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Bisect a plane figure.

  1. Jun 2, 2005 #1
    That L shaped thing reminded me of a puzzle. How do you draw a line that bisects an L-shaped plane figure? In this case, bisect means cut into two pieces of the same area.
     
  2. jcsd
  3. Jun 2, 2005 #2
    L-shaped thing? What exactly do you mean--any rectangle with a rectangular hole cut out of a corner, or something more specific?
     
  4. Jun 3, 2005 #3
    That is precisely what I mean.
     
  5. Jun 3, 2005 #4
    For reference construct the L-shape in this way: Draw rectangle ABCD. On side CD append rectangle DCEF. On side CE append rectangle CEGH. (none of the rectangles overlap)

    Now assume WLOG segment EG is longer than segment AD. Draw point I on CH so that CI = AD. Bisect IH at point J. Draw points K and L on CH and FE so that CK = FL = IJ. Line KL bisects the shape.
     
  6. Jun 3, 2005 #5
    Or, after you draw point I, just draw point J on CH so that CJ = IH, and line FJ bisects the shape.
     
  7. Jun 3, 2005 #6
    You might be right, I don't know. You provide no proof that your construction actually bisects the shape. Please add that in. I know of a much simpler construction that comes with a simple proof that it works. It is provided below. Don't look if you still want to work on the problem.


    First, consider that the L shaped object can be described as BicycleTree did, namely, a larger rectangle with a smaller rectangular hole cut from one corner.
    Consider the line that connects the centers of these two rectangles. Any line through the center of a rectangle bisects the rectangle. The indicated line bisects both rectangles, and so bisects the L shaped piece.

    This construction scales to three dimensions. I don't know the name of this shape, but consider a three dimensional object of 6 faces, each face of which is a quadrangle and congruent with one other face which is parallel to it. A cube is an example, but in general, much more irregular shapes are allowed.

    Consider a cake whose shape is of this kind and which has two cavities in it (not counting the cavities it will put in your teeth), both also of this kind. The centers of the three shapes define a plane which will cut the cake into two pieces of equal volume.
     
  8. Jun 3, 2005 #7
    BicycleTree, your constructions do not seem to work. Any line that passes between one point on the line FE and another point on the line CH, cannot be the solution when DF is much smaller than AD, and EG is only slightly larger than AD just from a casual glance at the figure.
     
  9. Jun 3, 2005 #8
    Doh... I accidentally assumed that FECD was a square.
     
  10. Jun 9, 2005 #9
    I have given this a bit of thought... There should be 2 lines that can both bisect it. YEs or no?

    Edit:

    I figured I should enplane why... If you have 2 rectangles any line drawn threw the midpoint of both of them bisects them both. If the two rectangles are attached creating a L shape. You can create 4 different rectangles (2pair of different rectangles) with 4 different midpoints. Therefore, two different bisecting lines. Someone tell me if that’s right?
     
    Last edited by a moderator: Jun 9, 2005
  11. Jun 9, 2005 #10
    xJuggleboy, you are right that there is more than one way to create an L shaped area from two rectangles. However, the problem does not ask you to do so. The problem starts with an L shaped figure and thus the rectangles are determined. Since the rectangles are determined, there is only one line that intersects the centers of both rectangles (if the rectangles have the same center, then the original shape is not an L.)

    Let me add that there are infinitely many lines that bisect the L shaped figure. Consider any boundary point of the L shape that is on the outer edge of the larger rectangle. Consider a line that goes through that point and is parallel to the line that it is on. Then let that line sweep across the figure like a radar sweep. At some point it will sweep half the area.

    The method given as a solution to the problem is the only one I know of that is constructable with straight edge and compass.
     
    Last edited: Jun 9, 2005
  12. Jun 9, 2005 #11

    NateTG

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    Actually, it's not that difficult to construct a bisector given an aribitrary point that the bisector must go through, but it can be tedious.
     
  13. Jun 9, 2005 #12
    I have my doubts about this. Take a line that makes a 20 degree angle with one of the sides of the L shaped figure and pass that line through the figure until it passes half of the area. This line is the bisector through all of its points, but it is not constuctable.
     
  14. Jun 9, 2005 #13

    NateTG

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    Since 20 degrees is not (IIRC) a constructable angle, that's probably correct. However, it doesn't contradict my claim, since I was talking about an arbitrary point, not an arbitrary angle.
     
  15. Jun 9, 2005 #14
    But there is only one line passing through a given point on the boundary that bisects the figure. Since my 20 degree line is that line for at least two points, your tedious but not difficult program seems to me to be impossible.
     
  16. Jun 9, 2005 #15

    NateTG

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    Well, actually, there may be more than one line that does so since the figure is not convex. (For example, on an L formed by cutting a 3x3 square out of a 5x5 square, there are 2 perpendicular lines that go through the 'crook' of the L that both bisect the L's area.)

    If you provide a line that is at a 20 degree angle from the horizontal, or any line at all, then it can be translated so that it bisects the figure using a compass and straightedge in finitely many steps.

    If you can provide a point that is on that '20 degree' bisector, then it is possible to construct a 20 degree angle from it to bisect the figure.

    The problem with doing what you suggested has nothing to do with problems bisecting the figure, but with constructing a line at that angle.
     
    Last edited: Jun 9, 2005
  17. Jun 9, 2005 #16
    Good point. Of those two lines, one of them behaves as I said, it intersects the boundary at two other points and is the only bisector that passes through those points. The other line however passes through the inside and outside corners of the L and at the inside corner there are two bisectors, so this line does not behave as I said it would.

    However, the existence of such a line has nothing to do with the main point I was trying to make. What I am saying is that you cannot construct a bisector through an arbitrary point in an arbitrary L shaped figure as you had indicated. I am not 100% sure that I am correct, but I think the image of that 20 degree line and its intersection with at least one 'outside' boundary point is compelling.

    The issue with 20 degree angles is not that they don't exist, but that they are not constructable. I don't need to construct the 20 degree line (I can't), I merely need to note that it exists. Ergo, I do not 'provide a line that is at a 20 degree angle' and you have nothing to translate.
     
  18. Jun 10, 2005 #17
    Once again I am misunderstood.... But you have answerd my question anyway =-P Thanks =-D
     
  19. Jun 10, 2005 #18

    NateTG

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    You're asserting that it is impossible to construct a bisector through some (plane figure) and a given point because it is impossible to construct a bisector through the same (plane figure) at a 20 degree angle to the horizontal - or if you prefer, some other arbitrary line.

    However, it's quite easy to construct a bisector through a rectangle that includes a given point (you don't even need a compass), but it is impossible to construct a '20 degree' bisector of the same rectangle.
     
  20. Jun 10, 2005 #19
    You are leaving out an important part of the assertion, namely that the 20 degree angle line is the ONLY bisector that goes through that point. Since you can't construct the 20 degree angle line, you lose the only chance you might have had to construct a bisector through that point.

    Please carry out this program:

    1. Draw, freehand, an L shaped figure, not too symmetric, not too assymetric. Roughly is good enough, no need to be too finicky about it.

    2. Draw, freehand, a line through the figure so that it bisects it, and so that it makes a 20 degree angle with one of the sides. Roughly is good enough, no need to be too finicky about it.

    3. Mark, in red, a point where the bisector intersects with the boundary, but not on the two 'inside' lines.

    4. Note that there is only one bisector that passes through that red point. Why? Consider a radar sweep with the red point as the center. It will only get one chance to sweep out half the area of the figure.

    5. Get back to me. At that time I will explain the implications of this program.
     
  21. Jun 10, 2005 #20

    NateTG

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    Do you know what 'not constructable' means?
     
  22. Jun 10, 2005 #21
    Nate's right... another thing is, what if the L-shape is originally designed so that the angle through the centers of the two rectangles is 20 degrees? You wouldn't say that the method didn't work in that case.

    That is a neat trick with the centers of the rectangles though.
     
  23. Jun 10, 2005 #22
    Obviously, such an L shaped figure is itself not constructable.

    Constructable means with straight edge and compass. I am not going to give a complete description, but if you start with a finite line segment and call its length 1, then the only lengths (in comparison to this unit length) that are constructable are roots of irreducible polynomials with rational coefficients that have either degree 1 (the straight edge: y = mx + b) or of even degree (the compass: [itex]x^2 + y^2 = r^2[/itex]). The cosine of 20 degrees is the root of an irreducible cubic polynomial with rational coefficients and therefore cannot be constructed with straight edge and compass. Since the cosine of 20 degrees is not constructable, neither is an angle of 20 degrees. Since an angle of 60 degrees is constructable, this means that you cannot trisect an arbitrary angle. My high school geometry teacher broadly hinted that the trisection problem was an open question, but he was playing a prank.

    There are many figures that imply unconstructable lengths. For instance the L shaped figure you spoke of. Such figures are themselves not constructable. If you are presented with such a figure, you certainly can construct otherwise unconstructble lengths. Those lengths are still considered nonconstructable because in order to construct them you need to use another nonconstructable object.

    Therefore, the program that I set before you needs this small adjustment. Make sure that the figure you draw is made of line segments whose lengths are constructable. Since any rational number is constructable, and the figure is to be drawn freehand, this is not much of a restriction. None the less, be sure that your drawing represents a constructable one.

    Nate is right about a great many things and he has correctly pointed out an error in something I wrote. I only have issue with one thing. He implied that there is a constructable bisector passing through each point of each L shaped figure. I claim that my program displays a counter-example to this. Please carry it out.
     
  24. Jun 10, 2005 #23

    NateTG

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    If you provide the point on the 'outside' of the L that is part of a 20 degree bisector, then it can be constructed. It's not always going to be this easy, but here's an example.

    OK, so let's say we have a nice easy L, formed by cutting a 1x10 rectangle out of the upper right of a 2x11 rectangle.

    Now, pick any point on the line segment from the exterior corner to five units in. (It should be obvious that a 20-degree line bisecting the L would hit somwhere along this segment.)

    Then, measure the distance from that point, to five units along the base.
    Construct a perpendicular up from the point five units along the base, to the top of the leg.
    Measure an equal distance along the top of the leg away from the corner, and mark that point. Then the line from the originally selected point to the constructed point will bisect the L:

    It's easy to see this because the total area of the L is 12 square units, and the region under the line forms a trapezoid with altitude one, and base lengths 6+n and 6-n so it has an area of 6.
     
  25. Jun 10, 2005 #24
    Why are you drawing the bisector before you draw the point that you want to draw the bisector through? That doesn't make sense. Nate isn't claiming that you have the power to draw a 20 degree bisector, he's claiming that once you have drawn an L and a point, you can draw a bisector of the L that passes through the point. So you draw the L and the point first, and these must follow your stipulation that they are constructable. Then you draw the bisector. If you succeed at this and the bisector makes a 20 degree angle with anything, then perhaps it was your point that was not constructable, not the bisector.

    I'm not sure that you actually can always draw such a bisector, but the unconstructable number argument as you presented it doesn't disprove it.

    Nate, your recent post wasn't clear to me. Maybe jimmy can understand what you mean but in places I can't (for example, "the line segment from the exterior corner to five units in"). Could you post a diagram or something?
     
  26. Jun 10, 2005 #25

    NateTG

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    I was trying to come up with an easy example. I just realized that there is an even easier one:

    A rectangle is an L. Clearly any line that goes through the center of the rectangle (which is a point) will bisect it. Pick any point other than the center of the rectangle that is on the 20 degree bisector, and it's very easy to see that a bisector can be constructed.
     
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