How many vertices are on a pentegonal prism? Is it 10?
411393
Yes 10, think of vertices like the corners.
397805
This problem is certainly tractable, you just need to\nfind a consistent/methodical way of counting the\nrectangles -- without missing any and without counting\nany of them multiple times.\n\nThe way I chose is to define each rectangle as being\ndefined by the square in the upper left corner and the\nsquare in the lower right corner.  Consider the following\nsimplified 2x3 case (where each square is labelled by\nits X and Y coordinates:\n\n0,0     0,1   0,2\n1,0     1,1   1,2\n\n\nThis 2X3 example has the following rectangles starting\nat 0,0.  Remember, each rectangle is defined by the\nsquare in its upper left and the square in the lower right:\n0,0   0,0\n0,0   0,1\n0,0   0,2\n0,0   1,0\n0,0   1,1\n0,0   1,2\n\nThen it also has the set anchored in the upper left by 0,1:\n0,1   0,1\n0,1   0,2\n0,1   1,1\n0,1   1,2\n\nThen it has the set anchored in the upper left by 0,2:\n0,2   0,2\n0,2   1,2\n\nThen those anchored by 1,0:\n1,0   1,0\n1,0   1,1\n1,0   1,2\n\nThose anchored by 1,1:\n1,1   1,1\n1,1   1,2\n\nAnd finally those anchored by 1,2:\n1,2   1,2\n\n\nRewriting the matrix another way, showing how many\nrectangles are anchored by each location in the matrix:\n\n2x3    2x2   2,1\n1x3    1x2   1x1\n\n\nand then doing the math:\n\n6      4     2\n3      2     1\n\nfor a total of 6+4+2+3+2+1 = 18\n\n\nDoing the same analysys for a 4x5 gives:\n\n4x5   4x4   4x3   4x2  4x1\n3x5   3x4   3x3   3x2  3x1\n2x5   2x4   2x3   2x2  2x1\n1x5   1x4   1x3   1x2  1x1\n\nor rewriting after doing the multiplication:\n\n20    16    12    8    4\n15    12     9    6    3\n10     8     6    4    2\n 5     4     3    2    1\n\nAdding up the columns gives:\n\n50    40    30   20   10\n\nFor a total of 50+40+30+20+10 = 150 total rectangles
416560
You should be able to picture it. Or at least, find a D+D player and look at a 12-sided die.\n\nOne half of this is a pentagon, surrounded on each of the 5 edges by other equal-sized pentagons.  The pentagon has 5 edges, and from each of the five corners is another edge which reaches to the 'equator' of the thing.  The bottom half is identical, with anoth 10 edges just like the flipside. Finally, around the 'equator', is another 5 edges.\n\n25 edges.  Don't need formulas, just good 3-D visualization skills.\n\nPS  It's 'polyhedron', by the way.  Polyhedra is plural.  Saying 'a polyhedra' is just like saying 'a women'.
766854
28! I just did it on paper putting 8 dots and connecting each one with the rest. There is a math formula for it but I don't remember it!
1169447
The Cube has 43,252,003,274,489,856,000 different possible configurations.\n\nI cheated and looked up the answer, but to figure it out you'll want to use combinatorics.  I'll have to work on that part of it...\n\nIf you're taking the cube apart and counting how many different ways you can put it back together, this is a relatively simple problem (although it doesn't match the above answer).\n\nThe center cubes don't move at all, so there is only one possible configuration of them.\n\nThere are 12 edge cubes, which each have 2 outside edges and therefore two orientations for each location.  The number of possible combinations of these edges is (2*12)*(2*11)*...*(2*2)*(2*1) = (2^12)*12!\n\nThere are 8 corner cubes, which each have 3 outside edges and therefore three orientations for each location.  The number of possible combinations of these corners is (3*8)*(3*7)*...*(3*2)*(3*1) = (3^8)*8!\n\nMultiply the number of possible combinations of edge cubes by the number of possible combinations of corner cubes and you end up with 519,024,039,293,878,300,000 combinations.  \n\nThis is different than the answer I looked up because there are possible combinations when you take the cube apart that cause it to no longer be solvable.  I can't think of an easy way to find the number of combinations that keep solvability.
462771
They're polyhedrons.  The 5 types are:  Tetrahedron (a 3-d figure with 4 equilateral triangle faces, 4 vertices, & 6 edges); Cube (a 3-d figure with 6 square faces, 4 vertices, & 12 edges); Octahedron (a 3-d figure with 8 equilateral triangle faces, 6 vertices, & 12 edges)); Dodecahedron (a 3-d figure with 12 regular pentagonal faces, 20 vertices, & 30 edges); and Icosahedron (a 3-d figure with 20 equilateral triangle faces, 12 vertices, & 30 edges).
645334
for a cube with 6 faces, the edges would be 8
824935
With three points you form a triangle, with four points a quadrilateral. Regions "outside" the polygon are equal to the number of sides (the number of points). The number of regions inside the polygon is the interesting number.\n\nFor two points, there are no regions "inside" the polygon.\nFor three points, there is one region inside the polygon.\nFor four points, there are four regions inside.
1092713
triskaidecagon\n\nor a 13-sided polygon, haha.\n\nCheck the source for polygon names from 1-20 sides  :)
730795
Here are "drawings" or descriptions of all 11 trees:\n\n_ _ _ _ _ _\n\n_ _ _ _ L\n\n_ _ _ L _\n\n_ L _ L\n\n_ _ L L\n\nX _ _\n\n_ X _\n\nY  (where each of the three spokes consists of 2 edges)\n\n X.L\n\nA star with 5 spokes: one spoke consists of 2 edges:\n\n1-2 1-3 1-4 1-5 1-6 6-7\n\nA start with 6 spokes:\n\n1-2 1-3 1-4 1-5 1-6 1-7\n\nStart with the one tree with 0 edges. After drawing all the trees with n edges, add an extra edge to get all the trees with n+1 edges, removing any duplicates ("isomorphic" trees).
682954
pentagon-5\nhexagon-6\npolygon-many
922356
There are 88 or 44 depending on how you define what a unique diagonal is.  ( Sorry, not trying to be "egghead-y" here, just trying to give you a complete explanation )\n\nANSWER OF 88:\n\nLet's look at an n-sided figure.  For each point (where two sides meet), you can't draw a diagonal from the point to the same point itself, nor from the point to the two points next to it on either side.  But, you can draw a diagonal to every other point, or to n-3 points.  You can do this for each of the n points.\n\nSo, in an n-sided figure, you can have n(n-3) diagonals.\n\nSo, in an 11-sided figure, you can have 11(11-3) diagonals = 11*8 diagonals = 88 diagonals.\n\nNotice this answer assumes that the diagonal from point A to point B is different than the diagonal from point B to point A!\n\nANSWER OF 44:\n\nTo eliminate such "duplicates", we need to look at it this way for an n-sided figure:\n\nPoint 1 has n-3 diagonals (same reasoning as above)\nPoint 2 has n-3 diagonals (same reasoning as above)\nPoint 3 has n-4 diagonals (since 1 of its diagonals was already counted under Point 1)\nPoint 4 has n-5 diagonals (since 2 of its diagonals have already been counted)\n... etc ... till we get to ....\nPoint n-2 has 1 diagonal\nPoint n-1 has 0 diagonals\nPoint n has 0 diagonals\n\nSo an 11-sided figure would have 44 diagonals.\n\nThe general formula being n(n-3)/2.\n\nIf you try this general formula for something that you can easily draw and count the diagonals of, such as a square or a pentagon, you will see it holds true.  (A square has 2 diagonals = 4(4-3)/2 = 4/2 = 2.  A pentagon has 5 diagonals = 5(5-3)/2 = 10/2 = 5.)\n\n~ ♥ ~
991638
I don't believe it can.  This is an example of a unicursal problem and you can find some more details at the puzzle link in the reference below, you'll need to scroll down to puzzle number 51.\n\nFrom my days of O Level Maths I remember that we touched on these and I seem to recall that there was an equation to do with the number of vertices and nodes that a shape had which would determine if it was unicursal or not.\n\nBasically only unicursal shapes (those with a specific ratio of nodes and vertices) can be drawn without removing a pencil from the page and without tracing over an existing line.\n\nI'll leave further research to you.\n\nHave fun
956378
There are 35 triangles.  Count them this way:  All triangles whose vertices are vertices of the pentagon.  There are exactly 10 of hese, one for every different set of three vertices.  Next count the triangles using exactly two adjacent vertices of the pentagon, plus one vertex inside the figure.  Each pair of adjacent vertices has three such triangles, so we get 15 more.  Now count the triangles which use two NON-adjacent vertices of the pentagon plus an inside vertex.  There is just one triangle for each such pair, and there are five pairs, so 5 more triangles.  Now count the triangles using one pentagon vertex and two inside vertices.  There is only one for each vertex, so 5 more.  There are NO triangles using only inside vertices, so we are done, and it adds up to 35.\n\nI copied and pasted this directly from http://mathforum.org/library/drmath/view/54888.html
681069
Pentadecagon or Pentakaidecagon\nLet me give you a list:\n\nn polygon \n2 digon \n3 triangle (trigon) \n4 quadrilateral (tetragon) \n5 pentagon \n6 hexagon \n7 heptagon \n8 octagon \n9 nonagon (enneagon) \n10 decagon \n11 hendecagon (undecagon) \n12 dodecagon \n13 tridecagon (triskaidecagon) \n14 tetradecagon (tetrakaidecagon) \n15 pentadecagon (pentakaidecagon) \n16 hexadecagon (hexakaidecagon) \n17 heptadecagon (heptakaidecagon) \n18 octadecagon (octakaidecagon) \n19 enneadecagon (enneakaidecagon) \n20 icosagon \n30 triacontagon \n40 tetracontagon \n50 pentacontagon \n60 hexacontagon \n70 heptacontagon \n80 octacontagon \n90 enneacontagon \n100 hectogon \n10000 myriagon
94604
A quotation from\nhttp://www.ams.org/featurecolumn/archive/sylvester3.html\n\n"In retrospect, it is clear that one could easily see that there had to be at least three ordinary lines for any point-line configuration. This follows from Melchior's work. Later it was shown that the number of ordinary points had to grow larger as the number of points in a plane finite configuration grew. Both G. Dirac and T. Motzkin were among those who explicitly conjectured that there would be at least n/2 ordinary points for a set of n plane points. However, it should again be noted that for n =7 and 13, there are configurations for which this bound does not hold. The additional requirement that n be large enough has to be added. It was announced by Hansen that he had a proof that there were always at least n/2 (n > 13), but the proof, unfortunately, had an error. Despite this, there were many correct new results in Hansen's investigations related to Sylvester-Gallai results. The number of ordinary points is known to be at least 3n/7 (n > 2), as was shown by L. Kelly and W. Moser, and the best known result (due to J. Csima and E. Sawyer) is that the number of ordinary lines is at least 6n/13 (n > 6). "
376901
There is no rigorous definition of "corner" in geometry, so I assume you mean vertex, as pretending to have an answer by massaging the definitions is not appealing to me. Under peer review, a square pyramid has 5 vertices, period. It would be laughable to suggest otherwise.\nIf your figure is a convex polyhedron, then no such polyhedron exists, as the Euler sum 4 - 8 + 5 = 1 when it should be 2 for such a polyhedron.\nSince you did not specify the type of "figure" you are looking for, I can visualize many less rigorous shapes than convex polyhedra with the properties you need.