| Posted on Sunday, 05 January, 2003 - 12:45 pm: |
We fill a right cone (base-down) with salt (or any other substance...) up till half the height. We turn the cone up-side down. Until what percentage of the height will the cone be filled with salt?
It is not very hard to get the answer, the interesting thing about this riddle is that it is very not intuitional...
Yatir
| Posted on Sunday, 05 January, 2003 - 01:58 pm: |
Wow!! Yeah Yatir you're very right... There's intuition for you (I guessed it would be a lot but not that much) I won't say more so I don't spoil it for anyone who's doing it...
Marcos
| Posted on Sunday, 05 January, 2003 - 02:14 pm: |
If my answer's correct, then it's quite unbelievable.
Chris
| Posted on Sunday, 05 January, 2003 - 02:27 pm: |
Yeah, I think you need to actually do it to believe it (I think I'll go try it now... All I need is a cone)
| Posted on Sunday, 05 January, 2003 - 04:00 pm: |
Is it unbelievably high?
| Posted on Sunday, 05 January, 2003 - 04:21 pm: |
Once you give it a lot of thought it becomes more and more believable...
| Posted on Sunday, 05 January, 2003 - 04:21 pm: |
Yeah (I mean if it was unbelievably low then we must have done something wrong in the maths!)...
Marcos
| Posted on Sunday, 05 January, 2003 - 04:25 pm: |
By the way, I just tried it with a funnel and it seems correct. It's intuitive now! (See this is the good things with problems involving solids - you can always try it... I mean, if you find something in abstract algebra counter-intuitive I doubt it will be as easy to convince yourself)
Marcos
| Posted on Sunday, 05 January, 2003 - 05:44 pm: |
Since we are on the topic can I just ask a q that I always wondered (but somehow still havent found out due to general excuses): how can you prove the formula of a chopped headed cone (if ppl dont mind me calling it that and what is the 'official' name for it)?
I'm still suffering from shock on the result of this riddle!
| Posted on Sunday, 05 January, 2003 - 05:58 pm: |
formula of a chopped headed cone??
You mean general equation?
love arun
| Posted on Sunday, 05 January, 2003 - 06:08 pm: |
There are several ways. You can find it out with calculus (volume of revolution, integrals...). OR you can take a cone of height h and cut off it a cone of height h', and notice that the radius changes linearaly to the height.
Yatir
| Posted on Sunday, 05 January, 2003 - 06:20 pm: |
This picture might help:
Arun, I did mean the general equation, namely V=(1/3)×ph(R2+r2+R r). I never used to remember this equation because intuitively (that's another one!) I always thought it'd be something like (1/3)×p(R+r)2 h. Does anyone have a proof?
| Posted on Sunday, 05 January, 2003 - 06:23 pm: |
I've just proved it by calculus (volume of revolution)...
If you want I'll post it...
Marcos
P.S. I think it's called a frustrum of a cone
| Posted on Sunday, 05 January, 2003 - 06:26 pm: |
Well,when i said general equation i actually meant the x2 +y2 =z2 type of equation.Nevertheless,since u have mentioned volume .. there is no doubt now.
As for the formula for the volume.
Hint:
1.Let the height of the original cone be x.
2.Try to express x in terms R,r and h.(Sub-hint:Use the test for Similarity of two triangles)
3.The rest is simply evaluation of the volume that is volume of big cone - volume of small cone.
love arun
| Posted on Sunday, 05 January, 2003 - 06:26 pm: |
By the way my physics teacher seems to think that it's the 'intuitive' one you mention! She couldn't construct a proof of it though... Now, I can explain why.
Marcos
| Posted on Sunday, 05 January, 2003 - 06:29 pm: |
Arun's method is better than mine... I had tried that approach before calculus but I reached a dead end/mental block and couldn't progress to get it solely in terms of the height of the frustrum...
Seeing it again now I don't see why I reached the dead end...
Marcos
| Posted on Sunday, 05 January, 2003 - 06:31 pm: |
Angelina, I just gave you a proof above... :) Notice that the radius changes linearly with the height. That means that in half the height of the cone you have half the radius of the base of the cone. Or in other words: r/R=(H-h)/H The volume of the frustrum of the cone (Thanks to Marcos), is given by: p/3(H R2-(H-h)r2) Now use the fact that: H=R(H-h)/r
Yatir
| Posted on Sunday, 05 January, 2003 - 06:36 pm: |
Oh sorry Yatir (and others)... didn't notice you'd mentioned all the approaches before
This is the problem when we're all online at the same time and are posting like maniacs!
Marcos
| Posted on Sunday, 05 January, 2003 - 06:39 pm: |
Oh of course! Sorry, not having a good day mentally! Yatir, your proof makes sense and I wonder how I didnt think of it! Thanks!
| Posted on Sunday, 05 January, 2003 - 06:41 pm: |
Actually Yatir, I think it's...
r/R = (H-h)/H
giving H = hR/(R-r)
Marcos
| Posted on Sunday, 05 January, 2003 - 06:53 pm: |
Marcos, but that is exactly what I wrote! (My turn to use some magic...)
and...
I think you meant to write: H=R(H-h)/r
Yatir
| Posted on Sunday, 05 January, 2003 - 07:03 pm: |
Yatir,
what Marcos wrote was right.He simply had used Dividendo before simplificaion.
love arun
| Posted on Sunday, 05 January, 2003 - 07:08 pm: |
I used what?!?!? (All I did was rearrange to get H)
Marcos
| Posted on Sunday, 05 January, 2003 - 07:10 pm: |
Of course
| Posted on Sunday, 05 January, 2003 - 07:20 pm: |
If a/b = c/d then
1.(a+b)/b = (c+d)/d (This is called Componendo)
2.(a-b)/b = (c-d)/d (This is called Dividendo)
3. b/a = d/c (This is called Invertendo)
4. a/c = b/d (This is called Alternendo)
5 (a+b)/(a-b) = (c+d)/(c-d) (This is called Componendo-Dividendo)
love arun
| Posted on Sunday, 05 January, 2003 - 07:28 pm: |
Uhuh... You learn something new every day! (If using NRICH, I think it's more often.)
Thanks Arun
Marcos
| Posted on Monday, 27 January, 2003 - 02:41 pm: |
The answer is about 95%