Does anyone have any information about the Swiss born German Mathematician Jakob Steiner, and websites or information about his thrms, in particular his proof that all euclidean constructions can be made with a straightedge.  Thanks for any help you can give me.\n\n~matttlocke
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http://en.wikipedia.org/wiki/Jakob_Steiner\nhttp://mathcircle.berkeley.edu/BMC3/constr2/node6.html
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Start here:\nhttp://www.math.pitt.edu/~sussmanm/2071Spring05/lab07/lab07.html#TheGramSchmidtMethod
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PART (a)\nUsing Pythagoras’ Theorem in triangle ABC,\nAB^2 + BC^2 = AC^2 = 85 [since AC = sqrt(85)]\nBoth sides AB and BC has to be positive integers.\nBy trial we take AB = 6, BC = 7 [since 6^2 + 7^2 = 36 + 49 = 85]\n\nAgain, using Pythagoras’ Theorem in triangle CDE,\nCD^2 + DE^2 = CE^2 = 85 [since CE = AC = sqrt(85)]\nBoth sides CD and DE has to be positive integers.\nBy trial we take CD = 2, DE = 9 [since 2^2 + 9^2 = 4 + 81 = 85]\n\nThus we have the length of four sides of the hexagon ABCDEF,\nAB = 6, BC = 7, CD = 2, DE = 9.\n\nNow, let us concentrate on triangle AFE.\nIn any triangle, the sum of any two sides is always greater than the third side.\nSo, in triangle AFE, AF + FE > AE\ni.e. AF + FE > sqrt(85) = 9.22 [since AE = AC sqrt(85)]\n\nAs length of 4 sides of the hexagon are fixed so for minimum perimeter the sides of triangle AFE has got to be minimum. As AE = sqrt(85) is also fixed so the other two sides has to be minimum and hence their sum has to be minimum i.e. (AF + FE) has to be minimum. But AF + FE > 9.22 and also both AF and FE has to be positive integers, \nso (AF + FE)min = 10.\n\nNow, 10 = 1+9 = 2+8 = 3+7 = 4+6 = 5+5 \n\nSides AF and FE cannot attain any pair of these values (1,9; 2,8; 3,7; 4,6; 5,5) as length of each side of the hexagon has got to be different positive integers and we already have the nos. 2, 6, 7, 9 as length of 4 sides.\n \nSo, AF + FE = 11 [next positive integer to 10]\nAgain 11 = 1 + 10 = 2 + 9 = 3 + 8 = 4 + 7 = 5 + 6\n\nThus, AF and FE can attain 2 pairs of values, either 1 and 10 or 3 and 8. \nSo the minimum value of (AF + FE) of possible values of AF and FE is 11.\nHence the minimum value of perimeter of the hexagon is \nAB + BC + CD + DE + EF + (FE + AF)\n= 6 + 7 + 2 + 9 +11\n= 35 \n\nPART (b)\nTo find a minimum value of AC, AC has to be less than present value sqrt(85). Further there must be two pairs of positive integers whose sum equals AC^2 (which is less than 85). So those positive integers has got to be in the range 1 through 9. The squares of these positive integers are presented below. One cannot find two pairs from these numbers such that the sum of each pair is equal to any number less than 85. \n1 4 9 16 25 36 49 64 81\nSo the minimum value of AC is sqrt(85) and the corresponding perimeter of the hexagon is 35.\n\nTo get the figure of this hexagon give me your email id so that I can send it as attachment.
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The difficulty here is that the term 'metric' for differential geometry and for topology is slightly different. The metric described in the website is of the differential geometry type. Essentially, it gives a differential formulation that can be integrated to find the length of a curve. Then 'straight' lines are the curves with minimal length. I'll see if I can find an explicit descrition of the metric in the topological sense. It is rather more complicated as I recall.
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Well, I'm not active in the field, but have found some link for you...\n\nHopefully you will get some of the information you are looking for - otherwise do not hesitate to contact the Zurich university (second link).
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This will happen for only two irregular pentagons in general:\n\nAll triangles as the angle sum of the three angles is a straight angle so the the angles so placed will always make straight lines ... and thus tessellate\n\nAll quadrilaterals as the angle sum of the four angles is a straight angle so the the angles so placed will make a complete revolution ... and thus tessellate\n\nOther particular polygons MAY tessellate BUT no more general cases will occur
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This website looks like it's got a lot of proofs.\n\nhttp://www.cut-the-knot.org/pythagoras/index.shtml
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http://www-math.mit.edu/~sipser/book.html\n\nI hope this may be of some help.
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Prove that if in an isosceles triangle a line segment does one of the following things it also does the other two.\n\n      A\n       ^\n    /  |  &#92;\n   -------\nB    D    C\n\nAD goes from the vertex to the midpoint of the base \nAD is perpendicular to BC \nAD bisects /A \n\n\nConverse: If the same line does any two of the previous three things, then the triangle is isosceles.
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The curve traced out by a point on the rim of a rolling circle is called a cycloid,  \n\nBecause of the frequency with which it provoked quarrels among mathematicians in the 17th century, the cycloid became known as the "Helen of Geometers" \n\na.yes but we do not know what they are\n\nb.yes there must be but we dont know what it is\n\nc.in mathematics any thing is possible but improbable
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Since the question is relative to whether it is good or not:\nYes, it is.\nWould you like us to work it out as well? I appreciate that it must come right out of a quesiton from, perhaps, your homework? Otherwise, why the last sentence? "One hint is that it is related to chapter- 'CIRCLES'" Can you attach that entire chapter for us as well?\n\nThanks!
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For an arbitrary angle, no.  There is an interesting algebraic proof to show that it is not possible to trisect an angle with a straightedge (no markings) and compass alone through the construction of a field.  Now it is obviously possible for certain angles such as the right angle to be trisected.  However, the subset of angles which can be trisected has measure 0 in the set of all possible angles (countable in an uncountable set).\n\n(See source below for more.)
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Pierre Laurent Wantzel (June 5, 1814 in Paris – May 21, 1848 in Paris) was a French mathematician.\n\nIn a paper from 1837[1], Wantzel proved that\n\ndoubling the cube \ntrisecting the angle \nconstructing a regular polygon whose number of sides is not the product of a power of 2 and any number of distinct Fermat primes (i.e. that does not fulfill the same conditions proven to be sufficient by Carl Friedrich Gauss) \nare all impossible using only compass and straightedge.\nwww-groups.dcs.st-and.ac.uk/~history/Biographies/Wantzel.html \nen.wikipedia.org/wiki/Pierre_Wantzel
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You are on the right track.\nThis a geometrical problem that requires drawings.\nCheck this link http://departments.kings.edu/chemlab/chemlab_v2/packgeo.html\n\nI'll get back to you later \n\nso later it is\n\n3--------R=r(1+2/3^.5)\n7--------R=3r\n19------R=5r\n37------R=7r\n61------R=9r
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It was believed that he saw the shadows of the pillars on the tiled temple floor, that looked like a right angled triangle with squares attached. \nFrom there he developed his theory that the square drawn on the hypotenuse of a right angled triangle, is equal in area to the sum of the squares drawn on the other two sides.\nPythagoras put forward a proof, but said proof contained errors to do with transient numbers, however it seemed the theory was sound.\nSome two centuries later the Great Euclid himself proved it, with his brilliant proof using congruent triangles, true poetry of geometry.
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mikii, that's rude, though vinyl is rude and opportunistic as well. shut up if u dont have any answer.\n\nEuclide?\n\nAnyway, plane geometry is also called Euclidean geometry. this is a geometry where 2 parallel line never ever intercept and triangles alway have 180 degree of internal angles. u can imagine it as geometry on a flat plane as we know it. As for spherical geometry, 2 parallel lines do cross and the inside angle of a triangle is larger than 180 degree. how much larger depends on the curvature of the plane and the area of the triangle. U can imagine this geometry as one of the surface of a sphere. Spherical geometry is also called elliptic geometry.\n\nAnother popular geometry is called a hyperbolic. it's a geomtry of the pseudosphere. it's triangles have inside angles of less than 180 degree and 2 parallel line never intercept.
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There was a thing circling the Internet this past Christmas, incorporating snacks in the holiday. \n\nThe story goes, a German baker wanted to create something to show his faith, so he designed a rope of dough in the form of arms bent forward and clasped in prayer. It looked sort of like the letter B on its side. But when it baked, the 2 hands didn't stick together, so he twisted them and tucked the tips under the top. \n\nThe new result was even better than he'd hoped for, because he'd accidentally created 3 holes (apparently for the trinity). I hadn't heard that the twisting was supposed to represent the crucifixion, but whatever. Anyway, I never heard of it before, either, but that's what about 30 people emailed me in November and December. \n\nPersonally, I doubt it. I've been to Bavaria and seen those German pretzel makers. And pretzels. Nothing holy about them. Pretty good with mustard though. Chalk it up to another urban legend online.