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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:
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[*]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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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
The daily problem!
Leeoz   216
N a minute ago by kjhgyuio
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!
216 replies
Leeoz
Mar 21, 2025
kjhgyuio
a minute ago
Inequality in triangle
Nguyenhuyen_AG   3
N 43 minutes ago by Nguyenhuyen_AG
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]
3 replies
Nguyenhuyen_AG
Today at 6:17 AM
Nguyenhuyen_AG
43 minutes ago
Problem 1
randomusername   73
N an hour ago by ND_
Source: IMO 2015, Problem 1
We say that a finite set $\mathcal{S}$ of points in the plane is balanced if, for any two different points $A$ and $B$ in $\mathcal{S}$, there is a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that $\mathcal{S}$ is centre-free if for any three different points $A$, $B$ and $C$ in $\mathcal{S}$, there is no points $P$ in $\mathcal{S}$ such that $PA=PB=PC$.

(a) Show that for all integers $n\ge 3$, there exists a balanced set consisting of $n$ points.

(b) Determine all integers $n\ge 3$ for which there exists a balanced centre-free set consisting of $n$ points.

Proposed by Netherlands
73 replies
randomusername
Jul 10, 2015
ND_
an hour ago
x is rational implies y is rational
pohoatza   44
N an hour ago by ezpotd
Source: IMO Shortlist 2006, N2, VAIMO 2007, Problem 6
For $ x \in (0, 1)$ let $ y \in (0, 1)$ be the number whose $ n$-th digit after the decimal point is the $ 2^{n}$-th digit after the decimal point of $ x$. Show that if $ x$ is rational then so is $ y$.

Proposed by J.P. Grossman, Canada
44 replies
pohoatza
Jun 28, 2007
ezpotd
an hour ago
Multiplicative function
Tales   37
N an hour ago by ezpotd
Source: IMO Shortlist 2004, number theory problem 2
The function $f$ from the set $\mathbb{N}$ of positive integers into itself is defined by the equality \[f(n)=\sum_{k=1}^{n} \gcd(k,n),\qquad n\in \mathbb{N}.\]
a) Prove that $f(mn)=f(m)f(n)$ for every two relatively prime ${m,n\in\mathbb{N}}$.

b) Prove that for each $a\in\mathbb{N}$ the equation $f(x)=ax$ has a solution.

c) Find all ${a\in\mathbb{N}}$ such that the equation $f(x)=ax$ has a unique solution.
37 replies
Tales
Mar 23, 2005
ezpotd
an hour ago
NICE INEQUALITY
Kyleray   3
N an hour ago by sqing
Let's $a,b,c>0$. Prove:
$$(\frac{a}{b+c}+\frac{b}{c+a})(\frac{b}{c+a}+\frac{c}{a+b})(\frac{c}{a+b}+\frac{a}{b+c})\geq \frac{(a+b+c)^2}{3(ab+bc+ca)}$$$\text{P/S: No mapple, please :(}$
3 replies
Kyleray
Mar 11, 2021
sqing
an hour ago
Tough inequality
TUAN2k8   4
N an hour ago by cazanova19921
Source: Own
Let $n \ge 2$ be an even integer and let $x_1,x_2,...,x_n$ be real numbers satisfying $x_1^2+x_2^2+...+x_n^2=n$.
Prove that
$\sum_{1 \le i < j \le n} \frac{x_ix_j}{x_i^2+x_j^2+1} \ge \frac{-n}{6}$
4 replies
TUAN2k8
May 28, 2025
cazanova19921
an hour ago
Easy Diophantne
anantmudgal09   20
N 2 hours ago by Adywastaken
Source: India Practice TST 2017 D1 P2
Find all positive integers $p,q,r,s>1$ such that $$p!+q!+r!=2^s.$$
20 replies
anantmudgal09
Dec 9, 2017
Adywastaken
2 hours ago
SOLVE: CDR style problem quick algebra
ryfighter   1
N 2 hours ago by franklin2013
It takes 3 people 10 minutes to mow 2 lawns. How many minutes will it take for 2 people to mow 10 lawns? Express your answer in hours as a decimal.

$(A)$ $1.25$
$(B)$ $75$
$(C)$ $01.025$
$(D)$ $1.5$
$(E)$ $15.25$
1 reply
ryfighter
Today at 3:19 AM
franklin2013
2 hours ago
Converse of a classic orthocenter problem
spartacle   43
N 2 hours ago by ihategeo_1969
Source: USA TSTST 2020 Problem 6
Let $A$, $B$, $C$, $D$ be four points such that no three are collinear and $D$ is not the orthocenter of $ABC$. Let $P$, $Q$, $R$ be the orthocenters of $\triangle BCD$, $\triangle CAD$, $\triangle ABD$, respectively. Suppose that the lines $AP$, $BQ$, $CR$ are pairwise distinct and are concurrent. Show that the four points $A$, $B$, $C$, $D$ lie on a circle.

Andrew Gu
43 replies
spartacle
Dec 14, 2020
ihategeo_1969
2 hours ago
Symmetric points part 2
CyclicISLscelesTrapezoid   22
N 2 hours ago by ihategeo_1969
Source: USA TSTST 2022/6
Let $O$ and $H$ be the circumcenter and orthocenter, respectively, of an acute scalene triangle $ABC$. The perpendicular bisector of $\overline{AH}$ intersects $\overline{AB}$ and $\overline{AC}$ at $X_A$ and $Y_A$ respectively. Let $K_A$ denote the intersection of the circumcircles of triangles $OX_AY_A$ and $BOC$ other than $O$.

Define $K_B$ and $K_C$ analogously by repeating this construction two more times. Prove that $K_A$, $K_B$, $K_C$, and $O$ are concyclic.

Hongzhou Lin
22 replies
CyclicISLscelesTrapezoid
Jun 27, 2022
ihategeo_1969
2 hours ago
Combi counting
Caasi_Gnow   3
N 2 hours ago by franklin2013
Find the number of different ways to arrange seven people around a circular meeting table if A and B must sit together and C and D cannot sit next to each other. (Note: the order for A and B might be A,B or B,A)
3 replies
+1 w
Caasi_Gnow
Mar 20, 2025
franklin2013
2 hours ago
Periodicity of factorials
Cats_on_a_computer   0
2 hours ago
Source: Thrill and challenge of pre-college mathematics
Let a_k denote the first non zero digit of the decimal representation of k!. Does the sequence a_1, a_2, a_3, … eventually become periodic?
0 replies
Cats_on_a_computer
2 hours ago
0 replies
Geometry question !
kjhgyuio   1
N 3 hours ago by whwlqkd
........
1 reply
kjhgyuio
4 hours ago
whwlqkd
3 hours ago
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
462 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
900 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
1268 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
363 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
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Yiyj1
1268 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
1312 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
1268 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
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fruitmonster97
2506 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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