Easy questions

fetofs · 2005-09-04, 17:00

1- A man buys bottles of wine for $1000. After taking 4 bottles and lifting the price of the dozen up $100, he sells it again for the same price. How many bottles did he buy?

2- How many 10cm diameter balls can you fit into a cubic box (1m dimensions)?

fetofs · 2005-09-05, 13:05

Quote:

Originally Posted by fetofs

1- A man buys bottles of wine for $1000. After taking 4 bottles and lifting the price of the dozen up $100, he sells it again for the same price. How many bottles did he buy?

Looks like I was doing the wrong equation on this one... Considering b as the number of dozens of bottles and d as the dozen price...

$\Large bd=1000$

$\Large (12b-4)(d+100)=1000$

$\Large b=-\frac{5}{3}/2$

Guess it's the latter

wblipp · 2005-09-05, 13:34

Quote:

Originally Posted by fetofs

$\Large (12b-4)$

Thats the number of bottles, not the number of dozens of bottles.

Ken_g6 · 2005-09-05, 23:39

1.
For some reason, I went at it the long way:

p*n = 1000
n = 1000/p
(p+100/12)*(1000/p-4) = 1000

(p+100/12)*(1000/p-4) = 1000
1000+100000/(p*12)-4*p-400/12 = 1000
100000/(p*12)-4*p-400/12 = 0
100000/12-4*p^2-400*p/12 = 0
-48*p^2-400*p+100000 = 0

p = (400+/-sqrt(160000+4*48*100000))/96
p = 4800/96 = $50 per bottle
So the answer is:
n = 1000/50 = 20 bottles

2. First, the answer is >= 10*10*10 = 1000.

Method 1:

1st bottom row: 10 across
Now, the second row fits in spaces creating equilateral
triangles. The centers of these balls are sqrt(3)/2 from the
centers of the first row. Consider that at each end, half a
sphere is required, but the rest can be filled in by triangles.
Effectively, the first row takes 10cm, and the subsequent rows
take ~8.66cm. However, every other row can only hold 9 balls.
So: 90/8.66 ~= 10.39 > 10, so one can fit 11 rows.
That's 6*10+5*9 = 105 balls on the bottom.

If we stacked these vertically, one could get 10*105 = 1050
balls in the box. It is probably possible to do better, by
nestling balls in subsequent layers in the triangles created by
the layer below.

Method 2:
But first, let's place the balls in vertically equilateral
triangles. Then the second layer takes ~8.66cm but each row
must be the opposite size of the row below it. So this layer
holds only 6*9+5*10 = 104 balls. We know this way we can get 11
layers, totaling 6*105+5*104 balls = 1150 balls.

Method 3: Back to the balls in the triangles. Such a
configuration of 4 balls forms a tetrahedron. According to
Mathworld, the height of this

[url=http://mathworld.wolfram.com/Tetrahedron.html]tetrahedron[/

a] is:

1/3*sqrt(6)*10cm ~= 8.165cm. So 90/8.165 ~= 11.02 > 11.

So one can fit 12 layers in the box this way!

Let's find out if we could fit one more row on the end of the
second layer. The center of the balls on that last row would be
offset by only a small amount. That amount is d on the
Mathworld "bottom view" diagram.

bad diagram:
_
d|_\
5 30 degrees

d/5 = tan(30) => x = 5 tan 30 = 1/3*sqrt(3)*5 ~= 2.89cm. The

space available ~=
3.9 cm, so it works! This means each layer will be just like
the respective layer in case 2, only some will be shifted over
by 2.89cm.

12 layers = 6*105+6*104 balls = 1254 balls.

This has been proven to be the most efficient packing for an infinite size, but for a finite size there may be more efficient packings. I believe this is an open question.

Mystwalker · 2005-09-06, 00:17

Assuming the balls can be pressed into arbitrary shapes, we only have to take care of the volume. This is ½ d³ or 500cm³.
Thus, 2000 balls (well, cubes by now :wink: ) would fit into the box...

fetofs · 2005-09-06, 13:00

Quote:

Originally Posted by wblipp

Thats the number of bottles, not the number of dozens of bottles.

Oh, forgot to add the $\Large (12b-4)/12$ Seems like I solved it correctly, and didn't input that part.

fetofs · 2005-09-06, 13:17

Quote:

Originally Posted by Ken_g6

12 layers = 6*105+6*104 balls = 1254 balls.

That really wasn't what I expected to hear.

Quote:

Originally Posted by Ken_g6

Consider that at each end, half a
sphere is required, but the rest can be filled in by triangles.
Effectively, the first row takes 10cm, and the subsequent rows
take ~8.66cm.

I didn't understand how did you pack the spheres in equilateral triangles.... Maybe that way?

Code:

 O 
OO

And how did you pack into tetrahedrons? A drawing would be nice, because I don't seem to understand the logic behind the smaller layers of triangles and polyhedrals...

wblipp · 2005-09-07, 01:48

I agree with Ken's solution at at least 1254 balls are possible.
The additional wiggle room in every direction is small, so I doubt that
repacking a few boundary layers would allow you to squeeze in an additional ball.

I'll try to clarify the packing. Start with 10 balls in a row along one wall on the bottom of the box.

Next snug a row of 9 balls next to the 10.
Alternate rows of 9 and 10 balls.
Looking down on the box, the centers of the balls lie on
equilateral triangles, so each row is 5*sqrt(3), about 8.66 cm further along.
11 rows takes up 10 interrow distances from center to center, plus
a radius of 5 at each end, about 96.6 cm total.

The next layer starts with a row of 9 stacked on
the first two rows of the bottom layer.
The center of this row lies above the centriod of the
equilateral triangle of the balls on the bottom row.
The centroid is 1/3 of the way from the base.
So the center of this first row of the second layer is
5+5*sqrt(3)/3, about 7.887 cm from the wall.

The next rows on this layer have the same inter-row spacing, 8.66 cm.
Adding 10 interrow spacings plus a 5 cm radius for the last row,
we can fit in 11 rows on this layer, too - reaching to 99.49 cm.

The first, and all odd layers, have 6*10+5*9=105 balls.
The second, and all even layers, have 6*9+5*10=104 balls.

The only issue remaining is to verify that 12 layers are possible.

This is easily calculated by starting with the equilateral triangle from 3 balls
on the bottom, and observing that you can get from the center of any of
these balls to the center of the upper ball by going to the centroid then up.
Going straight up forms a right triangle, so we have a right triangle with
the hypoteneus of 10 cm (2 radii) and one of the legs 10*sqrt(3)/3.
Pythagorus tells us the third leg, which is the vertical distance from
the center of layer 1 to the center of layer 2, is 10*sqrt(6)/3.
about 8.165 cm.

Packing in 12 layers will use up 11 inter-layer distances
plus 5 cm on each end, totaling about 99.815 cm.

This is tight. We could shrink this box to 100 x 99.49 x 99.82.
I'd be surprised is anyone can wiggle enough to fit another ball.

fetofs · 2005-09-07, 16:45

Quote:

Originally Posted by wblipp

Pythagorus tells us the third leg, which is the vertical distance from
the center of layer 1 to the center of layer 2, is 10*sqrt(6)/3.
about 8.165 cm.

Packing in 12 layers will use up 11 inter-layer distances
plus 5 cm on each end, totaling about 99.815 cm.

Greatly clarified! But how is 8.165*12+10=99.815

wblipp · 2005-09-07, 17:26

Quote:

Originally Posted by fetofs

Greatly clarified! But how is 8.165*12+10=99.815

As I said,

Quote:

Originally Posted by wblipp

Packing in 12 layers will use up 11 inter-layer distances
plus 5 cm on each end, totaling about 99.815 cm.

Hence 8.165*11+10=99.815

THILLIAR · 2005-09-07, 21:53

Quote:

Originally Posted by wblipp

This is easily calculated by starting with the equilateral triangle from 3 balls
on the bottom, and observing that you can get from the center of any of
these balls to the center of the upper ball by going to the centroid then up.
Going straight up forms a right triangle, so we have a right triangle with
the hypoteneus of 10 cm (2 radii) and one of the legs 10*sqrt(3)/3.
Pythagorus tells us the third leg, which is the vertical distance from
the center of layer 1 to the center of layer 2, is 10*sqrt(6)/3.
about 8.165 cm.

Packing in 12 layers will use up 11 inter-layer distances
plus 5 cm on each end, totaling about 99.815 cm.

This is tight. We could shrink this box to 100 x 99.49 x 99.82.
I'd be surprised is anyone can wiggle enough to fit another ball.

This does not fly.

8.165 cm is less than the true inter-layer vertical distance.
Show us all your caculations to come up with that distance.

2005-09-04, 17:00	#1
fetofs Aug 2005 Brazil 2×181 Posts	Easy questions 1- A man buys bottles of wine for $1000. After taking 4 bottles and lifting the price of the dozen up $100, he sells it again for the same price. How many bottles did he buy? 2- How many 10cm diameter balls can you fit into a cubic box (1m dimensions)?

2005-09-05, 23:39	#4
Ken_g6 Jan 2005 Caught in a sieve 110001011₂ Posts	1. For some reason, I went at it the long way: pn = 1000 n = 1000/p (p+100/12)(1000/p-4) = 1000 (p+100/12)(1000/p-4) = 1000 1000+100000/(p12)-4p-400/12 = 1000 100000/(p12)-4p-400/12 = 0 100000/12-4p^2-400p/12 = 0 -48p^2-400p+100000 = 0 p = (400+/-sqrt(160000+448100000))/96 p = 4800/96 = $50 per bottle So the answer is: n = 1000/50 = 20 bottles 2. First, the answer is >= 101010 = 1000. Method 1: 1st bottom row: 10 across Now, the second row fits in spaces creating equilateral triangles. The centers of these balls are sqrt(3)/2 from the centers of the first row. Consider that at each end, half a sphere is required, but the rest can be filled in by triangles. Effectively, the first row takes 10cm, and the subsequent rows take ~8.66cm. However, every other row can only hold 9 balls. So: 90/8.66 ~= 10.39 > 10, so one can fit 11 rows. That's 610+59 = 105 balls on the bottom. If we stacked these vertically, one could get 10105 = 1050 balls in the box. It is probably possible to do better, by nestling balls in subsequent layers in the triangles created by the layer below. Method 2: But first, let's place the balls in vertically equilateral triangles. Then the second layer takes ~8.66cm but each row must be the opposite size of the row below it. So this layer holds only 69+510 = 104 balls. We know this way we can get 11 layers, totaling 6105+5104 balls = 1150 balls. Method 3: Back to the balls in the triangles. Such a configuration of 4 balls forms a tetrahedron. According to Mathworld, the height of this [url=http://mathworld.wolfram.com/Tetrahedron.html]tetrahedron[/ a] is: 1/3sqrt(6)10cm ~= 8.165cm. So 90/8.165 ~= 11.02 > 11. So one can fit 12 layers in the box this way! Let's find out if we could fit one more row on the end of the second layer. The center of the balls on that last row would be offset by only a small amount. That amount is d on the Mathworld "bottom view" diagram. bad diagram: _ d\|_\ 5 30 degrees d/5 = tan(30) => x = 5 tan 30 = 1/3sqrt(3)5 ~= 2.89cm. The space available ~= 3.9 cm, so it works! This means each layer will be just like the respective layer in case 2, only some will be shifted over by 2.89cm. 12 layers = 6105+6104 balls = 1254 balls. This has been proven to be the most efficient packing for an infinite size, but for a finite size there may be more efficient packings. I believe this is an open question. Last fiddled with by Ken_g6 on 2005-09-05 at 23:40 Reason: Removed column spacer

2005-09-06, 00:17	#5
Mystwalker Jul 2004 Potsdam, Germany 831₁₀ Posts	Assuming the balls can be pressed into arbitrary shapes, we only have to take care of the volume. This is ½ d³ or 500cm³. Thus, 2000 balls (well, cubes by now :wink: ) would fit into the box... Last fiddled with by Mystwalker on 2005-09-06 at 00:19

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2005-09-07, 01:48	#8
wblipp "William" May 2003 New Haven 3·787 Posts	I agree with Ken's solution at at least 1254 balls are possible. The additional wiggle room in every direction is small, so I doubt that repacking a few boundary layers would allow you to squeeze in an additional ball. I'll try to clarify the packing. Start with 10 balls in a row along one wall on the bottom of the box. Next snug a row of 9 balls next to the 10. Alternate rows of 9 and 10 balls. Looking down on the box, the centers of the balls lie on equilateral triangles, so each row is 5sqrt(3), about 8.66 cm further along. 11 rows takes up 10 interrow distances from center to center, plus a radius of 5 at each end, about 96.6 cm total. The next layer starts with a row of 9 stacked on the first two rows of the bottom layer. The center of this row lies above the centriod of the equilateral triangle of the balls on the bottom row. The centroid is 1/3 of the way from the base. So the center of this first row of the second layer is 5+5sqrt(3)/3, about 7.887 cm from the wall. The next rows on this layer have the same inter-row spacing, 8.66 cm. Adding 10 interrow spacings plus a 5 cm radius for the last row, we can fit in 11 rows on this layer, too - reaching to 99.49 cm. The first, and all odd layers, have 610+59=105 balls. The second, and all even layers, have 69+510=104 balls. The only issue remaining is to verify that 12 layers are possible. This is easily calculated by starting with the equilateral triangle from 3 balls on the bottom, and observing that you can get from the center of any of these balls to the center of the upper ball by going to the centroid then up. Going straight up forms a right triangle, so we have a right triangle with the hypoteneus of 10 cm (2 radii) and one of the legs 10sqrt(3)/3. Pythagorus tells us the third leg, which is the vertical distance from the center of layer 1 to the center of layer 2, is 10sqrt(6)/3. about 8.165 cm. Packing in 12 layers will use up 11 inter-layer distances plus 5 cm on each end, totaling about 99.815 cm. This is tight. We could shrink this box to 100 x 99.49 x 99.82. I'd be surprised is anyone can wiggle enough to fit another ball.