Bisect an L-Shaped Plane Figure

[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.

 

[BicycleTree]

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

 

[Jimmy Snyder]

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

That is precisely what I mean.

 

[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.

 

[BicycleTree]

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

 

[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.

 

[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.

 

[BicycleTree]

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

 

[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 that’s right?

 

[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.

 

[NateTG]

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.

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.

 

[Jimmy Snyder]

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.

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.

 

[NateTG]

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.

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.

 

[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.

 

[NateTG]

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.

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.

 

[Jimmy Snyder]

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.)

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.

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.

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.

 

[xJuggleboy]

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

 

[NateTG]

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.

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.

 

[Jimmy Snyder]

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.

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.

 

[NateTG]

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.

Do you know what 'not constructable' means?

 

[BicycleTree]

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.

 

[Jimmy Snyder]

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.

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: $x^2 + y^2 = r^2$). 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's right.

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.

 

[NateTG]

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.

 

[BicycleTree]

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.

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?

 

[NateTG]

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?

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.

 

[Jimmy Snyder]

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.

In fact, on my way home from work I realized this too.

The problem with my construction is that when you mark the point red, you introduce a point into the figure that may not be constructable and once you do that, the proof that a 20 degree angle can't be constructed falls apart.

My intuition still tells me that there are unconstructable bisectors on the L, but I can't prove it and besides Nate has already implied that he has a program (tedious, but not difficult) that constructs them. Can you share it with us? Until you do, I consider this an open question and I intend to pursue my intuition by looking for a counter-example.

 

[Jimmy Snyder]

So you draw the L and the point first, and these must follow your stipulation that they are constructable.

It's sort of beating a dead horse now, but for future reference please undertand this: I only need to show that a bad point exists, I do not need to show that the bad point is constructable. Nate said he could construct a bisector through any point. If he had said that he could construct a bisector through any constructable point, my intuition would fail me.

The statement I made that kicked off this discussion was that there are infinitely many bisectors of an L shaped figure, but the line that connects the centers of the two rectangles is the only constructable one I know of. That is still true in the general case.

 

[NateTG]

My intuition still tells me that there are unconstructable bisectors on the L, but I can't prove it and besides Nate has already implied that he has a program (tedious, but not difficult) that constructs them. Can you share it with us? Until you do, I consider this an open question and I intend to pursue my intuition by looking for a counter-example.

A little warm up (maybe a lemma):

Given a polygon, and a length it is possible to construct a rectangle that has that length as the length of one of its sides in a finite number of steps.

Step 1: Disect the polygon into triangles.
Step 2: For each triangle construct a retangle of equal area - bisect one side and take the height from that side for the dimentions.
Step 3: Using a reversible construction for the geometric mean (there are several options) allows for the creation of a rectangle with the specified dimension as one of its sides. For each triangle.
Step 4: Stack the rectangles.

It's easy to find a rectangle with an area equal to half the area of another rectangle.

Pick some vertex to start with, and measure the areas on either sides of the vertex, if the areas on either side of the region are equal, we're done and that is the bisector.

Otherwise, sweep through the vertices until which side has more area flips. Now we've got a sector of the L that contains no vertices and a bisector.

So now the goal is to find a line through the point P that cuts the sector into appropriate areas.

There are a number of cases for the region in the area.
The region in the sector can be:
1. A trapezoid
2. A triangle.
3. A trapezoid and a triangle
4. Two trapezoids
5. Two triangles.

So the trick is to show that the case cannot occur, or that there is a construction for cutting off a chosen area from it.

I believe the triangle and trapezoid case is the toughest of the lot.

 

[BicycleTree]

Nate, I can't understand what you're saying:

Given a polygon, and a length it is possible to construct a rectangle that has that length as the length of one of its sides in a finite number of steps.

Given the length, isn't it very easy to construct a rectangle that has that length as the length of one of its sides? What's the polygon for?

Using a reversible construction for the geometric mean (there are several options) allows for the creation of a rectangle with the specified dimension as one of its sides. For each triangle.

What does this mean?

Pick some vertex to start with, and measure the areas on either sides of the vertex, if the areas on either side of the region are equal, we're done and that is the bisector.

"Areas on either side of a vertex"?

Otherwise, sweep through the vertices until which side has more area flips.

 

[Jimmy Snyder]

Given a polygon, and a length it is possible to construct a rectangle that has that length as the length of one of its sides in a finite number of steps.

I assume that you mean the following:
Given a polygon, and a length it is possible to construct a rectangle that has that length as the length of one of its sides and an area equal to that of the polygon in a finite number of steps.

The steps that follow seem to be a proof of this statement, but I do not understand step 3.

I have not progessed beyond that yet.

 

[Hurkyl]

I wonder if it would not be easier to simply write down equations defining the bisector in terms of the various parameters, and see if it's solvable with square roots.

 

[NateTG]

I wonder if it would not be easier to simply write down equations defining the bisector in terms of the various parameters, and see if it's solvable with square roots.

Probably, but I hadn't thought of doing that.