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2025 Caucasus MO Seniors P2
BR1F1SZ   3
N an hour ago by X.Luser
Source: Caucasus MO
Let $ABC$ be a triangle, and let $B_1$ and $B_2$ be points on segment $AC$ symmetric with respect to the midpoint of $AC$. Let $\gamma_A$ denote the circle passing through $B_1$ and tangent to line $AB$ at $A$. Similarly, let $\gamma_C$ denote the circle passing through $B_1$ and tangent to line $BC$ at $C$. Let the circles $\gamma_A$ and $\gamma_C$ intersect again at point $B'$ ($B' \neq B_1$). Prove that $\angle ABB' = \angle CBB_2$.
3 replies
BR1F1SZ
Mar 26, 2025
X.Luser
an hour ago
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IMO Shortlist 2010 - Problem G1
Amir Hossein   130
N an hour ago by LeYohan
Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$

Proposed by Christopher Bradley, United Kingdom
130 replies
Amir Hossein
Jul 17, 2011
LeYohan
an hour ago
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Nordic 2025 P3
anirbanbz   7
N an hour ago by anirbanbz
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 $\vert OE \vert = \vert OF \vert$.
7 replies
anirbanbz
Mar 25, 2025
anirbanbz
an hour ago
J
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CGMO6: Airline companies and cities
v_Enhance   13
N an hour ago by Marcus_Zhang
Source: 2012 China Girl's Mathematical Olympiad
There are $n$ cities, $2$ airline companies in a country. Between any two cities, there is exactly one $2$-way flight connecting them which is operated by one of the two companies. A female mathematician plans a travel route, so that it starts and ends at the same city, passes through at least two other cities, and each city in the route is visited once. She finds out that wherever she starts and whatever route she chooses, she must take flights of both companies. Find the maximum value of $n$.
13 replies
v_Enhance
Aug 13, 2012
Marcus_Zhang
an hour ago
J
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nice problem
hanzo.ei   0
2 hours ago
Source: I forgot
Let triangle $ABC$ be inscribed in the circumcircle $(O)$ and circumscribed about the incircle $(I)$, with $AB < AC$. The incircle $(I)$ touches the sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. A line through $I$, perpendicular to $AI$, intersects $BC$, $CA$, and $AB$ at $X$, $Y$, and $Z$, respectively. The line $AI$ meets $(O)$ at $M$ (distinct from $A$). The circumcircle of triangle $AYZ$ intersects $(O)$ at $N$ (distinct from $A$). Let $P$ be the midpoint of the arc $BAC$ of $(O)$. The line $AI$ cuts segments $DF$ and $DE$ at $K$ and $L$, respectively, and the tangents to the circle $(DKL)$ at $K$ and $L$ intersect at $T$. Prove that $AT \perp BC$.
0 replies
hanzo.ei
2 hours ago
0 replies
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Find a given number of divisors of ab
proglote   9
N 2 hours ago by zuat.e
Source: Brazil MO 2013, problem #2
Arnaldo and Bernaldo play the following game: given a fixed finite set of positive integers $A$ known by both players, Arnaldo picks a number $a \in A$ but doesn't tell it to anyone. Bernaldo thens pick an arbitrary positive integer $b$ (not necessarily in $A$). Then Arnaldo tells the number of divisors of $ab$. Show that Bernaldo can choose $b$ in a way that he can find out the number $a$ chosen by Arnaldo.
9 replies
proglote
Oct 24, 2013
zuat.e
2 hours ago
J
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2025 TST 22
EthanWYX2009   1
N 2 hours ago by hukilau17
Source: 2025 TST 22
Let \( A \) be a set of 2025 positive real numbers. For a subset \( T \subseteq A \), define \( M_T \) as the median of \( T \) when all elements of \( T \) are arranged in increasing order, with the convention that \( M_\emptyset = 0 \). Define
\[ P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T. \]Find the smallest real number \( C \) such that for any set \( A \) of 2025 positive real numbers, the following inequality holds:
\[ P(A) - Q(A) \leq C \cdot \max(A), \]where \(\max(A)\) denotes the largest element in \( A \).
1 reply
EthanWYX2009
5 hours ago
hukilau17
2 hours ago
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Deriving Van der Waerden Theorem
Didier2   0
2 hours ago
Source: Khamovniki 2023-2024 (group 10-1)
Suppose we have already proved that for any coloring of $\Large \mathbb{N}$ in $r$ colors, there exists an arithmetic progression of size $k$. How can we derive Van der Waerden's theorem for $W(r, k)$ from this?
0 replies
Didier2
2 hours ago
0 replies
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Not so classic orthocenter problem
m4thbl3nd3r   6
N 2 hours ago by maths_enthusiast_0001
Source: own?
Let $O$ be circumcenter of a non-isosceles triangle $ABC$ and $H$ be a point in the interior of $\triangle ABC$. Let $E,F$ be foots of perpendicular lines from $H$ to $AC,AB$. Suppose that $BCEF$ is cyclic and $M$ is the circumcenter of $BCEF$, $HM\cap AB=K,AO\cap BE=T$. Prove that $KT$ bisects $EF$
6 replies
m4thbl3nd3r
Yesterday at 4:59 PM
maths_enthusiast_0001
2 hours ago
J
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Functional equations
hanzo.ei   1
N 2 hours ago by GreekIdiot
Source: Greekldiot
Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$
1 reply
hanzo.ei
3 hours ago
GreekIdiot
2 hours ago
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Geometric Inequality
sqing   15
N Sep 23, 2015 by sqing
Source: China
In $\triangle ABC$ , prove that\[\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\geq \frac{2}{\sqrt{3}}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}).\]
15 replies
sqing
Sep 10, 2015
sqing
Sep 23, 2015
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Geometric Inequality
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Reveal topic tags
inequalities
geometric inequality
geometry
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G H BBookmark kLocked kLocked NReply
Source: China
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sqing
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sqing
#1 h  Sep 10, 2015, 11:24 AM • 1 Y
Y by Adventure10
In $\triangle ABC$ , prove that\[\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\geq \frac{2}{\sqrt{3}}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}).\]
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arqady
30161 posts
arqady
#2 h  Sep 10, 2015, 12:33 PM • 2 Y
Y by Adventure10, Mango247
After using Ravi's substitution we need to prove that
$3(x+y+z)(x+y)^2(x+z)^2(y+z)^2\geq4\left(\sum_{cyc}(x^2+3xy)\right)^2xyz$.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$. Hence, we need to prove that $f(w^3)\geq0$,
where $f(w^3)=u(9uv^2-w^3)^2-4(3u^2+v^2)^2w^3$.
We see that $f'(w^3)=-2u(9uv^2-w^3)-4(3u^2+v^2)^2<0$.
Thus, $f$ gets a minimal value for a maximal value of $w^3$, which happens for equality case of two variables.
Let $y=z=1$. Hence, we need to prove $(x-1)^2(x+3)\geq0$. Done!
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xzlbq
15849 posts
xzlbq
#3 h  Sep 11, 2015, 12:01 AM • 2 Y
Y by Adventure10, Mango247
Stonger is:

\[ \left( {y}^{2}+10\,yz+{z}^{2} \right) \left( z+x \right) ^{2} \left( x+y \right) ^{2} \left( x+y+z \right) \geq 4\, \left( {x}^{2}+{y}^ {2}+{z}^{2}+3\,xy+3\,yz+3\,xz \right) ^{2}xyz\]
BQ
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xzlbq
15849 posts
xzlbq
#4 h  Sep 11, 2015, 12:20 AM • 2 Y
Y by Adventure10, Mango247
\[\left( {z}^{2}+2\,xz+2\,{x}^{2}+2\,xy+{y}^{2} \right) ^{2} \left( x+y +z \right) ^{2}\geq 2/3\, \left( {z}^{2}+2\,xz+2\,{x}^{2}+{y}^{2}+2\,xy-2 \,yz \right) \left( {x}^{2}+{y}^{2}+{z}^{2}+3\,xy+3\,yz+3\,xz \right) ^{2}\]
=>
In triangle,

\[\frac{1}{k_a} \geq \frac{2}{3}\sqrt{3}(\frac{a}{a+b+c})(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\]
$k_a=2bc/(b^2+c^2)m_a$
This post has been edited 1 time. Last edited by xzlbq, Sep 11, 2015, 12:22 AM
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xzlbq
15849 posts
xzlbq
#5 h  Sep 11, 2015, 2:58 AM • 2 Y
Y by Adventure10, Mango247
\[\frac{1}{w_a}\geq \frac{2\sqrt{3}}{3}\frac{(2h_c+2r_c+h_a+h_b)(h_c+h_a+2h_b+2r_b)(h_a+r_a)a}{\sum{((2h_c+2r_c+h_a+h_b)(h_c+h_a+2h_b+2r_b)(h_a+r_a)a)}}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\]
deg=10

BQ
This post has been edited 3 times. Last edited by xzlbq, Sep 11, 2015, 3:09 AM
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szl6208
2032 posts
szl6208
#6 h  Sep 11, 2015, 4:44 AM • 1 Y
Y by Adventure10
xzlbq wrote:
Stonger is:

\[ \left( {y}^{2}+10\,yz+{z}^{2} \right) \left( z+x \right) ^{2} \left( x+y \right) ^{2} \left( x+y+z \right) \geq 4\, \left( {x}^{2}+{y}^ {2}+{z}^{2}+3\,xy+3\,yz+3\,xz \right) ^{2}xyz\]
BQ

We have
\[(y^2+10yz+z^2)(z+x)^2(x+y)^2(x+y+z)-4(x^2+3xy+3xz+y^2+3yz+z^2)^2xyz=\]
\[(x+y+z)(x^2y+x^2z+xy^2-2xyz+xz^2-y^2z-yz^2)^2+4yz(2z+2y+x)(x^2-yz)^2\ge{0}\]
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szl6208
2032 posts
szl6208
#7 h  Sep 11, 2015, 5:15 AM • 1 Y
Y by Adventure10
xzlbq wrote:
\[\left( {z}^{2}+2\,xz+2\,{x}^{2}+2\,xy+{y}^{2} \right) ^{2} \left( x+y +z \right) ^{2}\geq 2/3\, \left( {z}^{2}+2\,xz+2\,{x}^{2}+{y}^{2}+2\,xy-2 \,yz \right) \left( {x}^{2}+{y}^{2}+{z}^{2}+3\,xy+3\,yz+3\,xz \right) ^{2}\]
=>
In triangle,

\[\frac{1}{k_a} \geq \frac{2}{3}\sqrt{3}(\frac{a}{a+b+c})(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\]
$k_a=2bc/(b^2+c^2)m_a$

We have
\[(z^2+2xz+2x^2+2xy+y^2)^2(x+y+z)^2-\frac{2}{3}(z^2+2xz+2x^2+y^2+2xy-2yz)(\sum{(x^2+3yz)})^2\]
\[=\frac{1}{3}(x^2+2yz)(y-z)^4+\frac{1}{3}(x^2z+y^3-2y^2z)^2+\frac{1}{3}(x^2y-2yz^2+z^3)^2+\frac{2}{3}x(y+z)(x^2+y^2-3yz+z^2)^2+\frac{1}{3}(8x^2+18xy+18xz+13y^2+20yz+13z^2)(x^2-yz)^2\ge{0}\]
This post has been edited 1 time. Last edited by szl6208, Sep 11, 2015, 5:17 AM
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xzlbq
15849 posts
xzlbq
#8 h  Sep 11, 2015, 5:25 AM • 2 Y
Y by Adventure10, Mango247
Stronger is:

\[\frac{1}{\sqrt{h_a}} \geq \frac{\sqrt{2}}{2}(\frac{h_b+h_c}{h_a+h_b+h_c})\sqrt{\sqrt{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}\]

Click to reveal hidden text

=>
$\left( {\frac {1}{\sqrt {{\it h_a}}}}+{\frac {1}{\sqrt {{\it h_b}}}}+{ \frac {1}{\sqrt {{\it h_c}}}} \right) ^{2}\geq 2\,\sqrt {3} \left( {a}^{-1} +{b}^{-1}+{c}^{-1} \right) $

BQ
This post has been edited 4 times. Last edited by xzlbq, Sep 11, 2015, 5:29 AM
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szl6208
2032 posts
szl6208
#9 h  Sep 11, 2015, 6:19 AM • 1 Y
Y by Adventure10
xzlbq wrote:
Stronger is:

\[\frac{1}{\sqrt{h_a}} \geq \frac{\sqrt{2}}{2}(\frac{h_b+h_c}{h_a+h_b+h_c})\sqrt{\sqrt{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}\]

Click to reveal hidden text

=>
$\left( {\frac {1}{\sqrt {{\it h_a}}}}+{\frac {1}{\sqrt {{\it h_b}}}}+{ \frac {1}{\sqrt {{\it h_c}}}} \right) ^{2}\geq 2\,\sqrt {3} \left( {a}^{-1} +{b}^{-1}+{c}^{-1} \right) $

BQ

We have
\[(z+x)^2(x+y)^2(x^2+y^2+z^2+3xy+3yz+3xz)^2-3(x+y+z)xyz(z+2x+y)^4=\]
\[3x^3(4x^3+y^3+z^3)(y-z)^2+yz(60x^2+7y^2+10yz+7z^2)(x^2-yz)^2+\]
\[\frac{1}{2}(x^2+z^2)(x^3-x^2z+y^3-y^2z)^2+\frac{1}{2}(x^2+y^2)(x^3-x^2y-yz^2+z^3)^2+\]
\[\frac{1}{2}z(18x^5+4x^4y+22x^4z+68x^3z^2+12x^2y^3+64x^2yz^2+44xy^2z^2+12xyz^3+10xz^4+z^5)(x-y)^2+\]
\[\frac{1}{2}y(18x^5+22x^4y+4x^4z+68x^3y^2+64x^2y^2z+12x^2z^3+10xy^4+12xy^3z+44xy^2z^2+y^5)(-z+x)^2+\]
\[\frac{47}{2}x^2z^2(xz-y^2)^2+\frac{47}{2}x^2y^2(xy-z^2)^2\ge{0}\]
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xzlbq
15849 posts
xzlbq
#10 h  Sep 11, 2015, 6:23 AM • 1 Y
Y by Adventure10
Szl,Very good!
and kill 12>=deg>=8
BQ
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arqady
30161 posts
arqady
#11 h  Sep 14, 2015, 7:12 AM • 1 Y
Y by Adventure10
The following inequality is also true.
For all triangle prove that:
$$\frac{1}{l_a}+\frac{1}{l_b}+\frac{1}{l_c}\geq \frac{2}{\sqrt{3}}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$
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xzlbq
15849 posts
xzlbq
#12 h  Sep 14, 2015, 7:34 AM • 2 Y
Y by arqady, Adventure10
arqady wrote:
The following inequality is also true.
For all triangle prove that:
$$\frac{1}{l_a}+\frac{1}{l_b}+\frac{1}{l_c}\geq \frac{2}{\sqrt{3}}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$

5#,give a sum,get it.

BQ
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nicusorz
1158 posts
nicusorz
#13 h  Sep 16, 2015, 5:14 PM • 2 Y
Y by Adventure10, Mango247
we have :
$ \frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{ca}}=\frac{1}{r} $, then
$\frac{1}{r}\geq \frac{2}{\sqrt{3}}\sum \frac{1}{a} $ (Petrovic's inequality )
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sandumarin
1093 posts
sandumarin
#14 h  Sep 22, 2015, 11:48 AM • 1 Y
Y by Adventure10
$ LHS=\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}=\frac{a}{ah_a}+\frac{b}{bh_b}+\frac{c}{ch_c}=\frac{a+b+c}{2A}=\frac{2s}{2sr}=\frac{1}{r} $ ;where A are triangle area
$ RHS=\frac{2}{\sqrt{3}}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{2}{\sqrt{3}}(\frac{h_a}{ah_a}+\frac{h_b}{bh_b}+\frac{c}{ch_c})=\frac{2}{\sqrt{3}}\frac{h_a+h_b+h_c}{2A}=\frac{h_a+h_b+h_c}{sr\sqrt{3}} $
Ineq returns to:
$ h_a+h_b+h_c\le s\sqrt{3} $ (1)
Because by RAVI's Substitutions result: $ h_a=\frac{2\sqrt{xyz(x+y+z)}}{y+z},h_b=\frac{2\sqrt{xyz(x+y+z)}}{z+x},h_c=\frac{2\sqrt{xyz(x+y+z)}}{x+y} $ where $ x,y,z>0 $
It remains to show that:
$ \frac{2}{y+z}+\frac{2}{z+x}+\frac{2}{x+y}\le\sqrt{3(\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xy})} $
This result using GM ineq and Cauchy-Schwarz:
$ y+z\ge 2\sqrt{yz},z+x\ge 2\sqrt{zx},x+y\ge 2\sqrt{xy}\Rightarrow LHS\le\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}+\frac{1}{\sqrt{xy}}\le(by Cauchy-Schwarz)\sqrt{3(\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xy})} $
The proof is ended!
_________
Marin Sandu
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Virgil Nicula
7054 posts
Virgil Nicula
#15 h  Sep 23, 2015, 1:02 AM • 2 Y
Y by Adventure10, Mango247
See here (proposed problem) PP6 $:$ the bilateral inequality $\boxed{\ \frac {\sqrt 3}{R}\ \le\ \frac 1a+\frac 1b+\frac 1c\ \le\ \frac {\sqrt 3}{2r}\ }$

what is equivalently with $\boxed{\frac {r\sqrt 3}R\le \frac {h_a+h_b+h_c}{a+b+c}\le \frac {\sqrt 3}2}$ .
This post has been edited 7 times. Last edited by Virgil Nicula, Sep 23, 2015, 2:29 AM
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sqing
41310 posts
sqing
#16 h  Sep 23, 2015, 4:41 AM • 1 Y
Y by Adventure10
sqing wrote:
In $\triangle ABC$ , prove that\[\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\geq \frac{2}{\sqrt{3}}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}).\]
$\sum_{cyc}\frac{1}{h_a}=\frac{1}{abc}\sum_{cyc}\frac{a^2}{sinA}\geq\frac{(\sum_{cyc}a)^2}{abc\sum_{cyc}sinA}\geq\frac{3\sum_{cyc}bc}{abc\sum_{cyc}sinA}\geq\frac{2}{\sqrt{3}}\sum_{cyc}\frac{1}{a}.$
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