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
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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)