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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
Euler Line Madness
raxu   75
N 31 minutes ago by lakshya2009
Source: TSTST 2015 Problem 2
Let ABC be a scalene triangle. Let $K_a$, $L_a$ and $M_a$ be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of $AK_aL_a$ intersects $AM_a$ a second time at point $X_a$ different from A. Define $X_b$ and $X_c$ analogously. Prove that the circumcenter of $X_aX_bX_c$ lies on the Euler line of ABC.
(The Euler line of ABC is the line passing through the circumcenter, centroid, and orthocenter of ABC.)

Proposed by Ivan Borsenco
75 replies
+1 w
raxu
Jun 26, 2015
lakshya2009
31 minutes ago
Own made functional equation
Primeniyazidayi   8
N 42 minutes ago by MathsII-enjoy
Source: own(probably)
Find all functions $f:R \rightarrow R$ such that $xf(x^2+2f(y)-yf(x))=f(x)^3-f(y)(f(x^2)-2f(x))$ for all $x,y \in \mathbb{R}$
8 replies
Primeniyazidayi
May 26, 2025
MathsII-enjoy
42 minutes ago
IMO ShortList 2002, geometry problem 7
orl   110
N an hour ago by SimplisticFormulas
Source: IMO ShortList 2002, geometry problem 7
The incircle $ \Omega$ of the acute-angled triangle $ ABC$ is tangent to its side $ BC$ at a point $ K$. Let $ AD$ be an altitude of triangle $ ABC$, and let $ M$ be the midpoint of the segment $ AD$. If $ N$ is the common point of the circle $ \Omega$ and the line $ KM$ (distinct from $ K$), then prove that the incircle $ \Omega$ and the circumcircle of triangle $ BCN$ are tangent to each other at the point $ N$.
110 replies
orl
Sep 28, 2004
SimplisticFormulas
an hour ago
Cute NT Problem
M11100111001Y1R   6
N an hour ago by X.Allaberdiyev
Source: Iran TST 2025 Test 4 Problem 1
A number \( n \) is called lucky if it has at least two distinct prime divisors and can be written in the form:
\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]where \( p_1, \dots, p_k \) are distinct prime numbers that divide \( n \). (Note: it is possible that \( n \) has other prime divisors not among \( p_1, \dots, p_k \).) Prove that for every prime number \( p \), there exists a lucky number \( n \) such that \( p \mid n \).
6 replies
M11100111001Y1R
May 27, 2025
X.Allaberdiyev
an hour ago
Possible values of determinant of 0-1 matrices
mathematics2004   4
N 3 hours ago by loup blanc
Source: 2021 Simon Marais, A3
Let $\mathcal{M}$ be the set of all $2021 \times 2021$ matrices with at most two entries in each row equal to $1$ and all other entries equal to $0$.
Determine the size of the set $\{ \det A : A \in M \}$.
Here $\det A$ denotes the determinant of the matrix $A$.
4 replies
mathematics2004
Nov 2, 2021
loup blanc
3 hours ago
ISI UGB 2025
Entrepreneur   1
N 3 hours ago by Knight2E4
Source: ISI UGB 2025
1.)
Suppose $f:\mathbb R\to\mathbb R$ is differentiable and $|f'(x)|<\frac 12\;\forall\;x\in\mathbb R.$ Show that for some $x_0\in\mathbb R,f(x_0)=x_0.$

3.)
Suppose $f:[0,1]\to\mathbb R$ is differentiable with $f(0)=0.$ If $|f'(x)|\le f(x)\;\forall\;x\in[0,1],$ then show that $f(x)=0\;\forall\;x.$

4.)
Let $S^1=\{z\in\mathbb C:|z|=1\}$ be the unit circle in the complex plane. Let $f:S^1\to S^1$ be the map given by $f(z)=z^2.$ We define $f^{(1)}:=f$ and $f^{(k+1)}=f\circ f^{(k)}$ for $k\ge 1.$ The smallest positive integer $n$ such that $f^n(z)=z$ is called period of $z.$ Determine the total number of points $S^1$ of period $2025.$

6.)
Let $\mathbb N$ denote the set of natural numbers, and let $(a_i,b_i), 1\le i\le 9,$ be nine distinct tuples in $\mathbb N\times\mathbb N.$ Show that there are $3$ distinct elements in the set $\{2^{a_i}3^{b_i}:1\le i\le 9\}$ whose product is a perfect cube.

8.)
Let $n\ge 2$ and let $a_1\le a_2\le\cdots\le a_n$ be positive integers such that $$\sum_{i=1}^n a_i=\prod_{i=1}^n a_i.$$Prove that $$\sum_{i=1}^n a_i\le 2n$$and determine when equality holds.
1 reply
Entrepreneur
May 27, 2025
Knight2E4
3 hours ago
Recurrence trouble
SomeonecoolLovesMaths   3
N 3 hours ago by Knight2E4
Let $0 < x_0 < y_0$ be real numbers. Define $x_{n+1} = \frac{x_n + y_n}{2}$ and $y_{n+1} = \sqrt{x_{n+1}y_n}$.
Prove that $\lim_{n \to \infty} x_n = \lim_{n \to \infty} y_n$ and hence find the limit.
3 replies
SomeonecoolLovesMaths
May 28, 2025
Knight2E4
3 hours ago
Trigo or Complex no.?
hzbrl   5
N Today at 9:20 AM by GreenKeeper
(a) Let $y=\cos \phi+\cos 2 \phi$, where $\phi=\frac{2 \pi}{5}$. Verify by direct substitution that $y$ satisfies the quadratic equation $2 y^2=3 y+2$ and deduce that the value of $y$ is $-\frac{1}{2}$.
(b) Let $\theta=\frac{2 \pi}{17}$. Show that $\sum_{k=0}^{16} \cos k \theta=0$
(c) If $z=\cos \theta+\cos 2 \theta+\cos 4 \theta+\cos 8 \theta$, show that the value of $z$ is $-(1-\sqrt{17}) / 4$.



I could solve (a) and (b). Can anyone help me with the 3rd part please?
5 replies
hzbrl
May 27, 2025
GreenKeeper
Today at 9:20 AM
Quadruple Binomial Coefficient Sum
P162008   3
N Today at 4:28 AM by pineconee
Source: Self made by my Elder brother
$\sum_{p=0}^{\infty} \sum_{r=0}^{\infty} \sum_{q=1}^{\infty} \sum_{s=0}^{p+q - 1} \frac{((-1)^{p+r+s+1})(2^{p+q-1}) \binom{p + q - s - 1}{p + q - 2s - 1}}{4^s(2p^2q + 2pqr + pq + qr)(2p + 2q + 2r + 3)}.$
3 replies
P162008
Yesterday at 8:04 PM
pineconee
Today at 4:28 AM
2023 Putnam A1
giginori   29
N Yesterday at 10:52 PM by kidsbian
For a positive integer $n$, let $f_n(x)=\cos (x) \cos (2 x) \cos (3 x) \cdots \cos (n x)$. Find the smallest $n$ such that $\left|f_n^{\prime \prime}(0)\right|>2023$.
29 replies
giginori
Dec 3, 2023
kidsbian
Yesterday at 10:52 PM
A MATHEMATICA E BONITA
P162008   0
Yesterday at 7:54 PM
Source: Self made by my Elder brother
Let $K = \sum_{i=0}^{\infty} \sum_{j=0}^{\infty}\sum_{m=0}^{\infty}\sum_{l=0}^{\infty} \frac{1}{(i+j+m+l)!}$ where $i,j,k$ and $l \in W.$

Now, consider the ratio $Z$ defined as
$Z = \frac{\sum_{r=0}^{\lfloor k \rfloor} \sum_{i=0}^{\lfloor k \rfloor} (-1)^r \binom{\lfloor k \rfloor}{r}(\lfloor k \rfloor - r)^i}{\sum_{r=0}^{\lfloor k \rfloor + 1}(-1)^r\binom{\lfloor k \rfloor + 1}{r}(\lfloor k \rfloor + 1 - r)^{\lfloor k \rfloor + 1}}.$

The summation function $S(n)$ is given by
$S(n) = \sum_{j=1}^{n} \left(\binom{n}{j} (j!)\left(\sum_{b=0}^{j} \frac{(-1)^b}{b!}\right)\right)$

Let $p$ denotes the number of points of intersection between the curves
$x^2 + y^2 - \tan(e^x) - \frac{|x|}{\sin y} = 0, (x\sin (a))^y + (y - x\cos(a))^x = |a|.$

Define $A(m)$ as
$A(m) = p\left(\sum_{k=0}^{m} \binom{2m + 1}{k} ((2m + 1) - 2k) (-1)^k\right).$

The value of $X$ is
$X = \lim_{n \to \infty} \frac{\sqrt{n}}{e^n} \text{exp} \left(\int_{0}^{\infty} \lfloor ne^{-x} \rfloor \text{dx}\right).$

And, $8Y =$ Number of subsets of $\left(1,2,3,\cdots,100\right)$ whose sum of elements is divisible by $5.$

Finally, compute the value of $\frac{1}{Z} +S(4) + 1 + e^{A(20)} + X\sqrt{8\pi} + Y.$
0 replies
P162008
Yesterday at 7:54 PM
0 replies
Ultra-hyper saddle with logarithmic weight
randomperson1021   0
Yesterday at 5:22 PM
Fix integers \(k\ge 3\) and \(1<r<k\), a parameter \(\lambda>0\), and a real log-exponent \(\beta\in\mathbb R\). For every real \(a\) define
$$
F_{a,\beta}^{(k,r)}(x)
  \;:=\;
  \sum_{n\ge 1}
       n^{\,a}\,(\log n)^{\beta}\,e^{\lambda n^{r}}\,x^{\,n^{k}},
  \qquad 0\le x<1.
$$
Put
$$
\Lambda_{k,r,\lambda}
   \;:=\;
   \lambda\!\left(1-\frac{r}{k}\right)
   \left(\frac{\lambda r}{k}\right)^{\!\frac{r}{\,k-r\,}},
   \qquad
   \gamma=\frac{r}{k-r}.
$$
(1) Show that there exists a real constant \(c=c(k,r)\) (independent of \(\lambda\) and of \(\beta\)) such that
$$
\lim_{x\to 1^{-}}
      F_{a,\beta}^{(k,r)}(x)\,
      e^{-\Lambda_{k,r,\lambda}\,(1-x)^{-\gamma}}
      \;=\;
      \begin{cases}
          0, & a<c,\\[6pt]
          \infty, & a>c.
      \end{cases}
$$
(2) Determine this critical value \(c\) explicitly and verify that it coincides with the classical case \(r=1\), namely \(c=-\tfrac12\).

(3) Evaluate the finite, non-zero limit that occurs at the borderline \(a=c\) (your answer may depend on \(k,r,\lambda\) but not on \(\beta\)).
0 replies
randomperson1021
Yesterday at 5:22 PM
0 replies
3rd AKhIMO for university students, P5
UzbekMathematician   1
N Yesterday at 3:53 PM by grupyorum
Source: AKhIMO 2025, P5
Show that for every positive integer $n$ there exist nonnegative integers $p, q$ and integers $a_1, a_2, ... , a_p, b_1, b_2, ... , b_q \ge 2$ such that $$ n=\frac{(a_1^3-1)(a_2^3-1)...(a_p^3-1)}{(b_1^3-1)(b_2^3-1)...(b_q^3-1)} $$
1 reply
UzbekMathematician
May 28, 2025
grupyorum
Yesterday at 3:53 PM
Sum of three squares
perfect_radio   9
N Yesterday at 1:36 PM by RobertRogo
Source: RMO 2004, Grade 12, Problem 4
Let $\mathcal K$ be a field of characteristic $p$, $p \equiv 1 \left( \bmod 4 \right)$.

(a) Prove that $-1$ is the square of an element from $\mathcal K.$

(b) Prove that any element $\neq 0$ from $\mathcal K$ can be written as the sum of three squares, each $\neq 0$, of elements from $\mathcal K$.

(c) Can $0$ be written in the same way?

Marian Andronache
9 replies
perfect_radio
Feb 26, 2006
RobertRogo
Yesterday at 1:36 PM
Easy IMO 2023 NT
799786   133
N Apr 24, 2025 by Maximilian113
Source: IMO 2023 P1
Determine all composite integers $n>1$ that satisfy the following property: if $d_1$, $d_2$, $\ldots$, $d_k$ are all the positive divisors of $n$ with $1 = d_1 < d_2 < \cdots < d_k = n$, then $d_i$ divides $d_{i+1} + d_{i+2}$ for every $1 \leq i \leq k - 2$.
133 replies
799786
Jul 8, 2023
Maximilian113
Apr 24, 2025
Easy IMO 2023 NT
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2023 P1
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799786
1052 posts
#1 • 8 Y
Y by mmkkll, feranjos, CrazyFok, Littlelame, emmelin, KhaiMathAddict, cubres, ItsBesi
Determine all composite integers $n>1$ that satisfy the following property: if $d_1$, $d_2$, $\ldots$, $d_k$ are all the positive divisors of $n$ with $1 = d_1 < d_2 < \cdots < d_k = n$, then $d_i$ divides $d_{i+1} + d_{i+2}$ for every $1 \leq i \leq k - 2$.
This post has been edited 7 times. Last edited by v_Enhance, Sep 18, 2023, 12:33 AM
Reason: minor latex pet peeve
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qwedsazxc
167 posts
#2 • 9 Y
Y by srijan30pq, math90, akliu, Mehrshad, combo_nt_lover, Math.1234, KhaiMathAddict, cubres, Want-to-study-in-NTU-MATH
Assume $p<q$ be the smallest two prime divisors of $n$. Then $d_{k-1}=\frac{n}{p}$.
Assume $d_m=\frac{n}{q}$ for some $m$, and $d_{m+1}=\frac{n}{p^{c+1}}$ and $d_{m+2}=\frac{n}{p^c}$ for some nonnegative integer $c$.
Then, since $d_m\mid d_{m+1}+d_{m+2}$, $p^{c+1}\mid q+pq$ and $p|q$ which is a contradiction.
Therefore $n$ does not have two distinct prime factors; $n$=$p^t$ for some prime $p$ and a positive integer $t\neq1$. It's easy to show that this suffices.
This post has been edited 4 times. Last edited by qwedsazxc, Jul 8, 2023, 1:46 PM
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Tintarn
9045 posts
#3 • 13 Y
Y by PNT, math90, IAmTheHazard, Assassino9931, combo_nt_lover, Joider, Math.1234, Jia_Le_Kong, dhfurir, Sedro, Acorn-SJ, aidan0626, H_Taken
Solution
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Seungjun_Lee
526 posts
#5 • 2 Y
Y by dgkim, Kingsbane2139
From $d_{k-2} \mid d_{k-1} + d_{k}$ one can get that $d_{k-2} \mid d_{k-1}$ since $d_{k-2} \mid n = d_k$
Then $d_2 = \frac{n}{d_{k-1}} \mid \frac{n}{d_{k-2}} = d_3$ so $d_2 \mid d_3$. Here, since $d_2 \mid d_3 + d_4$, we get that $d_2 \mid d_4$.
Again, we can easily show $d_{k-3} | d_{k-1}$ in the same way, which leads to $d_{k-3} \mid d_{k-2}$

By induction, $d_1 \mid d_2 \mid \cdots \mid d_k$

If $p \mid n$ and $q \mid n$ for prime $p > q$
$\frac{n}{p} \mid \frac{n}{q}$ so $q \mid p$ and this is contradiction

Hence, $n = p^{k-1}$
This indeed fits the condition
This post has been edited 3 times. Last edited by Seungjun_Lee, Jun 24, 2024, 1:57 AM
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ihatemath123
3449 posts
#6 • 4 Y
Y by Inconsistent, centslordm, channing421, Zhaom
The answer is prime powers, which obviously work.

Note that the three largest divisors of $n$ are either $\left \{ \tfrac{n}{q}, \tfrac{n}{p}, n \right \}$ for distinct prime divisors $p$ and $q$ of $n$, or $\left \{ \tfrac{n}{p^2}, \tfrac{n}{p}, n \right \}$ for some prime divisor $p$ of $n$.

In the former case we have a contradiction, since
\[\frac{\frac{n}{p} + n}{\frac{n}{q}} = \frac{q(p+1)}{p},\]obviously not an integer.

So, the three largest divisors of $n$ are $\frac{n}{p^2}, \frac{n}{p}$ and $n$.

With similar reasoning, now consider the fourth largest divisor. It is either $\tfrac{n}{q}$ or $\tfrac{n}{p^3}$; the former option fails, since
\[ \frac{\frac{n}{p} + \frac{n}{p^2}}{\frac{n}{q}} = \frac{q(p+1)}{p^2}, \]obviously not an integer either.

We repeat this reasoning to deduce that $n$ is just $p^{k-1}$.
This post has been edited 4 times. Last edited by ihatemath123, Jul 18, 2024, 11:33 PM
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GrantStar
821 posts
#7 • 4 Y
Y by centslordm, jeff10, OronSH, MulhamAgam
HUH how is this IMO level?
sol
This post has been edited 1 time. Last edited by GrantStar, Jul 8, 2023, 5:27 AM
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bobthegod78
2982 posts
#8 • 1 Y
Y by centslordm
The answer is $p^e$ for all primes $p$ and $e>1$, which obviously works. Let $p$ be the smallest prime factor of $n$, we claim that $p$ is the only prime factor of $n$. We use induction downward to prove $d_i = \frac{n}{p^{k-i}}$. The base case is obvious. For the inductive step, if we instead had a different prime $q$, then
\[
\frac{n}{q} \mid \frac{n}{p^{k-i}} + \frac{n}{p^{k-i-1}},
\]but this is obviously false. We are done.
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LoloChen
479 posts
#9 • 1 Y
Y by GeoKing
This NT seems familiar. Maybe it's an old problem?
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carefully
241 posts
#10
Y by
LoloChen wrote:
This NT seems familiar. Maybe it's an old problem?
The only related problem I can think of is IMO 2002 P4, but it's not even close.
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Seungjun_Lee
526 posts
#11
Y by
LoloChen wrote:
This NT seems familiar. Maybe it's an old problem?

Maybe JBMO 2002 is similar with this problem
but this is even easier than that
https://artofproblemsolving.com/community/c6h58610p358021
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Seungjun_Lee
526 posts
#12 • 1 Y
Y by ihatemath123
ihatemath123 wrote:
The answer is prime powers, which obviously work.

Note that the three largest divisors of $n$ are either $\left \{ \frac{n}{q}, \frac{n}{p}, n \right \}$ for distinct prime divisors $p$ and $q$ of $n$, or $\left \{ \frac{n}{p^2}, \frac{n}{p}, n \right \}$ for some prime divisor $p$ of $n$.

In the former case we have a contradiction, since
\[\frac{\frac{n}{p} + n}{\frac{n}{q}} = \frac{q(p+1)}{p},\]obviously not an integer.

So, the three largest divisors of $n$ are $\frac{n}{p^2}, \frac{n}{p}$ and $n$.

With similar reasoning, now consider the fourth largest divisor. It is either $\frac{n}{q}$ or $\frac{n}{p^3}$; the former option fails, since
\[ \frac{\frac{n}{p} + \frac{n}{p^2}}{\frac{n}{q}} = \frac{q(p+1)}{p^2}, \]obviously not an integer either.

We repeat this reasoning to deduce that $n$ is just $p^k$.

I think you have a very small mistake
isn't it $p^{k-1}$??
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Anzoteh
126 posts
#13
Y by
I'll only show the more technical part below.

Partial solution
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beansenthusiast505
26 posts
#14 • 1 Y
Y by Tellocan
For any $n$, the 3 largest divisors are $n,\frac{n}{p},\frac{n}{q}$ for primes $p<q$ or $n,\frac{n}{p},\frac{n}{p^2}$.

However, first case doesn't work because $\frac{\frac{n}{p}+n}{\frac{n}{q}}=\frac{q}{p}+q$ which is not an integer.

Repeat for further cases. All should not work for $q$ as you will get $\frac{q(p+1)}{p^k}$ which are not integers. Thus only $p^{k-1}$ work


well there goes 10M N prediction more like 0M now
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Supertinito
45 posts
#15 • 38 Y
Y by Tintarn, JingheZhang, khan.academy, steppewolf, pieater314159, math90, oolite, Filipjack, eibc, TSandino, Creeper1612, ike.chen, TheStrayCat, Aryan-23, Sagnik123Biswas, leonhard8128, akliu, rg_ryse, spiritshine1234, LolitaLaLolita, Packito, GreatKillaOE, ihatemath123, EpicBird08, Madyyy, Alex-131, jestrada, Reakniseb, starchan, Sedro, antimonio, ATGY, khina, Marcus_Zhang, MELSSATIMOV40, aidan0626, NicoN9, SteppenWolfMath
This problem was proposed by me, Santiago Rodriguez from Colombia. I hope that you enjoy it. :)
This post has been edited 2 times. Last edited by Supertinito, Jul 16, 2023, 7:45 PM
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yofro
3151 posts
#16
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Far too easy for the IMO. It's just prime powers, which trivially work. Suppose $n$ isn't a prime power. Clearly the list of divisors goes $\left(1,\cdots,\frac{n}{q}, \frac{n}{p^k},\cdots,\frac{n}{p},n\right)$ for some primes $p,q$ and some $k\in\mathbb{N}$. Thus $\frac{n}{q}\mid \frac{n}{p^k}+\frac{n}{p^{k-1}}$ and hence
$$\frac{q(p+1)}{p^{k-1}}\in\mathbb{Z}$$Which is impossible unless $k=1$, and for $k=1$ we need $\frac{n+n/p}{n/q}$ to be an integer, which is impossible.
This post has been edited 1 time. Last edited by yofro, Jul 8, 2023, 6:15 AM
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