Mathematics Question and Solution
Let b = 1 unit.
Therefore;
Side length of the ascribed regular triangle c is;
c = 4b
c = 4(1)
c = 4 units.
Therefore d, radius of the inscribed circle is;
3(4d) = 4√(4²-2²)
12d = 4√(12)
3d = 2√(3)
d = ⅓(2√(3)) units.
d = 1.15470053838 units.
Calculating a.
e²+1² = (⅓(2√(3)))²
e² = ⅓(4)-1
e = √(⅓)
e = ⅓√(3) units.
e = 0.57735026919 units.
It implies;
(⅓(2√(3)))² = (⅓√(3))²+a²-2*(⅓√(3))acos30
⅓(4) = ⅓+a²-(a)
4 = 1+3a²-3a
3a²-3a-3 = 0
a²-a-1 = 0
(a-½)² = 1+(-½)²
(a-½)² = ¼(5)
a = ½±½√(5)
It implies;
a ≠ ½(1-√(5)) units.
a = ½(1+√(5)) units.
The Prove
a÷b
= ½(1+√(5))÷1
= ½(1+√(5))
= Phi (golden ratio).
Proved.
