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.
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
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Posted by Bractals
on 2012-11-28 16:38:18 |
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