40 replies to this topic

[Tomk]

I think everyone has either played pool or has at least seen it played.

The game is played with 15 colored balls and a white cue ball. For our purposes... let's ignore that cue ball as it is slightly undersized from the rest of the balls.

If you were to arrange those 15 balls into some shape on the table (no stacking the balls) to cover the minimum amount of area:

What would the shape be?
What area would it contain if the balls were 2" in diameter?

[terry1966]

triangle does the area include the gaps or are you just talking about the total area of each 2" ball added together? i'll let someone else find the area.

[Tomk]

I'm talking about the total area of the "shape" that encloses the balls.

[oldman960]

How about a triangle? 43.3 in²

[Tomk]

Interesting... That's a great answer if you can show me how you do it.

[oldman960]

I think I'm out a bit. I'll get back to you.

[terry1966]

to be honest i think just putting them all in a straight line down the centre of the table would use up the least area because there would be no empty gaps between them, and the total table area used would be exactly the area of the 15 balls. so area of 1 ball is 3.1415 inches squared total area is 47.1225 inches squared with any shape you make your always going to enclose extra empty space that needs to be added to the area taken up by the balls.

[Tomk]

OK... maybe I'm unclear. You need to make a "frame" of your shape and put the balls inside it. I'm looking for the area of the inside of the frame. If you put the balls in a straight line... your frame will be a rectangle. What is the area of the rectangle?

[terry1966]

How about a triangle? 43.3 in²

Interesting...

That's a great answer if you can show me how you do it.

Area = a*b*sin( C )/2

5 balls on each side = 10 inches a side, so a and b both = 10, being equilateral the inside angle C is 60 Area = 10*10*sin( 60 )/2 = 43.3 inch squared.

or on my calculator 60sinx10x10/2=43.3 inch squared.

Edited by terry1966, 06 November 2012 - 09:20 PM.

[Tomk]

@terry, In your previous post you stated that the balls inhabit 47.123 sq. in. Yet now you say you can "put" them into a triangle that encompasses only 43.3 sq. in.???

[terry1966]

nope was just showing you how i think oldman got his answer.

[oldman960]

Actually I used this

sqrt(3) X side²
____________
4

But as I said I made an error. It's not a 10x10x10 triangle (think about it, you'll see why). The area for the triangle inquestion is actually 62.35 in²

The rectangle is smaller at 60 in²

[terry1966]

if you draw a triangle from the centre of each corner ball you'll have an 8x8x8 triangle with an area of 27.712 now draw a rectangle on each side of that triangle 8x1 total area for all 3 is 24 you'll see you have now covered all the balls except for the final bits of each corner ball to add to the total area, now seeing how i haven't actually worked out how much is left i'm going to guess a 1/3rd for each ball so will add 3.141 making my total area..54.855 inches square.

[Tomk]

Well... the shape is identifiable and the area can be calculated. I'm not really sure what else to say.

[terry1966]

you trying to say you have a shape with a smaller area than a triangle with curved corners? a circular shape might cover less area but i haven't looked at it, i just assumed a triangle shape was the smallest. the frame around the triangle covers an area of 54.855 inches square so is the answer your looking for smaller?

Edited by terry1966, 09 November 2012 - 02:44 AM.