Mathematics Question and Solution
Notice.
The polygon is regular hexagon.
The three circles are congruent.
The inscribed triangle is equilateral.
Let the side length of the regular hexagon be a.
Let the a the congruent circles' radius be b.
It implies;
a = b+2 --- (1).
(2b)² = (b+1)²+a²-2a(b+1)cos60
4b² = b²+2b+1+a²-ab-a
3b² = 2b+1+a²-ab-a --- (2).
Substituting (1) in (2) to get b, radius (r) of the three congruent circles.
3b² = 2b+1+(b+2)²-b(b+2)-(b+2)
3b² = 2b+1+b²+4b+4-b²-2b-b-2
3b² = 1+4b+4-b-2
3b² = 3b+3
b²-b-1 = 0
Resolving the above quadratic equation via completing the square approach to get b, radius (r) of the three congruent circles.
(b-½)² = 1+(-½)²
(b-½)² = (5/4)
b-½ = ±½(√(5))
b = ½±0.5√(5)
It implies;
b ≠ ½(1-√(5))
b = ½(1+√(5)) units.
And b = r, radius of the three congruent circles.
r = ½(1+√(5)) units.
r = 1.6180339887 units.
