perplexus.info :: Shapes : Harmonic mean and constant determination puzzle

Harmonic mean and constant determination puzzle (Posted on 2015-06-13)

In a triangle ABC , ∠BAC = 90^o. 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.

See The Solution Submitted by K Sengupta

Rating: 4.5000 (2 votes)

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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

AD=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

cos(∠BAD)=[(-d^3+3d+2)+(1-d^2)^2-(1+d)^2]/[2(1-d^2)*sqrt(-d^3+3d+2)]

cos(∠BAD)=sqrt(2-d)/2

Using the double angle formula for this last cosine

cos(2*∠BAD)=2[sqrt(2-d)/2]^2-1=-d/2

which is precisely cos(∠BDA)

so k=2

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

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