 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Phibobrick's dimensions (Posted on 2016-10-14) 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?

 No Solution Yet Submitted by Ady TZIDON No Rating Comments: ( Back to comment list | You must be logged in to post comments.) proposed solution Comment 1 of 1
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
= 4

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 Please log in:

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