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Thread: Circles in square...

  1. #1
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    Circles in square...

    A circle is inscribed in a square with sides = 40.

    A smaller (of course!) circle tangent to the above
    circle and 2 sides of the square is inscribed in
    one of the corners of the square.

    What is the diameter of this circle?
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  2. #2
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    Re: Circles in square...

    The radius of the large circle plus the diameter of the smaller corner circle is equal to the half the length of the diagonal of the square.

    $R + d = 20\sqrt{2}$

    $R$ of course is just half the length of the side of the square.

    $R=\dfrac{40}{2}=20$

    $20+d = 20\sqrt{2}$

    $d = 20(\sqrt{2}-1)\approx 8.3$
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  3. #3
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    Re: Circles in square...

    ..BUT ~8.3 is not the diameter of the inscribed circle: it is the distance
    from large circle's circumference to the corner of the square.
    In order for circle to be inscribed, its diameter must be shorter than ~8.3; no?
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  4. #4
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    Re: Circles in square...

    Quote Originally Posted by DenisB View Post
    ..BUT ~8.3 is not the diameter of the inscribed circle: it is the distance
    from large circle's circumference to the corner of the square.
    In order for circle to be inscribed, its diameter must be shorter than ~8.3; no?
    You're right

    $d = \dfrac{20(\sqrt{2}-1)}{\sqrt{2}} \approx 5.86$
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  5. #5
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    Re: Circles in square...

    Circles in square...-circles.png

    In triangle $OBC$

    $\displaystyle (20+r)^2=2(20-r)^2$

    so $\displaystyle r=20 \left(3-2 \sqrt{2}\right)$

    and the diameter $\displaystyle =40\left(3-2\sqrt{2}\right)\approx 6.86$
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  6. #6
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    Re: Circles in square...

    Thanks Idea; was hoping for that solution!!
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  7. #7
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    Re: Circles in square...

    there is an easier way

    $\displaystyle r\sqrt{2}+r+20=20\sqrt{2}$

    solve for $r$
    Thanks from DenisB
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