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

3 comments:

  1. 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

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  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.

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