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Concurrency with 10 lines
oVlad   1
N 3 hours ago by kokcio
Source: Romania EGMO TST 2017 Day 1 P1
Consider five points on a circle. For every three of them, we draw the perpendicular from the centroid of the triangle they determine to the line through the remaining two points. Prove that the ten lines thus formed are concurrent.
1 reply
oVlad
Today at 1:31 PM
kokcio
3 hours ago
Concurrence, Isogonality
Wictro   40
N 4 hours ago by CatinoBarbaraCombinatoric
Source: BMO 2019, Problem 3
Let $ABC$ be an acute scalene triangle. Let $X$ and $Y$ be two distinct interior points of the segment $BC$ such that $\angle{CAX} = \angle{YAB}$. Suppose that:
$1)$ $K$ and $S$ are the feet of the perpendiculars from from $B$ to the lines $AX$ and $AY$ respectively.
$2)$ $T$ and $L$ are the feet of the perpendiculars from $C$ to the lines $AX$ and $AY$ respectively.
Prove that $KL$ and $ST$ intersect on the line $BC$.
40 replies
Wictro
May 2, 2019
CatinoBarbaraCombinatoric
4 hours ago
Prove excircle is tangent to circumcircle
sarjinius   7
N 4 hours ago by markam
Source: Philippine Mathematical Olympiad 2025 P4
Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on side $BC$. Points $X$ and $Y$ are chosen on lines $BI$ and $CI$ respectively such that $DXIY$ is a parallelogram. Points $E$ and $F$ are chosen on side $BC$ such that $AX$ and $AY$ are the angle bisectors of angles $\angle BAE$ and $\angle CAF$ respectively. Let $\omega$ be the circle tangent to segment $EF$, the extension of $AE$ past $E$, and the extension of $AF$ past $F$. Prove that $\omega$ is tangent to the circumcircle of triangle $ABC$.
7 replies
sarjinius
Mar 9, 2025
markam
4 hours ago
Easy geo
oVlad   3
N 5 hours ago by Primeniyazidayi
Source: Romania EGMO TST 2019 Day 1 P1
A line through the vertex $A{}$ of the triangle $ABC{}$ which doesn't coincide with $AB{}$ or $AC{}$ intersectes the altitudes from $B{}$ and $C{}$ at $D{}$ and $E{}$ respectively. Let $F{}$ be the reflection of $D{}$ in $AB{}$ and $G{}$ be the reflection of $E{}$ in $AC{}.$ Prove that the circles $ABF{}$ and $ACG{}$ are tangent.
3 replies
oVlad
Today at 1:45 PM
Primeniyazidayi
5 hours ago
abc(a+b+c)=3, show that prod(a+b)>=8 [Indian RMO 2012(b) Q4]
Potla   28
N 5 hours ago by mihaig
Let $a,b,c$ be positive real numbers such that $abc(a+b+c)=3.$ Prove that we have
\[(a+b)(b+c)(c+a)\geq 8.\]
Also determine the case of equality.
28 replies
Potla
Dec 2, 2012
mihaig
5 hours ago
In triangle \( PQR \), points \( A, B, C, D, E, F \) are constructed as follows:
Jackson0423   0
5 hours ago
In triangle \( PQR \), points \( A, B, C, D, E, F \) are constructed as follows: Points \( A \) and \( B \) lie on the extension of side \( QR \) such that \( AP = BP = QR \). Points \( C \) and \( D \) lie on the extension of side \( PQ \) such that \( CR = DR = PQ \). Points \( E \) and \( F \) lie on the extension of side \( RP \) such that \( EQ = FQ = RP \).

The points are placed in a clockwise order around triangle \( PQR \).

Prove that:
\[
\angle ACE + \angle FBD + \angle EAC = 180^\circ.
\]
0 replies
Jackson0423
5 hours ago
0 replies
Similar triangles formed by angular condition
Mahdi_Mashayekhi   5
N Today at 2:02 PM by sami1618
Source: Iran 2025 second round P3
Point $P$ lies inside of scalene triangle $ABC$ with incenter $I$ such that $:$
$$ 2\angle ABP = \angle BCA , 2\angle ACP = \angle CBA $$Lines $PB$ and $PC$ intersect line $AI$ respectively at $B'$ and $C'$. Line through $B'$ parallel to $AB$ intersects $BI$ at $X$ and line through $C'$ parallel to $AC$ intersects $CI$ at $Y$. Prove that triangles $PXY$ and $ABC$ are similar.
5 replies
Mahdi_Mashayekhi
Apr 19, 2025
sami1618
Today at 2:02 PM
Easy Geometry
TheOverlord   33
N Today at 1:37 PM by math.mh
Source: Iran TST 2015, exam 1, day 1 problem 2
$I_b$ is the $B$-excenter of the triangle $ABC$ and $\omega$ is the circumcircle of this triangle. $M$ is the middle of arc $BC$ of $\omega$ which doesn't contain $A$. $MI_b$ meets $\omega$ at $T\not =M$. Prove that
$$ TB\cdot TC=TI_b^2.$$
33 replies
TheOverlord
May 10, 2015
math.mh
Today at 1:37 PM
Existence of a circle tangent to four lines
egxa   3
N Today at 1:36 PM by mathuz
Source: All Russian 2025 10.2
Inside triangle \(ABC\), point \(P\) is marked. Point \(Q\) is on segment \(AB\), and point \(R\) is on segment \(AC\) such that the circumcircles of triangles \(BPQ\) and \(CPR\) are tangent to line \(AP\). Lines are drawn through points \(B\) and \(C\) passing through the center of the circumcircle of triangle \(BPC\), and through points \(Q\) and \(R\) passing through the center of the circumcircle of triangle \(PQR\). Prove that there exists a circle tangent to all four drawn lines.
3 replies
egxa
Apr 18, 2025
mathuz
Today at 1:36 PM
Two very hard parallel
jayme   0
Today at 12:46 PM
Source: own inspired by EGMO
Dear Mathlinkers,

1. ABC a triangle
2. D, E two point on the segment BC so that BD = DE= EC
3. M, N the midpoint of ED, AE
4. H the orthocenter of the acutangle triangle ADE
5. 1, 2 the circumcircle of the triangle DHM, EHN
6. P, Q the second point of intersection of 1 and BM, 2 and CN
7. U, V the second points of intersection of 2 and MN, PQ.

Prove : UV is parallel to PM.

Sincerely
Jean-Louis
0 replies
jayme
Today at 12:46 PM
0 replies
Functional equations
hanzo.ei   20
N Apr 17, 2025 by hanzo.ei
Source: Greekldiot
Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$
20 replies
hanzo.ei
Mar 29, 2025
hanzo.ei
Apr 17, 2025
Functional equations
G H J
G H BBookmark kLocked kLocked NReply
Source: Greekldiot
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hanzo.ei
20 posts
#1
Y by
Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$
Z K Y
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GreekIdiot
174 posts
#2
Y by
Not my problem :D Havent made such a beautiful FE myself yet.
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Mathzeus1024
823 posts
#3
Y by
It works for $\textcolor{red}{f(x)=\frac{1}{x}}$.
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GreekIdiot
174 posts
#4
Y by
Yeah we established that in another post
Z K Y
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hanzo.ei
20 posts
#5
Y by
pco, can you solve it :omighty:
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GreekIdiot
174 posts
#6
Y by
$f$ seems to be an involution. I wonder if we are able to prove that. Then we can eliminate other solutions to the assertion very easily.
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hanzo.ei
20 posts
#7
Y by
bump!!!!
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MR.1
102 posts
#8 • 3 Y
Y by hanzo.ei, Akakri, giangtruong13
solved with GioOrnikapa if you guys want solution please give me $10$ likes :-D
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GreekIdiot
174 posts
#9
Y by
lol aint no way
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GioOrnikapa
76 posts
#10 • 1 Y
Y by MR.1
A lot of liars nowadays smh
Z K Y
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truongphatt2668
322 posts
#11
Y by
hanzo.ei wrote:
Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$

I can prove $f(x)$ is injective and $f(1)=1$ anyone continue please?
This post has been edited 3 times. Last edited by truongphatt2668, Apr 3, 2025, 4:52 PM
Z K Y
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GreekIdiot
174 posts
#12
Y by
truongphatt2668 wrote:
hanzo.ei wrote:
Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$

I can prove $f(x)$ is injective and $f(1)=1$ anyone continue please?

I noticed that there exists some homogenous-like function by isolating $y$ on the $RHS$. Can you post the claims you made with proof so that we can create a complete solution?
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truongphatt2668
322 posts
#13
Y by
GreekIdiot wrote:
truongphatt2668 wrote:
hanzo.ei wrote:
Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$

I can prove $f(x)$ is injective and $f(1)=1$ anyone continue please?

I noticed that there exists some homogenous-like function by isolating $y$ on the $RHS$. Can you post the claims you made with proof so that we can create a complete solution?

Just do it, and I will give a complete solution :D
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GreekIdiot
174 posts
#14
Y by
$f(x)>0$ for all $x \in \mathbb R_+$ so all $y \in \mathbb R_+$ can be written as $\dfrac {f(m)}{f(n)}$ for some $m,n \in \mathbb R_+$
Then there exists some homogenous-kinda function (lets call it $g$) such that $f(xf(y)+f(x))=y^{\ell +1} \cdot g(x)$ and also $f(x+yf(x))=y^{\ell} \cdot g(x)$ thats what I meant to say. Correct me if wrong lol. :oops:
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jasperE3
11229 posts
#15
Y by
GreekIdiot wrote:
$f(x)>0$ for all $x \in \mathbb R_+$ so all $y \in \mathbb R_+$ can be written as $\dfrac {f(m)}{f(n)}$ for some $m,n \in \mathbb R_+$
Then there exists some homogenous-kinda function (lets call it $g$) such that $f(xf(y)+f(x))=y^{\ell +1} \cdot g(x)$ and also $f(x+yf(x))=y^{\ell} \cdot g(x)$ thats what I meant to say. Correct me if wrong lol. :oops:

What's a homogenous kinda function
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GreekIdiot
174 posts
#16
Y by
I am not sure how to call it in english or even what it is. Hope you can understand what I am saying from the symbols :D Thats the important part anyways, not some random math definition.
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jasperE3
11229 posts
#17
Y by
GreekIdiot wrote:
I am not sure how to call it in english or even what it is. Hope you can understand what I am saying from the symbols :D Thats the important part anyways, not some random math definition.

I can't, can you explain?
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GreekIdiot
174 posts
#18
Y by
So basically I am trying to define a second function, g, which exists and satisfies both relations above. Then proving g must be constant will help in proving that the only sol we have found so far is unique. Hope that clears things up.
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jasperE3
11229 posts
#19
Y by
GreekIdiot wrote:
So basically I am trying to define a second function, g, which exists and satisfies both relations above. Then proving g must be constant will help in proving that the only sol we have found so far is unique. Hope that clears things up.

How is $g$ defined
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GreekIdiot
174 posts
#20
Y by
$g(x) > 0$ is a must for all positive $x$. Then it could be any function but we may be able to narrow it down. Just brainstorming, nothing rigorous. This FE has been unsolved for some time, I doubt that I of all people will be the one to solve.
This post has been edited 3 times. Last edited by GreekIdiot, Apr 4, 2025, 8:28 PM
Z K Y
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hanzo.ei
20 posts
#21
Y by
:blush: :blush:
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