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Two Questions on Geometric Harmonics

Date: 11/24/2005 at 15:38:44
From: amar
Subject: geometry

ABC is an acute triangle, and P is any point inside it.  Draw BP and
extend it to meet AC at E.  Similarly draw CP and extend it to meet AB
at F.  Join EF.  Join AP.  Let the point where AP intersects EF be D.
Draw the perpendicular from D to BC meeting it at K.

Prove that angle EKF is bisected by KD.

I tried to use coordinate geometry but could not get any good result
as it only led to very long calculations.  I am looking for a pure
geometry proof.



Date: 11/26/2005 at 08:05:13
From: Doctor Floor
Subject: Re: geometry

Hi Amar,

Thanks for your question.  A geometric proof requires some knowledge 
I hope you will find in the presented links.  If needed it is not 
difficult to find more sites on the Internet which explain the facts 
I will use.

Note that the lines AC, AB, BP, and CP form a complete quadrilateral. 
See for instance:

  Complete Quadrilateral
    http://mathworld.wolfram.com/CompleteQuadrilateral.html 

Now let G be the point where EF meets BC.  Since EF, AP and BC are 
the diagonals of the complete quadrilateral, the points D and G are 
harmonic conjugates.  In other words, ED:DF = EG:GF.  See for instance

  Harmonic Ratio
    http://www.cut-the-knot.org/pythagoras/HarmonicRatio.shtml 

The locus of all points X with EX=XF is an Apollonius circle, which 
has DG as diameter.  See for instance:

  Locus Circle
   http://www.cut-the-knot.org/Curriculum/Geometry/LocusCircle.shtml 

When you read this page, you will note that for any point Y on this 
Apollonius circle, YD bisects angle AYF.  Now note that the circle
meets BC in a second point, say L, apart from G.  As DG is a diameter
of this circle, angle DLG must be a right angle.  But that means that
L = K.  And we are done.

If you have more questions, just write back.

Best regards,

- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 12/28/2005 at 00:53:07
From: amar
Subject: geometry

Thank you, Dr. Floor.  Your answer was quite helpful.  I have another
question which may be similar.

Let 2 circles intersect each other at B and C.  Their common tangent
touches them at P and Q.  Any circle is drawn through B and C cutting
PQ at L and M.  Prove that {PQ:LM} is harmonic.



Date: 12/28/2005 at 07:40:35
From: Doctor Floor
Subject: Re: geometry

Hi, Amar,

Thanks for your question.

This is really a theorem from projective geometry.  The circles 
through B and C constitute an involution that changes points L and 
M as you described, and P and Q are double points or fixed points of 
this involution.  A pair L,M of such an involution divides the fixed 
points always harmonically.  But that is all quite advanced, so let's 
try to find a more basic proof.

To do this, let's start with a special case, that the line PQ ("an 
infinite circle") meets BC at the midpoint of B and C ("and at 
infinity").  You can find this in the Dr. Math library at

  Intersecting Circles
    http://mathforum.org/library/drmath/view/55125.html 

Now let's go to the general case.  Let's call the point where BC and 
PQ meet W.  From the theory explained in the above page, we know that 
WL*WM = WQ^2 = WB*WC.

   

Now if we use notation

  WL = x
  WM = y
  WP = WQ = z
  
WL*WM = WQ^2 can be rewritten as xy = z^2.

From this we find:

  xy - z^2 = z^2 - xy = 0
  xy - z^2 - xz + yz = z^2 - xy - xz + yz
  (z+x)(y-z) = (z-x)(z+y)

  z+x   z+y
  --- = ---
  z-x   y-z

Now we can retranslate this using the figure as

  |PL|   |PM|
  ---- = ----
  |QL|   |QM|

or, in other words, LM divides PQ harmonically.

If you have more questions, just write back.

Best regards,

- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Triangles and Other Polygons
High School Triangles and Other Polygons

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