Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the problem 728 details.
Saturday, February 11, 2012
Problem 728: Triangle, Altitude, Perpendicular, Circumcircle, Midpoint, Intersecting Circles, Concurrency
Subscribe to:
Post Comments (Atom)
Note that centers O and O’ will be on AC.
ReplyDeleteLet NG cut EF at L and HM cut EF at L’
NG Perpen. To AB & HM perpen. To BC
Quadrilaterals ANQL and CDL’M are cyclic
Power from point P to circle ANQL= BN.BA=BL.BD
Power from point P to circle CDL’M=BM.BC=BL’.BD
Power of point P to circles O and O’ =BN.BA=BE.BF=BM.BC
From above we have BL.BD=BL’.BD => BL=BL’ => L coincide to L’
So BF, GN and MH are concurrent
My solution
ReplyDeletehttp://lectiimatematice.blogspot.com/2012/02/problem-728-triangle-altitude.html
A beautiful application of Ceva's Theorem (Trigonometric Version)
ReplyDeleteSketch:
Let EAC = x, ECA = y
AE, BE, CE are concurrent Cevians satisfying
sin ABD.sin ECB.sin EAC
= sin EBC.sin ECA.sin EAB
which implies
cos A.sin(C-y).sin x = cos C.sin y.sin (A-x)...(#)
EF,GN,HM are Cevians thro'the vertices E,G,H of triangle EGH.
sin HED=sin HCE=sin y ;
sin EGN=sin EAN=sin(A-x);
sin CHM=sin(90-C)=cos C
while
sin DEG=sin GAE=sin x;
sin NGA=sin(90-A)=cos A;
sin MHE=sin MCE= sin(C-y)
Clearly in view of (#)
The product of the above first three ratios
= the product of the above next three ratios.
Hence by Ceva, EF,GN,HM are concurrent.
Problem 728
ReplyDeleteThe ΒF is the radical axis of centers of circles O and O ',then BN.BA=BE.BF=BM.BC. So
A,N,M,C are concyclic.Then <MNB=<BCA.But If I bring the altitudes ck and AL (T = orthocenter).Is <AKC=<ALC=90 then A,K,L,C are concyclic. So <LKB=<BCA.Therefore
KL//NM then BK/BN=BL/BM ,but if NG and HM intersect the BF to P, P ' respectively
(KT//PN,LT//MP’). Is BK/BN=BT/BP, BL/BM=BT/BP’. So BT/BP=BT/BP’ then BP=BP’.
Therefore NG,MH,BD are concyrrent.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE