# Bisect a plane figure.

1. Jun 2, 2005

### Jimmy Snyder

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. Jun 2, 2005

### BicycleTree

L-shaped thing? What exactly do you mean--any rectangle with a rectangular hole cut out of a corner, or something more specific?

3. Jun 3, 2005

### Jimmy Snyder

That is precisely what I mean.

4. Jun 3, 2005

### BicycleTree

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.

5. Jun 3, 2005

### BicycleTree

Or, after you draw point I, just draw point J on CH so that CJ = IH, and line FJ bisects the shape.

6. Jun 3, 2005

### Jimmy Snyder

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.

7. Jun 3, 2005

### Jimmy Snyder

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.

8. Jun 3, 2005

### BicycleTree

Doh... I accidentally assumed that FECD was a square.

9. Jun 9, 2005

### xJuggleboy

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 thatâ€™s right?

Last edited by a moderator: Jun 9, 2005
10. Jun 9, 2005

### Jimmy Snyder

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
11. Jun 9, 2005

### NateTG

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.

12. Jun 9, 2005

### Jimmy Snyder

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.

13. Jun 9, 2005

### NateTG

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.

14. Jun 9, 2005

### Jimmy Snyder

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.

15. Jun 9, 2005

### NateTG

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
16. Jun 9, 2005

### Jimmy Snyder

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.

17. Jun 10, 2005

### xJuggleboy

Once again I am misunderstood.... But you have answerd my question anyway =-P Thanks =-D

18. Jun 10, 2005

### NateTG

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.

19. Jun 10, 2005

### Jimmy Snyder

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.

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.

20. Jun 10, 2005

### NateTG

Do you know what 'not constructable' means?