Then 
But I am stuck. I came across this functional equation by trying to find a limit of a sequence, I can show that if it is helpful. 
has solutio
and use variation of parameters in each case to find the solution. Is there any short way to calculate solution?
Y by
Let 
a point in the space and 
the centroid of a tetrahedron 
. Prove that:
a point in the space and 
the centroid of a tetrahedron 
. Prove that:
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Cute property of Pascal hexagon config
Miquel-point 0
Source: KoMaL B. 5444
In cyclic hexagon 
let 
denote the intersection of diagonals 
and 
, and let 
denote the intersection of diagonals 
and 
. Prove that if 
and 
, then 
bisects angle 
.
Proposed by Géza Kós, Budapest
let 
denote the intersection of diagonals 
and 
, and let 
denote the intersection of diagonals 
and 
. Prove that if 
and 
, then 
bisects angle 
.Proposed by Géza Kós, Budapest
0 replies
II_a - r_a = R - r implies A = 60
Miquel-point 0
Source: KoMaL B. 5421
The incenter and the inradius of the acute triangle 
are 
and 
, respectively. The excenter and exradius relative to vertex 
is 
and 
, respectively. Let 
denote the circumradius. Prove that if 
, then 
.
Proposed by Class 2024C of Fazekas M. Gyak. Ált. Isk. és Gimn., Budapest
are 
and 
, respectively. The excenter and exradius relative to vertex 
is 
and 
, respectively. Let 
denote the circumradius. Prove that if 
, then 
.Proposed by Class 2024C of Fazekas M. Gyak. Ált. Isk. és Gimn., Budapest
0 replies
Number of roots of boundary preserving unit disk maps
Assassino9931 3
N
by bsf714
Source: Vojtech Jarnik IMC 2025, Category II, P4
Let 
be the open unit disk in the complex plane and let 
be a holomorphic function such that 
. Let the Taylor series of 
be 
. Prove that the number of zeroes of (counted with multiplicities) equals 
.
be the open unit disk in the complex plane and let 
be a holomorphic function such that 
. Let the Taylor series of 
be 
. Prove that the number of zeroes of (counted with multiplicities) equals 
.
3 replies
|A/pA|<=p, finite index=> isomorphism - OIMU 2008 Problem 7
Jorge Miranda 2
N
by pi_quadrat_sechstel
Let be an abelian additive group such that all nonzero elements have infinite order and for each prime number 
we have the inequality 
, where 
, 
(where the sum has summands) and 
is the order of the quotient group 
(the index of the subgroup 
).
Prove that each subgroup of of finite index is isomorphic to .
be an abelian additive group such that all nonzero elements have infinite order and for each prime number 
we have the inequality 
, where 
, 
(where the sum has summands) and 
is the order of the quotient group 
(the index of the subgroup 
).Prove that each subgroup of
of finite index is isomorphic to .
2 replies
Prove the statement
Butterfly 8
N
by oty
Given an infinite sequence ![$\{x_n\} \subseteq [0,1]$](//latex.artofproblemsolving.com/8/e/4/8e4206325387485d73765c9d7070ecb1ce18ff60.png)
, there exists some constant 
, for any 
, among the sequence 
and 
could be chosen to satisfy 
and 
.
![$\{x_n\} \subseteq [0,1]$](http://latex.artofproblemsolving.com/8/e/4/8e4206325387485d73765c9d7070ecb1ce18ff60.png)
, there exists some constant 
, for any 
, among the sequence 
and 
could be chosen to satisfy 
and 
.
8 replies
Functional equation from limit
IsicleFlow 1
N
by jasperE3
Is there a solution to the functional equation 
Such That 
is even?
Such That 
is even?1 reply
f(m+n)≤f(m)f(n) implies existence of limit
Etkan 2
N
by Etkan
Let 
satisfy 
for all 
. Prove that
satisfy 
for all 
. Prove that
2 replies
Collinearity in a Harmonic Configuration from a Cyclic Quadrilateral
kieusuong 0
Let 
be a fixed circle, and let 
be a point outside such that 
. A variable line through intersects the circle at two points 
and 
, such that the quadrilateral 
is cyclic, where 
are fixed points on the circle.
Define the following:
-
,
-
,
-
is the tangent from to the circle , and 
is the point of tangency.
**Problem:**
Prove that for all such configurations:
1. The points
, 
, and are collinear.
2. The line
is perpendicular to chord 
.
3. As the line through varies, the point traces a fixed straight line, which is parallel to the isogonal conjugate axis (the so-called *isotropic line*) of the centers 
and .
---
### Outline of a Synthetic Proof:
**1. Harmonic Configuration:**
- Since
lie on a circle, their cross-ratio is harmonic:
![\[
(ANMB) = -1.
\]](//latex.artofproblemsolving.com/b/d/5/bd5301ba76f493aec0821c454dffc46dea93c0f2.png)
- The intersection points , and form a well-known harmonic setup along the diagonals of the quadrilateral.
**2. Collinearity of, , :**
- The line is tangent to , and due to harmonicity and projective duality, the polar of passes through .
- Thus,, , and must lie on a common line.
**3. Perpendicularity:**
- Since is tangent at and is a chord, the angle between and chord is right.
- Therefore, line is perpendicular to .
**4. Quasi-directrix of:**
- As the line through varies, the point moves.
- However, all such points lie on a fixed line, which is perpendicular to 
, and is parallel to the isogonal (or isotropic) line determined by the centers and .
---
**Further Questions for Discussion:**
- Can this configuration be extended to other conics, such as ellipses?
- Is there a pure projective geometry interpretation (perhaps using polar reciprocity)?
- What is the locus of point, or of line , as varies?
*This configuration arose from a geometric investigation involving cyclic quadrilaterals and harmonic bundles. Any insights, counterexamples, or improvements are warmly welcomed.*
be a fixed circle, and let 
be a point outside such that 
. A variable line through intersects the circle at two points 
and 
, such that the quadrilateral 
is cyclic, where 
are fixed points on the circle.Define the following:
-
,-
,-
is the tangent from to the circle , and 
is the point of tangency.**Problem:**
Prove that for all such configurations:
1. The points
, 
, and are collinear.2. The line
is perpendicular to chord 
.3. As the line through
varies, the point traces a fixed straight line, which is parallel to the isogonal conjugate axis (the so-called *isotropic line*) of the centers 
and .---
### Outline of a Synthetic Proof:
**1. Harmonic Configuration:**
- Since
lie on a circle, their cross-ratio is harmonic:![\[
(ANMB) = -1.
\]](http://latex.artofproblemsolving.com/b/d/5/bd5301ba76f493aec0821c454dffc46dea93c0f2.png)
- The intersection points , and form a well-known harmonic setup along the diagonals of the quadrilateral.**2. Collinearity of
, , :**- The line
is tangent to , and due to harmonicity and projective duality, the polar of passes through .- Thus,
, , and must lie on a common line.**3. Perpendicularity:**
- Since
is tangent at and is a chord, the angle between and chord is right.- Therefore, line
is perpendicular to .**4. Quasi-directrix of
:**- As the line through
varies, the point moves.- However, all such points
lie on a fixed line, which is perpendicular to 
, and is parallel to the isogonal (or isotropic) line determined by the centers and .---
**Further Questions for Discussion:**
- Can this configuration be extended to other conics, such as ellipses?
- Is there a pure projective geometry interpretation (perhaps using polar reciprocity)?
- What is the locus of point
, or of line , as varies?*This configuration arose from a geometric investigation involving cyclic quadrilaterals and harmonic bundles. Any insights, counterexamples, or improvements are warmly welcomed.*
0 replies
Find solution of IVP
neerajbhauryal 2
N
by Mathzeus1024
Show that the initial value problem ![\[y''+by'+cy=g(t)\]](//latex.artofproblemsolving.com/7/8/8/7882dc954ab6dba2445b68d370d630ce1c66a6db.png)
with 
, where 
are constants has the form ![\[y(t)=\int^{t}_{t_0}K(t-s)g(s)ds\,\]](//latex.artofproblemsolving.com/6/3/c/63c34df1c0013bf659bc76506a93873a5f879070.png)
![\[y''+by'+cy=g(t)\]](http://latex.artofproblemsolving.com/7/8/8/7882dc954ab6dba2445b68d370d630ce1c66a6db.png)
with 
, where 
are constants has the form ![\[y(t)=\int^{t}_{t_0}K(t-s)g(s)ds\,\]](http://latex.artofproblemsolving.com/6/3/c/63c34df1c0013bf659bc76506a93873a5f879070.png)
2 replies
Sets on which a continuous function exists
Creativename27 1
N
by alexheinis
Source: My head
Find all 
that exist function 
such continuous on 
and discontinuous on 
that exist function 
such continuous on 
and discontinuous on 
1 reply
Japanese Olympiad
parkjungmin 6
N
by mathNcheese_aops
It's about the Japanese Olympiad
I can't solve it no matter how much I think about it.
If there are people who are good at math
Please help me.
I can't solve it no matter how much I think about it.
If there are people who are good at math
Please help me.
6 replies
3D geometry theorem
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Y by KAME06
Just use the vectors (like in Leibniz's formula for a triangle).
Or, see this: https://vixra.org/pdf/1003.0187v1.pdf
Or, see this: https://vixra.org/pdf/1003.0187v1.pdf
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