605
   

NeoPets Riddles (Lenny Conundrums) and Answers Here

 
 
DeviatedUnderworld
 
  1  
Reply Wed 14 Jan, 2009 05:37 pm
@eilonwy42,
Me neither so both our answers are messed
0 Replies
 
Dandy Warhols
 
  1  
Reply Wed 14 Jan, 2009 05:44 pm
@DeviatedUnderworld,
i've got 718.5
there is 2 r for a icos-whatever and i'm not sure a use the good one to find the volume
0 Replies
 
jjbuttcrack
 
  1  
Reply Wed 14 Jan, 2009 05:44 pm
@eilonwy42,
yeah -- i got 726.5 as well... and that's without taking into account the 3cm.

but part of me is hoping we don't have to... the question just asks what the volume IS -- not how much liquid it can hold.
if you want to account for the thickness the answer i'm getting is 717.1
DeviatedUnderworld
 
  1  
Reply Wed 14 Jan, 2009 05:50 pm
@jjbuttcrack,
Even if I dont take into account the 3cm the end formula is 2.18169499 * 6.59 ^ 3 right. My calulator gives me 624.3818614... if i round that up 624.4 Not sure how you're getting higher
DeviatedUnderworld
 
  1  
Reply Wed 14 Jan, 2009 05:54 pm
@DeviatedUnderworld,
because when calculating the volume of an icos-thingy you use the radius not the diameter other because the icos-thingy works with a literka = radius not a diameter...or am I REALLY wrong?!
jjbuttcrack
 
  1  
Reply Wed 14 Jan, 2009 06:00 pm
@DeviatedUnderworld,
If the edge length of a regular icosahedron is a, the radius of a circumscribed sphere (one that touches the icosahedron at all vertices) is 0.9510565163*a

so, the length of the side is slightly larger than the radius
stinkypops
 
  1  
Reply Wed 14 Jan, 2009 06:06 pm
@jjbuttcrack,
Pretty sure you have to account for the 3cm...
0 Replies
 
DeviatedUnderworld
 
  1  
Reply Wed 14 Jan, 2009 06:09 pm
@jjbuttcrack,
ah man I knew I wasn't paying enough attention. So do you add that extra bit onto the radius? And also does the surface area still fit within the sphere?
ro67
 
  1  
Reply Wed 14 Jan, 2009 06:11 pm
@DeviatedUnderworld,
I subtracted from the radius.
0 Replies
 
shaneybobo
 
  1  
Reply Wed 14 Jan, 2009 06:13 pm
wow i'm completely stumped at this point...
0 Replies
 
hfjai
 
  1  
Reply Wed 14 Jan, 2009 06:28 pm
@jjbuttcrack,
Quote:
If the edge length of a regular icosahedron is a, the radius of a circumscribed sphere (one that touches the icosahedron at all vertices) is 0.9510565163*a

so, the length of the side is slightly larger than the radius


I don't think you use the formula for a circumscribed sphere. I think you have to use the formula for an inscribed sphere (tangent to each of the icosahedron's faces), which is .7557613141*a.

Also, I'm pretty sure that after you calculate the volume of the icosahedron, THEN you subtract the volume taken up by its 20 surfaces given that each is 3 cm thick. At least, that's the way I did it. Of course, there would be some overlap at the joints (vertices) but I'm hoping it would be small enough not to affect the answer.
darkiaia
 
  1  
Reply Wed 14 Jan, 2009 06:45 pm
@hfjai,
3 cm doesnt affecct 6.59 meters
0 Replies
 
Garie
 
  1  
Reply Wed 14 Jan, 2009 06:47 pm
How did you guys calculate the radius of a sphere? I got 16.926 m as the radius.
I Am Not Bob
 
  1  
Reply Wed 14 Jan, 2009 07:42 pm
@Garie,
I have gotten so far as realizing the answer can be found using the formula in Euclid's Elements: Book 13: Proposition 16. The main problem I am having is identifying the length os line segment BD which would be used to figure the size of a pentagonal facet of the icosahedron. If anyone can figure that measurement for me, I can solve the rest of it, but I can't get that part.
0 Replies
 
PeckyP
 
  1  
Reply Wed 14 Jan, 2009 07:46 pm
Hi guys, just to try and provide more than an answer, I'll provide my calculations to help everyone understand what's going on.

We start with V = 1200 m^3 as given in the question.
V = 4/3*pi*r^3
r = 6.5922076505088... (I went to 31 decimals since that's what my calculator gives me)

We know the material used for the icosahedron is 3cm thick. This means that the interior radius is 3cm less than the exterior. Id est, the radius we need is 3cm less than what we found.
r = 6.5622076505088...

Now, we need to find the edge length of the icosahedron because that is the length that we use to calculate volume. We need the formula of the edge length where the radius describes a circumscribed sphere (where all the vertices touch the sphere. Otherwise that sphere won't encase the icosahedron.). This formula is r = 0.9510565163*a (where a is the edge length; approximately).

We get, a = 6.8999134520822672705910139163207

Now the volume of an icosahedron with edge length a is approximately V = 2.18169499*a^3.

So, we end up with:
V = 716.67947048211927207653509403511

According to submission guidelines, we would enter 716.6.

- PeckyP -
PeckyP
 
  1  
Reply Wed 14 Jan, 2009 07:53 pm
@hfjai,
Quote:
I don't think you use the formula for a circumscribed sphere. I think you have to use the formula for an inscribed sphere (tangent to each of the icosahedron's faces), which is .7557613141*a.

Also, I'm pretty sure that after you calculate the volume of the icosahedron, THEN you subtract the volume taken up by its 20 surfaces given that each is 3 cm thick. At least, that's the way I did it. Of course, there would be some overlap at the joints (vertices) but I'm hoping it would be small enough not to affect the answer.


If you use an inscribed sphere that means the sphere is contained COMPLETELY by the icosahedron. In other words, the icosahedron extends beyond the space we have allocated to it as the radius to the vertex of an icosahedron is larger than the radius to the middle of the face of a triangle.
hfjai
 
  1  
Reply Wed 14 Jan, 2009 07:57 pm
@PeckyP,
Quote:
Quote:

Quote:
I don't think you use the formula for a circumscribed sphere. I think you have to use the formula for an inscribed sphere (tangent to each of the icosahedron's faces), which is .7557613141*a.

Also, I'm pretty sure that after you calculate the volume of the icosahedron, THEN you subtract the volume taken up by its 20 surfaces given that each is 3 cm thick. At least, that's the way I did it. Of course, there would be some overlap at the joints (vertices) but I'm hoping it would be small enough not to affect the answer.


If you use an inscribed sphere that means the sphere is contained COMPLETELY by the icosahedron. In other words, the icosahedron extends beyond the space we have allocated to it as the radius to the vertex of an icosahedron is larger than the radius to the middle of the face of a triangle.


Yeah, I think you're right and I had it backwards. Grrr. I hate geometry.
Razz
PeckyP
 
  1  
Reply Wed 14 Jan, 2009 08:01 pm
@hfjai,
Quote:
Quote:
Quote:
I don't think you use the formula for a circumscribed sphere. I think you have to use the formula for an inscribed sphere (tangent to each of the icosahedron's faces), which is .7557613141*a.

Also, I'm pretty sure that after you calculate the volume of the icosahedron, THEN you subtract the volume taken up by its 20 surfaces given that each is 3 cm thick. At least, that's the way I did it. Of course, there would be some overlap at the joints (vertices) but I'm hoping it would be small enough not to affect the answer.


If you use an inscribed sphere that means the sphere is contained COMPLETELY by the icosahedron. In other words, the icosahedron extends beyond the space we have allocated to it as the radius to the vertex of an icosahedron is larger than the radius to the middle of the face of a triangle.


Yeah, I think you're right and I had it backwards. Grrr. I hate geometry.


You don't become a University of Waterloo Math student if you don't love Geometry! =P *hugs geometry* (Ouch! My prism poked me. lolz)
0 Replies
 
eversoclever
 
  0  
Reply Thu 15 Jan, 2009 03:31 am
@Jen Aside,
As with a number of these problems there is a missing piece of information, and a bit of ambiguity in the question.

Firstly, we do not know the thickness of the sphere Dr. Sloth intended to be built. As the spherical area set aside had to contain the sphere and it's contents, the area would have to have been greater than that calculated from 1200 cubic metres (unless he was using advanced Virtupets materials which allowed such volumes to be contained using a thickness of 0.0(lots more zeros)1) metres.

Secondly, the question asks for the volume of the tank, but is unclear as to whether this is the volume of the tank including or excluding its contents.
PeckyP
 
  1  
Reply Thu 15 Jan, 2009 05:47 am
@eversoclever,
Quote:
As with a number of these problems there is a missing piece of information, and a bit of ambiguity in the question.

Firstly, we do not know the thickness of the sphere Dr. Sloth intended to be built. As the spherical area set aside had to contain the sphere and it's contents, the area would have to have been greater than that calculated from 1200 cubic metres (unless he was using advanced Virtupets materials which allowed such volumes to be contained using a thickness of 0.0(lots more zeros)1) metres.

Secondly, the question asks for the volume of the tank, but is unclear as to whether this is the volume of the tank including or excluding its contents.


Okay, to attempt to answer your question, I'll try to provide my reasoning.
First, the question says Sloth set aside a spherical area, not an area for a spherical tank that could contain a volume of water. The actual area he left available was able to contain the 1200 cubic metres of water. So, there is no thickness because we are not considering any tank in that area.
Second, when asked for the volume of a tank of water, say the water that we buy for our watercoolers, our being those of us that have one, we use the volume of liquid that is contained inside. Or, for another example, with gasoline containers, it is stated what the container holds when it is full in litres or gallons. This refers to the volume of the interior space contained by the container, not the volume of space the container were to occupy if it was stored in my closet. (And we all know litres is just fancy talk for 1000 cubic centimetres or 0.001 cubic metres.)

I hope that helps clear up some of your queries.

- PeckyP -
 

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