Some obviousness about harmonic divisions
by math_explorer, Mar 14, 2011, 10:29 AM
If
are on a line (in that order), then these statements are pairwise equivalent:
-
is a harmonic division
-
and
are inverses w.r.t. the circle with diameter 
-
and
are inverses w.r.t. the circle with diameter 
- the circle with diameter
is a circle of Apollonius with foci 
- the circle with diameter
is a circle of Apollonius with foci 
If
is a point outside
:
is right iff
is bisected (internally/externally) by
and/or
;
is right iff
is bisected (internally/externally) by
and/or
;

iff
![\[ \frac{\sin AOB / \sin OAB}{\sin BOC / \sin OCB} = \frac{\sin AOD / \sin OAD}{\sin COD / \sin OCD} \]](//latex.artofproblemsolving.com/e/c/0/ec0e34ea53d974f4a27d839e3145b3ea55d58566.png)
But
and
so the equation becomes
![\[ \sin AOB \sin COD = \sin AOD \sin BOC \]](//latex.artofproblemsolving.com/7/e/c/7ec2844397b8003fdc4b67d2ff8fd76f9727af36.png)
depending on only the angles between
,
,
,
.
This is why we can say
is a harmonic pencil.
Suppose
is a triangle and the tangents through
,
to the circumcircle of
intersect at
. Then
is the
-symmedian of
.
Proof: Let the perpendicular bisector of
(also passing through the center of the circumcircle and through
) intersect the circle at
, with
closer to
; let the midpoint of
be
.
Then since
and
are inverses w.r.t. the circumcircle,
is a harmonic division. Since
is right because
is a diameter of the circumcircle,
is bisected by
. Since
is the midpoint of arc
,
also bisects
. Thus
and
are isogonals w.r.t.
. Since
is the
-median,
is the
-symmedian.

-

-



-



- the circle with diameter


- the circle with diameter


If











iff
![\[ \frac{\sin AOB / \sin OAB}{\sin BOC / \sin OCB} = \frac{\sin AOD / \sin OAD}{\sin COD / \sin OCD} \]](http://latex.artofproblemsolving.com/e/c/0/ec0e34ea53d974f4a27d839e3145b3ea55d58566.png)
But


![\[ \sin AOB \sin COD = \sin AOD \sin BOC \]](http://latex.artofproblemsolving.com/7/e/c/7ec2844397b8003fdc4b67d2ff8fd76f9727af36.png)
depending on only the angles between




This is why we can say

Suppose








Proof: Let the perpendicular bisector of







Then since


















This post has been edited 1 time. Last edited by math_explorer, Aug 19, 2011, 9:26 AM