Wednesday, September 28, 2016

Geometry Problem 1267 Triangle, Incircle, Excircle, Circle, Tangency Points, Perpendicular, 90 Degrees, Parallelogram

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

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Geometry Problem 1267 Triangle, Incircle, Excircle, Circle, Tangency Points, Perpendicular, 90 Degrees, Parallelogram.
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3 comments:

  1. Peter TranSeptember 29, 2016 at 12:39 AM

    https://goo.gl/photos/1anav7EpeN3Uvv8e9

    Let ED meet AC at P and E1D1 meet AC at P1
    Note that triangles BED and BE1D1 are isosceles and ED//E1D1
    CF=AF1= p- AB where p= semi perimeter of triangle ABC
    1. Per the result of problem 1265 and 1266 we have
    FC/FA= PC/PA= k
    We can calculate FP in term of k and AC
    FP= ( (k/1+k) + k/(1-k)). AC
    Similarly F1A/F1C=P1A/P1C= k and F1P1= ( (k/1+k) + k/(1-k)). AC
    So FP=F1P1
    Triangle PGF congruent to P1G1F1 ( case ASA) => FG=F1G1
    We also have FG//F1G1 => FGF1G1 is a parallelogram

    2. we have triangle AFG congruent to CF1G1 ( case SAS) and triangle CGF1 congruent to AG1F1 ( case SAS)
    So AG= CG1 and CG= AG1
    And AGCG1 is a parallelogram

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  2. APOSTOLIS MANOLOUDISSeptember 29, 2016 at 3:06 AM

    Problem 1267
    Is AF_1=CF(O=incircle, O_1=excircle corresponding to AC) or AE_1=AF_1=CF=CD. But triangleAE_1G_1 is similar with triangle G_1D_1C then E_1G_1/G_1D_1=AE_1/CD_1=DC/CD_1 so than vice versa Thales' theorem we have
    E_1D//G_1C. Also triangle AEG is similar with triangle GDC so EG/GD=AE/DC=AE/AE_1
    then AG//DE_1.Terefore AG//CG_1.Similar CG//AG_1 so AG_1CG is parallelogram.
    If AC Intersects GG_1 in M then G_1M=MG and F_1M=MF (AF_1=FC) so F_1G_1FG is
    parallelogram.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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  3. Sumith PeirisSeptember 29, 2016 at 4:18 AM

    With the usual notation, 2s = a+b+c,

    AE1= AF1 = CF =CD = s-c and
    AE = AF = CF1 =CD1 = s-a

    From Problems 1265 and 1266 we have

    <E1AG = < G1CD1 and < EAG = < DCG, so < GAG1 = GCG1
    Further AG1/CG1 = AF1/CF1 = (s-c)/(s-a) = CF/AF

    Hence Tr.s GAG1 and GCG1 are similar. However the side GG1 is common to
    these 2 triangles. Hence they must also be congruent and therefore AGCG1 is a
    parallelogram.

    Hence Tr.s AF1G1 ≡ CGF (SAS),
    so AG1 = GC
    Similarly Tr.s AF1G ≡ CFG1 (SAS),
    so AG = CG1

    So in quadrilateral FGF1G1, the opposite sides are equal, hence the same
    is a parallelogram.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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