# Elementary proof that there is no paradoxical decomposition using triangular pieces

I am teaching a geometry course and I am trying to understand two definitions in the textbook ("Geometry with Geometry Explorer" by Michael Hvidsten.)

Definition: The area of a rectangle is its base times its height.

Definition: If two figures can be made equivalent, we will say that they have the same area.

Here we say that two figures can be made equivalent if each can be split into the same finite number of polygons (without loss of generality, triangles) such that corresponding pairs are congruent. Note that "split" does not exactly mean "partitioned" here, because we allow the edges of the triangles to overlap.

To me it seems that the first definition is defining "area" in a special case, and the second definition is defining "has the same area." However, we are clearly meant to infer that if two rectangles "have the same area" in the second sense then they have the same "area" in the first sense. Is this obvious?

Of course one can prove it using analytic methods because triangles are Lebesgue measurable. However, the course takes a synthetic approach to geometry, so it would be better to avoid this. So my question is the following:

Is there a proof in elementary synthetic geometry that two rectangles $R$ and $R'$ with different values for "base times height" (e.g. $1 \times 1$ and $2 \times 1$) cannot be split into finitely many triangles $T_i,\ldots,T_n$ and $T'_1,\ldots,T'_n$ respectively, with $T_i \cong T'_i$ for all $i \le n$?

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Is the Wallace–Bolyai–Gerwien theorem what you're looking for? – Daron Apr 11 '14 at 20:52
@Daron The Wallace–Bolyai–Gerwien theorem (applied to rectangles) appears to be the converse of what I am looking for. But thanks for pointing out that theorem to me; it answers a question of one of my students. – Trevor Wilson Apr 11 '14 at 20:54
For the question in the title, one can show there is no Banach-Tarski paradox in the plane. – André Nicolas Apr 11 '14 at 21:07
I hope that you mentioned that in $L(\Bbb R)$ the Paradox is false, at least. (After all, last time I asked you implicitly assumed a proper class of Woodin cardinals... :-)) – Asaf Karagila Apr 11 '14 at 21:10
@AndréNicolas A good point. But my hope is that if the pieces are restricted to be triangles, there is a proof that does not use advanced notions like "solvable group". – Trevor Wilson Apr 11 '14 at 21:12

I think what you want is in Chapter 5 of Hartshorne's book "Geometry: Euclid and Beyond", which covers Hilbert's synthetic theory of area. The theorem you want is a slight rewording of the statement in Andre's post, namely: there is a finitely additive function defined on all figures in the plane that extends the standard "length times height" area formula of a rectangle, and that is invariant under rigid motions.

Here is an outline of the proof of this theorem. One first derives the standard "one half base times height" formula for a triangle. Then one subdivides each figure into triangles and adds their areas. Finally one proves that the result is well-defined independent of subdivision; this is Lemma 23.5 in Harshorne. I've skipped some steps of the outline in which one proves special cases of well-definedness, on the way to the general proof of well-definedness.

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This proof that the sum of the "one half base times height" figures is independent of the subdivision sounds like exactly what I want. I will take a look. Thanks. – Trevor Wilson Apr 11 '14 at 23:49
More precisely, what I want is equivalent to this well-definedness for subdivisions of rectangles (which I hope is one of the special cases; when I get access to the book I will check this.) – Trevor Wilson Apr 12 '14 at 0:12

Edit: The answer refers to the old title, "Why is there no Banach-Tarski paradoxical decomposition using triangles?"

We address only the question in the title, which is different from the question in the body.

There is a finitely additive "measure" in the plane that (i) extends Lebesgue measure and (ii) is invariant under rigid motions. So suppose that $A$ and $B$ $A$ and $B$ have distinct measures, with the measure of $A$ less than the measure of $B$. Then we cannot decompose $A$ into a finite number of (arbitrary) subsets and reassemble these to form $B$.

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A good point. But now I can't change the title without making your answer obsolete. – Trevor Wilson Apr 11 '14 at 21:25
No problem, if the title changes I can modify the answer, or delete it. As to the question in the body, as you know a rectangle can be decomposed into triangles, and the parts reassembled to form a square. If we obtain two non-congruent rectangles, this violates the (vague) Euclidean axiom about the whole not being equal to one of its parts. – André Nicolas Apr 11 '14 at 21:32
Assuming that by non-congruent rectangles you mean rectangles such that one is (or is congruent to) a proper subset of the other, I see what you mean. I had not thought to apply Euclid's common notion "the whole is greater than the part" to areas. I changed the title, by the way. – Trevor Wilson Apr 12 '14 at 0:04
Sorry, "typo." I meant if we obtain non-congruent squares. – André Nicolas Apr 12 '14 at 0:13

The best presentation I've seen of this, at an elementary level, is in the book Geometry. A metric approach with models, by Richard S. Millman and George D. Parker. (There was actually a thread on Mathoverflow on a similar topic, years ago, which is how I found this reference. Unfortunately, I couldn't locate the thread today. But you should look for it, as it includes sketches and other references.)

In Millman-Parker, you want to look at Chapter 10, Area. The book develops the theory in a way that allows the authors to treat areas (for polygonal regions) for both Euclidean and Hyperbolic geometry, essentially simultaneously, before specializing in each case. This way, they first identify the features an area function should satisfy, and then proceed to show in each case that there is indeed essentially a unique such a function.

For the Bolyai-Gerwein theorem itself, once one knows that areas are well defined, the presentation I like best is the one in Wagon's book on the Banach-Tarski paradox. The sketch there leaves a few details out, and for these, the nicest presentation I know of is in Howard Eves's A survey of geometry.

(Unfortunately, although elementary, the argument in the Millman-Parker book takes a while to present carefully -- and more so if one also includes the Bolyai-Gerwein theorem, which should be the case. It would be nice to start a discussion motivating the need for the development of Lebesgue measure with these results, and then a discussion of Hilbert's third problem, and Dehn's solution. I tried a sketch of this once, but not in full detail.)

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Hi Andres, thanks. Section 10.2 seems to contain everything I was hoping for. I'm glad to see that it is a bit simpler than the solution in Andre Nicolas's answer. – Trevor Wilson Apr 12 '14 at 3:10