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Math Help - Cyclic quad

  1. #1
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    Cyclic quad

    Had "fun" with this:

    Circle radius 65 has center M(65,65), thus tangent to both x and y positive axis.

    ABCD is a cyclic quad, it's vertices on the circumference of circle M.

    It's area is 6072 square units.

    Coordinates of A are (2,81), coordinates of C are (98,9).

    What are the coordinates of B and D?

    In case some of you suspect I'm "fishing" for an answer:
    the x and y coordinates of B and D add up to 260 (that's from my solution).

    I'm posting this for your pleasure(!), hoping nobody finds more than 1 solution
    Thanks from Shakarri
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  2. #2
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    Re: Cyclic quad

    Are the coordinates positive integers or any real numbers is acceptable?
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  3. #3
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    Re: Cyclic quad

    Positive integers.
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  4. #4
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    Re: Cyclic quad

    Answer is:
    A(2,81)
    B(2,49)
    C(98,9)
    D(128,81)
    But as you can switch B and D there are two answers to this question.
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  5. #5
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    Re: Cyclic quad

    Thanks; mine was:
    A(2,81), B(98,121), C(98,9), D(32,9) ..... clockwise !
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  6. #6
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    Re: Cyclic quad

    OK I did a quick calculations on your answer.
    So with your points:
    Distance between A and B: a=104
    Distance between B and C: b=112
    Distance between C and D: c=66
    Distance between D and E: d=78
    Using Brahmagupta's formula:
     A=\sqrt{(s-a)(s-b)(s-c)(s-d)}
    Where semiperimeter  s is defined as s=\frac{a+b+c+d}{2}
    So with your numbers, semiperimeter s would be:
    s=\frac{104+112+66+78}{2}=180
    And Area would be:
     A=\sqrt{(180-104)(180-112)(180-66)(180-78)}=7752
    Am I doing something wrong or did you miss something somewhere?
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  7. #7
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    Re: Cyclic quad

    Oh oh...7752 is also what I have for that one; my bad, copied wrong one;
    1000 apologies of which you may have 1
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  8. #8
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    Re: Cyclic quad

    No problem.
    At least it had one answer... and I had fun solving it anyways.
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