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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.

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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
A glass is filled with milk. Two-thirds of its content is poured out and replace
Vulch   0
33 minutes ago
A glass is filled with milk. Two-thirds of its content is poured out and replaced with water. If this process of pouring out two-thirds the content and replacing with water is repeated three more times, then the final ratio of milk to water in the glass, is
0 replies
Vulch
33 minutes ago
0 replies
2014 JBMO Shortlist G2
parmenides51   6
N an hour ago by tilya_TASh
Source: 2014 JBMO Shortlist G2
Acute-angled triangle ${ABC}$ with ${AB<AC<BC}$ and let be ${c(O,R)}$ it’s circumcircle. Diameters ${BD}$ and ${CE}$ are drawn. Circle ${c_1(A,AE)}$ interescts ${AC}$ at ${K}$. Circle ${{c}_{2}(A,AD)}$ intersects ${BA}$ at ${L}$ .(${A}$ lies between ${B}$ and ${L}$). Prove that lines ${EK}$ and ${DL}$ intersect at circle $c$ .

by Evangelos Psychas (Greece)
6 replies
parmenides51
Oct 8, 2017
tilya_TASh
an hour ago
MATHCOUNTS
ILOVECATS127   26
N an hour ago by ILOVECATS127
Hi,

I am looking to get on my school MATHCOUNTS team next year in 7th grade, and I had a question: Where do the school round questions come from? (Sprint, Chapter, Team, Countdown)
26 replies
ILOVECATS127
May 7, 2025
ILOVECATS127
an hour ago
Disjoint Pairs
MithsApprentice   42
N an hour ago by endless_abyss
Source: USAMO 1998
Suppose that the set $\{1,2,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}$ ($1\leq i\leq 999$) so that for all $i$, $|a_i-b_i|$ equals $1$ or $6$. Prove that the sum \[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|  \] ends in the digit $9$.
42 replies
MithsApprentice
Oct 9, 2005
endless_abyss
an hour ago
FE with gcd
a_507_bc   8
N an hour ago by Tkn
Source: Nordic MC 2023 P2
Find all functions $f: \mathbb{N} \to \mathbb{N}$ such that $$\gcd(f(x),y)f(xy)=f(x)f(y)$$for all positive integers $x, y$.
8 replies
a_507_bc
Apr 21, 2023
Tkn
an hour ago
2014 JBMO Shortlist G1
parmenides51   19
N 2 hours ago by tilya_TASh
Source: 2014 JBMO Shortlist G1
Let ${ABC}$ be a triangle with $m\left( \angle B \right)=m\left( \angle C \right)={{40}^{{}^\circ }}$ Line bisector of ${\angle{B}}$ intersects ${AC}$ at point ${D}$. Prove that $BD+DA=BC$.
19 replies
parmenides51
Oct 8, 2017
tilya_TASh
2 hours ago
Stars and bars i think
RenheMiResembleRice   1
N 2 hours ago by NicoN9
Source: Diao Luo
Solve the following attached with steps
1 reply
RenheMiResembleRice
2 hours ago
NicoN9
2 hours ago
Sequence
Titibuuu   1
N 2 hours ago by Titibuuu
Let \( a_1 = a \), and for all \( n \geq 1 \), define the sequence \( \{a_n\} \) by the recurrence
\[
a_{n+1} = a_n^2 + 1
\]Prove that there is no natural number \( n \) such that
\[
\prod_{k=1}^{n} \left( a_k^2 + a_k + 1 \right)
\]is a perfect square.
1 reply
Titibuuu
Today at 2:22 AM
Titibuuu
2 hours ago
2013 Japan MO Finals
parkjungmin   0
2 hours ago
help me

we cad do it
0 replies
parkjungmin
2 hours ago
0 replies
Calculate the distance AD
MTA_2024   1
N 2 hours ago by MTA_2024
A semi-circle is inscribed in a quadrilateral $ABCD$. The center $O$ of the semi-circle is the midpoint of segment $AD$. We have $AB=9$ and $CD=16$.
Calculate the distance $AD$.
1 reply
MTA_2024
Yesterday at 3:50 PM
MTA_2024
2 hours ago
IMO ShortList 1999, algebra problem 2
orl   11
N 2 hours ago by ezpotd
Source: IMO ShortList 1999, algebra problem 2
The numbers from 1 to $n^2$ are randomly arranged in the cells of a $n \times n$ square ($n \geq 2$). For any pair of numbers situated on the same row or on the same column the ratio of the greater number to the smaller number is calculated. Let us call the characteristic of the arrangement the smallest of these $n^2\left(n-1\right)$ fractions. What is the highest possible value of the characteristic ?
11 replies
orl
Nov 14, 2004
ezpotd
2 hours ago
Coolabra
Titibuuu   2
N 2 hours ago by no_room_for_error
Let \( a, b, c \) be distinct real numbers such that
\[
a + b + c + \frac{1}{abc} = \frac{19}{2}
\]Find the maximum possible value of \( a \).
2 replies
Titibuuu
Today at 2:21 AM
no_room_for_error
2 hours ago
Hard centroid geo
lucas3617   0
2 hours ago
Source: Revenge JOM 2025 P5
A triangle $A B C$ has centroid $G$. A line parallel to $B C$ passing through $G$ intersects the circumcircle of $A B C$ at $D$. Let lines $A D$ and $B C$ intersect at $E$. Suppose a point $P$ is chosen on $B C$ such that the tangent of the circumcircle of $D E P$ at $D$, the tangent of the circumcircle of $A B C$ at $A$ and $B C$ concur. Prove that $G P = P D$.
0 replies
lucas3617
2 hours ago
0 replies
9 What is the best way to learn math???
lovematch13   95
N 3 hours ago by Craftybutterfly
On the contrary, I'm also gonna try to send this to school admins. PLEASE DO NOT TROLL!!!!
95 replies
lovematch13
May 22, 2023
Craftybutterfly
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
575 posts
#2
Y by
w problem i cannot lie
113(do i even need to write a solution)
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GallopingUnicorn45
376 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
3749 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
851 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
1266 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
353 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
1266 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
1289 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
1266 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
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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
2492 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
14 posts
#17
Y by
In the sum at the ... there are not present all the missing numbers?
Z K Y
N Quick Reply
G
H
=
a