OK.  Square is NOT the right word.  Cuboidal polyhedra, or something technical like that.\n\nBasically, it is possible to get a bubble with multiple flat sides.  How?  \n\nPepe made one at the weekend.  I never knew he was so clever.  He's only just turned 2.  I'll get him to explain the answer.  (except you'd have to wait 3 years, till he can write.)
137883
Bubbles are good examples of minimal surfaces. Meaning that they will always attain a shape which gives them the smallest surface area, given the contraints upon them. No contraints (just floatin' free) will always create spherical bubbles for this reason. If you make bubbles stick to some sort of shape, or stick bubbles together, they will create more complex shapes, including some with flat sides (imagine, for instance, two bubbles stuck together, they intersect in a flat circle). So, with the right bubbles and a lot of luck, you would be able to create some rather interesting shapes in the middle of a cluster. But free-floating non-spherical bubbles are an impossibility.
540824
Square and regular hexagon together \n\nEach decagon is surrounded on each alternate side by a square and a hexagon\n\nAlternately instead of the regular hexagon you could put 6 equilateral triangles tesselated into a regular hexagon or a combination of hexagons and equilateral triangles.
454989
Does it have to be a triangle??\n\nA cylinder is what comes to mind for me.  Like a soda can sliced in three equal parts.
14346
Yes, it is possible.\n\nDenote the number of black dots belonging to a red dot by N.\n\nThe present problem is formulated for N=10.\n\nLet us first solve a simpler problem with N=2.\n\nStep 1. \nTake a square, that is a four-sided regular polygon, with the side equal to R. Put the two red dots at the opposite [i.e., connected by a diagonal] vertices of the square. Put the two black dots at the remaining two vertices. If you draw a circle of radius R around each of the two red dots, you will find exactly 2 black dots lying precisely on the circle line. The problem is almost solved, but the number of red dots (2) is equal to the number of black dots (2).\n\nStep 2. \nTake another square and repeat the above procedure. \nNow we have the two square compounds. In total we have 4 red dots and 4 black dots.\n\nStep 3. \nPut the 5th red dot anywhere in the plane and draw a circle of radius R around it. \nDrag each of the above compounds as a whole closer to this circle. \nPlace one of the black dots belonging to the first compound and one of the black dots belonging to the second compound on the circle line surrounding the 5th red dot.\n\nWhile doing so, make sure that you move and rotate each compound as a whole, keeping all the sides of the squares equal to R. \n\nThe problem with N=2 has been solved: in total we have 5 red dots and 4 black dots satisfying the condition of the problem, 5>4. \n\nLet us solve the same problem for N=3.\n\nIn this case, one has to extend the Step 1.\n\nStep 1a. \nHaving constructed the square compound according to the Step 1 above, draw parallel vectors of length R starting at the vertices of the square. Make sure that the vectors do NOT coincide with the sides of the square. \nIf the vector starts at the red vertex, put the black dot at the end of the vector. If the vector starts at the black vertex, put the red dot at the end of the vector. \n\nNow we have one compound consisting of the two shifted squares containing 4 red dots and 4 black dots. If you draw a circle of R around each of the four red dots, you will find exactly 3 black dots lying precisely on the circle line.\n\nStep 2. Repeat the procedure 1+1a.\nStep 2a. Repeat the procedure 1+1a.\n\nNow that we have 3 compounds with equal numbers of red and black dots (in total 12 red dots and 12 black dots), put the 13th red dot anywhere in the plane and follow the instructions from Step 3 above.\n\nThe problem with N=3 has been solved: in total we have 13 red dots and 12 black dots satisfying the condition of the problem, 13>12. \n\nFor N=4, take the compound constructed for N=3 at Step 1a and translate ALL 8 vertices by a vector of length R.\nAgain, if the vector starts at the red vertex, put the black dot at the end of the vector and vice versa. You will have 8 red dots and 8 black dots, etc.\n\nThe above procedure applies to an arbitrary N, in particular to N=10.\n\n\nAn interesting question is: what is the minimal number of dots needed to solve the problem for a given N.
540583
I'm not sure, but I have an idea.  The surface area of a sphere is 4 pi r^2.  You have 40 points spread out over that area, so each point is "entitled" to 1/40th of the total area, or (pi r^2)/10 square units.\n\nNow think about how these points, or actually, small areas around points, are going to be packed.  They won't be circles ... if you put several pennies together on a table, there's be small gaps wherever three pennies join together.\n\nHexagons are more efficient.  Draw a regular hexagon, and you'll see that you can surround it with six more hexagons, completely filling the space without any gaps.\n\nNow consider just one hexagon -- the space surrounding just one point, which is located in the center.  This hexagon, which has an area of (pi r^2)/10, is composed of six equilateral triangles.\n\nSuppose the side of the hexagon is "a" (which is also the side of one of the six small equilateral triangles).  Then the height of the small triangle is (a/2) sqrt(3), and the area of the small triangle is 1/2 bh = (a/2)^2 sqrt(3).  Since there are six small triangles in the hexagon, the area of the hexagon is 6(a/2)^2 sqrt(3) = 3/2 a^2 sqrt(3).\n\nBut that must equal (pi r^2)/10, so\n\n3/2 a^2 sqrt(3) = (pi r^2)/10\n\nSolving for a, we get a = sqrt[pi sqrt(3)/45] r = 0.3477 r\n\n[One cautionary note here:  I'm doing plane geometry, and all these hexagons are plane figures.  This assumes the sphere is circumscribed around all the vertices of the 40 hexagons ... something like a soccer ball.\n\n[Total area, though, will be equal to the sphere because we divided the sphere's total surface area by 40.  Although I don't want to do the work, I think it's possible to include the curvature into this analysis.]\n\nNow we want to know how far apart two adjacent points are located.  Each point is at the center of a hexagon, and the distance between two hexagon centers is twice the height of any one of the small triangles.\n\nThis height is (a/2) sqrt(3), so the distance between points is a sqrt 3 = sqrt[pi sqrt(3)/15] r = 0.6023 r (and all the angles are 60 degrees).  (And that 0.6023 r is probably the answer to your question.)\n\nNow let's get the curvature in there and figure out the central angles.  Assume the hexagons are all stretched so that they form part of the sphere.  Draw a great circle passing through two adjacent points.  We want the central angle of the sector where the arc length is 0.6023 r.  But because arc length is just r theta (in radians), the central angle is 0.6023 radians, or 34.5 degrees apart.
43539
Yes, definitely multiple postings... \n\nwell... a fourth dimension is time... so anything that changes over time is a fourth dimensional object. But that isn't what you are really asking.\n\nWe can only perceive four dimensions with our senses, so perceiving it is difficult. To us it would look as if it were a regular three-dimensional object, perhaps changing over time in some subtle way. Jack L. Chalker did a good job of describing such a structure in his "Quintarra Marathon" books. This is similar in idea to how you would describe a three-dimensional object such as a pencil to a person living in a two-dimensional world, at best they could only see a cross-section of the pencil, or if the world they were living on were curved they would see multiple cross-sections and not realize they were the same object (put a pencil through a folded piece of paper).\n\nIf you are looking for a way to visualize it to help with problem solving I recommend thinking about it in terms of computer arrays, or a group of three dimensional objects, each can be looked at seperately.\n\nGood luck with your conundrum :-)
140027
One possibility is that the sphere is symmetrical and has 6 faces in each hemisphere with a 5-sided pole face with 5 adjacent faces.  This makes 10 edges in each hemisphere and the 5 edges that connect the 2 hemispheres makes 25 edges total.
350038
eclipze Big Big news for you!\n★http://www.osoq.com/funstuff/extra/extra02.asp?strName=eclipze
84253
There is a delightful article by the late Dr. John Kenny on that, back a few years ago, looking at the formulae for surface area and volumes of spheres; I'll see if I can find it for you--it's in my files SOMEWHERE--and give you the reference, and the formulae he generated. It turns out that they're not precisely the same as what you've got--but not terribly far from it.\n\nAs I said, I'll see if I can find it, and I'll get you the reference and the formulae.\n\nAbout 9pm CST...\n\nOK, here is the formulae, done the best I can on this thing...\n\n\nFor a sphere of dimension 2*k, radius R\n\nVolume: (pi^k)*(R^(2*k))/k!\n\nFor a sphere of dimension 2*k + 1 and radius R\n\nVolume:\n(2^(2*k+1))*k!*(pi^k)* (R^(2*k+1))/(2*k+1)!\n\n\nTo get the surface areas for each of these, all you do is differentiate with respect to R...\n\nAs I said, it was a little more complex than you thought--the surface areas come out a little bit differently for odd and even numbered dimensions.  I do not recall why, frankly.
558440
Try the 2-2 pattern. it's the best way. The 3 is done with that criss-cross pattern, but the 2-2 allows you to continue the circle, just make it have a big radius
257012
U ripped dis off from da AISSE 2006 sample que paper didn't u. Here's da value points for it\nLet ABCD be the bucket which is the frustrum\nof a cone with vertex O (as in fig.) ½\nLet ON = x\nΔ ONB ~ Δ OMC ∴ = Ü x = 4 1\n∴ height of cone = 8 + 4 = 12 cm ½\nVolume of bucket = [π(9)².12 — π(3)².4] cm³ 1\n= 312 π cm³\nSlant height of cone of radius 9cm = 9² + 12² cm ½\n∴ L = 15 cm ½\nSlant height of cone of radius 3cm = 3² + 4² cm\n = 5 cm ½\nArea of the copper sheet used to form bucket\n= [π(9) (15) - π(3) (5) + π(3)² cm² 1\n129π cm²
550054
OK, I think what you're saying is that all the corners have to be colored and that if you can rotate a square to superimpose it on another square, or if they're mirror images, then they're the same.  You didn't say that all colors have to be used, so I'll assume they don't.  That said...\n\nStart with squares in which all four corners are the same color.  There are obviously 4 of these.\n\nNow take squares in which three corners are one color and the fourth is different.  There are 12 of these (three squares can be any of four colors, and the odd one can be any of the other three.)\n\nNext, take squares in which two corners are one color and two are another.  There are six color combinations (red-white, red-blue, red-green, white-blue, white-green, blue-green) and each one can have two patterns:  either the same-colored corners are next to each other, or they're diagonally opposite.  So there are 12 of these.\n\nNext, consider square in which two corners are the same and the others are two different colors.  For each combination of a major (2-corner) color and two minor colors, there are only two patterns:  same colors adjacent and same colors opposite.  Whichever way you arrange the odd colors, it will be the mirror image of the other way and therefore identical.  So for each major color, you can pick any two of three minor colors.  This is the same as leaving out any one, so there are three options for minor colors.  4*3*2 patterns is 24 different squares.\n\nFinally, consider squares in which each corner is a different color.  Now, pick a color and a corner.  Two of the other colors will be adjacent to that corner, and one will be diagonally opposite.  For any choice of opposite colors, either arrangement of adjacent colors will just make another mirror image.  So the only choice here is which color will be opposite.  That means there are three different 4-color squares.\n\nNow, add it all up: 4 one-color squares, (12 + 12 =) 24 two-color squares, 24 three-color squares, and 3 four-color squares, equals 55 unique squares.
120690
1st of all: "angles". \n\nAnd then, let's work on the questions\n\n- first, remember that any pentagon has the sum of the internal angles equal to 3*180° \n(n-2)*180° is the general rule for an n-agone\nso:\n\ni) none: otherwise you would have a degenerating pentagon (the last angle would be 180° so the polygon would be a rectangle)\n\nh) infinite: if you only fix the angles and you can choose the sides lenght, you can get infinite similar polygons.\n\ng) infinite: you have even more choices on the other three angles\n\nf) infinite: take a square and "open" it by one vertex, to make room for another side: you have infinite chances to choose the fifth side's lenght.\n\ne) infinite: even more generical than f\n\nd) infinite: cut a regular pentagon with a line parallele to one side and intersecting two adiacent sides\n\nc)this is possible for shapes which are symmetrical with respect to some of their diagonals; for the pentagon, it is impossible: you should "stretch" one side in order to change the angles of a regular pentagon\n\nb) infinite: see answer d)\n\na) one: the regular pentagon
331738
I can't see your image, but if it is like any of these other pictures (see link), the problem is with the slopes of the triangles.  They may appear the same, but they aren't and that is where you lose a fraction of a square.  These small differences eventually add up to a full square if you rearrange things differently.\n\nThe key is in the slopes.\n\nI wish I could see the picture because I like area paradox puzzles like these.\n\n-----\n\nI'm back and can see your picture now... it is just like the first paradox in my link.\n\nThe answer, as I said is the slopes of the triangles.  One triangle has a rise of 2 for a run of 5 (slope = 0.40) and the other has a rise of 3 for a run of 8 (slope = 0.375).  So despite how it looks in the picture, that is not a straight line along the top.\n\nIn the top case, the slope causes a slight bend inward on that line.  In the bottom case, the slope causes a slight bend outward on that line.  If you take the difference between these two you get 1 sq. unit which accounts for the "missing" square.  The second link shows this visually.
169534
Disagree.  You can have a square that has sides of 8 inches and another with sides of 3 inches.  They are similar because the angles are congruent, and the sides are in proportion, but they are NOT congruent figures.  (Same SIZE and shape).  If two polygons are congruent, then their sides are in a scale factor of 1:1 and their angles are congruent, thus they are similar also.
270277
That would be an octahedron, and it would look like a double square pyramid (two square pyramids put base to base).
615061
You need to get a little more specific in terms of what kind of "3d equivalent" you want.  Two kinds of 3d solids you can form easily from 2d polygons (closed shapes made of line segments) are prisms and pyramids.  You make a prism of any 2d shape by taking 2 copies of the flat shape, one above the other, and connecting the corresponding vertices (corners) with edges (lines).  From this perspective, a cube, is a square prism.  A triangular prism looks a little like a tent, with two triangles (top and bottom) and three rectangles or squares around the sides.  To make a pyramid, choose any 2d shape as the base, and then a point above the base as the top, and radiate edges down from the top to all the base's corners.  There are other "3d equivalents", so if you're thinking of something else, let me/us know.\n\nYou also need to be more specific about what kind(s) of sides interest you: edges (1d lines) or faces (2d flat surfaces).  The "sides" of a 2d shape are all edges, but 3d shapes like cubes have both edges and faces (a cube has 12 edges and 6 faces).  Any prism made from a polygon with N "sides" (edges) will have 3N edges (N on the top, N on the bottom, and N more connecting top and bottom) and N+2 faces (N around the sides plus top and bottom).  Any pyramid made from a polygon with N "sides" (edges) will have 2N edges (N around the base and N more reaching up to the top point) and N+1 faces (N around the sides plus the base).\n\nDoes this do it?