|
A program tries a range of arc lengths, given the radius is 11 cm. It finds the excess of the sum of the angles of the polygons (solved via breaking them down into right triangles; 10 for the pentagons, 12 for the hexagons) over 180°. The total excess over the appropriate total degrees for a plane figure isproportional to the area of the total sphere that is covered. When the total excess is 720° the whole sphere is covered, so we look for the case where the total excess is this value.
% 12 pentagons 20 hexagons
clearvars,clc
hexcentral=60; pentcentral=72;
r=11;
for edge= 4.469:.00001:4.47
circ=2*pi*r;
arc=edge/r*180/pi;
% halve the isocesles to get a right triangle
a=arc/2; A=30; B=90;
b=asind(sind(a)*sind(B)/sind(A));
C=2*acotd(tand((A-B)/2) * sind((a+b)/2)/sind((a-b)/2));
excess1=(A+B+C-180)*12; % 12 right triangles in a hexagon
a=arc/2; A=pentcentral/2; B=90;
b=asind(sind(a)*sind(B)/sind(A));
C=2*acotd(tand((A-B)/2) * sind((a+b)/2)/sind((a-b)/2));
excess2=(A+B+C-180)*10; % 10 right triangles in a pentagon
excess=excess1*20+excess2*12;
disp([edge,excess ])
end
The 720° figure is easy to remember as a sphere is completely covered by 8 90°-90°-90° triangles, an excess of 90° for each, making 720°,
This final stage of the program shows:
4.4697 719.994315361315
4.46971 719.997696436486
4.46972 720.001077520747
4.46973 720.004458614106
so the answer is 4.46972 cm.
|
