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Math Help - Inscribe a semicircle into a square

  1. #1
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    Inscribe a semicircle into a square

    How do you inscribe a semicircle into a square? And the semicircle's diameter can not be the one of the diagonals of the square. It has to look like the picture below.
    Attached Thumbnails Attached Thumbnails Inscribe a semicircle into a square-example.jpg  
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  2. #2
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    Very messy unless someone else has a solution. I don't even really count this as a solution. However, if you use the following diagram, then from the triangle inside the lower right sector, you can find x in terns of r. Then you have sides of a right triangle that will relate r and d. That is, you can form the ratio of r and d. However, I did a little of the math, and it is VERY messy [I might easily have missed something.] Further, that does not at all imply a drafting solution, but it's the best I can come up with at this time and under my present circumstances.
    Attached Thumbnails Attached Thumbnails Inscribe a semicircle into a square-circle-square.jpg  
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  3. #3
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    Hello, arrowhead566!

    How do you inscribe a semicircle into a square?
    And the semicircle's diameter can not be the one of the diagonals of the square.
    I believe that the original problem asked for the semicircle of maximum area.
    Code:
                    C
        -   * - - * o * - - - *
        :   | *:::::|:::::*   |
        : D o:::::::|:::::::* |
        :   | *:::::|R:::::::*|
        :   |  R*:::|:::::::::|
        x   |     * |::::R::::*
        :   |       o:-:-:-:-:o B
        :   |      O: *:::::::*
        :   |       :   *R::::|
        :   |       :  45*::*|
        - E o - - - * - - - o *
                    F       A

    Let x = side of the square.
    Let R = radius of the semicircle.
    Let O = center of the semicircle.

    The semicircle is tangent to the square at A,B,C,D.
    . . Then: . OA = OB = OC = OD = R

    In right triangle OFA\!:\;\angle OAF = 45^o
    . . Hence: . OF = \frac{R}{\sqrt{2}}

    Since CO + OF \:=\:x, we have: . R + \frac{R}{\sqrt{2}} \:=\:x

    . . Hence: . R\left(1 + \frac{1}{\sqrt{2}}\right) \:=\:x \quad\Rightarrow\quad R\left(\frac{\sqrt{2}+1}{\sqrt{2}}\right) \:=\:x \quad\Rightarrow\quad R \;=\;\frac{\sqrt{2}}{\sqrt{2}+1}\,x


    Therefore: . R \;=\;(2-\sqrt{2})x

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  4. #4
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    This is the exact problem.
    Inscribe a semicircle into the square given below.
    And then my teacher put a picture of what it was supposed to look like, which I have attached in my previous post.
    I see what you're doing, but how did you get the semicircle's center? Or what would be the first time in constructing this after the square?
    O yes, also when you say what R is equal to, how would you go about constructing a segment that multiplies to side x?
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  5. #5
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    Hello again, arrowhead566!

    When you say: . R \:=\:(2-\sqrt{2})x
    how would you go about constructing a segment that multiplies to side x ?
    I've come up with a construction; maybe someone can improve on it.

    On a horizontal line, mark off points A,B,C,D
    . . so that: . AB \,=\,BC\,=\,CD\,=\,1
    Code:
              1       1       1
        - o - - - o - - - o - - - o -
          A       B       C       D


    At C, erect perpendicular CE so that: . CE = 1
    Code:
                          E
                          o     _
                          | *  √2
                         1|   *
                          |     *
        - o - - - o - - - o - - - o - -
          A       B       C   1   D
    Draw ED\!:\;\;ED = \sqrt{2}




    Using D as center and DE as radius,
    draw an arc cutting BC at P.
    Code:
                          E
                        . o
                      .   | * 
                     .    |   *
                    .     |     *
        - * - - - * o - - * - - - o -
          A       B P     C       D
    Hence: . BP = 2-\sqrt{2}




    Through A, draw a line to the upper-right.
    On the line, measure off AQ = x
    Code:
                          *
                        *
                      *
                 Q  *
                  o
             x  *  \
              *     \
            *        \
        - o - - - - - o - - o -
          A           B     P
    Draw QB.




    Code:
                          *
                     R  *
                      o
                 Q  *  \
                  o     \
             x  *  \     \
              *     \     \
            *        \     \
        - o - - - - - o - - o -
          A           B     P

    Through P, construct  PR \parallel BQ.


    Then: . QR \:=\:(2-\sqrt{2})x

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