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Math Help - An unusual problem: probability meets geometry

  1. #1
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    An unusual problem: probability meets geometry

    A good mixture of probability and plane geometry:
    Creative Unusual Geometry Problems: Fox 198
    Attached Thumbnails Attached Thumbnails An unusual problem: probability meets geometry-8foxes.com_198.gif  
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  2. #2
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    Quote Originally Posted by oswaldo View Post
    A good mixture of probability and plane geometry:
    Creative Unusual Geometry Problems: Fox 198
    Why is 0 not an option?

    CB
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  3. #3
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    It must be "any triangle" or "within the overall picture there should be one triangle".

    Expecting C collide with O has a zero chance, but there should be other ways of getting "a" triangle -as part of the final figure.
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  4. #4
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    Hello, oswaldo!

    A random process starts from point O,
    Point A is obtained in a uniformly selected direction, 1 unit from O.
    The same process is repeated at point A to obtain point B,
    and at point B to obtain point C.

    What is the probability that a triangle is obtained when point C is generated?

    . . (A)\;\frac{1}{8} \qquad (B)\;\frac{1}{9} \qquad (C)\;\frac{1}{12} \qquad (D)\;\frac{1}{16} \qquad (E)\;\frac{1}{18}

    Code:
                    B
                    o
                   *  *
                  *     *
                 *        *
        O o * * * * * * * * o A
               * 
              o
             C
    I got: . {\color{blue}(E)\;\frac{1}{18}}


    Segment OA can be any direction.


    Code:
                B
                o
                 \
                  \
                   \
                    \
                 60 \
        O o * * * * * o A
                 60 /
                    /
                   /
                  /
                 /
                o
                B
    Segment AB must be within 60 on either side of OA.
    Otherwise, OB > 1 and we will have no triangle.

    P(\angle OAB \leq 60^o) \:=\:\frac{120^o}{360^o} \:=\:\frac{1}{3}



    Code:
                B
                o
               / *
              /60*
             /     *
            /       *
                     *
        O o * * * * * o A
    Segment BC must be within 60 of AB.

    P(\angle ABC \leq 60^o) \:=\:\frac{60^o}{360^o} \:=\:\frac{1}{6}


    Therefore: . P(OABC\text{ forms a triangle}) \;=\;\frac{1}{3}\cdot\frac{1}{6} \;=\;\frac{1}{18}

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, oswaldo!

    I got: . {\color{blue}(E)\;\frac{1}{18}}


    Segment OA can be any direction.

    Code:
                B
                o
                 \
                  \
                   \
                    \
                 60 \
        O o * * * * * o A
                 60 /
                    /
                   /
                  /
                 /
                o
                B
    Segment AB must be within 60 on either side of OA.
    Otherwise, OB > 1 and we will have no triangle.

    P(\angle OAB \leq 60^o) \:=\:\frac{120^o}{360^o} \:=\:\frac{1}{3}


    Code:
                B
                o
               / *
              /60*
             /     *
            /       *
                     *
        O o * * * * * o A
    Segment BC must be within 60 of AB.

    P(\angle ABC \leq 60^o) \:=\:\frac{60^o}{360^o} \:=\:\frac{1}{6}


    Therefore: . P(OABC\text{ forms a triangle}) \;=\;\frac{1}{3}\cdot\frac{1}{6} \;=\;\frac{1}{18}
    Ahh .. different question from my reading which required O and C be coincident, not that segment BC intersect segment OA

    CB
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  6. #6
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    As Captain was objecting, assuming "that segment BC intersect segment OA" the answer is NOT 1/18. I think it is because angle OAB can be bigger than 60 degrees. Assume it is 70. A triangle is still possible. I think AOB can have values upto 90 degrees. That will change the solution with new combinations.

    P( 0 \leq \angle OAB \leq 90^o) \:=\: \ldots{}
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  7. #7
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    Actually this can be simulated in Excel. And you should have an answer.
    Here is my approach:
    a: first angle randomly generated
    b: second angle randomly generated

    a in [0,60]
    Then 0 < b < (180-a)/2 forms a triangle.

    a in [60,90]
    Then 0 < b < (180-2a) forms a triangle.
    Compute probabilities in both case.
    Then multiple by 2 (why?)
    You should get the answer.
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  8. #8
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    Quote Originally Posted by oswaldo View Post
    Actually this can be simulated in Excel. And you should have an answer.
    Here is my approach:
    a: first angle randomly generated
    b: second angle randomly generated

    a in [0,60]
    Then 0 < b < (180-a)/2 forms a triangle.

    a in [60,90]
    Then 0 < b < (180-2a) forms a triangle.
    Compute probabilities in both case.
    Then multiple by 2 (why?)
    You should get the answer.
    That is correct. Working in radians rather than degrees, the probability is given by \frac1\pi\left(\int_0^{\pi/3}\frac{\pi-\theta}{4\pi}d\theta + \int_{\pi/3}^{\pi/2}\frac{\pi-2\theta}{2\pi}d\theta\right), which works out as 1/12.
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