The arc length of the parabolic segment
(1)
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illustrated above is given by
(2)
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(3)
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(4)
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and the area is given by
(5)
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(6)
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(Kern and Bland 1948, p. 4). The weighted mean of is
(7)
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(8)
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so the geometric centroid is then given by
(9)
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(10)
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The area of the cut-off parabolic segment contained between the curves
(11)
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(12)
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can be found by eliminating ,
(13)
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so the points of intersection are
(14)
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with corresponding -coordinates . The area is therefore given by
(15)
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(16)
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(17)
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The maximum area of a triangle inscribed in this segment will have two of its polygon vertices at the intersections and , and the third at a point to be determined. From the general equation for a triangle, the area of the inscribed triangle is given by the determinant equation
(18)
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Plugging in and using gives
(19)
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To find the maximum area, differentiable with respect to and set to 0 to obtain
(20)
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so
(21)
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Plugging (21) into (19) then gives
(22)
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This leads to the result known to Archimedes in the third century BC, namely
(23)
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