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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
2023 Japan Mathematical Olympiad Preliminary
parkjungmin   0
7 minutes ago
Please help me if I can solve the Chinese question
0 replies
parkjungmin
7 minutes ago
0 replies
Polynomial divisible by x^2+1
Miquel-point   1
N an hour ago by luutrongphuc
Source: Romanian IMO TST 1981, P1 Day 1
Consider the polynomial $P(X)=X^{p-1}+X^{p-2}+\ldots+X+1$, where $p>2$ is a prime number. Show that if $n$ is an even number, then the polynomial \[-1+\prod_{k=0}^{n-1} P\left(X^{p^k}\right)\]is divisible by $X^2+1$.

Mircea Becheanu
1 reply
Miquel-point
Apr 6, 2025
luutrongphuc
an hour ago
the same prime factors
andria   5
N an hour ago by bin_sherlo
Source: Iranian third round number theory P4
$a,b,c,d,k,l$ are positive integers such that for every natural number $n$ the set of prime factors of $n^k+a^n+c,n^l+b^n+d$ are same. prove that $k=l,a=b,c=d$.
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andria
Sep 6, 2015
bin_sherlo
an hour ago
A sharp estimation of the product
mihaig   0
an hour ago
Source: VL
Let $n\geq4$ and let $a_1,a_2,\ldots, a_n\geq0$ be reals such that $\sum_{i=1}^{n}{\frac{1}{2a_i+n-2}}=1.$
Prove
$$a_1+\cdots+a_n+2^{n-1}\geq n+2^{n-1}\cdot\prod_{i=1}^{n}{a_i}.$$
0 replies
mihaig
an hour ago
0 replies
Integration Bee Kaizo
Calcul8er   63
N 3 hours ago by MS_asdfgzxcvb
Hey integration fans. I decided to collate some of my favourite and most evil integrals I've written into one big integration bee problem set. I've been entering integration bees since 2017 and I've been really getting hands on with the writing side of things over the last couple of years. I hope you'll enjoy!
63 replies
Calcul8er
Mar 2, 2025
MS_asdfgzxcvb
3 hours ago
Japanese high school Olympiad.
parkjungmin   1
N 3 hours ago by GreekIdiot
It's about the Japanese high school Olympiad.

If there are any students who are good at math, try solving it.
1 reply
parkjungmin
Yesterday at 5:25 AM
GreekIdiot
3 hours ago
Already posted in HSO, too difficult
GreekIdiot   0
4 hours ago
Source: own
Find all integer triplets that satisfy the equation $5^x-2^y=z^3$.
0 replies
GreekIdiot
4 hours ago
0 replies
Square on Cf
GreekIdiot   0
4 hours ago
Let $f$ be a continuous function defined on $[0,1]$ with $f(0)=f(1)=0$ and $f(t)>0 \: \forall \: t \in (0,1)$. We define the point $X'$ to be the projection of point $X$ on the x-axis. Prove that there exist points $A, B \in C_f$ such that $ABB'A'$ is a square.
0 replies
GreekIdiot
4 hours ago
0 replies
Japanese Olympiad
parkjungmin   4
N Today at 8:55 AM by parkjungmin
It's about the Japanese Olympiad

I can't solve it no matter how much I think about it.

If there are people who are good at math

Please help me.
4 replies
parkjungmin
Saturday at 6:51 PM
parkjungmin
Today at 8:55 AM
ISI UGB 2025 P3
SomeonecoolLovesMaths   11
N Today at 8:21 AM by Levieee
Source: ISI UGB 2025 P3
Suppose $f : [0,1] \longrightarrow \mathbb{R}$ is differentiable with $f(0) = 0$. If $|f'(x) | \leq f(x)$ for all $x \in [0,1]$, then show that $f(x) = 0$ for all $x$.
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SomeonecoolLovesMaths
Yesterday at 11:32 AM
Levieee
Today at 8:21 AM
D1020 : Special functional equation
Dattier   4
N Today at 7:57 AM by Dattier
Source: les dattes à Dattier
1) Are there any $(f,g) \in C(\mathbb R,\mathbb R_+)$ increasing with
$$\forall x \in \mathbb R, f(x)(\cos(x)+3/2)+g(x)(\sin(x)+3/2)=\exp(x)$$?

2) Are there any $(f,g) \in C(\mathbb R,\mathbb R_+)$ increasing with
$$\forall x \in \mathbb R, f(x)(\cos(x)+3/2)+g(x)(\sin(x)+3/2)=\exp(x/2)$$?
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Dattier
Apr 24, 2025
Dattier
Today at 7:57 AM
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Tricky123   1
N Today at 6:57 AM by navier3072
X is continuous random variable having spectrum
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$E(X)=\int_{0}^{\infty}(1-F(x)-F(-x))dx$ and find the expression of $V(x)$

Ans:- $V(x)=\int_{0}^{\infty}(2x(1-F(x)+F(-x))dx-m^{2}$

How to solve help me
1 reply
Tricky123
Yesterday at 9:51 AM
navier3072
Today at 6:57 AM
Tough integral
Martin.s   0
Today at 4:00 AM
$$\int_0^{\pi/2}\ln(\tan(\theta/2))
\;\frac{4\cos\theta\cos(2\theta)}{4\sin^4\theta+1}\,d\theta.$$
0 replies
Martin.s
Today at 4:00 AM
0 replies
Minimum value
Martin.s   3
N Yesterday at 5:24 PM by Martin.s
What is the minimum value of
$$
\frac{|a + b + c + d| \left( |a - b| |b - c| |c - d| + |b - a| |c - a| |d - a| \right)}{|a - b| |b - c| |c - d| |d - a|}
$$over all triples $a, b, c, d$ of distinct real numbers such that
$a^2 + b^2 + c^2 + d^2 = 3(ab + bc + cd + da).$

3 replies
Martin.s
Oct 17, 2024
Martin.s
Yesterday at 5:24 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]
import graph; size(10cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -14.241848214741799, xmax = 11.317889835069966, ymin = -5.96603640835851, ymax = 9.940734075836872;  /* image dimensions */
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); 
 /* draw figures */
draw((1.7619209254634167,8.7237821962927)--(-7.69687926965853,-0.48362131590244306), linewidth(1)); 
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
8864 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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