Question

Six boys stood equally spaced on a circle of radius 40 feet. Each boy walked to all of the other non-adjacent persons on the circle, shook their hands and then returned to his original spot on the circle before the next boy started his trip to shake hands with all of the other non-adjacent boys on the circle. After all six boys had done this, what is the least distance in feet that could have been traveled

Answer 1

The lest distance in feet that could have been traveled is 480 + 480·√3 feet or approximately 1311.38 feet

Explanation:

The parameters given are;

Six boys equally spaced round a circle

Radius of circle = 40 feet

Angle subtended by the arc between 2 boys = 360°/6 = 60°

Path of motion of each boy = To the other non adjacent boys = Inscribed Kite

∴ Path of motion of each boy = Right kite

Interior angles of the right kite = 90°, 90°, 120°, 60° (Angles subtended at the center = 2 × angle at the circumference)

Hence distance traveled by each boy = Perimeter of the right kite

Therefore, the distance traveled by each boy = 2 × (Long diagonal × sin((largest angle)/2) + Long diagonal × cos((largest angle)/2)

The distance traveled by each boy = 2 × (80 × sin(60) + 80 × cos(60))

`The\, distance\, traveled\, by \, each\, boy = 2* \left (80\cdot (√(3))/(2)+80* (1)/(2) \right )`

`The\, distance\, traveled\, by \, each\, boy = 80 + 80\cdot √(3)}`

Hence we have;

The distance traveled by the six boys = 6 × (80 + 80·√3) = 480 + 480·√3 feet.

The lest distance in feet that could have been traveled = 480 + 480·√3 feet = 1311.38 feet.