Imagine a brick with sides of lengths 1, Phi=1·61803... and phi=1/Phi=0·61803.
Clearly, the longest side is the sum of the other two lengths since 1 + phi = Phi
i. Show that the largest face (area C=1 x Phi) is the sum of the other two face's areas (area A =1 x phi and area B=phi x Phi) ?
ii. Evaluate the surface area S of the brick.
iii. Find the diagonal across the brick.
iv. What is the ratio of the surface areas of
the "Phi-bonacci brick" and its surrounding sphere?
i. A = 1*phi = 1/Phi; B = phi*Phi = 1
Phi = phi + 1
1 * Phi = 1*phi + phi*Phi
C = A + B
ii. S = 2*(phi + Phi + 1) = 4 * Phi = 2*(1 + sqrt(5))
iii. Phi^2 = 1+Phi
phi^2 = 2-Phi
d^2 = 1 + Phi^2 + phi^2
= 1 + 3
d = 2
iv. r = d/2 = 1
A = 4 * pi
S/A = (1+sqrt(5))/(2*pi) = Phi/pi ~= .5150362148004845
Posted by Charlie
on 2016-10-14 11:45:47