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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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0 replies
jlacosta
May 1, 2025
0 replies
Factorial Divisibility
Aryan-23   47
N a minute ago by ezpotd
Source: IMO SL 2022 N2
Find all positive integers $n>2$ such that
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47 replies
Aryan-23
Jul 9, 2023
ezpotd
a minute ago
2-var inequality
sqing   3
N 25 minutes ago by sqing
Source: Own
Let $ a,b> 0 ,a^3+ab+b^3=3.$ Prove that
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1 viewing
sqing
an hour ago
sqing
25 minutes ago
Infinite number of sets with an intersection property
Drytime   8
N 42 minutes ago by math90
Source: Romania TST 2013 Test 2 Problem 4
Let $k$ be a positive integer larger than $1$. Build an infinite set $\mathcal{A}$ of subsets of $\mathbb{N}$ having the following properties:

(a) any $k$ distinct sets of $\mathcal{A}$ have exactly one common element;
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Drytime
Apr 26, 2013
math90
42 minutes ago
Factorials divide
va2010   37
N an hour ago by ND_
Source: 2015 ISL N2
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va2010
Jul 7, 2016
ND_
an hour ago
IMC 1994 D2 P1
j___d   13
N Yesterday at 11:20 PM by krigger
Let $f\in C^1[a,b]$, $f(a)=0$ and suppose that $\lambda\in\mathbb R$, $\lambda >0$, is such that
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j___d
Mar 6, 2017
krigger
Yesterday at 11:20 PM
D1039 : A strange and general result on series
Dattier   0
Yesterday at 10:33 PM
Source: les dattes à Dattier
Let $f \in C([0,1];[0,1])$ bijective, $f(0)=0$ and $(a_k) \in [0,1]^\mathbb N$ with $ \sum \limits_{k=0}^{+\infty} a_k$ converge.

Is it true that $\sum \limits_{k=0}^{+\infty} f(a_k)\times f^{-1}(a_k)$ converge?
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Dattier
Yesterday at 10:33 PM
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Aproximate ln(2) using perfect numbers
YLG_123   5
N Yesterday at 8:55 PM by ei_killua_
Source: Brazilian Mathematical Olympiad 2024, Level U, Problem 1
A positive integer \(n\) is called perfect if the sum of its positive divisors \(\sigma(n)\) is twice \(n\), that is, \(\sigma(n) = 2n\). For example, \(6\) is a perfect number since the sum of its positive divisors is \(1 + 2 + 3 + 6 = 12\), which is twice \(6\). Prove that if \(n\) is a positive perfect integer, then:
\[
\sum_{p|n} \frac{1}{p + 1} < \ln 2 < \sum_{p|n} \frac{1}{p - 1}
\]where the sums are taken over all prime divisors \(p\) of \(n\).
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YLG_123
Oct 12, 2024
ei_killua_
Yesterday at 8:55 PM
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P162008   4
N Yesterday at 8:40 PM by vmene
Source: Self made by my Elder brother
$\sum_{p=0}^{\infty} \sum_{r=0}^{\infty} \sum_{q=1}^{\infty} \sum_{s=0}^{p+q - 1} \frac{((-1)^{p+r+s+1})(2^{p+q-1}) \binom{p + q - s - 1}{p + q - 2s - 1}}{4^s(2p^2q + 2pqr + pq + qr)(2p + 2q + 2r + 3)}.$
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P162008
Thursday at 8:04 PM
vmene
Yesterday at 8:40 PM
IMC 1994 D1 P5
j___d   5
N Yesterday at 5:39 PM by krigger
a) Let $f\in C[0,b]$, $g\in C(\mathbb R)$ and let $g$ be periodic with period $b$. Prove that $\int_0^b f(x) g(nx)\,\mathrm dx$ has a limit as $n\to\infty$ and
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j___d
Mar 6, 2017
krigger
Yesterday at 5:39 PM
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giginori   21
N Yesterday at 3:32 PM by pie854
Let $n$ be an even positive integer. Let $p$ be a monic, real polynomial of degree $2 n$; that is to say, $p(x)=$ $x^{2 n}+a_{2 n-1} x^{2 n-1}+\cdots+a_1 x+a_0$ for some real coefficients $a_0, \ldots, a_{2 n-1}$. Suppose that $p(1 / k)=k^2$ for all integers $k$ such that $1 \leq|k| \leq n$. Find all other real numbers $x$ for which $p(1 / x)=x^2$.
21 replies
giginori
Dec 3, 2023
pie854
Yesterday at 3:32 PM
Putnam 2019 A1
awesomemathlete   33
N Yesterday at 3:25 PM by cursed_tangent1434
Source: 2019 William Lowell Putnam Competition
Determine all possible values of $A^3+B^3+C^3-3ABC$ where $A$, $B$, and $C$ are nonnegative integers.
33 replies
awesomemathlete
Dec 10, 2019
cursed_tangent1434
Yesterday at 3:25 PM
IMC 1994 D1 P2
j___d   5
N Yesterday at 3:11 PM by krigger
Let $f\in C^1(a,b)$, $\lim_{x\to a^+}f(x)=\infty$, $\lim_{x\to b^-}f(x)=-\infty$ and $f'(x)+f^2(x)\geq -1$ for $x\in (a,b)$. Prove that $b-a\geq\pi$ and give an example where $b-a=\pi$.
5 replies
j___d
Mar 6, 2017
krigger
Yesterday at 3:11 PM
A Construction in Multivariable Analysis
MrOrange   0
Yesterday at 2:11 PM
Source: Garling's A COURSE IN MATHEMATICAL ANALYSIS
Construct a continuous real valued function \( f \) on \( \mathbb{R}^2 \) for which
\[
\lim_{R \to \infty} \int_{\|x\|_2 \leq R} f(x) \, dx = 0
\]and for which
\[
\lim_{R \to \infty} \int_{\|x\|_\infty \leq R} f(x) \, dx \text{ does not exist.}
\]
0 replies
MrOrange
Yesterday at 2:11 PM
0 replies
Possible values of determinant of 0-1 matrices
mathematics2004   4
N Yesterday at 1:56 PM 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
Yesterday at 1:56 PM
A problem about equal areas
Henry_2001   9
N Sep 5, 2022 by HamstPan38825
Source: 2000 China Second Round Olympiad P1
In acute-angled triangle $ABC,$ $E,F$ are on the side $BC,$ such that $\angle BAE=\angle CAF,$ and let $M,N$ be the projections of $F$ onto $AB,AC,$ respectively. The line $AE$ intersects $ \odot (ABC) $ at $D$(different from point $A$).
Prove that $S_{AMDN}=S_{\triangle ABC}.$
9 replies
Henry_2001
Aug 13, 2019
HamstPan38825
Sep 5, 2022
A problem about equal areas
G H J
G H BBookmark kLocked kLocked NReply
Source: 2000 China Second Round Olympiad P1
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Henry_2001
165 posts
#1 • 2 Y
Y by Adventure10, Mango247
In acute-angled triangle $ABC,$ $E,F$ are on the side $BC,$ such that $\angle BAE=\angle CAF,$ and let $M,N$ be the projections of $F$ onto $AB,AC,$ respectively. The line $AE$ intersects $ \odot (ABC) $ at $D$(different from point $A$).
Prove that $S_{AMDN}=S_{\triangle ABC}.$
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coldheart361
26 posts
#2 • 1 Y
Y by Adventure10
Can you give me small hint
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Limerent
39 posts
#3 • 1 Y
Y by Mango247
We start with mentioning that $\Delta MNF \sim \Delta CBD.$
proof
Lemma 1: $\frac{AB\cdot CD}{BD}+AC=\frac{AD\cdot BC}{BD}$
proof
We want $$S_{ABC}=S_{AMDN}$$$$S_{ABF}+S_{AFC}=S_{AMD}+S_{ADN}$$$$\frac{1}{2}AB\cdot AFsin(\angle BAF)+\frac{1}{2}AF\cdot ACsin(\angle FAC)=\frac{1}{2}AM\cdot ADsin(\angle FAC)+\frac{1}{2}AD\cdot ANsin(\angle BAF)$$$$AF(AB\cdot \frac{MF}{AF}+AC\cdot\frac{NF}{AF})=AD(AM\cdot\frac{NF}{AF}+AN\cdot\frac{MF}{AF})$$$$AF(AB\cdot MF+AC\cdot NF)=AD(AM\cdot NF+AN\cdot MF)$$Divide by $FN$ and use $\frac{MF}{NF}=\frac{CD}{BD}$.
$$AF(AB\cdot \frac{CD}{BD}+AC)=AD(AM+AN\cdot \frac{CD}{BD})$$Using Lemma 1: $$\frac{AF\cdot BC}{BD}=AM+AN\cdot\frac{CD}{BD}$$This is equivalent to $$AF=AM\cdot \frac{NF}{NM}+AN\cdot\frac{NF}{NM}$$Which is just Ptolemy theorem for $AMFN$.

[asy]
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pen dotstyle = black; /* point style */ 
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draw((-7.69687926965853,-0.48362131590244306)--(4.904379462048781,-0.5778950720000042), linewidth(1)); 
draw((4.904379462048781,-0.5778950720000042)--(1.7619209254634167,8.7237821962927), linewidth(1)); 
draw(circle((-1.3737046562445991,2.482789896605576), 6.984420711370107), linewidth(1) + rvwvcq); 
draw((1.7619209254634167,8.7237821962927)--(-3.6152839988300443,-0.5141569413699879), linewidth(1) + dtsfsf); 
draw((1.7619209254634167,8.7237821962927)--(2.4172067581662664,-3.3833096014855637), linewidth(1) + dtsfsf); 
draw((-5.616382898360874,1.5415794707760124)--(4.01227895316257,2.0627224343031854), linewidth(1) + rvwvcq); 
draw((-5.616382898360874,1.5415794707760124)--(-3.6152839988300443,-0.5141569413699879), linewidth(1) + rvwvcq); 
draw((-3.6152839988300443,-0.5141569413699879)--(4.01227895316257,2.0627224343031854), linewidth(1) + rvwvcq); 
 /* dots and labels */
dot((-7.69687926965853,-0.48362131590244306),dotstyle); 
label("$B$", (-7.625027712498727,-0.3227630623057912), NE * labelscalefactor); 
dot((4.904379462048781,-0.5778950720000042),dotstyle); 
label("$C$", (4.965082120946319,-0.40526705859311163), NE * labelscalefactor); 
dot((-3.6152839988300443,-0.5141569413699879),dotstyle); 
label("$F$", (-3.549330295905114,-0.3557646608207194), NE * labelscalefactor); 
dot((2.4172067581662664,-3.3833096014855637),linewidth(4pt) + dotstyle); 
label("$D$", (2.4899622323267168,-3.243404530876935), NE * labelscalefactor); 
dot((-5.616382898360874,1.5415794707760124),linewidth(4pt) + dotstyle); 
label("$M$", (-5.54592700605826,1.6738336478473634), NE * labelscalefactor); 
dot((4.01227895316257,2.0627224343031854),linewidth(4pt) + dotstyle); 
label("$N$", (4.074038961043263,2.2018592240862143), NE * labelscalefactor); 
dot((2.2642971136055023,-0.5581438324854661),linewidth(4pt) + dotstyle); 
label("$E$", (2.324954239752077,-0.42176785785057574), NE * labelscalefactor); 
dot((1.761920925463417,8.7237821962927),linewidth(4pt) + dotstyle); 
label("$A$", (1.8299302620281561,8.851681324844241), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]
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Hayalim
27 posts
#4 • 3 Y
Y by Mango247, Mango247, Mango247
The second line to the right of the last line | NF | not | MF | should be. Also congratulations on the very clever solution "Limerent"
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Jufri
68 posts
#5
Y by
$S_{AMDN}$ what is it?
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JG666
287 posts
#6
Y by
Jufri wrote:
$S_{AMDN}$ what is it?

It's the area of the quadrilateral $AMDN$.
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Muke_Yang
13 posts
#7 • 1 Y
Y by Slim.Shady
notice that $MN$ is perpendicular to $AD$so we know $S_{AMDN} = \frac{1}{2} AD \cdot MN$
set $\angle BAD=\angle CAF=\theta$,so we have $\frac{AD}{sin(C+\theta)}=\frac{c}{sinC}=2R$,$AD=2R\cdot sin(C+\theta)$
and we know $\frac{MN}{sinA}=\frac{AM}{sin\angle ANM}=\frac{AM}{sin\angle AFM}=AF$,$\frac{AF}{sinC}=\frac{b}{sin(C+\theta)}$
so,$MN=AF\cdot sinA=\frac{b\cdot sinC \cdot sinA}{sin(C+\theta)}$ ,$S_{AMDN}=\frac{1}{2} AD \cdot MN=\frac{1}{2}(2RsinC) \cdot bsinA=\frac{1}{2} bcsinA=S_{ABC}$

$Q.E.D.$
Attachments:
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Slim.Shady
14 posts
#8 • 1 Y
Y by Muke_Yang
Denote $\alpha = \angle BAE =\angle CAF$.
$S_{AMDN}=S_{\triangle AMD}+S_{\triangle AND}=\frac{1}{2}AD\cdot AM\cdot \sin\alpha +\frac{1}{2}AD\cdot AN\cdot \sin(A-\alpha)=\frac{1}{2}AD\cdot (AF\cos(A-\alpha))\cdot \sin\alpha+\frac{1}{2}AD\cdot (AF\cos\alpha)\cdot \sin(A-\alpha)=\frac{1}{2}AD\cdot AF\cdot \sin A$.
Since $\angle BAE =\angle CAF$, it's easy to know that $\triangle ABD \sim \triangle AFC$, from which we can get $AB\cdot AC=AD\cdot AF$.
So the expression above equals to $\frac{1}{2}AB\cdot AC\cdot \sin A$, which is the area of $\triangle ABC$, as desired.
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bestmonitors
1 post
#9
Y by
dinh? ban la nhat
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HamstPan38825
8869 posts
#10
Y by
Let $\theta = \angle BAD = \angle FAC$. We have
\begin{align*}
2[AMDN] &= 2([AMD]+[AND]) \\
&= AD(AM \sin \theta + AN \sin(A-\theta)) \\
&= AD(AF \sin \theta \cos(A-\theta) + \cos \theta \sin(A-\theta)) \\
&= AD \cdot AF \cdot \sin \theta.
\end{align*}So it suffices to show that $AD \cdot AF = AB \cdot AC$, but this is simply true because $\triangle ABD \sim \triangle AFC$.
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