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.

Click the figure below to view more details of problem 1267.

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

Posted by Antonio Gutierrez at 8:50 PM
Labels: , , , , , ,

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

    ReplyDelete
  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

    ReplyDelete
  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

Subscribe to: Post Comments (Atom)