Billiard Ball Brainteaser
Given a billiards table that has a playing surface measuring 100" x 50", with no pockets, how many standard sized billiard balls (2.25" diameter) can be placed on the table so that they rest freely on the playing surface?
That means no tricks like piling them up or compressing the cushions! 100" x 50" is the measurement inside the cushions!
That means no tricks like piling them up or compressing the cushions! 100" x 50" is the measurement inside the cushions!
Comments
What do I win, besides your respect and admiration?
So I'll guess that the real answer is probably closer 1188.
968
1012
1056
1100
1144
1188
1232
1276
1350
1394
1438.
One of those should be correct.
[ ] impressed with mathematical analysis
2.25" diameter ball goes into 100" length just under 44.5 times.
2.25" diameter ball goes into 50" width just under 22.25 times.
So the max. length of a straight row of balls is 44.
Minimum # of rows is 22 without a stagger.
Only alternative is to stagger beginning on the short side . . .
Too hot here at work, please PM or post answer.
0 0 0 0 0 0 0
And another layer on top
0 0 0 0 0 0
0 0 0 0 0 0
This is the best close packing algorithm, as the size of the container gets large.
Imagine a equilateral triangle of 3 balls.
0
0 0
If we bisect the triange from the top to the bottom.
We have
0
|
|
0--
We know the distance between any two centres of balls is 2*(2.25)
So the Hypot side is 4.5"
Now we know the angle on the left is 60 deg.
So we can solve for the height of the triangle ...
Looks like a winner to me!
44 balls may be placed along the long rail! If we now place 43 balls snuggled in tight in the spaces between balls in the first row we see that we have a whole bunch of equilateral triangles and the distance between rows, as measured through the center of the balls will be 2.25*sin60.
Through this method it is easy to get more than 22 rows. In fact you can fit 25. 13 rows will contain 44 balls, and 12 will contain 43.
This is NOT the answer, but does lead you along the path!
vroom vroom!
Please forgive my mood! gg Spain!
actually I like the 968 but I guess that's too obvious
13*44 + 12*43 = 1088
The answer is ABOVE 1088! <-- that is an exclamation point.
diagram or gtfo
'bout
Continue to alternate between 22 and 21 balls down the length of the table.
How many rows can you fit?
This is given as 97.75/(2.25*sin60) +1 = ~51.16 or 51 rows.
26 rows of 22 balls + 25 rows of 21 balls = 1097 balls on the table.
It is possible to squeeze 1 more in for a total of 1098. Much respect and admiration for any posted solution!
k i got 1097 but i did it without trig...is that good? just working on the extra ball now.