Collinearity in a Harmonic Configuration from a Cyclic Quadrilateral
by kieusuong, May 15, 2025, 2:26 PM
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.*
that exist function 
such 
continuous on 
and discontinuous on 
satisfy 
for all 
.
![$\{x_n\} \subseteq [0,1]$](http://latex.artofproblemsolving.com/8/e/4/8e4206325387485d73765c9d7070ecb1ce18ff60.png)
,
,
,
and 
could be chosen to satisfy 
and 
.
be the open unit disk in the complex plane and let 
be a holomorphic function such that 
.
.
.
Such That 
is even?
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
![\[y''+by'+cy=g(t)\]](http://latex.artofproblemsolving.com/7/8/8/7882dc954ab6dba2445b68d370d630ce1c66a6db.png)
with 
,
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)
and use variation of parameters in each case to find the solution. Is there any short way to calculate solution?
be an abelian additive group such that all nonzero elements have infinite order and for each prime number 
we have the inequality 
,
,
(where the sum has 
is the order of the quotient group 
(the index of the subgroup 
)
May 2025
April 2025