring MathCounts Question? 2025 MathCounts National Sprint Round #29
ilikemath247365   28
N 13 minutes ago by tikachaudhuri
What is the value of the expression below?

$\frac{(1! + 2! + 3!)(2! + 3! + 4!)(3! + 4! + 5!)...(98! + 99! + 100!)}{(1! - 3(2!) + 3!)(2! - 3(3!) + 4!)(3! - 3(4!) + 5!)...(98! - 3(99!) + 100!)}$.
28 replies
ilikemath247365
Yesterday at 1:16 AM
tikachaudhuri
13 minutes ago
find question
mathematical-forest   0
an hour ago
Are there any contest questions that seem simple but are actually difficult? :-D
0 replies
mathematical-forest
an hour ago
0 replies
Inspired by qrxz17
sqing   2
N an hour 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
an hour ago
D,E,F are collinear.
TUAN2k8   1
N an hour ago by Beelzebub
Source: Own
Help me with this:
1 reply
TUAN2k8
Yesterday at 1:07 AM
Beelzebub
an hour 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
sqing
3 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 2 hours 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
2 hours ago
Stop Projecting your insecurities
naman12   53
N 2 hours 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
2 hours ago
9 How many squares do you have memorized
LXC007   107
N 2 hours ago by SapphireKitty
How many squares have you memorized. I have 1-20

Edit: to clarify i mean positive squares from 1 so if you say ten you mean you memorized the squares 1,2,3,4,5,6,7,8,9 and 10
107 replies
LXC007
May 17, 2025
SapphireKitty
2 hours ago
Roots of unity
Henryfamz   1
N 2 hours 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
2 hours ago
Shortest number theory you might've seen in your life
AlperenINAN   11
N 2 hours 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
2 hours ago
9 Mock AMC 8
Leeoz   2
N 6 hours ago by Yihangzh
This is just your average mock AMC 8 on AoPS... nothing sus at all...
Just PM me your answers if you actually want to try it and be on the leaderboard...

Leaderboard

This test is for entertainment purposes only...

Also if you all want, I can post the solutions in June.
2 replies
Leeoz
Today at 4:46 AM
Yihangzh
6 hours ago
Overly wordy problems
ZMB038   4
N Today at 3:31 AM by CJB19
Hey everyone, here we can post questions with way to many extraneous words, that are actually easy.
Try to solve the one above yours.
I'll start:
Click to reveal hidden text
4 replies
ZMB038
Yesterday at 11:14 PM
CJB19
Today at 3:31 AM
Tricky summation
arfekete   12
N Apr 3, 2025 by KevinKV01
If $\dots = 7$, what is the value of $1 + 2 + 3 + \dots + 100$?
12 replies
arfekete
Apr 2, 2025
KevinKV01
Apr 3, 2025
Tricky summation
G H J
G H BBookmark kLocked kLocked NReply
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arfekete
259 posts
#1 • 3 Y
Y by eg4334, aidan0626, lpieleanu
If $\dots = 7$, what is the value of $1 + 2 + 3 + \dots + 100$?
Z K Y
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nitride
582 posts
#2
Y by
w problem i cannot lie
113(do i even need to write a solution)
Z K Y
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GallopingUnicorn45
460 posts
#3
Y by
how is this even in middle school math category

$1 + 2 + 3 + ... + 100 = 1 + 2 + 3 + 7 + 100 = \boxed{113}$
Z K Y
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MathPerson12321
3796 posts
#4
Y by
GallopingUnicorn45 wrote:
how is this even in middle school math category

$1 + 2 + 3 + ... + 100 = 1 + 2 + 3 + 7 + 100 = \boxed{113}$

There's no elementary math school category
Z K Y
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DhruvJha
899 posts
#5
Y by
MathPerson12321 wrote:
GallopingUnicorn45 wrote:
how is this even in middle school math category

$1 + 2 + 3 + ... + 100 = 1 + 2 + 3 + 7 + 100 = \boxed{113}$

There's no elementary math school category

I think there's a user created one
Z K Y
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Yiyj1
1267 posts
#6
Y by
DhruvJha wrote:
MathPerson12321 wrote:
GallopingUnicorn45 wrote:
how is this even in middle school math category

$1 + 2 + 3 + ... + 100 = 1 + 2 + 3 + 7 + 100 = \boxed{113}$

There's no elementary math school category

I think there's a user created one

never heard of it, doubt the op would know
Z K Y
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blueprimes
362 posts
#7 • 4 Y
Y by aidan0626, lpieleanu, arfekete, eg4334
GallopingUnicorn45 wrote:
how is this even in middle school math category

$1 + 2 + 3 + ... + 100 = 1 + 2 + 3 + 7 + 100 = \boxed{113}$

Clearly sir, you are deeply mistaken.

Solved with resources, greendivisors, eg4334, lpieleanu, SigmaPiE, Arcticturn, and CoolJupiter.

Here, having several continguous characters as a variable name is absurd! A clear counterexample is in programming, a variable name is invalid if it contains spaces. Thus, the only reasonable explanation is a multiplication using the $\cdot$ symbol standard. We want to solve:
\[ \cdot \cdot \cdot = 7. \]But this is just $\cdot^2 = 7 \implies \cdot = \sqrt{7}$. Since we are in Middle School Math, we will not consider the case of $\cdot = -\sqrt{7}$ as surely outrage will spark. Now if you are not experienced in the dark arts, a feeble-minded individual would simply plug in $\dots = 7$ and sum it up. How absurd! Instead, we explore the more reasonable path of multiplying the "normal" sum of $5050$ by $\sqrt{7}$, as every unit in the sum is replaced by the embedded $\cdots$ within the sequence, clearly the intended path of the creator.

Now suppose it is thousands of years ago and we do not have a calculator. We instead use the approximation $\sqrt{7} \approx 2.64575131106$ written by Euclid himself on a humble rock. Multiplying with our fingers, we obtain
\[ 5050 \cdot \sqrt{7} \approx 13361.0441209. \]Since $5050$ has $3$ significant figures, we round our answer accordingly to scientific procedure to obtain $\boxed{1.34 \times 10^4}$.
This post has been edited 2 times. Last edited by blueprimes, Apr 2, 2025, 2:37 AM
Z K Y
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Yiyj1
1267 posts
#8
Y by
blueprimes wrote:
GallopingUnicorn45 wrote:
how is this even in middle school math category

$1 + 2 + 3 + ... + 100 = 1 + 2 + 3 + 7 + 100 = \boxed{113}$

Clearly sir, you are deeply mistaken.

Solved with resources, greendivisors, eg4334, lpieleanu, SigmaPiE, Arcticturn, and CoolJupiter.

Here, having several continguous characters as a variable name is absurd! A clear counterexample is in programming, a variable name is invalid if it contains spaces. Thus, the only reasonable explanation is a multiplication using the $\cdot$ symbol standard. We want to solve:
\[ \cdot \cdot \cdot = 7. \]But this is just $\cdot^2 = 7 \implies \cdot = \sqrt{7}$. Since we are in Middle School Math, we will not consider the case of $\cdot = -\sqrt{7}$ as surely outrage will spark. Now if you are not experienced in the dark arts, a feeble-minded individual would simply plug in $\dots = 7$ and sum it up. How absurd! Instead, we explore the more reasonable path of multiplying the "normal" sum of $5050$ by $\sqrt{7}$, as every unit in the sum is replaced by the embedded $\cdots$ within the sequence, clearly the intended path of the creator.

Now suppose it is thousands of years ago and we do not have a calculator. We instead use the approximation $\sqrt{7} \approx 2.64575131106$ written by Euclid himself on a humble rock. Multiplying with our fingers, we obtain
\[ 5050 \cdot \sqrt{7} \approx 13361.0441209. \]Since $5050$ has $3$ significant figures, we round our answer accordingly to scientific procedure to obtain $\boxed{1.34 \times 10^4}$.

wait why am i able to edit ur post
Z K Y
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Yrock
1311 posts
#9
Y by
#8 nah don't mind it it won't work its just a weird glitch
Z K Y
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Yiyj1
1267 posts
#10
Y by
Yrock wrote:
#8 nah don't mind it it won't work its just a weird glitch

oh aight chat

edit: one more post away from 1200!
This post has been edited 1 time. Last edited by Yiyj1, Apr 2, 2025, 2:43 AM
Z K Y
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arfekete
259 posts
#11 • 1 Y
Y by Amkan2022
blueprimes wrote:
GallopingUnicorn45 wrote:
how is this even in middle school math category

$1 + 2 + 3 + ... + 100 = 1 + 2 + 3 + 7 + 100 = \boxed{113}$

Clearly sir, you are deeply mistaken.

Solved with resources, greendivisors, eg4334, lpieleanu, SigmaPiE, Arcticturn, and CoolJupiter.

Here, having several continguous characters as a variable name is absurd! A clear counterexample is in programming, a variable name is invalid if it contains spaces. Thus, the only reasonable explanation is a multiplication using the $\cdot$ symbol standard. We want to solve:
\[ \cdot \cdot \cdot = 7. \]But this is just $\cdot^2 = 7 \implies \cdot = \sqrt{7}$. Since we are in Middle School Math, we will not consider the case of $\cdot = -\sqrt{7}$ as surely outrage will spark. Now if you are not experienced in the dark arts, a feeble-minded individual would simply plug in $\dots = 7$ and sum it up. How absurd! Instead, we explore the more reasonable path of multiplying the "normal" sum of $5050$ by $\sqrt{7}$, as every unit in the sum is replaced by the embedded $\cdots$ within the sequence, clearly the intended path of the creator.

Now suppose it is thousands of years ago and we do not have a calculator. We instead use the approximation $\sqrt{7} \approx 2.64575131106$ written by Euclid himself on a humble rock. Multiplying with our fingers, we obtain
\[ 5050 \cdot \sqrt{7} \approx 13361.0441209. \]Since $5050$ has $3$ significant figures, we round our answer accordingly to scientific procedure to obtain $\boxed{1.34 \times 10^4}$.

Best solution so far but this makes a slight assumption which seems trivial but is actually incorrect. However, this would probably still get partials.

Intended sol (according to some moppers): Click to reveal hidden text

Remark: I don't know how it would be expected in contest for anyone to actually be able to evaluate $1 + 2 + 3 + 100$ within a reasonable timing even after finding the (already hard) cruxes of considering $G$ and finding $\cdot = \sqrt{7}$, so this problem is probably best just to be posted here for us to speculate and not used within a timed contest.
This post has been edited 6 times. Last edited by arfekete, Apr 2, 2025, 2:53 AM
Z K Y
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fruitmonster97
2505 posts
#12
Y by
ithinkaboveiswrong?

For clarity, we will write any " " in our math as "space". Then space$\cdot$space$\cdot$space$\cdot$space$=7,$ so space=$\sqrt[4]{7}.$

We aim to compute space$1$space$+$space$2$space$+$space$3$space$+$space$\cdot$space$\cdot$space$\cdot$space$+$space$100.$ This is simply:
\[\sqrt[4]{7}1\sqrt[4]{7}+\sqrt[4]{7}2\sqrt[4]{7}+\sqrt[4]{7}3\sqrt[4]{7}+\sqrt[4]{7}\cdot\sqrt[4]{7}\cdot\sqrt[4]{7}\cdot\sqrt[4]{7}+\sqrt[4]{7}100=\sqrt[4]{71}\sqrt[4]{7}+\sqrt[4]{72}\sqrt[4]{7}+\sqrt[4]{73}\sqrt[4]{7}+\sqrt[4]{7+\sqrt[4]{7+\sqrt[4]{7+\sqrt[4]{7}}}}+\sqrt[4]{7100}.\]We will now estimate to the nearest integer, because every number in the problem is an integer. we have 1.6^4=6.5536<7 but 1.7^4=8.3521 so $\sqrt[4]{7}\approx1.61.$ Similarly, $\sqrt[4]{71}\sqrt[4]{72}\sqrt[4]{73}\approx3\sqrt[4]{72}\approx8.7.$ Thus, the first part is $8.7\cdot1.61\approx14.$

for the second part, finitely many nested roots bad. infinitely many better. assume infinitely many. let it be $x.$ then $x=\sqrt[4]{7+x},$ so $x^4=x+7.$ Now, use newton's method on $f(x)=x^4-x-7.$ Guess $x_0=2.$ Then $x_1=x_0-\tfrac{f(x_0)}{f'(x_0)}=2-\tfrac{7}{31}=\tfrac{55}{31}.$ Close enough.

Finally, $\sqrt[4]{7100}\approx\sqrt[4]{6561}=9.$ Our sum is $14+9+\tfrac{55}{31}\approx\boxed{25},$ which fittingly enough is the last two digits of the year. Also, the sum of the first two parts and the last part are, when rounded, are the two squares that when combined with the three in the date, make the first five squares, which is a beautiful easter egg in memorium for easter being in (last two digits of year)-(month number) days. $\blacksquare$
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KevinKV01
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In the sum at the ... there are not present all the missing numbers?
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