A tennis ball can with radius r holds a certain number of tennis balls also with the same radius. The amount of space in the tennis ball can that is not occupied by the tennis balls equals at most the volume of one tennis ball. How many tennis balls does the can hold?
Barry sent me this problem months ago & I've been avoiding it because geometry scares me.
I finally shamed myself into attempting it just now & got an answer of 2. Unfortunately, I typed up my solution, loaded it to flickr, but flickr is on the blink so I can't post.
I solved it (assuming I did solve it) algebraically, then resorted to "logic and reasoning" to check.
Unfortunately, I'm confused by logic and reasoning at the moment.
I was thinking that because a sphere is 2/3 of a cylinder of same radius, with each ball you put inside a same-radius cylinder you end up with 1/3 of a ball's worth of empty space....which now implies to me that the answer should be 3 balls, not 2.
sigh
update (1):
I'm mixing things up
The 1/3 that's left over isn't 1/3 of a tennis ball. It's 1/3 of a cylinder with the same radius as the tennis ball.
I better forget the logic and reasoning & stick to algebra.
Assuming I didn't screw up the algebra, that is.
update (2):
OK, so in between dealing with screaming autistic youths, loading the dishwasher, & microwaving a taco for Jimmy, I realized that I don't need to know "how much of a tennis ball-sized volume is left over."
I just need to know how much empty volume is left over, period, then figure out how many multiples of that empty space add up to the volume of 1 tennis ball.
volume of tennis ball with radius r: 4/3πr^3
height of cylinder that fits just one tennis ball: 2r
volume of cylinder w/height of 2r: πr^2h = 2πr^3
vol. of cylinder - vol. of 1 tennis ball = vol. of empty space
2πr^3 - 4/3πr^3 = 2/3πr^3 empty space left over when 1 tennis ball is in cylinder
2 tennis balls leaves 2 empty spaces, each 2/3πr^3 in volume:
2/3πr^3 + 2/3πr^3 = 4/3πr^3, which is the volume of 1 tennis ball
so: 2 tennis balls
update (3):
Barry says the problem comes from Dolciani's Algebra 2! (I include the exclamation point because I'm happy to discover I am able to solve a problem from that book. cool.)
original wording:
A tennis ball can in the shape of a cylinder with a flat top and bottom of the same radius as the tennis balls is designed so that the space inside the can that is NOT occupied by the balls has volume at most equal to the volume of one ball. What is the largest nubmer of balls the can will contain?