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 MAX octagonal area (Posted on 2012-11-28)
The octagon ABCDEFGH is inscribed in a circle, with the vertices around the circumference in the given order. Given that the polygon ACEG is a square of area 5 and the polygon BDFH is a rectangle of area 4, find the maximum possible area of the octagon.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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 Solution | Comment 1 of 2
`Let O and r be the center and radius of the circle.Let [KL...ST] denote the area of polygon KL...ST and <XYZ denote the  angle XYZ.`
`  5 = [ACEG] = 2*r^2   r^2 = 5/2                              (1)                                     Let 2*z = m(<BOD) with |BD| < |BH|.`
`  4 = [BDFH] = 4*r^2*cos(z)*sin(z)  1 = r^2*cos(z)*sin(z)  1 = (5/2)*cos(z)*sin(z)  1 = (25/4)*cos(z)^2*sin(z)^2  4 = 25*cos(z)^2*(1-cos(z)^2)  25*cos(z)^4-25*cos(z)^2+4 = 0`
`Therefore,`
`  cos(z)^2 = 4/5  cos(z) = 2/sqrt(5)                     (2)  sin(z)^2 = 1-cos(z)^2 = 1-4/5 = 1/5  sin(z) = 1/sqrt(5)                     (3)`
`  <AOE = <AOB + <BOD + <DOE   180 =   x  +  2*z +   y               (4)`
`  <AOE = <AOB +  <BOC  +  <COD  + <DOE   180 =   x  + (90-x) + (90-y) +   y    (5)`
`After constructing the polygon A-H withGeometer's Sketchpad it was noted thatA-H is symmetrical about the center point O. Using this and (5),`
`  [A-H] = 2*[ABCDE]        = 2*([AOB]+[BOC]+[COD]+[DOE])        = r^2*[sin(x)+sin(90-x)+               sin(90-y)+sin(y)]        = r^2*[cos(x)+sin(x)+               cos(y)+sin(y)]            (6)`
`From the symmetry of x and y in (4) and (5) we know that [A-H] has a maximum orminimum when x = y. From Geometer's Sketchpad we know that it is a maximum.Therefore, (4) and (6) give us`
`  x = y = 90-z                           (7)  max[A-H] = 2*r^2*[cos(z)+sin(z)]       (8)`
`Plugging in the values from (1), (2),and (3) into (8) gives`
`  max[A-H] = 2*(5/2)*[2/sqrt(5)+1/sqrt(5)]                 = 3*sqrt(5) ~= 6.708204                                                          This agrees with Geometer's Sketchpad.`
`QED`

Edited on November 28, 2012, 4:42 pm
 Posted by Bractals on 2012-11-28 16:38:18

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