|
In order for the inscribed octagon have all its angles equal, any two sticks together must subtend the same length arc, and so there can be no more than two different lengths represented. Therefore the right triangle must be isosceles. The ratio of the legs must be sqrt(2):1. For an area of 50, the shorter legs should be 10 units long, making the hypotenuse-length stick equal 10*sqrt(2).
That equal arc that the two sticks form on the circumscribed circle must be a 90-degree arc as there are 4 long-short pairs of sticks altogether. If the radius of the circumscribing circle is r, then the chord connecting the outer endpoints of one pair of sticks has length r*sqrt(2). As the two sticks' ends mark a chord of a 90-degree arc, the internal angle is 270/2 = 135 degrees.
The area of the octagon can be considered as the area of a square formed by four such chords, plus 4 times the area of a triangle formed by such a chord and the two stick lengths.
First we need to find this chord length, r*sqrt(2). We can use the law of cosines on the triangle it makes with the two stick lengths:
2 r^2 = 100 + 200 - 200 sqrt(2) cos 135 = 300 + 200 = 500
r^2 = 500/2
r = sqrt(500/2) = sqrt(250)
So the side of the central square in that circumcircle is sqrt(250) * sqrt(2) = sqrt(500). Thus its area is 500 square units.
Each of the four triangles completing the area of the octagon has side lengths 10, 10*sqrt(2) and 10*sqrt(5). Heron's formula can be used:
s = 10*(1+sqrt(2)+sqrt(5))/2
a = 10
b=10*sqrt(2)
c=10*sqrt(5)
triangle area = sqrt(s*(s-a)*(s-b)*(s-c))
This area comes out to 50.
The total area of the octagon is then 500 + 4 * 50 = 700.
From Enigma No. 1528, "Ad hoctagon", by Susan Denham, New Scientist, 17 January 2009, page 44. |
