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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

we have a pyramid which has got a base with area of 64 cm ² and a high 3 cm ... we want to put inside it balls with 1 cm radius . how many balls can we put inside it ?

my answer : 9 balls

can i see your answers please ?

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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

anyone here ?

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

well first of all, is it a triangle or square based pyramid, if triangle, equaliteral i suppose?

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I would imagine square, 8x8.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

I guess 5 balls, but I can't do the calculations now.

3-D is very hard trig work.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

I go 10 (9+1), with no reason other than it hasn't been taken yet. What do I win if I am right?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

well i went through a load of big calculations, on the base of the triangle

the base is 8x8, and the range of placements you can have for a ball to fit, resting on the base

is a square of 3.73125*3.73125

so id estimate you could fit... about 4 balls inside of the pyramid, you couldnt fit 4 on the base, but allowing different altitudes in arrangement, you can probably fit 4 balls

radius of 1 = diameter of 2.

lol to get that square, i found the distance from the corner, along the diagonal (under the diagonal elevation edge), the precise distance, at which you could put a ball, and it would fit under the diagonal edge, this value was √41 / 3 = 2.134...

and then i found the distance along the middle of each base edge, to the other, (i.e. under the faces of elevation) at which it would just touch the face, this value was 10/6 = 1.6666....

the first distance value is bigger, so you have to take that one, 8 - 2*2.134... = 3.73125...

if you want me to scan in the bit of paper i worked this out on, i will

*Last edited by luca-deltodesco (2006-06-22 19:38:41)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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Yeah ... I just did a rough cross-section: width-wise 2 balls just don't quite fit side-by-side, but *may* fit if placed in the diagonal corners, so my previous estimate was gloriously optimistic.

4 looks more reasonabubble.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

MathsIsFun wrote:

Yeah ... I just did a rough cross-section: width-wise 2 balls just don't quite fit side-by-side, but *may* fit if placed in the diagonal corners, so my previous estimate was gloriously optimistic.

4 looks more reasonabubble.

i was kind of thinking in some sort of triangle arrangemewnt for 3 of them and then the fourth in the centre raised from the base a little bit were it is a bit higher

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

suppose θ is the angle between waist surface and base surface.

tanθ=3/4, sinθ=3/5 and cosθ=4/5

radius/nearest ball center position=tan(θ/2)=sinθ/(1+cosθ)=1/3

nearest ball center position=3

let's see how many balls can we mostly have in one row or in one column on the base

two side balls take up distance 3+1+1+3=8

so at most 2 balls in one column and 2 balls in one row.

the best way to put balls on base seems to be 2 by 2 arrangement.

then can we put other balls on the 4 balls?

you can only put 1 ball on them, I'm afraid. More balls will not be contained under the narrower space between side surfaces.

so together **5** balls at most.

**X'(y-Xβ)=0**

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

actually

taking into the account the fact that the area in which a ball could be placed isnt actually square

its more of a bloated square (since corner distance is smaller than mid edge distance)

you probably could fit 4 balls on base, this is actually, an accurate model (or atleast, the the displacement of the mid part of edges are at an accurate distance out, i wouldnt know the exact shape of the curve)

*Last edited by luca-deltodesco (2006-06-22 20:09:03)*

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

George,Y wrote:

you can only put 1 ball on them, I'm afraid. More balls will not be contained under the narrower space between side surfaces.

so together5balls at most.

i disagree

the height of the pyramid is 3, ball diameter is 2, so you would need to sink the ball down from the top by about 1.2 taking into consideration the slope of the pyramid lengths, and there just isnt enough room on the base to make that kind of room

just fitting 4 balls in the first place is risky

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

actually infact, i think you may be able to fit 6

i forgot that the way i had calculated it, that little drawing i had, the red balls dont have to be contained by the blacklines, the CENTER of the balls has to be contained by it, so you can fit 5 comfortably and possible one more making more space using the vertical space still left at centre

so infact, looking again, i think it is 5

*Last edited by luca-deltodesco (2006-06-22 20:15:58)*

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

so you mean you can virturely fit 5 balls on the base.

But I insist a square arrangement.

when you kick a ball into the very corner under two side surfaces, you can imagine make it two and roll them away under the two surfaces. The paths would be orthogonal to each other and the new two balls now have no difference than two balls kissing their own one side and the base.

*Last edited by George,Y (2006-06-22 20:26:44)*

**X'(y-Xβ)=0**

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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

umm ... okay i'll explain how did i say 9 balls

64 cm ^ 2 >>> 8 * 8

we cant use the hight because 3 cm is not enough for a ball ! ... the diameter is 2 !

if we calculate it we will get that we can put 4 balls in the 8 cm ! but in fact we cant do that because the pyramid has got a problem in the corner we cant put 4 fit ! so lets say 3

3 * 3 = 9

sorry guys my english is not that much ... my language is arabic

thanks for help ... try more ... i need a certain answer

where is ganesh ? help us with it

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

RauLiTo wrote:

sorry guys my english is not that much ... my language is arabic

Don't say sorry, because:

a) This is "math is fun", not "english must be perfect".

b) Your English beats my Arabic 100-0

c) Arabic-speakers have had a massive influence on mathematics. If my memory serves me: Algebra; the concept of zero; (shared glory with India) our decimal numbers, just to name a few.

d) Mathematics is the "universal language", and English should not be a barrier

So ... can you translate these postings?: http://www.mathsisfun.com/forum/search. … =952435847

(Silly me: I though "RauLiTo" was a hispanic name)

Anyway, back on topic, this is a really great puzzle!

There will no doubt be an exact answer if we do the math properly ... it has just been more fun to make educated guesses!

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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

you are right mathisfun ... thanks for saying that

yah raulito is a hispanic name my real name is yousif '' arabic one '' mu username is raulito because i lo e raul gonzalez

math properly is the right way to think you are right

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

I'm still just guessing, but there are five vertices (pointy corners) to the pyramid, so I say put the 5 balls as close as you can get them to each corner as they hit the inner walls and try to figure out if the upper ball hits one of the four lower balls and do the four lower balls stay inside their quadrants so they don't interfere with each other?? The lower 4 balls would touch 3 walls each and the upper 5th ball would touch 4 walls on its upper half of ball.

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Okay RauLiTo , why don't you get some hard-paper and some ping-pang balls to get it yourself? That's the most correct way...

**X'(y-Xβ)=0**

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