All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Harmonic mean and constant determination puzzle (Posted on 2015-06-13)
In a triangle ABC , ∠BAC = 90o. Point D lies on the side BC, and satisfies
∠BDA = k*∠BAD, where k is a real constant.

Find the value of k, given that length of AD is the harmonic mean between the respective lengths of BD and CD.

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution | Comment 1 of 2
B=(-1,0), C=(1,0), D=(d,0), O=(0,0)
BD=1+d, CD=1-d, OD=d
The harmonic mean is 2(1+d)(1-d)/((1+d)+(1-d))=1-d^2
OA = 1

Using the law of cosines on triangle ADO find
cos(ODA)=(d^2+(1-d^2)^2-1^2)/(2(d^2)(1-d^2))
=-d/2=cos(∠BDA)

Using the law of cosines on triangle ABD find
AB^2=(d+1)^2+(1-d^2)^2-2(d+1)(1-d^2)*-d/2=-d^3+3d+2

Using the law of cosines on triangle ABD again

Using the double angle formula for this last cosine
which is precisely cos(∠BDA)

so k=2

 Posted by Jer on 2015-06-15 12:36:54

 Search: Search body:
Forums (0)