Prof. Templeton Posted February 23, 2009 Report Share Posted February 23, 2009 Nine Octahedrons of equal size are packed inside a 1x1x1 unit cube such that no more can fit inside. What is the maximum length of the edges of the octahedrons? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 23, 2009 Report Share Posted February 23, 2009 i think the answer is .3536 units my thought is, i pack the octahedron in by putting 2 rows of 2 on top of one another with one in the midde of all of them. Thinking of it as four square pyramids on top of one another, the height of one of those pyramids would be .25 units. The side of the pyramid would be x. Then I drew a line from the center of the square where the height bisects to the edge of the pyramid. This creates a right triangle, so solving for x: .25^2 + (.5*x*sqrt2)^2 = x^2 x=sqrt 0.125 x=.3536 Quote Link to comment Share on other sites More sharing options...
0 HoustonHokie Posted February 23, 2009 Report Share Posted February 23, 2009 I think I could stack them in layers of 4, 1, & 4 octahedrons, where the apexes of the top layer touch the apexes of the bottom layer. The single octahedron in the middle layer could fit in the gap in between the other 8. It doesn't come out exactly cubic, so I'm guessing there's some more optimization to be done, but I get a side length of 1 / (2 * sqrt(2)) = 0.3536. My overall volume used only comes out to 9/48, so I imagine there's a more compact arrangement possible. Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted February 23, 2009 Author Report Share Posted February 23, 2009 i think the answer is .3536 units my thought is, i pack the octahedron in by putting 2 rows of 2 on top of one another with one in the midde of all of them. Thinking of it as four square pyramids on top of one another, the height of one of those pyramids would be .25 units. The side of the pyramid would be x. Then I drew a line from the center of the square where the height bisects to the edge of the pyramid. This creates a right triangle, so solving for x: .25^2 + (.5*x*sqrt2)^2 = x^2 x=sqrt 0.125 x=.3536 I think I could stack them in layers of 4, 1, & 4 octahedrons, where the apexes of the top layer touch the apexes of the bottom layer. The single octahedron in the middle layer could fit in the gap in between the other 8. It doesn't come out exactly cubic, so I'm guessing there's some more optimization to be done, but I get a side length of 1 / (2 * sqrt(2)) = 0.3536. My overall volume used only comes out to 9/48, so I imagine there's a more compact arrangement possible. Those 9 will surely fit inside the cube. But as HH noticed there's alot of room left inside. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 23, 2009 Report Share Posted February 23, 2009 Nine Octahedrons of equal size are packed inside a 1x1x1 unit cube such that no more can fit inside. What is the maximum length of the edges of the octahedrons? Someone has a bunch of role-playing dice... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 24, 2009 Report Share Posted February 24, 2009 Since you didn't say they have to be regular octahedrons, you could squash them down so that their height was 1/9 unit. the four edges around the middle could be 1unit each Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted February 24, 2009 Author Report Share Posted February 24, 2009 Since you didn't say they have to be regular octahedrons, you could squash them down so that their height was 1/9 unit. the four edges around the middle could be 1unit each They are quite regular, as the picture indicates. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 24, 2009 Report Share Posted February 24, 2009 could it be as big as you want since 1 could equal 1000 or more or less Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 24, 2009 Report Share Posted February 24, 2009 I'm gonna go for 3 rows of 3 in my solution: Front and Side view So with this arrangment i think the limiting factor is depth, but i'm not too sure on that. If it is then the max side length i found was .408, this gives us a packing factor of around .289 which is still fairly low, so there's probly some better way. Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted February 24, 2009 Author Report Share Posted February 24, 2009 could it be as big as you want since 1 could equal 1000 or more or less They can certainly be as big as you want. As long as they fit inside the 1x1x1 cube. I'm gonna go for 3 rows of 3 in my solution: Front and Side view So with this arrangment i think the limiting factor is depth, but i'm not too sure on that. If it is then the max side length i found was .408, this gives us a packing factor of around .289 which is still fairly low, so there's probly some better way. That's an improvement alright. Here's a hint. The arrangement I'm thinking of takes up more then half of the volume of the cube, and like some of the earlier packings has one octahedron in the center of the cube. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 25, 2009 Report Share Posted February 25, 2009 Well this kind of has one in the centre, but its in the same plane as the others So i noticed that when you cut a cube as shown you get a hexagon, and when you arange the octahedrons like so, you get a hexagonal arangement which happens to fit quite nicely in the box along the plane. With the remaining 2 i put them in the corners. With this arrangement i get a side length of .4714, and a packing of .44, which is still less than the correct answer, but closer. Am i on anywhere near the right track? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 25, 2009 Report Share Posted February 25, 2009 Actually, that last one i posted could fit upto 15, so there has to be a different packing arrangement... Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted February 25, 2009 Author Report Share Posted February 25, 2009 I'll give you the packing, but you'll have to figure out the maximum length of the sides. five are drawn in, you can see where the other four would go. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 25, 2009 Report Share Posted February 25, 2009 Ugh took me a while to figure out that it wasnt an impossible box you'd drawn, but that you'd actually used perspective Thanks for the hint, i would never have gotten it otherwise side length: .53033 packing factor: .6328 Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted February 26, 2009 Author Report Share Posted February 26, 2009 (edited) Ugh took me a while to figure out that it wasnt an impossible box you'd drawn, but that you'd actually used perspective Thanks for the hint, i would never have gotten it otherwise side length: .53033 packing factor: .6328 Yeah, you got it. Looking back I should have given you the length and had you try to figure out the packing. It may have been more challenging. Care to post how you arrived at your answer, so I don't have to? Edit: I can't take credit for the drawing. Edited February 26, 2009 by Prof. Templeton Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 26, 2009 Report Share Posted February 26, 2009 Well i was hoping to avoid writing all this out, but since you asked so nicely: Just a warning, i usually go about things in the most complicated way, so no doubt there will be an easier way to do this calculation Let x = side length of octahedron The height of a square pyramid with side length x can be shown to be x*SQRT(2)/2 Using this we can find out that the angles alpha and beta of the projection of the side view of the octahedron to find alpha we use the one of the right angled triangle rules: tan(alpha) = (x*SQRT(2)/2)/(.5*x) alpha = arctan(SQRT(2)) The angles in a triangle add up to 180 degrees so: beta = 180 - 2*arctan(SQRT(2)) If we look at the cube from the plane shown in red above, we get: gamma = 360 - 3*beta gamma = 360 - 3*(180 - 2*arctan(SQRT(2))) gamma = 6*arctan(SQRT(2)) - 180 theta = (180 - gamma)/2 theta = 180 - 3*arctan(SQRT(2)) lambda = theta + alpha lambda = 180 - 3*arctan(SQRT(2)) + arctan(SQRT(2)) lambda = 180 - 2*arctan(SQRT(2)) sin(lambda) = 0.5/x x = 0.5/sin(lambda) x = 0.5/sin(180 - 2*arctan(SQRT(2))) x = 0.53033 Quote Link to comment Share on other sites More sharing options...
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Prof. Templeton
Nine Octahedrons of equal size are packed inside a 1x1x1 unit cube such that no more can fit inside. What is the maximum length of the edges of the octahedrons?
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