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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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29 Mar 2019, 00:35
29 Mar 2019, 00:35
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Difficulty:
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Question Stats:
50% (03:10) correct
50%
(02:56)
wrong
based on 92
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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) \(\frac{\sqrt{3}}{2}\)
(A) \(\frac{\sqrt{2}}{3}\)
(B) \(\frac{1}{2}\)
(C) \(\frac{\sqrt{3}}{3}\)
(D) \(\frac{\sqrt{2}}{2}\)
(E) \(\frac{\sqrt{3}}{2}\)
Re: Points A and C lie on a circle centered at O, each of BA and BC are ta
[#permalink]
18 Apr 2019, 05:47
18 Apr 2019, 05:47
2
Kudos
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
Posted from my mobile device
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
Posted from my mobile device
Intern
Joined: 14 Mar 2018
Posts: 20
Given Kudos: 249
Location: India
Concentration: Finance, Marketing
GMAT 1: 760 Q50 V42
GPA: 3.48
Re: Points A and C lie on a circle centered at O, each of BA and BC are ta
[#permalink]
27 May 2019, 04:49
27 May 2019, 04:49
m1033512 wrote:
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
Posted from my mobile device
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
Posted from my mobile device
Could you please provide a diagram for the solution?
Thanks!
