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Geometry
gggzul   2
N 32 minutes ago by gggzul
In trapezoid $ABCD$ segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle A$ and $\angle C$ meet at $P$. Angle bisectors of $\angle B$ and $\angle D$ meet at $Q$. Prove that $ABPQ$ is cyclic
2 replies
gggzul
2 hours ago
gggzul
32 minutes ago
thank you !
Piwbo   2
N 37 minutes ago by Piwbo
Given positive integers $a,b$ such that $a$ is even , $b$ is odd and $ab(a+b)^{2023}$ is divisible by $a^{2024}+b^{2024}$ .Prove that there exists a prime number $p$ such that $a^{2024}+b^{2024}$ is divisible by $p^{2025}$
2 replies
Piwbo
an hour ago
Piwbo
37 minutes ago
Inequality involving square root cube root and 8th root
bamboozled   2
N 39 minutes ago by bamboozled
If $a,b,c,d,e,f,g,h,k\in R^+$ and $a+b+c=d+e+f=g+h+k=8$, then find the maximum value of $\sqrt{ad^3 g^4} +\sqrt[3]{be^3 h^4} + \sqrt[8]{cf^3 k^4}$
2 replies
bamboozled
5 hours ago
bamboozled
39 minutes ago
find positive n so that exists prime p with p^n-(p-1)^n$ a power of 3
parmenides51   12
N an hour ago by n-k-p
Source: JBMO Shortlist 2017 NT5
Find all positive integers $n$ such that there exists a prime number $p$, such that $p^n-(p-1)^n$ is a power of $3$.

Note. A power of $3$ is a number of the form $3^a$ where $a$ is a positive integer.
12 replies
parmenides51
Jul 25, 2018
n-k-p
an hour ago
hard problem
Cobedangiu   5
N 2 hours ago by KhuongTrang
$a,b,c>0$ and $a+b+c=7$. CM:
$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+abc \ge ab+bc+ca-2$
5 replies
Cobedangiu
Yesterday at 4:24 PM
KhuongTrang
2 hours ago
Nordic 2025 P3
anirbanbz   9
N 2 hours ago by Tsikaloudakis
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
9 replies
anirbanbz
Mar 25, 2025
Tsikaloudakis
2 hours ago
Aime type Geo
ehuseyinyigit   1
N 2 hours ago by ehuseyinyigit
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
1 reply
ehuseyinyigit
Yesterday at 9:04 PM
ehuseyinyigit
2 hours ago
Arbitrary point on BC and its relation with orthocenter
falantrng   34
N 2 hours ago by Mamadi
Source: Balkan MO 2025 P2
In an acute-angled triangle \(ABC\), \(H\) be the orthocenter of it and \(D\) be any point on the side \(BC\). The points \(E, F\) are on the segments \(AB, AC\), respectively, such that the points \(A, B, D, F\) and \(A, C, D, E\) are cyclic. The segments \(BF\) and \(CE\) intersect at \(P.\) \(L\) is a point on \(HA\) such that \(LC\) is tangent to the circumcircle of triangle \(PBC\) at \(C.\) \(BH\) and \(CP\) intersect at \(X\). Prove that the points \(D, X, \) and \(L\) lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus
34 replies
falantrng
Apr 27, 2025
Mamadi
2 hours ago
Which numbers are almost prime?
AshAuktober   5
N 3 hours ago by Jupiterballs
Source: 2024 Swiss MO/1
If $a$ and $b$ are positive integers, we say that $a$ almost divides $b$ if $a$ divides at least one of $b - 1$ and $b + 1$. We call a positive integer $n$ almost prime if the following holds: for any positive integers $a, b$ such that $n$ almost divides $ab$, we have that $n$ almost divides at least one of $a$ and $b$. Determine all almost prime numbers.
original link
5 replies
AshAuktober
Dec 16, 2024
Jupiterballs
3 hours ago
If $b^n|a^n-1$ then $a^b >\frac {3^n}{n}$ (China TST 2009)
Fang-jh   16
N 3 hours ago by Aiden-1089
Source: Chinese TST 2009 6th P1
Let $ a > b > 1, b$ is an odd number, let $ n$ be a positive integer. If $ b^n|a^n-1,$ then $ a^b > \frac {3^n}{n}.$
16 replies
Fang-jh
Apr 4, 2009
Aiden-1089
3 hours ago
Confusing inequality
giangtruong13   1
N 3 hours ago by Natrium
Source: An user
Let $a,b,c>0$ such that $a+b+c=3$. Find the minimum: $$P= \sum_{cyc} \frac{a}{b} + \sum_{cyc} \frac{1}{a^3+b^3+abc}$$
1 reply
giangtruong13
Yesterday at 8:04 AM
Natrium
3 hours ago
n-gon function
ehsan2004   10
N Apr 4, 2025 by Zany9998
Source: Romanian IMO Team Selection Test TST 1996, problem 1
Let $ f: \mathbb{R}^2 \rightarrow \mathbb{R} $ be a function such that for every regular $ n $-gon $ A_1A_2 \ldots A_n $ we have $ f(A_1)+f(A_2)+\cdots +f(A_n)=0 $. Prove that $ f(x)=0 $ for all reals $ x $.
10 replies
ehsan2004
Sep 13, 2005
Zany9998
Apr 4, 2025
n-gon function
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G H BBookmark kLocked kLocked NReply
Source: Romanian IMO Team Selection Test TST 1996, problem 1
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ehsan2004
2238 posts
#1 • 3 Y
Y by Adventure10, centslordm, Mango247
Let $ f: \mathbb{R}^2 \rightarrow \mathbb{R} $ be a function such that for every regular $ n $-gon $ A_1A_2 \ldots A_n $ we have $ f(A_1)+f(A_2)+\cdots +f(A_n)=0 $. Prove that $ f(x)=0 $ for all reals $ x $.
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perfect_radio
2607 posts
#2 • 2 Y
Y by Adventure10, Mango247
ehsan2004 wrote:
Prove that $f(x)=0$ for all reals $x$.

You wanted to say "$f(A)=0$ for all points $A$", right?
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ehsan2004
2238 posts
#3 • 2 Y
Y by Adventure10, Mango247
perfect_radio wrote:
ehsan2004 wrote:
Prove that $f(x)=0$ for all reals $x$.

You wanted to say "$f(A)=0$ for all points $A$", right?

excuse me, my meant was $f(x)\equiv 0$
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perfect_radio
2607 posts
#4 • 2 Y
Y by Adventure10, Mango247
Take $A \neq B$. Let $\ell$ be the perpendicular bisector of $AB$. Construct a rhombus $ACBD$, with $C,D \in \ell$ and $\measuredangle DAC = \measuredangle DBC = \dfrac{\pi}{3}$. This yields $f(A)+f(C)+f(D)=0=f(B)+f(C)+f(D)$, so $f(A)=f(B)=t$, $\forall A \neq B$.

Therefore $nt=0$, so $t=0$.

Have I done something wrong? It looks too good to be true :blush: (because I used the condition given only for $n=3$)
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enescu
741 posts
#5 • 2 Y
Y by Adventure10, Mango247
Actually, $n$ is fixed in the original statement.
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perfect_radio
2607 posts
#6 • 2 Y
Y by Adventure10, Mango247
enescu wrote:
Actually, $n$ is fixed in the original statement.
Oops... sorry :( . do you know the solution for $n \geq 4$?
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enescu
741 posts
#7 • 7 Y
Y by Batominovski, wateringanddrowned, Adventure10, Upwgs_2008, Mango247, and 2 other users
Yes. Let $A$ be an arbitrary point. Consider a regular $n-$gon $AA_{1}A_{2}\ldots A_{n-1}.$ Let $k$ be an integer, $0\leq k\leq n-1.$ A rotation with center $A$ of angle $\dfrac{2k\pi}{n}$ sends the polygon $AA_{1}A_{2}\ldots A_{n-1}$ to $A_{k0}A_{k1}\ldots A_{k,n-1},$ where $A_{k0}=A$ and $A_{ki}$ is the image of $A_{i}$, for all $i=1,2,\ldots,n-1.$

From the condition of the statement, we have
\[ \sum_{k=0}^{n-1} \sum_{i=0}^{n-1}{f(A_{ki})}=0.  \]
Observe that in the sum the number $f(A)$ appears $n$ times, therefore
\[ nf(A)+\sum_{k=0}^{n-1} \sum_{i=1}^{n-1}{f(A_{ki})}=0.  \]
On the other hand, we have
\[ \sum_{k=0}^{n-1} \sum_{i=1}^{n-1}{f(A_{ki})}=\sum_{i=1}^{n-1} \sum_{k=0}^{n-1}{f(A_{ki})}=0,  \]
since the polygons $A_{0i}A_{1i}\ldots A_{n-1,i}$ are all regular $n-$gons. From the two equalities above we deduce $f(A)=0,$ hence $f$ is the zero function.
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TheUltimate123
1740 posts
#8 • 1 Y
Y by MS_asdfgzxcvb
We may assume \(n\) is even, since for \(n\) odd, the sum of the vertices in any \(2n\)-gon is zero.

Now let \(A_1\cdots A_n\) be a regular \(n\)-gon. For each \(i\) and \(j\), let \(M_{ij}\) be the midpoint of \(\overline{A_iA_j}\) (so in particular, \(M_iM_i=A_i\)), and let \(O\) be the center of the \(n\)-gon.

We know since \(M_{i1}M_{i2}\cdots M_{in}\) and \(M_{1,1+i}M_{2,2+i}\cdots M_{n,n+i}\) are regular \(n\)-gons that \begin{align*}     0=\sum_i\sum_jf(M_{ij})     =n\cdot f(O)+\sum_j\sum_{\substack{i<n\\ i\ne n/2}}f(M_{j,j+i})     &=n\cdot f(O) \end{align*}
This post has been edited 1 time. Last edited by TheUltimate123, Jun 30, 2021, 8:40 AM
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jasperE3
11293 posts
#9
Y by
ehsan2004 wrote:
Let $ f: \mathbb{R}^2 \rightarrow \mathbb{R} $ be a function such that for every regular $ n $-gon $ A_1A_2 \ldots A_n $ we have $ f(A_1)+f(A_2)+\cdots +f(A_n)=0 $. Prove that $ f(x)=0 $ for all reals $ x $.

The claim for just $n=4$:
https://aops.com/community/p1703551
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AshAuktober
1004 posts
#10
Y by
For $n$ even, draw a lot of $n$-gons with diametres the segments through the respective vertices and the centre of some $n$-gon, and the calculation works out to give $f(\text{ centre }) = 0$, so we're done.
For $n$ odd, notice that the statement then holds for $2n$ as well, so we're done.
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Zany9998
11 posts
#11
Y by
I wonder if there exists a coloring proof. i.e. label all negative points red, all 0 points yellow and all positive points green. I’ve proved that both red and green are dense in R^2 if whole board is not yellow. Is this sufficient to prove that there exists a regular n-gon whose vertices have at least one red and no green or vice versa? The condition seems to be strong enough.
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