Points A and C lie on a circle centered at O, each of BA and BC are ta

Question

Points A and C lie on a circle centered at O, each of BA and BC are tangent to the circle, and triangle ABC is equilateral. The circle intersects BO at D. What is BD/BO ?

(A)  \frac{\sqrt{2}}{3}

(B)  \frac{1}{2}

(C)  \frac{\sqrt{3}}{3}

(D)  \frac{\sqrt{2}}{2}

(E)  \frac{\sqrt{3}}{2}

m1033512 · Answer

IMO B

extend OB so that it cut circle at M

now by circle property

BD*BM = AB^2

BD*(BO+OM) = AB^2

let the raidus B r

then
BD*(BO+OM) = AB^2

find the value of BO and AB in terms of r

in right angle triangle AOB , AB = √3r

and OB = 2r

we get BD = r

so BD/BO = r/2r = 1/2

award kudos if helpful

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ankitamundhra28 · Answer

Could you please provide a diagram for the solution?

Thanks!

nick1816 · Answer

Let AB=BC=CA=a
and OA=OC=OD=r

a/sin150=r/sin30
 a=\sqrt{3}r

AC^2=BD*BE 
 a^2= BD*(BD+2r) 
 3r^2=BD^2+2r*BD 
 BD^2+2r*BD-3r^2=0 
BD=r or -3r 
BD can't be -ve, Hence BD=r

BD/BO=r/(r+r)=1/2

GMATGuruNY · Answer

A radius drawn to a tangent line forms a RIGHT ANGLE.
Thus, triangle OAB is a right triangle, as shown in the figure below:
 https://i.ibb.co/cwdCz24/kite-and-circle.png 
In the figure above:
The circle has a radius of 2.
BO = 1+3 = 4
BD = BO - radius OD = 4-2 = 2
Thus:
 \frac{BD}{BO} = \frac{2}{4} = \frac{1}{2}

B

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Points A and C lie on a circle centered at O, each of BA and BC are ta

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Points A and C lie on a circle centered at O, each of BA and BC are ta

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