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Interesting problem from a friend
v4913   11
N an hour ago by YaoAOPS
Source: I'm not sure...
Let the incircle $(I)$ of $\triangle{ABC}$ touch $BC$ at $D$, $ID \cap (I) = K$, let $\ell$ denote the line tangent to $(I)$ through $K$. Define $E, F \in \ell$ such that $\angle{EIF} = 90^{\circ}, EI, FI \cap (AEF) = E', F'$. Prove that the circumcenter $O$ of $\triangle{ABC}$ lies on $E'F'$.
11 replies
v4913
Nov 25, 2023
YaoAOPS
an hour ago
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Curious inequality
produit   2
N an hour ago by RainbowNeos
Positive real numbers x_1, x_2, . . . x_n satisfy x_1 + x_2 + . . . + x_n = 1.
Prove that
1/(1 −√x_1)+1/(1 −√x_2)+ . . . +1/(1 −√x_n)⩾ n + 4.
2 replies
produit
May 10, 2025
RainbowNeos
an hour ago
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Inspired by an old problem
RainbowNeos   0
an hour ago
Given $n\geq 3$ and $a_i \geq 0, i=1,2,...,n$. Show that
\[\sum_{i=1}^n (a_i-a_{i+1})^3\leq \frac{1}{2\sqrt{3}} (\sum_{i=1}^n a_i )^3,\]where $a_{n+1}=a_1$.
0 replies
RainbowNeos
an hour ago
0 replies
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Random walk
EthanWYX2009   3
N 2 hours ago by maromex
As shown in the graph, an ant starts from $4$ and walks randomly. The probability of any point reaching all adjacent points is equal. Find the probability of the ant reaching $1$ without passing through $6.$
3 replies
EthanWYX2009
Yesterday at 9:58 AM
maromex
2 hours ago
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Diophantine
TheUltimate123   32
N 3 hours ago by Kempu33334
Source: CJMO 2023/1 (https://aops.com/community/c594864h3031323p27271877)
Find all triples of positive integers \((a,b,p)\) with \(p\) prime and \[a^p+b^p=p!.\]
Proposed by IndoMathXdZ
32 replies
TheUltimate123
Mar 29, 2023
Kempu33334
3 hours ago
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Parallel lines in incircle configuration
GeorgeRP   4
N 3 hours ago by VicKmath7
Source: Bulgaria IMO TST 2025 P1
Let $I$ be the incenter of triangle $\triangle ABC$. Let $H_A$, $H_B$, and $H_C$ be the orthocenters of triangles $\triangle BCI$, $\triangle ACI$, and $\triangle ABI$, respectively. Prove that the lines through $H_A$, $H_B$, and $H_C$, parallel to $AI$, $BI$, and $CI$, respectively, are concurrent.
4 replies
GeorgeRP
May 14, 2025
VicKmath7
3 hours ago
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Geometry
shactal   1
N 3 hours ago by ricarlos
Two intersecting circles $C_1$ and $C_2$ have a common tangent that meets $C_1$ in $P$ and $C_2$ in $Q$. The two circles intersect at $M$ and $N$ where $N$ is closer to $PQ$ than $M$ . Line $PN$ meets circle $C_2$ a second time in $R$. Prove that $MQ$ bisects angle $\widehat{PMR}$.
1 reply
shactal
Yesterday at 3:04 PM
ricarlos
3 hours ago
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IMO ShortList 2001, combinatorics problem 6
orl   15
N 3 hours ago by awesomeming327.
Source: IMO ShortList 2001, combinatorics problem 6
For a positive integer $n$ define a sequence of zeros and ones to be balanced if it contains $n$ zeros and $n$ ones. Two balanced sequences $a$ and $b$ are neighbors if you can move one of the $2n$ symbols of $a$ to another position to form $b$. For instance, when $n = 4$, the balanced sequences $01101001$ and $00110101$ are neighbors because the third (or fourth) zero in the first sequence can be moved to the first or second position to form the second sequence. Prove that there is a set $S$ of at most $\frac{1}{n+1} \binom{2n}{n}$ balanced sequences such that every balanced sequence is equal to or is a neighbor of at least one sequence in $S$.
15 replies
orl
Sep 30, 2004
awesomeming327.
3 hours ago
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Number Theory Chain!
JetFire008   64
N 4 hours ago by Primeniyazidayi
I will post a question and someone has to answer it. Then they have to post a question and someone else will answer it and so on. We can only post questions related to Number Theory and each problem should be more difficult than the previous. Let's start!

Question 1
64 replies
JetFire008
Apr 7, 2025
Primeniyazidayi
4 hours ago
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Serbian selection contest for the IMO 2025 - P1
OgnjenTesic   3
N 4 hours ago by MR.1
Source: Serbian selection contest for the IMO 2025
Let \( p \geq 7 \) be a prime number and \( m \in \mathbb{N} \). Prove that
\[\left| p^m - (p - 2)! \right| > p^2.\]Proposed by Miloš Milićev
3 replies
OgnjenTesic
Thursday at 4:01 PM
MR.1
4 hours ago
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Nice "if and only if" function problem
ICE_CNME_4   0
5 hours ago
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[ f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0. \](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )

Please do it at 9th grade level. Thank you!
0 replies
ICE_CNME_4
5 hours ago
0 replies
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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
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n-gon function
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Source: Romanian IMO Team Selection Test TST 1996, problem 1
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ehsan2004
2238 posts
ehsan2004
#1 h  Sep 13, 2005, 10:11 AM • 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
perfect_radio
#2 h  Oct 1, 2005, 1:42 PM • 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
ehsan2004
#3 h  Oct 1, 2005, 1:51 PM • 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
perfect_radio
#4 h  Oct 1, 2005, 1:57 PM • 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
enescu
#5 h  Oct 3, 2005, 3:36 PM • 2 Y
Y by Adventure10, Mango247
Actually, $n$ is fixed in the original statement.
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perfect_radio
2607 posts
perfect_radio
#6 h  Oct 3, 2005, 6:10 PM • 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
enescu
#7 h  Oct 3, 2005, 6:27 PM • 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
TheUltimate123
#8 h  Jun 30, 2021, 8:39 AM • 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
11379 posts
jasperE3
#9 h  Aug 24, 2021, 7:01 PM
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
1009 posts
AshAuktober
#10 h  Apr 2, 2025, 10:09 AM
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
Zany9998
#11 h  Apr 4, 2025, 3:31 PM
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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