The sport scene is tense and exciting in Boston this time of year. Celtics playing Cavaliers, Bruins playing Flyers, Red Sox breaking our hearts every night. Sport fans could hardly sleep. When we do fall asleep, we imagine ourselves playing basketball, then growing bats and chewing gum, then the bats morph into hockey sticks and the baseballs turn into packs flying around with a swooshing sound. In a nightmare, you could imagine becoming smaller and smaller and... turning into a baseball ball. You turn your head and see a giant and heavy basketball chasing you. Attempting to escape the orange giant's jumping steps, you (as a baseball), are darting closer and closer to the wall, looking for a corner. Could a baseball hide safely from a basketball in a corner without being "hurt"?
Basketball diameter is 9.39". Baseball diameter is 2.866".
( puzzle adapted from B. Kordemsky book)
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2 comments:
No. I found it easiest to visualize in 2D, then generalize to 3D. Take a small circle of radius r, and a big circle of radius R (i.e. r < R). Place them in a corner (i.e. a right angle) so that they are tangent to the vertical and horizontal (a drawing here would help, but my right angle is oriented like the letter "L" or the positive x and y axes, and the circles are tangent to the horizontal [bottom] line segment and the vertical line segment). For the small circle to be able to "hide" it needs to fit in the gap in the corner under the large circle. We now have to calculate some distances. From the corner to the center of the big circle is R*sqrt(2), and from the corner to the center of the small circle is r*sqrt(2) (put two squares at the corner, one with side length r and one with side length R -- the desired distances are the lengths of the diagonals and the vertices of the squares are the corner, the points where the circles touch the horizontal and vertical and the centers of the circles). Imagine the small circle to be as big as possible -- it will be tangent to the big circle at the point where the line segment connecting their centers and the corner cross the big circle. Now you can see that the distance from the corner to that point of tangency is r + r*sqrt(2), and the distance from the corner to the center of the big circle is R*sqrt(2). That distance also has to be R*sqrt(2) - R. So for the little circle to be able to hide, r*(sqrt(2) + 1) < R*(sqrt(2) - 1)
Now we generalize to 3D. We place cubes at the corner, with radii of r and R. The diagonals of the cubes will be r*sqrt(3) and R*sqrt(3) respectively. Now r*(sqrt(3) + 1) < R*(sqrt(3) - 1) for the small ball to be able to hide. With r = 2.866" and R = 9.39" this is not possible.
2.866*(sqrt(3) + 1) = 7.83
9.39*(sqrt(3) - 1) = 6.874
That is correct, kj!
Unfortunately this dream is a real nightmare: no escape from the basketball. Wake up!
And you are right, graphical demonstration would work best here. Click to see my version of it.
A second puzzle point for kj!
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