In showing that this parallelogram
Figure 13.1. Parallelogram.
has the same area as a rectangle with the same base and height (altitude), we can easily cut the parallelogram into two pieces and rearrange them to form this rectangle
.
Figure 13.2. Equivalent by dissection to a rectangle.
We say that two figures (F and G) are equivalent by dissection (F =d G) if one can be cut up into a finite number of pieces and the pieces rearranged to form the other.
QUESTION: If two planar polygons have the same area, are they equivalent by dissection?
ANSWER: Yes! If the polygon is bounded on either the plane or on a sphere or on a hyperbolic plane.
You will prove these results about dissections in this chapter and the next and use them to look at the meaning of area. In this chapter you will show how to dissect any triangle or parallelogram into a rectangle with the same base. Then you will do analogous dissections on spheres and hyperbolic planes after first defining an appropriate analog of parallelograms and rectangles. After that you will show that two polygons on a sphere or on a hyperbolic plane which have the same area are equivalent by dissection to each other. The analogous result on the plane must wait until the next chapter.
The proofs and solutions to all the problems can be done using "=d", but if you wish you can use the weaker notion of "=s": We say that two figures (F and G) are equivalent by subtraction (F =s G) if there are two other figures, S and S', such that S =d S' and FÈS =d GÈS', where F & S and G & S' intersect at most in their boundaries. Saying two figures are equivalent by subtraction means that they can be arrived at by removing equivalent parts from two initially equivalent figures, like this:
Figure 13.3. Equivalent by subtraction.
If we cut out the two small squares as shown, we can see that the shaded portions of the rectangle and the parallelogram are equivalent by subtraction, but it is not at all obvious that one can be cut up and rearranged to form the other.
Equivalence by dissection is the stronger of the two methods of proof, and is generally preferable — it is much more obvious and in many ways, more convincing. All of the problems presented here can be proved by dissection, and we would urge you to use this method over equivalent by subtraction whenever you can.
Some of the dissection problems ahead are very simple, while some are rather difficult. If you think that a particular problem was so easy to solve that you may have missed something, chances are you hit the nail right on the head. For almost all of the problems, it is very helpful to make paper models and actually cut them up and fit the pieces together. Most of the dissection proofs will consist of two parts: First show where to make the necessary cuts, and then prove that your construction works, i.e., that all the pieces do in fact fit together as you say they do.
Problem 13.1. Dissect Plane Triangle & Parallelogram
a. Show that on the plane every triangle is equivalent by dissection to a parallelogram with the same base no matter which base of the triangle you pick.
Part a is fairly straightforward, so don't try anything complicated. You only have to prove it for the plane — a proof for spheres and hyperbolic planes will come in a later problem after we find out what to use in place of parallelograms. Make paper models, and make sure your method works for all possible triangles with any side taken as the base. In particular, make sure that your proof works for triangles whose heights are much longer than their bases. Also, you need to show that the resulting figure actually is a parallelogram.
b. Show that on a plane every parallelogram is equivalent by dissection to a rectangle with the same base and height.
This problem is also rather simple, and a partial proof of this was given in the introduction at the beginning of this chapter. But for this problem, your proof must also work for tall, skinny parallelograms for which the given construction doesn't work:
Figure 13.4. Tall, skinny parallelogram.
You may say that you can simply change the orientation of the parallelogram, but, as for Part a, we want a proof that will work no matter which side you choose as the base. Again, don't try anything too complicated, and you only have to work on the plane.
Dissection Theory on Spheres & Hyperbolic Planes
The above statements take on a different flavor when working on spheres and hyperbolic planes since one cannot construct parallelograms and rectangles, as such, on these spaces. We can define two types of polygons on spheres and hyperbolic spaces, and then restate the above two problems for these spaces. The two types of polygons are the Khayyam quadrilateral and the Khayyam parallelogram. These definitions were first put forth by the Persian geometer-poet Omar Khayyam in the eleventh century AD.2 Through a bit of Western chauvinism, geometry books generally refer to these quadrilaterals as Saccheri quadrilaterals after the Italian monk who translated into Latin and extended the works of Khayyam and others.
Figure 13.5. Khayyam parallelograms.
A Khayyam quadrilateral (KQ) is a quadrilateral such that AB @ CD and ÐBAD @ ÐADC @ p/2. A Khayyam parallelogram (KP) is a quadrilateral such that AB @ CD and AB is a parallel transport of DC along AD. In both cases, BC is called the base and the angles at its ends are called the base angles. One can check that a KQ on the plane is a rectangle and that a KP on the plane is a parallelogram.
Problem 13.2. Khayyam Quadrilaterals
a. Prove that the base angles of a KQ are congruent. and that they are more than a right angle on the sphere.
b. Prove that the perpendicular bisector of the top of a KQ is also the perpendicular bisector of the base.
c. Show that the base angles are greater than a right angle on a sphere and less than a right angle on a hyperbolic plane.
To begin this problem, note that the definitions of KP and KQ make sense on the plane as well as on spheres and hyperbolic planes. The pictures in Figure 13.5 are deliberately drawn with a curved line for the base to emphasize the fact that the base angles are not necessarily congruent to the right angles. You should think of these quadrilaterals and parallelograms in terms of parallel transport instead of parallel lines. Everything you have learned about parallel transport and triangles on spheres and hyperbolic planes can be helpful for this problem. Symmetry can also be useful.
Now we are prepared to modify Problem 13.1 so that it will apply to spheres.
Problem 13.3. Dissect Spherical & Hyperbolic Triangles and Khayyam Parallelograms
a. Show that every hyperbolic triangle and every small spherical triangle is equivalent by dissection to a Khayyam parallelogram with the same base.
Try your proof from Problem 13.1, as a first stab at the problem. You only need to look at a sphere. You should also look at the different proofs given for Problem 13.1. The only difference between the plane and spheres and hyperbolic planes as far as this problem is concerned is that you must be more careful on spheres and hyperbolic planes because there are no parallel lines; there is only parallel transport. Some of the proofs for Problem 29 work well on a sphere, and others do not. Remember that the base of a KP is the side opposite the given congruent angles.
b. Prove that every Khayyam parallelogram is equivalent by dissection to a Khayyam quadrilateral with the same base.
As with Part a, start with your planar proof and work from there. As before, your method must work for tall, skinny KPs. Once you have come up with a construction, you must then show that the pieces actually fit together as you say they do, and prove that the angles at the top are right angles.
*Problem 13.4. Spherical Polygons Dissect to Biangles
In the next chapter you will show that every polygon on the plane is equivalent by dissection to a square, and then we will use this and the Pythagorean Theorem to show that any two polygons with the same area are equivalent by dissection. This does not apply to spheres and hyperbolic planes because there are no squares on these surfaces. However, we have already shown in Problems 7.1 and 7.4 that two polygons (or triangles) on the same sphere have the same area if they have the same holonomy. Thus every polygon on a sphere must have the same area as some biangle (a region of a sphere bounded by two great circles, same as a lune) with the same holonomy. Now we can show that not only do they have the same area but they are also equivalent by dissection.
a. Show that every simple small polygon on a sphere is equivalent by dissection to a biangle with the same holonomy. That is, the angle of the biangle is equal to (½)(2p - sum of the exterior angles of the polygon).]
Consequently, two simple small polygons on a sphere with the same area are equivalent by dissection.
Outline of Proof
The proof of this result can be completed by proving the following steps (or lemmas). (This proof was first suggested to me by my daughter, Becky, now Rebecca Wynne.)
Theorem. On a hypersphere, two simple polygons with the same area are equivalent by dissection.
Two published proofs in English are in [NE: Millman & Parker (1981)], page 267, and [Z: Boltyanski (1978)], page 62. These proofs are similar, and both use the first four steps above and use the completeness of the reals (in the form of a version of the intermediate value theorem). You can check that Becky's proof above does not use completeness. In addition, the proof of the same result on the plane (see the discussion between Problems 14.2 and 14.3) also does not need the use of completeness axiom.
The first four steps above will also work (with essentially the same proofs) on a hyperbolic plane. But there is no clear replacement for the biangles (which do not exist on a hyperbolic sphere). There is a proof of:
1
From the translation by Edward Fitzgerald.
2 See [A: Khayyam, 1931].