Monday, October 8, 2018

Geometry Problem 1394: Three Circles and Three Pair of Common Internal Tangents, Concurrency

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

Details: Click on the figure below.

Geometry Problem 1394: Three Circles and Three Pair of Common Internal Tangents, Concurrency, Tutoring.

Posted by Antonio Gutierrez at 1:24 PM
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4 comments:

  1. AnonymousOctober 9, 2018 at 6:01 PM

    circle A, radius is Ra;
    connect AB, ab go through D,
    same for AC,BC; E on AC, F on BC;
    so AE/EC= Ra/Rc, CF/FB=Rc/Rb; BD/DA=Rb/Ra;
    so AE/EC * CF/FB * BD/DA =1,
    so AF,BE,CD concurrent

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  2. c.t.e.oOctober 10, 2018 at 1:05 PM

    By ceva's theorem: (AE)/(EC).(CF/CB).(BD/DA)=1 (Tr AET similar to tr ECP T,P tg points)
    (Ra/Rc).(Rc/Rb).(Rb/Ra)=1

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  3. AnonymousOctober 12, 2018 at 5:46 AM

    Fie triunghiul format de ABC. Se arată ușor că punctele E , F și D se găsesc pe laturile AC, CB și AB.
    Notând cu M punctul de tangență cu cercul A și cu N punctul de tangență comună cu cercul C, din asemănarea triunghiurilor AME și CNE ( AA) se obține raportul AE/EC = r_A/r_C si analog CF/FB = r_c/r_B respectiv BD/DA = r_B/r_A . De aici AE/EC ∙ CF/FB ∙ BD/DA= r_A/r_C ∙ r_A/r_C ∙ r_B/r_A = 1 și conform reciproca teoremei Ceva rezultă că dreptele AF, CD, BE sunt concurente.
    Florin Popa , Comănești, România.

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  4. Peter TranOctober 16, 2018 at 7:31 PM

    https://photos.app.goo.gl/bEcZvA4JTfREGsin9

    Let rA, rB, rC are radius of circles A, B and C
    Note that B,F,C are colinear
    Triangle BB’F similar to CC’F ( case AA)
    So BF/FC= BB’/CC’= rB/rC
    Similarly we also have CE/AE= rC/rA and DA/DB= rA/rB
    Multiply these expressions side by side and simplify it we get
    BF/BC x EC/EA x DA/DB= 1
    So AF, BE, CD are concurrent per Ceva’s theorem

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