Proposed Problem

See complete Problem 278 at:

gogeometry.com/problem/p278_tangent_circle_common_external_tangent_chord.htm

Level: High School, SAT Prep, College geometry

## Thursday, April 2, 2009

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

Labels:
chord,
common tangent,
external,
tangent

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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.

ReplyDeleteThus 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

We can prove that

ReplyDelete(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.

good job

ReplyDelete