Join the centre of the regular dodecahedron to one of its edges. What are the angkes of the triangle thus obtained.
217600
69.0948, 69.0948, 41.8104\n\nMore precisely, one of the base angles is equal to\n\narccos( 0.5 * sin(36) * ( 4 - sec(18)^2 )^0.5 / ( 1 + sin(36)^2 + sin(36) * sec(18) )^0.5 )
408661
a) 4 "things" attached to central atom and no lone pairs then: sp3-> tetrahedral e- arrangement, tetrahedral shape, 109.5°\nb) 4 "things" attached to central atom one lone pair then: sp3 -> tetrahedral e- arrangement, trigonal pyramidal shape, < 109.5° (about 107°)\nc) 3 "things" attached to central atom one lone pair then: sp2 -> trigonal planar e- arrangement, bent shape, < 120° (guess about 115°?). This is most challanging, try to figure out how many e- in total you have (18), then arrange them around the central atom using usual valances.
393801
Your first Question:\n\nNow,\n\nImagine a Triangle with vertices ABC. \nAny line from A to BC is AD. D is the point of intersection on BC.\n\nLet E and F be the midpoints of Sides AB anc AC respectively.\n\nJoin E and F to meet AD in G.\n\nNow for the proof:\nTake Triangles AGE and ADB these two are similar.\n\nReason being that EF would be parallel to BC from Mid Point theorem. Therefore, the two triangles will be similar as Angle E and Angle B are equal and Angle A is common. \n\nNow take Triangle ADB, E is the mid point of AB and EG is parallel to BD there for from Converse of Midpoint Theorem G becomes the Midpoint of AD.\n\nHence proved.
10829
milkyway: \nI think your answer is incomplete. Unfortunately, I don't have my own proof, but let me criticize yours. \nSorry.:-)\n\nWhat is definitely proven by induction in your post is that you are able to construct an acceptable set of N points. However, in my view, your reasoning does not prove the statement for a given set.\n\n1. From the very beginning, you have N points. How can one realize the induction scheme in this case? Perhaps, you mean that you can temporarily erase N-3 points (keeping in mind their positions) and start with 3 points forming a triangle, or "3 couples" in your terms. Then you restore the erased points one by one at their positions.  I believe this should be explained in your proof. \n\n2. Note that if you start with 3 aligned points, by "restoring" the 4th point, you can generate an allowed set of N+1 points out of the prohibited set of N points. This is missing in your proof, where you deal only with allowed N-sets.  \n\nBut this is not the most serious flaw there.\n\n3. You write: "now, this way there always are the old two couples, plus two new couples (new point-non aligned point of each of such two couples). So going from N to N+1, the number of acceptable couples is always >=3."  \n\nI disagree with this statement. \n\nIndeed, imagine that the three couples, which work at the stage N, form a triangle. Aligning the newly "restored" point with one side of the triangle, you kill one working couple. Still you have the two remaining old couples and get a possibly working new couple (new point -- non aligned point of the triangle). However, it is not at all guaranteed that there is no other point lying on the line going through this new possibly working couple. If such point exists, the number of working couples in this case gets reduced from 3 to 2 when going from N to N+1. At the next step, this way I will reduce the number of working couples from 2 to 1 and so on.\n\nThe transition from N to N+1 is therefore not justified.\n\n4. A general question. For N=infinity the statement is false for points located at nodes of a regular \n(say, square) lattice. Doesn't it contradict to the very idea of proving this statement by induction?\n\n\n------------------------------\n\n\n\n\nHere is my proof.\n\nGiven a set of N points, the number of all possible couples is M=N(N-1)/2.\n\nFor each couple draw a straight line going through it. We get M lines.\n\nLet us assume that the statement we are going to prove is false: i.e., none of those M lines goes through ONLY TWO points.\n\nThen, of course, each of those M lines coincides with at least the two other lines, so the actual number of distinguishable lines K < M.\n\nWe have N points and K different lines.\n\nEach of the NK pairs (point + line) is characterised by a distance S(n,k) between  the point n and the line k. \n\nThis distance S(n,k) is defined as a length of a segment perpendicular to the line k with one of the ends lying on the line k and the other end coinciding with the point n.\n\nLet us list all the NK distances S(n,k). Since NK is a finite integer number, one can find the shortest value (or several equal shortest values) of S(n,k). \n\nTake a closer look at the point P and the line L with the shortest S(n=P,k=L).\n\nBy assumption, there are at least three points (L1,L2, and L3) lying on the line L. Therefore, at least two of these three points (say, L1 and L2) can be found to the one side (say, to the right) of the point A where the perpendicular to the line L starting at the point P hits the line L. The case when one of the three points (say, L1) coincides with the point A is treated in exactly the same way.\n\nNow we have a triangle (P-L1-L2), with two vertices L1 and L2 lying on the line L to the right of A in the order A,L1,L2 from left to right. \n\nSince L1 lies to the right of A (or coincides with A), the angle between P-L1 and L1-L2 is greater than (or equal to) pi/2.\n\nTherefore t
291952
Not an answer, but a suggestion. Do some research on "Close Packing" of spheres, especially hexagonal sphere packing. You may find what you need.\n\nWhat you need is an angle.\n\nThe answer is going to be d = r + (2r)cos(theta) \n\nwhere theta is defined as follows:\n\nDraw line 1 connecting connecting the center of the 4th sphere with the center of one of the spheres in the bottom layer. Draw line 2 so that it goes through the center of the bottom sphere and through the point where it is tangent to the table top. Theta is the angle between those two lines. \n\nYou may be able to get theta from one of the sites dealing with sphere packing.
540122
a triangle with angles 30, 60,90\nhas sides in ratio 1: (SR 3) : 2\nThe equilateral triangle can be formed from\ntwo 30,60,90 triangles.\neach of those can be formed from 3 smaller 30,60,90 triangles\nso the equilateral triangle can be formed from six 30,60,90 triangles.\nIf radius of circle is 1 unit,\neach of the six small 30,60,90 triangles has \nsides (1/2) ((SR 3)/2) 1\nso the equilateral triangle has side SR(3) units\nso if the equilateral triangle has side x, the circle has radius x /(SR(3)),\nand the circle circumference is 2 pi x /((SR 3))
477328
Here's the answer you want:  With a compass set to the radius of the circle, start marking off radii around the circle, dividing it into 6 equal parts.  Connect any 3 alternate points and you'll have your triangle.\n\nNO MATH REQUIRED!
614924
Its been a long time since I took Geometry but maybe I can give you a small pointer.\n\nTry dividing the rectangle into 3 segments lengthwise and 3 segments widthwise\n\nA--B--C--D\nE--F--G--H\nJ---K--L--M\nN--O--P--Q\n\nDraw the diagonal through NKGD\n\nProve that the segments NK, KG, and DG are equal (prove that the triangles are the same)\n\nNow draw the segments from the vertex Q to the mid point of AD and AN.  Prove that the mid point of AD is the same as the mid point of BC and the same for AN and EJ.\n\nNow we have to prove that the line segment we drew in the last step instersect points K and G. You can probably do this with similar triangles (the triangle NQ and mid point of EJ is similar to JK and mid point of EJ therefore the point K is in the triangle)\n\nSo if you prove that the lines drawn from Q to the midpoints intersect K and G and we proved that NK KG and GD are congruent we have proved the original hypothesis.\n\nYou can turn this gabably gook into a real mathematical proof.\n\nGood Luck
363766
what are you supposed to prove?\n\nedit:\nyou mean why the three vertices of an equilateral triangle are concyclic? i am unable to make out theproblem
511264
Draw lines from the top of the vertical lines in the H to the opposite lower corners. There you go.\n\nIn response to the comment below:\nBTW\nI guess the middle line of the H would have to be not directly centered between the lines.  It has to be a little highter or lower so the 2 lines do not intersect on the center line.  Then you come out with 5 individual triangles and not triangles within triangles.
153521
Let the two sides be formed from three points, (a,b), (c,d), (e,f). So: Side A is (a,b) to (c,d) and Side B is (c,d) to (e,f).\n\nMidpoint of A: ([a+c]/2, [b+d]/2)\nMidpoint of B: ([c+e]/2, [d+f]/2)\nThe vector formed is:\n([c+e]/2 - [a+c]/2, [d+f]/2 - [b+d]/2) =\n([e-a]/2, [f-b]/2)\nThe third side is formed from (e,f) to (a,b), or\n([e-a], [f-b])\nIt's easy to see then that they are parallel.
306101
Okay, I got something really different, but it might be wrong, I used tangents and stuff to figure it out, and I assumed a dodecagon had 12 sides, but I might be wrong. Anyways, I got 38.4 for the perimeter, and 115.2 for the area.
159133
Take the average between the side of an inscribed square\nand that of an inscribed hexagon. With a hexagon you\nget an isosceles triangle with base as a side and a 60\ndegree angle at the apex, leaving 120 degrees to be split\nbetween the equal pair which are then also each 60 degrees.\nSo you have an equilateral triangle whose base is equal\nto the radius of the circle and of course the base is a side\nwhich has a length of 1. \n  Now inscribe a square. There are now 4 isosceles\ntriangles with a 90 degree angle at the apex and a pair\nof 45's at the base. Draw an altitude from the apex, \nperpendicular to the base, thus bisecting the 90 at\nthe apex making another 90, 45, 45 triangle whose\nside is 1/2 the side of that of the square. The hypotenuse\nis a radius of the circle so the side of the last mentioned\ntriangle is 1/2^2 (.707). This makes the side of the\nsquare 1.414. So averaging 1 and 1.414 you get 1.207.\n Using trig we find the real answer to be 1.1755.\nSo the percentage of error is 2.67 percent from 1.207.\nI wouldn't win an engineering prize with that kind of\naccuracy, but I think I could build a pretty good\ncatapult.
631140
What is a point-fill set?\n\nIf you are referring to tiling a planar area with regular polygons, the 60 degree interior angles of the equilateral triangle and the 108 degree interior angles of the regular polygon do not play nicely together ... no sum 60 n + 108 m (n, m integers >= 0) adds up to 360 degrees, so you can't join a corner of several at a point without gaps and/or overlaps.
167513
The centroid of a polyhedron is simply the average of the respective coordinates of all the vertices of the polyhedron. For example, if the coordinates are (x1,y1,z1) (x2,y2,z2) ... (xN,yN,zN), then the centroid would be, \n\nxC = (x1+x2+...+xN)/N\nyC = (y1+y2+...+yN)/N\nzC = (z1+z2+...+zN)/N
310773
a) isoceles\nb) hexagon\nc) hemisphere\nD) equilateral\ne) \nf)\nG) sphere or spherical\nh)\ni)\nj) scalene\n\nthese are the ones i know. next time, do your own homework!
489140
I find this an interesting question, and I suspect there are several possible methods for solving it.  However, I recommend we deal with it straightforwardly by solving for, say, the base length, height ... any 'variables' found within these polygons' area formulas.\n\nThe understanding here is that both polygons are regular polygons (the triangle is therefore equilateral, the hexagon has 6 equal-length sides, etc.).\n\nThe most creative solutions (and least mathematical) would probably involve cutting up one of the shapes and "re-gluing" it to form the other, thereby gaining some hint at the relationship between their perimeters.\n\nI normally prefer straight math, and would recall:\n\n A(triangle) = (1/2) base * height\n A(hexagon) = ([3sqrt(3)]/2) (side)^2\n\nBut let's "logic" our way through this one first.  Think of the hexagon as being composed of 6 equilateral triangles, each containing 1/6 of the overall area.  Because the original hexagon had an area in 1:1 ratio with the given triangle, we now can state:\n\n Area (SmallTriangle) = (1/6) (BigTriangle)\n\nRecalling a particular theorem concerning similar polygons ("The ratio of their areas will be the square of the ratio of their sides"), we can take this further:\n\n Ratio of BigTri:SmallTri Areas = 6:1\n\nSo,\n\n Ratio of BigTri:SmallTri Sides = sqrt(6):1\n\nPicture now how the small triangles would be "glued together" to form their parent hexagon.  Only the "outside side" would contribute to the hexagon's perimeter ... SIX such sides would be needed.  In short:\n\n Ratio of hexagon perimeter:BigTri perimeter = 6sqrt(6):1\n\nHope this (is correct and) helps!