Thursday, April 2, 2009

Problem 278: Tangent Circles, Common External Tangent, Chord

Proposed Problem

Problem 278: Tangent Circles, Common External Tangent, Chord.

See complete Problem 278 at:
gogeometry.com/problem/p278_tangent_circle_common_external_tangent_chord.htm

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 5:16 PM
Labels: , , ,

5 comments:

  1. AjitApril 4, 2009 at 9:08 PM

    Extend AB on either side to meet the gircle with radius a at F and at G on the other side. Also let FD and GE when extended meet in H. Now FHG is a rt. angled triangle with CD & CE perpendicular to FH & GH. Let angle FCD = θ . Therefore angle CGE = θ . Now cos(θ)=CD/CF=x/2a--(1). In tr. HCG we've cos(θ)= GC/GH = (GC/GF)/(GH/GF). But from tr. HGF we've cos(θ) = GH/GF.
    Thus cos(θ)=(GC/GF)/cos(θ) or [cos(θ)]^2=GC/GF But GC/GF=2b/(2a+2b)= b/(a+b). Thus using equation (1) (x/2a)^2=b/(a+b) or
    x = 2a*(b/(a+b))^(1/2). QED
    Ajit: ajitathle@gmail.com

    ReplyDelete
  2. Mrudul M. ThatteJune 15, 2009 at 7:41 AM

    We can prove that
    (DE)*(DE) = 4*a*b
    In triangle BED by Pythagoras Theorem,
    BD * BD = 4*a*b + b*b

    In Triangle ABD by Stewart's Theorem,
    BC * AD * AD + AC * BD * BD
    = AB * CD * CD + AB * AC * BC

    So,
    a*a*b + 4*a*a*b + a*b*b
    = (a+b)* x*x + (a+b)*a*b

    So,
    5*a*a*b + a*b*b = x*x (a+b) + a*a*b + a*b*b

    Thus,
    4a*a*b / (a+b) = x*x

    So,
    x = 2*a * [Square root of b/(a+b)]

    Hence the proof.

    ReplyDelete
  3. AnonymousApril 26, 2011 at 3:14 PM

    good job

    ReplyDelete
  4. ArifJanuary 19, 2018 at 4:15 AM

    Draw the tangent line common to the circles A and B . The tangent line meets DE in X.
    We see that

    DE =2CX

    From problem 277 we know that DE =2(ab)^(1/2) and

    CX=(ab)^(1/2)

    Then we have

    AX^2=a^2+ab
    AX=(a^2+ab)^(1/2)

    We connect DA and ACXD is concylical so we aplly ptolemy:

    x(a^2+ab)^(1/2)=a(ab)^(1/2)+a(ab)^(1/2)
    x(a^2+ab)^(1/2)=2a(ab)^(1/2)
    x=2a(ab)^(1/2)/(a^2+ab)^(1/2)
    x=2a(b)^(1/2)/(a+b)^(1/2)

    ReplyDelete
  5. Sumith PeirisJanuary 3, 2021 at 10:16 AM

    Let the common tangent at C meet DE at F. Let DE = 2t and let CE = y

    FD = FC = FE so < DCE = 90 and x^2 + y^2 = 4t^2 …… (1)

    From similar Triangles, x/t = y/b ……(2) and y/t = x/a ….. (3)

    (1) X (2) gives xy/t^2 = xy/ab hence t^2 = ab ….. (4)

    From (3), (x^2 + y^2)/(a^2 + t^2) = x^2/a^2 and so from (1),
    4t^2 / (a^2 + t^2) = x^2/a^2

    Therefore x = 2at/√(a2 + t2) = 2a/√{ab/(a2 + ab)} = 2a/√{b/(a+b)}

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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