Saturday, February 11, 2012

Problem 728: Triangle, Altitude, Perpendicular, Circumcircle, Midpoint, Intersecting Circles, Concurrency

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 728 details.

Online Geometry Problem 728: Triangle, Altitude, Perpendicular, Circumcircle, Midpoint, Intersecting Circles, Concurrency.

4 comments:

  1. Note that centers O and O’ will be on AC.
    Let 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

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  2. My solution

    http://lectiimatematice.blogspot.com/2012/02/problem-728-triangle-altitude.html

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  3. A beautiful application of Ceva's Theorem (Trigonometric Version)
    Sketch:
    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.

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  4. Problem 728
    The 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

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