Two Questions on Geometric HarmonicsDate: 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/ |
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