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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
Four tangent lines concur on the circumcircle
v_Enhance   35
N an hour ago by bin_sherlo
Source: USA TSTST 2018 Problem 3
Let $ABC$ be an acute triangle with incenter $I$, circumcenter $O$, and circumcircle $\Gamma$. Let $M$ be the midpoint of $\overline{AB}$. Ray $AI$ meets $\overline{BC}$ at $D$. Denote by $\omega$ and $\gamma$ the circumcircles of $\triangle BIC$ and $\triangle BAD$, respectively. Line $MO$ meets $\omega$ at $X$ and $Y$, while line $CO$ meets $\omega$ at $C$ and $Q$. Assume that $Q$ lies inside $\triangle ABC$ and $\angle AQM = \angle ACB$.

Consider the tangents to $\omega$ at $X$ and $Y$ and the tangents to $\gamma$ at $A$ and $D$. Given that $\angle BAC \neq 60^{\circ}$, prove that these four lines are concurrent on $\Gamma$.

Evan Chen and Yannick Yao
35 replies
v_Enhance
Jun 26, 2018
bin_sherlo
an hour ago
tangent and tangent again
ItzsleepyXD   0
an hour ago
Source: holder send me this
Given an acute non-isosceles triangle $ABC$ with circumcircle $\Gamma$. $M$ is the midpoint of segment $BC$ and $N$ is the midpoint of $\overarc{BC}$ of $\Gamma$ (the one that doesn't contain $A$). $X$ and $Y$ are points on $\Gamma$ such that $BX \parallel CY \parallel AM$. Assume there exists point $Z$ on segment $BC$ such that circumcircle of triangle $XYZ$ is tangent to $BC$. Let $\omega$ be the circumcircle of triangle $ZMN$. Line $AM$ meets $\omega$ for the second time at $P$. Let $K$ be a point on $\omega$ such that $KN \parallel AM, \omega_b$ be a circle thaat passes through $B$, $X$ and tangents to $BC$ and $\omega_C$ be a circle that passes through $C,Y$ and tangents to $BC$. Prove that circle with center $K$ and radias $KP$ is tangent to circles $\omega_B,\omega_C$ and $\Gamma$.
0 replies
ItzsleepyXD
an hour ago
0 replies
find angle
TBazar   0
2 hours ago
Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$
0 replies
TBazar
2 hours ago
0 replies
Geometry
Lukariman   10
N 2 hours ago by Captainscrubz
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
10 replies
Lukariman
Tuesday at 12:43 PM
Captainscrubz
2 hours ago
Simple inequality
sqing   21
N 3 hours ago by Bexultan
Source: JBMO 2011 Shortlist A3
$\boxed{\text{A3}}$If $a,b$ be positive real numbers, show that:$$ \displaystyle{\sqrt{\dfrac{a^2+ab+b^2}{3}}+\sqrt{ab}\leq a+b}$$
21 replies
sqing
May 15, 2016
Bexultan
3 hours ago
Primes p such that p and p^2+2p-8 are primes too
mhet49   44
N 3 hours ago by MITDragon
Source: Albanian National Math Olympiad 2012
Find all primes $p$ such that $p+2$ and $p^2+2p-8$ are also primes.
44 replies
mhet49
Apr 1, 2012
MITDragon
3 hours ago
Integer polynomial w factorials
Solilin   1
N 3 hours ago by Tkn
Source: 9th Thailand MO
Let $a_1, a_2, ..., a_{2012}$ be pairwise distinct integers. Show that the equation $(x -a_1)(x - a_2)...(x - a_{2012}) = (1006!)^2$ has at most one integral solution.
1 reply
Solilin
Yesterday at 2:12 PM
Tkn
3 hours ago
Combo NT
a_507_bc   4
N 3 hours ago by Namura
Source: Silk Road 2024 P1
Let $n$ be a positive integer and let $p, q>n$ be odd primes. Prove that the positive integers $1, 2, \ldots, n$ can be colored in $2$ colors, such that for any $x \neq y$ of the same color, $xy-1$ is not divisible by $p$ and $q$.
4 replies
a_507_bc
Oct 20, 2024
Namura
3 hours ago
Old hard problem
ItzsleepyXD   2
N 4 hours ago by ItzsleepyXD
Source: IDK
Let $ABC$ be a triangle and let $O$ be its circumcenter and $I$ its incenter.
Let $P$ be the radical center of its three mixtilinears and let $Q$ be the isogonal conjugate of $P$.
Let $G$ be the Gergonne point of the triangle $ABC$.
Prove that line $QG$ is parallel with line $OI$ .
2 replies
ItzsleepyXD
Apr 25, 2025
ItzsleepyXD
4 hours ago
Polynomial Factors
somebodyyouusedtoknow   1
N 4 hours ago by luutrongphuc
Source: San Diego Honors Math Contest 2025 Part II, Problem 2
Let $P(x)$ be a polynomial with real coefficients such that $P(x^n) \mid P(x^{n+1})$ for all $n \in \mathbb{N}$. Prove that $P(x) = cx^k$ for some real constant $c$ and $k \in \mathbb{N}$.
1 reply
somebodyyouusedtoknow
Apr 26, 2025
luutrongphuc
4 hours ago
I need the technique
DievilOnlyM   15
N 6 hours ago by sqing
Let a,b,c be real numbers such that: $ab+7bc+ca=188$.
FInd the minimum value of: $5a^2+11b^2+5c^2$
15 replies
DievilOnlyM
May 23, 2019
sqing
6 hours ago
Linear colorings mod 2^n
vincentwant   1
N 6 hours ago by vincentwant
Let $n$ be a positive integer. The ordered pairs $(x,y)$ where $x,y$ are integers in $[0,2^n)$ are each labeled with a positive integer less than or equal to $2^n$ such that every label is used exactly $2^n$ times and there exist integers $a_1,a_2,\dots,a_{2^n}$ and $b_1,b_2,\dots,b_{2^n}$ such that the following property holds: For any two lattice points $(x_1,y_1)$ and $(x_2,y_2)$ that are both labeled $t$, there exists an integer $k$ such that $x_2-x_1-ka_t$ and $y_2-y_1-kb_t$ are both divisible by $2^n$. How many such labelings exist?
1 reply
vincentwant
Apr 30, 2025
vincentwant
6 hours ago
sqrt(n) or n+p (Generalized 2017 IMO/1)
vincentwant   1
N 6 hours ago by vincentwant
Let $p$ be an odd prime. Define $f(n)$ over the positive integers as follows:
$$f(n)=\begin{cases}
\sqrt{n}&\text{ if n is a perfect square} \\
n+p&\text{ otherwise}
\end{cases}$$
Let $p$ be chosen such that there exists an ordered pair of positive integers $(n,k)$ where $n>1,p\nmid n$ such that $f^k(n)=n$. Prove that there exists at least three integers $i$ such that $1\leq i\leq k$ and $f^i(n)$ is a perfect square.
1 reply
vincentwant
Apr 30, 2025
vincentwant
6 hours ago
Flight between cities
USJL   5
N Today at 2:19 AM by Photaesthesia
Source: 2025 Taiwan TST Round 1 Mock P5
A country has 2025 cites, with some pairs of cities having bidirectional flight routes between them. For any pair of the cities, the flight route between them must be operated by one of the companies $X, Y$ or $Z$. To avoid unfairly favoring specific company, the regulation ensures that if there have three cities $A, B$ and $C$, with flight routes $A \leftrightarrow B$ and $A \leftrightarrow C$ operated by two different companies, then there must exist flight route $B \leftrightarrow C$ operated by the third company different from $A \leftrightarrow B$ and $A \leftrightarrow C$ .

Let $n_X$, $n_Y$ and $n_Z$ denote the number of flight routes operated by companies $X, Y$ and $Z$, respectively. It is known that, starting from a city, we can arrive any other city through a series of flight routes (not necessary operated by the same company). Find the minimum possible value of $\max(n_X, n_Y , n_Z)$.

Proposed by usjl and YaWNeeT
5 replies
USJL
Mar 8, 2025
Photaesthesia
Today at 2:19 AM
Do not try to case bash lol
ItzsleepyXD   4
N May 1, 2025 by straight
Source: Own , Mock Thailand Mathematic Olympiad P3
Let $n,d\geqslant 6$ be a positive integer such that $d\mid 6^{n!}+1$ .
Prove that $d>2n+6$ .
4 replies
ItzsleepyXD
Apr 30, 2025
straight
May 1, 2025
Do not try to case bash lol
G H J
G H BBookmark kLocked kLocked NReply
Source: Own , Mock Thailand Mathematic Olympiad P3
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ItzsleepyXD
133 posts
#1
Y by
Let $n,d\geqslant 6$ be a positive integer such that $d\mid 6^{n!}+1$ .
Prove that $d>2n+6$ .
Z K Y
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Haris1
77 posts
#2 • 2 Y
Y by ItzsleepyXD, Rayanelba
We get $2^{v_{2}(n!)+1}|\phi(d)$,
bounding it we get $d>\phi(d)\geq2^{v_{2}(n!)+1}$, it is easy to show that the expression with power of $2$ is bigger than $2n+6$ for $n\geq6$
Z K Y
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cursed_tangent1434
623 posts
#3 • 2 Y
Y by ehuseyinyigit, reni_wee
Extremely trivial. Same essential idea as the above post but with more details.

Note that if $p \mid d$ is a prime, then clearly $p \not \in \{2,3\}$ since the left hand side is relatively prime to $6$. Thus,
\begin{align*}
6^{n!} &\equiv -1 \pmod{p}\\
6^{2n!} & \equiv 1 \pmod{p}
\end{align*}Hence, $\text{ord}_p(6)\mid 2n!$ but $\text{ord}_p(6) \nmid n!$ implying that $\nu_2(\text{ord}_p(6)) = \nu_2(n!)+1$. But then, since $\text{ord}_p(6) \mid p-1$,
\[\nu_2(p-1) \ge \nu_2(\text{ord}_p(6)) \ge \nu_2(n!)+1\]Thus, $p-1 \ge 2^{\nu_2(n!)+1}$. But note that this implies,
\[d \ge p \ge 2^{\nu_2(n!)+1}+1 \ge 2^{1+\lfloor \frac{n}{2}\rfloor+ \lfloor \frac{n}{4} \rfloor + \dots }+1 \ge 2^{2+\lfloor\frac{n}{2}\rfloor}+1\]by Legendre's Formula since $n \ge 6$ (so $\frac{n}{4}>1$).

However,
\[d \ge 2^{2+\lfloor \frac{n}{2} \rfloor}+1 > 2^{\frac{n+3}{2}}> 2(n+3)\]since it is easy to see that $2^{\frac{n+3}{2}} > 2(n+3)$ for all $n \ge 6$ which is indeed the desired result.
Z K Y
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reni_wee
45 posts
#4
Y by
Let $p \mid d \mid 6^{n!} +1$; be a prime.
First of all $p = 2,3$ obviously doesn't work.
When $p = 5$, $6^{n!} + 1 \equiv 2(\text{ mod }5 )$. $\therefore p\neq 5$ as well.
$$6^{n!} \equiv -1 (\text{mod }p)$$$$6^{2n!} \equiv 1 (\text{mod }p)$$Let $ord_{p}6 = a. \implies a\mid 2n!,  a\nmid n!$. $\implies v_2(a) = v_2(n!) + 1$.
From Fermat's Little theorem, $a \mid p-1$
$$v_2(p-1) \geq v_2(a) = v_2(n!) +1,  $$Thus eliminating $a$. Let $v_2(n!) = x ; x\in\mathbb{Z^{+}}$. Then $p-1\geq 2^{x+1}y$; $y\in\mathbb{Z^+}$
From Legendre's
$$ v_2(n!) = \sum_{i = 1}^{\infty} \lfloor{\frac{n}{2^i}}\rfloor$$$$\therefore p-1 \geq 2^{\sum_{i = 1}^{\infty} \lfloor{\frac{n}{2^i}}\rfloor} y$$As $n\geq 6$
$$p-1 \geq 2^{1+\lfloor{\frac{n}{2}}\rfloor + \lfloor{\frac{n}{4}}\rfloor}y$$$$           \geq 2^{\frac{3n-1}{4}}y $$As $\lfloor{\frac{n}{4}}\rfloor \geq \frac{n-3}{4}, \lfloor{\frac{n}{2}}\rfloor \geq \frac{n-1}{2}.$

We can see that when $n\geq6, 2^{\frac{3n-1}{4}}\geq 2n+6$
$$\therefore p-1 \geq 2^{\frac{3n-1}{4}}y > 2n+6$$$$d \geq p > 2n +6 $$
Z K Y
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straight
414 posts
#5
Y by
Cute problem
Z K Y
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