In need of a throughout explanation of the following problem:

bobbym · 2011-10-10 07:59:33

Do not eat up the profits. That chocolate is a killer.

SmellyMan · 2011-10-10 08:12:06

bobbym wrote:

Do not eat up the profits. That chocolate is a killer.

Hehe.

Anyway, thank you for everything today! You deserve a golden cupcake award for the best forum companion.

I have a big day coming up, I'll get some rest now, thank you again!

Good night!

bobbym · 2011-10-10 08:29:43

Okay, see you later and good luck with the big day.

SmellyMan · 2011-10-17 03:16:41

Well, didn't check out these forums for quite a while now, I am coming in with another interesting problem I had.

So, my gym teacher got a new volleyball ball, and the ball has holes in it, probably for better grip.
So he, as the joker he is, said, how many holes are there in they volleyball?

So I looked up some information, and have no idea how to solve this problem:

r=11cm
There are 9 dots per square centimetre

This is all I know, I presume some more information can be given, but I'm trying to figure it out, the numbers I get don't seem reliable as they don't follow the logic.

Using some unknown methods to man, the first result I got was around 1600, the second was around 4000.

Could anyone care to explain this? The idea behind it seems interesting and I really want to answer my gym teacher, just to make him go 'WOW'.

Best regards!

Bob · 2011-10-17 03:18:52

hi SmellyMan

Answer in preparation if you are able to stay on-line a while.

Bob

SmellyMan · 2011-10-17 03:20:29

bob bundy wrote:

hi SmellyMan
Answer in preparation if you are able to stay on-line a while.
Bob

I am actually trying to solve it right now, and just got stuck at some point!

I am more than willing to wait.

Bob · 2011-10-17 03:21:29

hi again,

I guess I'd work out the surface area in square cm and times by 9.

How does that seem?

Bob

Bob · 2011-10-17 03:24:17

Obviously, there are some assumptions here. Is it possible to fit exactly 9 holes in each sq cm? It's a sphere so there might be odd bits of surface we cannot cover exactly. The calculation will give a decimal result so will need rounding. etc etc.

Bob

SmellyMan · 2011-10-17 03:26:22

bob bundy wrote:

hi again,
I guess I'd work out the surface area in square cm and times by 9.
How does that seem?
Bob

Well let me tell you how I tried to solve the problem, without numbers.

First I calculated the surface area and the diameter, after that I imagined the ball was actually a spread out rectangle!

And by the surface I had put down, I would simply see how long it is, by the diameter I got beforehand and after that just simply multiplay the X axis with an Y axis to get the number of dots!

EDIT: Also, as far as I see it, you calculated how many dots would actually come onto the whole surface, without any space in between them.

The dots were like this:

http://imageshack.us/photo/my-images/266/unledvk.png/

This ofcourse isn't very accurate, but I don't need an exact result at all, just to actually see how many holes there actually are on the friggin ball .

Last edited by SmellyMan (2011-10-17 03:33:40)

Bob · 2011-10-17 03:33:31

First I calculated the surface area and the diameter, after that I imagined the ball was actually a spread out rectangle!

Do you mean "circumference and diameter" ? Making it into a rectangle is tricky because of the curvature. The formula I used is exact for the surface area as it is derived using calculus.

I haven't actually plugged in the figures for a value yet ..... waits a while ....... I get 13 685 rounded to the nearest whole number. But, due to the formerly mentioned factors, I would round that some more and I'd probably say over 13 000 holes.

Bob

SmellyMan · 2011-10-17 03:35:49

bob bundy wrote:

First I calculated the surface area and the diameter, after that I imagined the ball was actually a spread out rectangle!
Do you mean "circumference and diameter" ? Making it into a rectangle is tricky because of the curvature. The formula I used is exact for the surface area as it is derived using calculus.
I haven't actually plugged in the figures for a value yet ..... waits a while ....... I get 13 685 rounded to the nearest whole number. But, due to the formerly mentioned factors, I would round that some more and I'd probably say over 13 000 holes.
Bob

Yes, I did mean circumference and diamater, sorry for that.

But still, you're overlooking the fact that dots appear only on 9 per every square centimetre, and that they are evenly spaced out.

Bob · 2011-10-17 03:40:03

Yes< I've just got to that bit.

I had pictured like below.

I'll need to re-calculate. Hang on a mo.

Bob

SmellyMan · 2011-10-17 03:43:10

bob bundy wrote:

Yes< I've just got to that bit.
I had pictured like below.
I'll need to re-calculate. Hang on a mo.
Bob

Oh you, oh you...

But that would work too, I don't see the reason why not

EDIT: I'll need to run out into the bakery, I forgot we were out of bread, don't hate me if I don't reply anything for the next 20 minutes or so!

Last edited by SmellyMan (2011-10-17 03:45:47)

Bob · 2011-10-17 03:54:27

hi

If you've gotta have bread, then that's cool. Eat it while it's fresh.

The problem with holes on the border of two or more squares means that, at 9 per square I'm counting some more than once. The actual number per square is less than that.

Say a hole is "in a square" only if it is either (i) in it and not on a border, or (ii) if it is on the left hand border but not on the left border of the square above, or (iii) if it is on the bottom border but not on the bottom border of the next right square.

That way every hole is assigned to one square only. Out of the original 9, two are along the top border so they belong to the square above; two more are along the right border and so belong to the square on the right; and one is in the top right corner and so belongs to the square diagonally up and right from the first square.

Only four holes are really "in the square" so we should say it is 4 holes per square cm. Change the calc. to

which I make a little over 6 000.

Bob

Last edited by Bob (2011-10-17 04:02:48)

SmellyMan · 2011-10-17 04:33:14

bob bundy wrote:

hi
If you've gotta have bread, then that's cool. Eat it while it's fresh.
The problem with holes on the border of two or more squares means that, at 9 per square I'm counting some more than once. The actual number per square is less than that.
Say a hole is "in a square" only if it is either (i) in it and not on a border, or (ii) if it is on the left hand border but not on the left border of the square above, or (iii) if it is on the bottom border but not on the bottom border of the next right square.
That way every hole is assigned to one square only. Out of the original 9, two are along the top border so they belong to the square above; two more are along the right border and so belong to the square on the right; and one is in the top right corner and so belongs to the square diagonally up and right from the first square.
Only four holes are really "in the square" so we should say it is 4 holes per square cm. Change the calc. to
which I make a little over 6 000.
Bob

Well, I'm back, and as easy as it seems, I didn't understand any of it.

But the problem here is, I actually measured that information by my own hands. Care to explain?

EDIT: I see what you tried to show me there, but I simply can't see how 4 dots are in a square centimeter?

Or maybe the edge dots don't belong in the same square?

Last edited by SmellyMan (2011-10-17 04:37:21)

Bob · 2011-10-17 05:24:33

hi SmellyMan

The trick is to find a way of only counting each hole once.

So I've assigned holes to squares. My latest diagram is colour coded to show which hole belongs to which square. I hope you can see all the different colours here.

You should be able to see that every hole has a colour. None are left out and because you cannot give a hole more than one colour, that means no hole is counted twice. I knew what rule I wanted but had a lot of trouble putting it into words. While you were out at the bakers, I had several edit attempts and it sounds like I still haven't made it clear.

But, a picture speaks a thousand words. That should make it clear why it is 4 holes per square. Any others on the border of that square belong to another square and are coloured for that.

Bob

SmellyMan · 2011-10-17 06:54:49

bob bundy wrote:

hi SmellyMan
The trick is to find a way of only counting each hole once.
So I've assigned holes to squares. My latest diagram is colour coded to show which hole belongs to which square. I hope you can see all the different colours here.
You should be able to see that every hole has a colour. None are left out and because you cannot give a hole more than one colour, that means no hole is counted twice. I knew what rule I wanted but had a lot of trouble putting it into words. While you were out at the bakers, I had several edit attempts and it sounds like I still haven't made it clear.
But, a picture speaks a thousand words. That should make it clear why it is 4 holes per square. Any others on the border of that square belong to another square and are coloured for that.
Bob

Oh I see, so you take 4 points, so you can cover ALL of the surface area?

SmellyMan · 2011-10-18 05:55:15

And, I have another problem...

I was watching this video:
http://www.youtube.com/watch?v=VoGSkYLA … ideo_title

And there is something that doesn't go in my mind.

I do agree you have to turn only 2 cards, but I think you should turn the cards D and 3, and not D and 7.

It makes sense, that there is no reason to pick D and 3, as if 3 has a D behind it the statement could still be either false or true.

The statement was:' Every card with a D on one side has a 3 on the other.'

But I question myself, how do you know, without turning all 4 of the cards, you can't know what's on the other side unless you check it.

Someone wiser than me, care to explain?

Best regards!

bobbym · 2011-10-18 06:08:01

Hi;

Great video! I do not know about his answer but the allure of being smarter than 57% of all mathematicians is too strong. I am picking 2 cards, D and 7!

SmellyMan · 2011-10-18 06:11:23

bobbym wrote:

Hi;
Great video! I do not know about his answer but the allure of being smarter than 57% of all mathematicians is too strong. I am picking 2 cards, D and 7!

That is correct!

Would you care to explain the logic behind your great mind?

I must also thank you for the help the last day, it helped us immensely!

Regards!

bobbym · 2011-10-18 06:17:05

Would you care to explain the logic behind your great mind?

I see you have not been paying attention and have not been reading all my stuff. I hope you are being sarcastic.

I wonder how many moderators got it right?

I guess programmers have a built in edge here.

The statement is an if then statement.

If D then 3

Of course we must turn over the D. If it does not have a 3 on the other side then the statement is false.

The statement does not say that only a D has a 3 on the other side. K could have a 3 on the other side too. We do not know. Turning over the 3 provides us with no new information.

SmellyMan · 2011-10-18 06:22:57

bobbym wrote:

Would you care to explain the logic behind your great mind?
I see you have not been paying attention and have not been reading all my stuff. I hope you are being sarcastic.
I wonder how many moderators got it right?
I guess programmers have a built in edge here.
The statement is an if then statement.
If D then 3
Of course we must turn over the D. If it does not have a 3 on the side then the statement is false.
The statement does not say that only a D has a 3 on the other side. K could have a 3 on the other side too. We do not know. Turning over the 3 provides us with no new information.

Yes, I was being a bit sarcastic, and I do not follow you completely, 30.000 posts are quite a handfull.

Well I explained it to myself like this:

The statement:'Every card with a D on one side has a 3 on the other.'

First, D is necessary, it is the first condition you have to check.

Checking the 3, I agree, it doesn't give us any new information.

But there could be a 3 behind the K, and a D behind the 7, so it would only make sense that you could turn over any of these cards? And you can't be sure there's no 3 behind the K if you turn around the 7?

EDIT: I thought about it, and it does make sense!

He implied at the start that each card has a number and a letter, so checking the K is not relevant as even if there is a 3 on the other side you don't prove the statement is correct.

It is a one way statement, starting with D.

Makes sense now, after thinking about it!

Thank you kindly!

Last edited by SmellyMan (2011-10-18 06:30:43)

bobbym · 2011-10-18 06:31:38

Hi;

Did you read the comments after the video? They will provide you with a clue!

SmellyMan · 2011-10-18 06:33:26

bobbym wrote:

Hi;
Did you read the comments after the video? They will provide you with a clue!

I edited my previous post, I got it after thinking about it!

And reading a bit of comments, they gave me a clue.

Thank you again!

Last edited by SmellyMan (2011-10-18 06:34:29)

bobbym · 2011-10-18 06:38:59

Hi;

Remember before ever solving a word problem of any type: Read the problem, again and again. Then read it one more time. Copy every relevant piece of information onto paper.

Math Is Fun Forum

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