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Thread: Two Sides of a Square Divided by a Circle

  1. #1
    Apr 2008

    Two Sides of a Square Divided by a Circle

    Here's a problem I'm having that involves a square and a circle. The square ABCD is partially within the circle so that the side AB of the square is also a chord of the circle and the side DC is a tangent of the circle. Here's an illustration of the problem in case my description doesn't convey it clearly enough:

    The circumference of the circle divides the sides AD and BC by a certain ratio which looking at the image might be 1 : 3. How do I determine for certain what this ratio is?
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  2. #2
    Senior Member
    Dec 2007
    Anchorage, AK
    Let point O be the center of the circle, M the midpoint of line segment AB, and N the midpoint of CD (the point of tangency). Draw the line segment MN; O will lie on this line. Let us define s as the length of the sides of the square, and r as the radius of the circle. Then the length of MN is s, and the lengths of AO, BO, and NO are all r.

    Now, define x to be the length of MO. Thus as MO+NO=MN, x=s-r. However, looking at the right triangle AMO, we see $\displaystyle r^2=\left(\frac{s}{2}\right)^2+x^2$
    Substituting in x=s-r:
    $\displaystyle r^2=\left(\frac{s}{2}\right)^2+(s-r)^2$
    $\displaystyle r^2=\left(\frac{s}{2}\right)^2+s^2-2rs+r^2$
    $\displaystyle 0=\frac{5}{4}s^2-2rs$
    $\displaystyle r=\frac{5}{8}s$
    And so $\displaystyle x=s-r=\frac{3}{8}s$
    Now, note that the length of the portion of AD (or BC) inside the circle is 2x, and thus the portion outside is s-2x. The result above then gives them to be $\displaystyle 2x=\frac{3}{4}s$, $\displaystyle s-2x=\frac{1}{4}s$, for the 3:1 ratio you observed in your drawing.

    --Kevin C.
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