B=(1,0), C=(1,0), D=(d,0), O=(0,0)
BD=1+d, CD=1d, OD=d
The harmonic mean is 2(1+d)(1d)/((1+d)+(1d))=1d^2
AD=1d^2
OA = 1
Using the law of cosines on triangle ADO find
cos(∠ODA)=(d^2+(1d^2)^21^2)/(2(d^2)(1d^2))
=d/2=cos(∠BDA)
Using the law of cosines on triangle ABD find
AB^2=(d+1)^2+(1d^2)^22(d+1)(1d^2)*d/2=d^3+3d+2
Using the law of cosines on triangle ABD again
cos(∠BAD)=[(d^3+3d+2)+(1d^2)^2(1+d)^2]/[2(1d^2)*sqrt(d^3+3d+2)]
cos(∠BAD)=sqrt(2d)/2
Using the double angle formula for this last cosine
cos(2*∠BAD)=2[sqrt(2d)/2]^21=d/2
which is precisely cos(∠BDA)
so k=2

Posted by Jer
on 20150615 12:36:54 