if a mobius strip is cut along the middle, how many separate strips are obtained? is this/ are these mobius?
469366
I dont think the answer above sufficiently answers your question so I'll put in my $0.02. \n\nIf you cut a mobius strip once along in the middle, you will get 2 separate strips (by logic, try it yourself). These 2 strips will also be mobius strips, so long as you have not done any other modifications to the original mobius strip. \n\nAgain the easiest way to test this is to do it yourself. Wont take longer than 5 minutes! :)
956315
The simplest way is to cut it into five slices, each with an arc of 72 degrees (or 2*pi/5 radians). You could probably also figure out a way to cut it into strips so that each strip has equal area, but that would be significantly more complicated.
196942
wow, that's pretty amazing.  i was convinced that was wrong until i tried drawing the "counterexample" i had in my head.  other than reading through that proof (props to the guy who posted that link, btw) i guess the best way to realize this is true is to show that for any five closed shapes in two dimensions, there will always be at least two shapes that do not touch one another.  because of this, you can color those two shapes the same color, so you will have a maximum of four colors.\n\nthis problem is reducible to a graph with five nodes, where each node represents a shape and each edge represents a shared boundary.  (showing that this is reducible to this is probably the difficult part of the proof, which is why i'm glossing over it).  you cannot connect each node to every other node without there being at least one set of intersecting edges.  if edges intersect, it indicates that the shapes (or countries) represented by the two nodes connected by one of the intersecting edges do not share a border.\n\nhere's my suggested proof (assuming we accept the reduction to this graph problem):\nstart with a graph G with three nodes, {A, B, C}.  There are also edges {AB, BC, AC}.  So, G is a closed graph with three nodes and three edges where each node is connected to each other node in the graph (i.e. G is a triangle).  If our reduction of the coloring problem is correct, then there exists a map reducible to this graph that requires three colors to be colored in the described way.\n\nNext, construct a graph H by adding a node D to graph G.  There are two possible ways to add D: 1) D can be on the interior of closed graph G; or 2) D can be outside the interior of G.\n\nIn case 1, let the new graph with nodes A,B,C,D be called H.  We need to show that we can make H such that D is connected to A, B, and C without any intersecting edges.  This is simple enough; straight edges to A, B, and C from D do not intersect each other or the existing edges of the graph G.  So, H demonstrates that a map can require four colors.  H now has four separate enclosed interior regions: triangles ABD, ACD, BCD, and ABC.\n\nFrom H, construct a graph I by adding another node E.  Repeating the above procedure, E can be placed in four places: a) in the interior of ABD; b) in the interior of ACD; c) in the interior of BCD; or d) on the exterior of ABC.\n\nCASE a: As with constructing H, edges can be drawn from E to A, B, and D.  However, C is on the exterior of ABD, so any edge from E to C must cross either AB, BD, or AD.  So, in case a, there must be an intersection, so four colors is still an option.\nCASES b and c follow similarly.\nCASE d: E is on the exterior of ABC, but D is on the interior of ABC.  Any edge between E and D must intersect AB, BC, or AC.  So, coloring with four colors is possible in this case.  Since it is possible in all cases, it is possible for any map with five regions with a sub-map reducible to a graph like H.\n\nCase 2) Instead of placing D in the interior of G, make a graph H' by placing D on the exterior of triangle ABC.  Any quick drawing will show that D is easily connected to A, B, and C.  So there is a map with four regions that requires four colors.  To make the rest of this easy, note that H' also produces four enclosed regions: ABD, ABC, BCD, and ADC.  I won't go through the trouble to prove this, but clearly H' is isomorphic to H.  That being said, whatever is true of the general graph H is also true of the general graph H'.  Therefore, a graph I' made by adding a node to H' cannot be complete without there being at least one set of intersecting edges.\n\nSince this is true for all cases, any map reducible to a graph with five nodes is colorable with just four colors.  QED\n\nSorry this is so long.  I hope it's helpful.
860033
First off, a perpendicular bisector (by definition) is a line that cuts something in half.\n\nWe know that a triangle is composed of three lines and therefore (as I’m sure you’ve already presumed) has three perpendicular bisectors.\n\nSo lets try to find one.  Its rather difficult without a triangle to show you on but we’ll make do with words.\n\nFirst, find the midpoint of any given line of your triangle.  You can do this by measuring it (if it’s to scale) or by using a midpoint formula (you can only do this if your triangle is on a graph).  If measuring, find the length of the side and divide by two for the distance from the start of the line to the middle of it.  If it is on a graph you can use the formula: ((x1+x2)/2),((y1+y2)/2).  You’ve guessed it, this formula does the exact same thing as the first method but with numbers instead of with a ruler.  In case you’re wondering; x1 refers to the x value of the first point of the line, y1 refers to the y value of the first point of the line, x2 refers to the x value of the second point of the line, and y2 refers to the y value of the second point of the line.  You will end up with an answer that looks like: (x,y).  This is the coordinate of your midpoint!\n\nNow that you have the midpoint you must connect it to the opposite vertex of your triangle.  This is easily done by hand and with a ruler.  You’ll notice how this has divided the triangle into two exact halves.  If you’re doing this in the mathematically sound way (i.e. with many formulas and calculations) please keep reading, your task is much more laborious!  You need to find the slope between the midpoint (M) and your vertex (V).  So we must use the slope formula: slope=(y2-y1)/(x2-x1).  Where (x1,y1) are the coordinates of the midpoint and (x2,y2) are the coordinates of the vertex.  Now that your slope is achieved you need to find the y-intercept.  This is found by using the equation: y=mx+b.  Sub in your midpoint coordinates for x and y (since they are a point on this line) and rearrange the equation so you have an answer for b (the y-intercept).  Now that you have an equation for this bisector, draw it in and give yourself a pat on the back!\n\nRepeat this process for the next two lines in your triangle.  Here’s a hint though, you don't actually need to find the third bisector because it will just confirm what the last two have told you!  Besides, you only need to have two bisectors to find the intersection of the bisectors.\n\nSpeaking of which, now we are going to find the intersection.  If you’ve been doing it all by hand and ruler you simply just need to put a point where your bisectors have met and show that you have, indeed, found the intersection.  If you have to do it the arduous method (in which case I applaud you), please read on.  Since you have you have the equations for each bisector, this will be easy.  You are simply going to find where these two lines intersect.  If you have a Texas Instruments Graphing Calculator (or similar device), now would be the time to whip it out and graph these two lines.  If you don’t have this device, have no fear; algebra is here!\n\nIf you have a pretty good understanding of algebra or even just pre-algebra you’ll understand this part fairly quickly.  We have two equations and two unknowns.  We have to solve for x and y in these two equations.  This will give us the x-value and y-value that these two lines share.  I’m going to use the two following equations to aid both my instruction and your understanding.\n\nLets say the two lines are y=2x-3 and y=x+1.  We know that y=2x-3 in the first equation so we’ll sub it into the second equation.  Therefore 2x-3=x+1.  From here you need to move the xs to one side and the numbers to another, so: 2x-x=1+3 which become x=4 (don’t forget that the number’s sign (+ or -) is inversed when you change sides (change side, change sign).  Now that we know x, we can sub x’s value into the first equation.
654251
No, because the charges will have to distribute to opposite sides of the strip (as if it was a loop rather than a mobius strip), however since we are dealing with a one dimensional object (essentially), the charges cannot do this.
278253
5 lines?? must they be straight lines or can they be curved or another circle inside??  If you put the four lines thru like 2 x's then draw a small circle it is 16 pieces.
450313
10 percoset should do the trick\nThis is a joke....just in case you are dumb enough to try it.lol
789773
let us understand it in a simple way:\n   Suppose you have a magnet as shown below:\n                                [ N ]\n                                [ S ]\n\nBy cutting the magnet centrally, if you  mean cutting it vertically, then we would obtain the following configuration after cutting the magnet:\n                              [N] [N]\n                              [S] [S]\nThat is to say we have resulted in two magnets with polarities as shown above.Now as we all know that like polarities(N-N, S-S) repel and unlike polarities(N-S,S-N) attract, the result follows that the two magnet in above configuration repel each other while they attract when one of them is rotated by 180.\n[When one of the magnet is  rotated (say the second one) the following configuration results:\n                            [N] [S]\n                            [S] [N]\nand hence the result.]
38235
---------Y cm----------\n==========\n    1 cm\n_______________________\n          "           "   '\n    1.5cm "    B      "   '\n         _"__x cm_____"   ' \n  A       "           "   '3.5 cm\n      2cm "           "   ' \n__________"___________"   '\n\n\n7/2= 3.5\n\n3.5 cm- 2 cm= 1.5 cm\n\nY cm- X cm = 1 cm (from statement "The length in the hole is 1 centimeter less than the length of metal sheet. Length of metal sheet is Y cm, therefore, Y cm- 1 cm= X cm or Y cm - X cm= 1 cm).\n\nIf you compute the area of Rectangle A, that would be..\n\nlength x width = 3.5 cm x 1 cm = 3.5 cm sq.\n\n"After the whole is cut, are of remaining metal is 11 cm squared"... therefore\n\nArea of Rectangle A + Area of Rectangle B = 11 cmsq.\n\n11 cm sq. - 3.5 cm sq= 7.5 cm sq.\n\n\nGiven: Area of Rectangle B is 7.5 cm sq.\n       Length x width  (rect. B) = 7.5 cm sq.\n\nWe know width is 1.5 cm (from 3.5 cm- 2 cm).\n\nSolve: Length x width = Area\n       Length x 1.5 = 7.5\n       Length = 7.5 / 1.5\n       Length = 5\n\nTherefore:\n\nLength of x (hole) = 5 cm.\nLength of Y (metal sheet) = 5 cm + 1 cm. = 6 cm.\n\nI hope this helps. Let me know if you have other questions.
434240
Generally, midpoints divide the lengths into half. So then, the whole perimeter of LMN is 1/2 of ABC. So the ratio is 1/2.\n\nYou can pick two CORRESPONDING sides, and their ratio is still 1/2. To find the area, there's a formula:\n\n(Side 1/Side 2)^2\n\n(One length of LMN/One length of ABC)^2\n= (1/2)^2\n= (1^2)/(2^2)\n= (1)/(2*2)\n= 1/4\n\nNote: one length of LMN doesn't have to be 1 and another length of ABC doesn't have to be 2. They can share common multiples, but you end up simplifying to ratio 1/2.\n\nratio of perimeter of triangles LMN to ABC is 1/2\nratio of areas of triangles LMN to ABC is 1/4
134473
The answer is obviously (A). The bar may shrink a little, but it can't disappear completely, as the options B, C and D imply!!\nHence, by process of elimination, the answer is (A). :-)
941718
Geometry Session 2, Part B: Linkage-Strip Constructions  \n... (or the hands-on version) to try to build quadrilaterals with the given lengths. ... Can a set of four lengths make two different quadrilaterals? ...learner.org/channel/courses/learningmath/.../part_b/constructing.html \nEasy Construction of Bicentric Quadrilateral II  \nEasy Construction of Bicentric Quadrilateral ... By construction, four quadrilaterals MPND, MPLC, LPKB, and KPNA are also cyclic. ...www.cut-the-knot.org/Curriculum/Geometry/BicentricQuadri2.shtml \nBrahmagupta Quadrilaterals  \nIn this paper we give a construction of Brahmagupta. quadrilaterals from rational Heron triangles. ... Construction of Brahmagupta quadrilaterals ...forumgeom.fau.edu/FG2002volume2/FG200221.pdf \nEasy Construction of Bicentric Quadrilateral  \nEasy Construction of Bicentric Quadrilateral ... appears as a consequence of another construction of bicentric quadrilaterals. ...www.cut-the-knot.org/Curriculum/Geometry/BicentricQuadri.shtml \nOnline Lesson On Geometrical Construction Using iMovies  \n... On Geometrical Construction Using iMovies. LESSON ... Module 3 - Construction of Quadrilaterals. After this module, the students should be able to. 1. ...ali.apple.com/ali_sites/sgli/exhibits/1001179/Lesson_Objectives.html
1062497
What everyone has said is correct, but look at it this way: Consider a square. We know from the theorem of pythagoras that the ratio of the hypotenuese to the length of one side = sqrt(2) or, if you like, the ratio of one side to the hypotenuese = 1/sqrt(2). These constant ratios are independent of the size of the square. We have the ratio of the perimeter of the square to the hypotenuese = 4/sqrt(2) = 2*sqrt(2), which is independent of the size of the square.  Similarly the ratio of the circumference of a given circle to the longest chord(the diameter) has an exact value, and similarly this value is independent of the size of the circle. As has been mentioned, for any regular n-gon(square, pentagon, hexagon etc) the ratio of the perimeter to the length of the longest line segment that can be drawn within the n-gon will have a specific value depending on n, and this value can be determined. As we let n approach infinity, the n-gon approaches a circle and the ratio of the perimeter of the n-gon to the longest "chord" approaches a specific limit and this limit is pi, the ratio of the circumference to the diameter.  And, as mentioned above, analytic techniques can (and have) determined the value of the constant pi to a trillion or so digits.\n\nFootnote: SK, determining the limit of a series or even that the limit exists does not necessarily require calculus.\nFor example, the sum of a geometric series such as Sum[1/(2^n)] = 1/2 + 1/4 + 1/8+....+1/2^n is proved without calculus to be 1- 1/2^n and the limit of it (without calculus) as n---> inf. = 1. Check out the below link regarding pi.\n\n http://mathworld.wolfram.com/ArchimedesAlgorithm.html
708197
no thats the hyperbola. a hyperbola has two parts. the parobola is when the conic section is cut parrallel to the conic section so you only ever get one continuous curve. if you cut directly downwards you will get two, unless you go through the intersection when the two parts of the hyperbola then intersect.
131179
No.\n\nA midpoint is the direct middle of two points.\n\nMany segments can be connected. These connected segments will have a midpoint that will differ from the midpoints of the individual segments but... yeah... the answer is NO.
815746
It looks as if there's one big triangle, but in fact the smallest angle of the green triangle isn't the same as the smallest angle of the red triangle, and so although it looks as if there's one big triangle cut up, it isn't so.  And rearranging it makes the hole appear.  I can do a diagram showing just where the extra area unit is, but not on here.\n\nDo you know trigonometry? If so, notice the tangents of the two angles I mentioned are different.  One is 2/5 and the other is 3/7.\n\nIf you don't know trigonometry, just notice that the green triangle has the sides about the rightangle in the ratio 2:5, the red triangle has them in the ratio 3:7.  It's the swapping of the positions of those two triangles that takes up the extra area.  The green triangle has the smaller angle, so when it's the lower triangle the "hypotenuse" of the big triangle has an inward kink.  When the red triangle is the lower one it has an outward kink.  The space between these two "hypotenuses" is one square unit of area.
1114562
twice the length from one end to the middle