## Thursday, April 2, 2009

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

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

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

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,
= 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.

3. good job

4. 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)