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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.

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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
ISI UGB 2025 P4
SomeonecoolLovesMaths   4
N 25 minutes ago by Ghoshadi
Source: ISI UGB 2025 P4
Let $S^1 = \{ z \in \mathbb{C} \mid |z| =1 \}$ be the unit circle in the complex plane. Let $f \colon S^1 \longrightarrow S^2$ be the map given by $f(z) = z^2$. We define $f^{(1)} \colon = f$ and $f^{(k+1)} \colon = f \circ f^{(k)}$ for $k \geq 1$. The smallest positive integer $n$ such that $f^{(n)}(z) = z$ is called the period of $z$. Determine the total number of points in $S^1$ of period $2025$.
(Hint : $2025 = 3^4 \times 5^2$)
4 replies
SomeonecoolLovesMaths
Yesterday at 11:24 AM
Ghoshadi
25 minutes ago
The Return of Triangle Geometry
peace09   9
N an hour ago by mathfun07
Source: 2023 ISL A7
Let $N$ be a positive integer. Prove that there exist three permutations $a_1,\dots,a_N$, $b_1,\dots,b_N$, and $c_1,\dots,c_N$ of $1,\dots,N$ such that \[\left|\sqrt{a_k}+\sqrt{b_k}+\sqrt{c_k}-2\sqrt{N}\right|<2023\]for every $k=1,2,\dots,N$.
9 replies
1 viewing
peace09
Jul 17, 2024
mathfun07
an hour ago
Set Partition
Butterfly   0
an hour ago
For the set of positive integers $\{1,2,…,n\}(n\ge 3)$, no matter how its elements are partitioned into two subsets, at least one of the subsets must contain three numbers $a,b,c$ ($a=b$ is allowed) such that $ab=c$. Find the minimal $n$.
0 replies
Butterfly
an hour ago
0 replies
Points Lying on its Cevian Triangle's Thomson Cubic
Feuerbach-Gergonne   1
N an hour ago by golue3120
Source: Own
Given $\triangle ABC$ and a point $P$, let $\triangle DEF$ be the cevian triangle of $P$ with respect to $\triangle ABC$. Let $H$ be the orthocenter of $\triangle ABC$, and denote the isotomic conjugate of $H, P$ with respect to $\triangle ABC$ by $X, Q$, respectively. Let the centroid of $\triangle DEF$ be $M$, and denote the isogonal conjugate of $P$ with respect to $\triangle DEF$ by $R$. Prove that
$$
P, Q, X \text{ are collinear} \iff P, R, M \text{ are collinear}. 
$$or in brief
$$
P \in \text{ K007 of } \triangle ABC \iff P \in \text{ K002 of } \triangle DEF. 
$$
1 reply
Feuerbach-Gergonne
Jul 19, 2024
golue3120
an hour ago
No more topics!
Prime Numbers
TRcrescent27   7
N Apr 30, 2025 by Goutamioqmtopper
Source: 2015 Turkey JBMO TST
Let $p,q$ be prime numbers such that their sum isn't divisible by $3$. Find the all $(p,q,r,n)$ positive integer quadruples satisfy:
$$p+q=r(p-q)^n$$
Proposed by Şahin Emrah
7 replies
TRcrescent27
Jun 22, 2016
Goutamioqmtopper
Apr 30, 2025
Prime Numbers
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Source: 2015 Turkey JBMO TST
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TRcrescent27
32 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $p,q$ be prime numbers such that their sum isn't divisible by $3$. Find the all $(p,q,r,n)$ positive integer quadruples satisfy:
$$p+q=r(p-q)^n$$
Proposed by Şahin Emrah
This post has been edited 4 times. Last edited by TRcrescent27, Jul 25, 2017, 8:05 PM
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Wave-Particle
3690 posts
#2 • 1 Y
Y by Adventure10
A push in the right direction

I'll finish up my solution later because i have to go right now..or someone else can finish it up for me if they wish.
This post has been edited 1 time. Last edited by Wave-Particle, Jun 22, 2016, 8:15 PM
Reason: LaTeX
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RagvaloD
4913 posts
#3 • 3 Y
Y by Bariskoyuncufan, Adventure10, Mango247
$3|pq(p-q)(p+q)$ for every $p,q$
If $3|p-q \to 3|p+q$-contradiction, so $3|pq$

Case 1: $p=3$
$3+q=r(3-q)^n$
$(3-q) | 6$
$3-q =(-6,-3,-2,-1,1,2) \to q=2,5$
$q=2$: $5=r 1^n \to r=5$

$q=5$: $8=r(-2)^n \to (r,n)=(2,2)$

Case 2: $q=3$
$3+p=r(p-3)^n$
$(p-3)|6$
$p-3=(-1,1,2,3,6) \to p=2,5$

$p=2$: $5=r(-1)^n \to (r,n)=(5,2k)$
$p=5$: $8=r 2^n \to (r,n)=(1,3),(2,2),(4,1)$

Answer: $(p,q,r,n) = (3,2,5,k),(3,5,2,2),(2,3,5,2k),(5,3,1,3),(5,3,2,2),(5,3,4,1)$
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IsaacJimenez
14 posts
#4 • 1 Y
Y by Adventure10
As $n$ is a positive integer, then $p-q$ divides $r{ (p-q) }^{ n }=p+q$. Now, see that the $gcd(p+q,p-q)=gcd(p+q,2p)=gcd(2q,2p)$ which is $1$ or $2$; but, as $p-q$ divides $p+q$ then $p-q$ divides $2$.

If $p-q=\pm 1$ then is clear than $p=2$, $q=3$, $r=5$ and $n$ is even, or $p=3$, $q=2$, $r=5$ and $n$ positive integer.

If $p-q=\pm 2$, then $p$ or $q=3$; otherwise, $p$ and $q$ are $1$ and $-1$ $mod$ $6$ and $3$ divides $p+q$. So, if $q=3$ and $5=5$, sp $p+q=8$. So $n=3$ and $r=1$, $n=2$ and $r=2$ or $n=1$ and $r=8$. If $p=3$ and $q=5$ then $n=2$ and $r=2$.
This post has been edited 5 times. Last edited by IsaacJimenez, Jun 28, 2016, 4:11 PM
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Namisgood
30 posts
#5
Y by
nice question
This post has been edited 1 time. Last edited by Namisgood, Apr 18, 2025, 8:07 AM
Reason: who cares?
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Pseudo_Matter
4 posts
#6 • 1 Y
Y by Namisgood
I have attached the solution
Attachments:
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Namisgood
30 posts
#7
Y by
Pseudo_Matter wrote:
I have attached the solution

so orz
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Goutamioqmtopper
3 posts
#8
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I checked it it is correct @Pseudo_Matter
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