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Checking a summand property for integers sufficiently large.
DinDean   3
N 32 minutes ago by DinDean
For any fixed integer $m\geqslant 2$, prove that there exists a positive integer $f(m)$, such that for any integer $n\geqslant f(m)$, $n$ can be expressed by a sum of positive integers $a_i$'s as
\[n=a_1+a_2+\dots+a_m,\]where $a_1\mid a_2$, $a_2\mid a_3$, $\dots$, $a_{m-1}\mid a_m$ and $1\leqslant a_1<a_2<\dots<a_m$.
3 replies
DinDean
Yesterday at 5:21 PM
DinDean
32 minutes ago
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Interesting inequalities
sqing   2
N 39 minutes ago by lbh_qys
Source: Own
Let $ a,b> 0 $ and $ a+b\leq 2ab . $ Prove that
$$\frac{ 9a^2- ab +9b^2 }{ a^2(1+b^4)}\leq\frac{17 }{2}$$$$\frac{a- ab+b }{ a^2(1+b^4)}\leq\frac{1 }{2}$$$$\frac{2a- 3ab+2b }{ a^2(1+b^4)}\leq\frac{1 }{2}$$
2 replies
sqing
an hour ago
lbh_qys
39 minutes ago
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Set: {f(r,r):r in S}=S
Sayan   7
N 41 minutes ago by kamatadu
Source: ISI (BS) 2007 #6
Let $S=\{1,2,\cdots ,n\}$ where $n$ is an odd integer. Let $f$ be a function defined on $\{(i,j): i\in S, j \in S\}$ taking values in $S$ such that
(i) $f(s,r)=f(r,s)$ for all $r,s \in S$
(ii) $\{f(r,s): s\in S\}=S$ for all $r\in S$

Show that $\{f(r,r): r\in S\}=S$
7 replies
Sayan
Apr 11, 2012
kamatadu
41 minutes ago
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26 or 30 coins in a circle
NO_SQUARES   0
an hour ago
Source: Kvant 2025 no. 2 M2833
There are a) $26$; b) $30$ identical-looking coins in a circle. It is known that exactly two of them are fake. Real coins weigh the same, fake ones too, but they are lighter than the real ones. How can you determine in three weighings on a cup scale without weights whether there are fake coins lying nearby or not??
Proposed by A. Gribalko
0 replies
NO_SQUARES
an hour ago
0 replies
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f(x,y)=0 iff (x,y) \in S, where |S|=2024
NO_SQUARES   0
an hour ago
Source: Kvant 2025 no. 2 M2832
There are $2024$ points of general position marked on the coordinate plane (i.e., points among which there are no three lying on the same straight line). Is there a polynomial of two variables $f(x,y)$ a) of degree $2025$; b) of degree $2024$ such that it equals to zero exactly at these marked points?
Proposed by Navid Safaei
0 replies
NO_SQUARES
an hour ago
0 replies
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Tangents forms triangle with two times less area
NO_SQUARES   0
an hour ago
Source: Kvant 2025 no. 2 M2831
Let $DEF$ be triangle, inscribed in parabola. Tangents in points $D,E,F$ forms triangle $ABC$. Prove that $S_{DEF}=2S_{ABC}$. ($S_T$ is area of triangle $T$).
From F.S.Macaulay's book «Geometrical Conics», suggested by M. Panov
0 replies
NO_SQUARES
an hour ago
0 replies
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IMO ShortList 2002, number theory problem 1
orl   76
N an hour ago by NerdyNashville
Source: IMO ShortList 2002, number theory problem 1
What is the smallest positive integer $t$ such that there exist integers $x_1,x_2,\ldots,x_t$ with \[x^3_1+x^3_2+\,\ldots\,+x^3_t=2002^{2002}\,?\]
76 replies
orl
Sep 28, 2004
NerdyNashville
an hour ago
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Number of lucky numbers
NO_SQUARES   0
an hour ago
Source: Kvant 2025 no. 2 M2830
There are coins in denominations of $a$ and $b$ doubloons, where $a$ and $b$ are given mutually prime natural numbers, with $a < b < 100$. A non-negative integer $n$ is called lucky if the sum in $n$ doubloons can be scored with using no more than $1000$ coins. Find the number of lucky numbers.
From the folklore
0 replies
NO_SQUARES
an hour ago
0 replies
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Woaah a lot of external tangents
egxa   3
N 2 hours ago by NO_SQUARES
Source: All Russian 2025 11.7
A quadrilateral \( ABCD \) with no parallel sides is inscribed in a circle \( \Omega \). Circles \( \omega_a, \omega_b, \omega_c, \omega_d \) are inscribed in triangles \( DAB, ABC, BCD, CDA \), respectively. Common external tangents are drawn between \( \omega_a \) and \( \omega_b \), \( \omega_b \) and \( \omega_c \), \( \omega_c \) and \( \omega_d \), and \( \omega_d \) and \( \omega_a \), not containing any sides of quadrilateral \( ABCD \). A quadrilateral whose consecutive sides lie on these four lines is inscribed in a circle \( \Gamma \). Prove that the lines joining the centers of \( \omega_a \) and \( \omega_c \), \( \omega_b \) and \( \omega_d \), and the centers of \( \Omega \) and \( \Gamma \) all intersect at one point.
3 replies
egxa
Apr 18, 2025
NO_SQUARES
2 hours ago
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2023 Hong Kong TST 3 (CHKMO) Problem 4
PikaNiko   3
N 2 hours ago by lightsynth123
Source: 2023 Hong Kong TST 3 (CHKMO)
Let $ABCD$ be a quadrilateral inscribed in a circle $\Gamma$ such that $AB=BC=CD$. Let $M$ and $N$ be the midpoints of $AD$ and $AB$ respectively. The line $CM$ meets $\Gamma$ again at $E$. Prove that the tangent at $E$ to $\Gamma$, the line $AD$ and the line $CN$ are concurrent.
3 replies
PikaNiko
Dec 3, 2022
lightsynth123
2 hours ago
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Prove that line TA',OY,MX are concurrent
Bunrong123   1
N Dec 30, 2019 by hectorraul
Let the incircle and the $A$-mixtilinear incircle of a triangle ABC touch $AC, AB$ at $E, F$ and $K, J$ resp.$EF$ and $JK$ meet BC at $X, Y$ resp. The $A$-mixtilinear incircle touches the circumcircle of $ABC$ at $T$ and the reflection of $A'$ in $O$, the circumcenter is $A'$.The midpoint of arc $BAC$ is $M$.Prove that the lines $TA', OY, MX$ are concurrent.
1 reply
Bunrong123
Dec 30, 2019
hectorraul
Dec 30, 2019
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Prove that line TA',OY,MX are concurrent
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Bunrong123
79 posts
Bunrong123
#1 h  Dec 30, 2019, 5:29 AM • 1 Y
Y by Adventure10
Let the incircle and the $A$-mixtilinear incircle of a triangle ABC touch $AC, AB$ at $E, F$ and $K, J$ resp.$EF$ and $JK$ meet BC at $X, Y$ resp. The $A$-mixtilinear incircle touches the circumcircle of $ABC$ at $T$ and the reflection of $A'$ in $O$, the circumcenter is $A'$.The midpoint of arc $BAC$ is $M$.Prove that the lines $TA', OY, MX$ are concurrent.
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hectorraul
363 posts
hectorraul
#2 h  Dec 30, 2019, 9:28 AM • 3 Y
Y by Bunrong123, Adventure10, Mango247
I will skip many details,

Let $N=A'I\cap (O)$, $D$ where the incircle touches $BC$, $Z=A'T\cap MN$ and $M_A, M_B, M_C$ the midpoints of the arcs $BC, CA, AB$.

1- $NAFIE$ is cyclic and there is a spiral similarity centered at $N$ sending $(E,B)$ to $(F,C)$. Actually, $XNEB$ and $XNFC$ are cyclics and then $M,N,X$ are collinear.

2- Homotecy at $T$ sending $A-mix$ to $(O)$ shows that $T,J, M_B$ and $T,K,M_C$ are collinear.

3- Pascal theorem shows that $K,I,J$ are collinear and then I is the midpoint of $KJ$.

4- In $\triangle TKJ$, $TA$ is symmedian and $TI$ is median, then $\angle JTI=\angle ATK$ with some angle working we can deduce that $T,I,M$ are collinear.

5- Inversion at $M_A$ with radious $M_AB$. $T$ is the intersection of $(O)$ and the circle of diameter $IM_A$, then the image of $T$ is the intersection of $BC$ and $JK$ which is $Y$. Conclusion $M_A, T, Y$ are collinear.

6- With angle working we get $\angle DIM_A=\angle M_AAA'=\angle M_ANA'$, then with the previous inversion $M_A, D, N$ are collinear and also $INYTD$ is cyclic, then $A,N,Y$ are collinear.

7- Finally Pascal on $(O)$. The intersections $Z=MN\cap A'T$, $Y= AN\cap M_AT$ and $O=AA'\cap MM_A$ are collinear and we are done.
This post has been edited 2 times. Last edited by hectorraul, Dec 30, 2019, 9:33 AM
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