Proposed Problem
See complete Problem 277 at:
gogeometry.com/problem/p277_tangent_circle_common_external_tangent.htm
Level: High School, SAT Prep, College geometry
Thursday, April 2, 2009
Problem 277: Tangent Circles, Common External Tangent
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Join BE and draw AF perpendicular to BE meeting the latter in F. Now DE=AF and in tr. ABF: AB^2=AF^2+BF^2 or BF^2= AF^2 = BF^2 or DE^2 = AF^2 - BF^2. But AF=a+b and BF=a-b. Hence, x^2 =DF^2=(a+b)^2-(a-b)^2 or x^2 =4ab or x = 2*(ab)^(1/2)
ReplyDeleteQED
Ajit: ajitathle@gmail.com
(a+b)^2-(b-a)^2=x^2
ReplyDeletea^2+2ab+b^2-(b^2-2ab+a^2)=x^2
4ab=x^2
√(4ab)=x
2√(ab)=x
Draw the tangent line common to the circles A and B at the point C. The tangent line meets DE at the point X .
ReplyDeleteWe have:
DX=CX=XE= y
And
x=2y
The points ACXD and BCXE are concylical. Therefore ∠CXD =∠CBE
If we draw the line AX it bisects the angle ∠CXD , and the line BX bisects ∠CBE
Therefore triangles ACX and XCB are similar and we get :
y/b=a/y
y^2=ab
(x/2)^2=ab
x^2=4ab
x=2(ab)^(1/2)