You guys rock–it makes me sad that I don’t have enough spare copies to give to the great entries I’ve been getting. I will be back momentarily with this week’s puzzle. First:
Four bugs sit 10 inches apart in a square. They move toward each other, forming a logarithmic spiral.
How far do they move until they hit the center?
Does this address any confusion? Guha and Mikey, if nothing else, demonstrate that upside-down sunglasses really do look like bug-eyes. A relativity game indeed.
They answer correctly, and are this week’s winners!
Several people did solve the puzzle with calculus–and it looks something like this (thanks B. Li):
‘Suppose the beetles always move with velocity v towards the next beetle. The radial component of v will be v*cos(pi/4)=v/sqrt(2) for the entire path.
‘The radial distance from each beetle’s starting point to the centre of the square is 10/sqrt(2). Distance travelled radially = (time taken along entire path)* (radial speed). So each beetle will take a time of (10/sqrt(2))/(v/sqrt(2)) = 10/v to reach the centre of the square.
‘Since the velocity of each beetle is v and distance=velocity*time, each beetle will travel a distance of v*10/v=10 inches before reaching the centre of the square.’
However, Gardner made sure to say that it could be done without calculus. The simplest expression I got was this (thanks, D. Graham):
‘The motion of each target bug is such that from the chaser’s perspective their distance doesn’t change (other than through the motion of the chaser), only the angle at which they must walk, due to the continuous perpendicularity of the paths. Therefore, each bug will travel 10 inches before meeting the next bug (less the length of a bug!’
I got two good poems this week, and they deserve special mention. They’ll go in the Hall of Fame page later today.
THE FINAL PUZZLE, after the jump.
A cylindrical hole 6-inches long has been drilled straight through the center of a solid sphere. What is the volume remaining in the sphere?
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