Three pentagramshaped stars (the stars formed from the diagonals of a regular pentagon) are stacked up so that the bottom two ends of the tips touch the middle ends of the tips of the star below.
(See diagram.)
The distance from the top of the stack to the floor (where the bottom star's "feet" rest) is 4 feet.
What is the distance between the bottom two ends of the tips of the stack that touch the floor?
I extended the left side 'leg' of the top pentagram to the level of the base of the structure. I am considering this line a the hypotenuse of a right triangle with its top angle being half of the apex angle of the structure. Its length will be 4/cos(18¨¬).
I then joined the left side endpoint of the horizontal line of the top pentagram to the left side 'foot'.
I proceeded to extend legs of the lower two pentagrams, which are parallel to it, upwards through the structure and crossing intersections of certain other line sections.
Now enclosing the lower pentagram in a pentagon I noticed that the right most oblique side is parallel and equal to the middle section of the leg of the top pentagram; I called this length A.
I considered the horizontal line of the lowest pentagram as having two parts. One, made of two segments equals A. I defined the remaining one as B.
Now I note that this line, of length A+B is equal in length to each of the longer segments in the top pentagram.
I now begin to map A and B values to the hypotenuse starting at the base of the structure.
The oblique height to the first horizontal is A, to the next is A+B, and the leg of the top pentagram is has the values of A+B, A and A+B.
Thus the hypotenuse has a length of 5A + 3B which equals 4/cos(18¨¬).
Now the ratio of the Golden Section, which I will not go into:
B:A = (¡î51)/2
As B = A(¡î51)/2, the hypotenuse formula becomes:
5A + 3A(¡î51)/2 = 4/cos(18¨¬).
This reduces to A = 8/{(7+3¡î5)cos(18¨¬) or .613625.
I concur with Nikki

Posted by brianjn
on 20040503 23:40:37 