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First one is 60.
First you could count how many vertices there are, that will tell you how many pairs you could join in total then you need to subtract the number of segments that correspond to edges and diagonals across faces.
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for the second:
if x and y are two of the angles then x/2 + y/2 = 90 (because if the angle bisectors were perpendicular it would form a triangle with angles 90,x/2,y/2) so x + y = 180 so the third angle would have to be 0.
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for the third one:
im not gonna write it all out cause typing math is annoying :D however, go about it by drawing the trapizoid formed by the area between R and r (the radius that you are trying to find, bisect the trough for this*), then draw the line to form a rectangle and a triangle, r is equal to R minus the top bit of the triangle. Hope that makes sense to you.
* itll look like this
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(ascii diagrams are hard  )
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For #1, Divinek has the right idea (but I think he has the wrong answer). Count all the vertices in your dodecahedron and find out how many ways you can connect a line between them. The subtract the number of ways that connecting to verices will make that line on a face or an edge.
A dodecahedron has twenty vertices and thirty edges. You should make sure you can figure that out yourself. Then you count all the ways you can connect the 20 vertices - thats just 20 choose 2 = 190.
Then how many of those lines are actually edges or on the faces of the dodecahedron. A pentagon will have 5 choose 2 = 10 lines connecting the vertices. 5 will be edges of the pentagon, so 5 lines per pentagon interior. 5*12 = 60 lines. Add that to the 30 edges = 90 lines on the edges.
So there should be 100(?) interior diagonals. You should double check my math and make sure you understand it.
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Oh crap mine was for face diaganols. Whoops. Yeah it's 100 interior.
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3 Lions
United States3705 Posts
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United States24753 Posts
I just tried doing 3 on my own and I got an ugly answer haha...
A/B where
A = R( cos(x/2)*cot(x/2) + sin(x/2) - 1)
B = (cos(x/2) / sin(x/2)) + 1
No idea if it's right :p
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On January 20 2009 13:19 micronesia wrote:
I just tried doing 3 on my own and I got an ugly answer haha...
A/B where
A = R( cos(x/2)*cot(x/2) + sin(x/2) - 1)
B = (cos(x/2) / sin(x/2)) + 1
No idea if it's right :p
what are A and B?
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United States24753 Posts
On January 20 2009 13:24 ThaddeusK wrote:Show nested quote +On January 20 2009 13:19 micronesia wrote:
I just tried doing 3 on my own and I got an ugly answer haha...
A/B where
A = R( cos(x/2)*cot(x/2) + sin(x/2) - 1)
B = (cos(x/2) / sin(x/2)) + 1
No idea if it's right :p what are A and B?
My answer is just the result that you get when you divide the expression A by the expression B.
edit: it was kinda a rush job though so I wouldn't be surprised if there was an error...
I mainly relied on the law of sines.
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i got r = (R(1-tan(x/2))/(1+tan(x/2))
the difference between R and r is equal to (R+r)tan(x/2). (the triangle i mentioned in my earlier post has R+r as its vertical side and delta r as its horizontal side) so r = R - (R+r)tan(x/2) which should equal (R(1-tan(x/2))/(1+tan(x/2)) if i didnt mess anything up.
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Weird. I got same as thaddeusK just with sines instead of tangents. Here is my work:
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I came expecting a thread about boring mass vulture battles.
Thread did not deliver.
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On January 20 2009 14:07 oxidized wrote:
Weird. I got same as thaddeusK just with sines instead of tangents. Here is my work:
it should be r = (z+r)sin(x/2) and R = (z + 2r + R)sin(x/2) because sinx/2 = r/z+r, however i think your method is correct, i think my assumption that the radius to where the circle intersects with the trough is perpendicular to the vertical is false.
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EPT
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