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Iran TST Starter
M11100111001Y1R   2
N 2 hours ago by sami1618
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[ a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i} \]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[ a_n < \frac{1}{1404} \]
2 replies
1 viewing
M11100111001Y1R
Tuesday at 7:36 AM
sami1618
2 hours ago
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Twin Prime Diophantine
awesomeming327.   23
N 3 hours ago by HDavisWashu
Source: CMO 2025
Determine all positive integers $a$, $b$, $c$, $p$, where $p$ and $p+2$ are odd primes and
\[2^ap^b=(p+2)^c-1.\]
23 replies
awesomeming327.
Mar 7, 2025
HDavisWashu
3 hours ago
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Troublesome median in a difficult inequality
JG666   2
N 3 hours ago by navid
Source: 2022 Spring NSMO Day 2 Problem 3
Determine the minimum value of $\lambda\in\mathbb{R}$, such that for any positive integer $n$ and non-negative reals $x_1, x_2, \cdots, x_n$, the following inequality always holds:
$$\sum_{i=1}^n(m_i-a_i)^2\leqslant \lambda\cdot\sum_{i=1}^nx_i^2,$$Here $m_i$ and $a_i$ denote the median and arithmetic mean of $x_1, x_2, \cdots, x_i$, respectively.

Duanyang ZHANG, High School Affiliated to Renmin University of China
2 replies
JG666
May 22, 2022
navid
3 hours ago
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Concurrent lines, angle bisectors
legogubbe   0
3 hours ago
Source: ???
Hi AoPS!

Let $ABC$ be an isosceles triangle with $AB=AC$, and $M$ an arbitrary point on side $BC$. The internal angle bisector of $\angle MAB$ meets the circumcircle of $\triangle ABC$ again at $P \neq A$, and the internal angle bisector of $\angle CAM$ meets it again at $Q \neq A$. Show that lines $AM$, $BQ$ and $CP$ are concurrent.
0 replies
legogubbe
3 hours ago
0 replies
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Fractional Inequality
sqing   33
N 3 hours ago by Learning11
Source: Chinese Girls Mathematical Olympiad 2012, Problem 1
Let $ a_1, a_2,\ldots, a_n$ be non-negative real numbers. Prove that
$\frac{1}{1+ a_1}+\frac{ a_1}{(1+ a_1)(1+ a_2)}+\frac{ a_1 a_2}{(1+ a_1)(1+ a_2)(1+ a_3)}+$ $\cdots+\frac{ a_1 a_2\cdots a_{n-1}}{(1+ a_1)(1+ a_2)\cdots (1+ a_n)} \le 1.$
33 replies
sqing
Aug 10, 2012
Learning11
3 hours ago
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Geometry angle chasing olympiads
Foxellar   1
N 3 hours ago by Ianis
Let \( \triangle ABC \) be a triangle such that \( \angle ABC = 120^\circ \). Points \( X, Y, Z \) lie on segments \( BC, CA, AB \), respectively, such that lines \( AX, BY, \) and \( CZ \) are the angle bisectors of triangle \( ABC \). Find the measure of angle \( \angle XYZ \).
1 reply
Foxellar
4 hours ago
Ianis
3 hours ago
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Iran Inequality
mathmatecS   17
N 3 hours ago by Learning11
Source: Iran 1998
When $x(\ge1),$ $y(\ge1),$ $z(\ge1)$ satisfy $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2,$ prove in equality.
$$\sqrt{x+y+z}\ge\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$$
17 replies
mathmatecS
Jun 11, 2015
Learning11
3 hours ago
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Problem 4
codyj   87
N 4 hours ago by ezpotd
Source: IMO 2015 #4
Triangle $ABC$ has circumcircle $\Omega$ and circumcenter $O$. A circle $\Gamma$ with center $A$ intersects the segment $BC$ at points $D$ and $E$, such that $B$, $D$, $E$, and $C$ are all different and lie on line $BC$ in this order. Let $F$ and $G$ be the points of intersection of $\Gamma$ and $\Omega$, such that $A$, $F$, $B$, $C$, and $G$ lie on $\Omega$ in this order. Let $K$ be the second point of intersection of the circumcircle of triangle $BDF$ and the segment $AB$. Let $L$ be the second point of intersection of the circumcircle of triangle $CGE$ and the segment $CA$.

Suppose that the lines $FK$ and $GL$ are different and intersect at the point $X$. Prove that $X$ lies on the line $AO$.

Proposed by Greece
87 replies
codyj
Jul 11, 2015
ezpotd
4 hours ago
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IMO96/2 [the lines AP, BD, CE meet at a point]
Arne   47
N 5 hours ago by Bridgeon
Source: IMO 1996 problem 2, IMO Shortlist 1996, G2
Let $ P$ be a point inside a triangle $ ABC$ such that
\[ \angle APB - \angle ACB = \angle APC - \angle ABC. \]
Let $ D$, $ E$ be the incenters of triangles $ APB$, $ APC$, respectively. Show that the lines $ AP$, $ BD$, $ CE$ meet at a point.
47 replies
1 viewing
Arne
Sep 30, 2003
Bridgeon
5 hours ago
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A sharp one with 3 var (3)
mihaig   4
N 5 hours ago by aaravdodhia
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$a^2+b^2+c^2+5abc\geq8.$$
4 replies
mihaig
Tuesday at 5:17 PM
aaravdodhia
5 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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