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Thread: geometry an olympics question(quadrilateral inequality)

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    geometry an olympics question(quadrilateral inequality)

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    Re: geometry an olympics question(quadrilateral inequality)

    You can start be calling angle ADC = theta; then since |DA| = x you have |AC| = x sin(theta), and angle BAC = theta. Draw a line from C to be perpendicular to AB - this is the height of triangle ABC, which is |AC|sin (theta) = x sin^2(theta). Given the area of triangle ABC = 24, we get 24 = (1/2)x^2sin^2(theta). To minimize x you want to maximize sin^2(theta). Take it away...
    Last edited by ChipB; Mar 29th 2017 at 12:50 PM.
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    Re: geometry an olympics question(quadrilateral inequality)

    Many Thanks.
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