Proposed Problem

See complete Problem 270 at:

gogeometry.com/problem/p270_tangent_circles_fractional_exponent.htm

Level: High School, SAT Prep, College geometry

## Tuesday, March 17, 2009

### Problem 270. Tangent Circles, Common external tangent, Fractional exponents

Labels:
circle,
common tangent,
diameter,
right triangle

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Extend FD & GE to meet in H. HFG is a rt. angled triangle where HC is perpendicular to FG. Now by the previous problem (no. 269) we've:

ReplyDeletea = FH^3/x^2 & b = GH^3/x^2.

Thus, a^(2/3)+ b^(2/3)=(FH^3/x^2)^(2/3)+ (GH^3/x^2)^(2/3)=(FH^2+GH^2)/x^(4/3)

= x^2/x^(4/3) = x^(2/3)

or a^(2/3) + b^(2/3)= x^(2/3)

Ajit: ajitathle@gmail.com

Let

ReplyDeleteAC =r

BC =R

From problem 278 we can find

DC^2 =4r^2R/(r+R)

CE^2=4R^2r/(r+R)

Then

a^2=FC^2-DC^2

a^2=4r^2-4r^2R/(r+R)

a^2=4r^3/(r+R)

And similarly for b we find

b^2=CG^2-CE^2

b^2=4R^2-4R^2r/(r+R)

b^2=4R^3/(r+R)

Than

a^(2/3)+b^(2/3)=(2/(r+R))^(1/3)r+R(2/(r+R))^(1/3)

a^(2/3)+b^(2/3)=(2/(r+R))^(1/3)(r+R)

a^(2/3)+b^(2/3)=(2(r+R))^(2/3)

And 2(r+R)=c therefore :

a^(2/3)+b^(2/3)=c^(2/3)