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MAX octagonal area (Posted on 2012-11-28) Difficulty: 4 of 5
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    
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Solution 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 with
Geometer's Sketchpad it was noted that
A-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 or
minimum 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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