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Problem Of The Week 11

with 7 comments


Triangle ABC is right angled at C. m<ABC=60 and AB=10. Let P be a point randomly chosen inside triangle ABC, such that BP extended meets CA at D. What is the probability that BD>5root(2)?

Written by Implex

September 11, 2008 at 3:46 pm

7 Responses

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  1. Is the answer to this 0.18?
    If it is, i’ll post my approach.

    milind

    September 11, 2008 at 5:05 pm

  2. No, the answer is much larger than that

    outtimed

    September 11, 2008 at 5:15 pm

  3. oops.. misunderstood the problem. BD>5root2.
    We get the triangle with sides 5root(3) and 5
    Now, for BD=5root(2), we get an isosceles triangle with sides 5 each. For this the angle DBC=45deg.
    So for the small 15 deg part, we get BD>5root(2)
    so, prob =15/60=.25 (much larger to nahi hai par)

    milind

    September 11, 2008 at 5:25 pm

  4. The answer is 1- root(3)/3

    outtimed

    September 11, 2008 at 5:42 pm

  5. I think we are comparing sides here. For BD to be 5sqrt2 CD = 5.

    For all values of 5<CD<5root3 the condition is satisfied .Hence probability = (root3-1)/root3

    Amit

    September 11, 2008 at 7:18 pm

  6. yeah right @ Amit, but we will find it easier if we compare areas and not sides

    outtimed

    September 11, 2008 at 7:24 pm

  7. Yes, thats the correct way, So if we fix CD=5 then for the condition to be satisfied P should lie in tri(ABD) out of possible tri(ABC). Nice method🙂

    Amit

    September 11, 2008 at 7:40 pm


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