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Math Help - Semicircle & Chords (Urgent)

  1. #1
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    Semicircle & Chords (Urgent)

    Hey guys. I have no idea how to even start this question. Any ideas?? Thanks!

    "A semi-circle is dawn outwardly on a chord 'AB' of a circle with center 'O' and units raidus. The point 'C' on this semi-circle which sticks out of the given circle the furthest is on the radius 'ODC' which is perpendcular to 'A'. Of course the Length of 'OC' is dependant on the choice of chord 'AB' in the first place. Determine 'AB' so that 'OC' is maximum"
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  2. #2
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    Hello, furnis1!

    First, make a sketch . . .
    [I'll use Calculus to maximize OC.]


    A semi-circle is dawn outwardly on a chord AB of a circle with center O and raidus 1.
    The point C on this semi-circle is the furthest from center O.
    Of course, the length of OC is dependant on the choice of chord AB.
    Determine AB so that OC is maximum.
    Code:
                  * * *
              *           *  A
            *               o
           *              / :*   o
                       1/   :      o
          *           / θ  D: *
          *       O * - - - + * - - o C
          *           \     : *
                       1\   :      o
           *              \ :*  o
            *               o
              *           *  B
                  * * *

    The circle with center O has radii OA = OB = 1.

    A semicircle is constructed (outward) with center D and diameter AB.
    . . It has radii: DA = DC = DB.

    Let \angle AOC = \theta,\;\;OC = d

    In right triangle ADO\!:\;\;OD = \cos\theta\,\text{ and }\,DA = \sin\theta = DC

    Then: . d \:=\:OD + DC \:=\:\cos\theta + \sin\theta


    To maximize d, set d' = 0

    We have: . d'\:=\:\sin\theta + \cos\theta \:=\:0\quad\Rightarrow\quad \sin\theta \:=\:\cos\theta \quad\Rightarrow\quad \tan\theta\:=:1

    . . Hence: . \theta \:=\:45^o

    OC is maximum when \theta = 45^o


    Then \angle AOB = 90^o and \Delta AOB is an isosceles right triangle.

    Since its legs are length 1, its hypotenuse is: . AB \:=\:\sqrt{2}

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  3. #3
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    great answer prof! thanks for the help - that really cleared it up.
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