## "Extra-geometric" proofs of the Pythagorean Theorem

These pages have recently mentioned two proofs of the Pythagorean Theorem which seem to draw their inspiration from outside the usual subject matter of "synthetic" geometry - I call them "extra-geometric" proofs. The purpose of this note is to examine these arguments with an eye to whether they truly bring new insight into our understanding of the Pythagorean Theorem, or merely obscure its essence in an unfamiliar context, perhaps by introducing (either explicitly or implicitly) additional assumptions which are tantamount to assuming the Pythagorean Theorem itself. In this setting, it is particularly important to look closely at the assumptions necessary to make the "connection" between the geometry and the extra-geometric context.

As a simple example of an unsatisfactory extra-geometric proof, consider this "hammer and tongs" approach via Cartesian coordinates:

We begin by assuming the Cartesian formula for the distance between two points, with coordinates (u, v) and (w, x): distance2 = (w - u)2 + (x - v)2. We also need the well-known fact that the slopes of perpendicular lines are negative reciprocals (that is, their product is -1).

Let the coordinates be given as follows: A = (s, t); B = (u, v); C = (m, n), where angle ACB is right. Apply the distance formula to compute the squares of the lengths of the sides, a, b, c. Compute the difference a2 + b2 - c2 (you initially get 18 terms; 8 of them cancel immediately). Compute the slopes of segments AC and BC. Since ACB is right, the product of these slopes is -1. Multiply this out. Insert this expression (8 terms) into the formula for the difference a2 + b2 - c2. If you haven't dropped a minus sign, everything cancels, and you find that a2 + b2 - c2 = 0, or a2 + b2 = c2. (If the sides of the original triangle are horizontal and vertical, we get into trouble computing slopes, but in this case the distance formula is precisely the Pythagorean Theorem.)

What does this proof tell us? Not much. The Cartesian formula for distance is simply a special case of the Pythagorean Theorem, in 17th Century clothing. This formula, which is necessary to link the geometry to the algebra, essentially assumes what we are trying to prove. Perhaps the most charitable thing that can be said for this proof is that we have picked ourselves up a short way by our bootstraps: we have assumed the Pythagorean Theorem for right triangles with horizontal and vertical sides, and extended it to right triangles with any orientation. Certainly all the algebraic technique provides no additional geometric insight.

Can we do any better using Vector Algebra instead of the old-fashioned kind? Suppose vectors A and B are perpendicular. Then their vector sum C = A + B forms the hypotenuse of a right triangle. It is traditional to introduce a "scalar product" X·Y, with the following properties:

Commutative Law: U·V = V·U

Distributive Law: U·(V + W) = U·V + U·W

The connection to the geometry resides in two additional assumptions:

Length Formula: |U|2 = U·U

Orthogonality: if U and V are perpendicular, then U·V = 0.

 |C|2 = (A + B)·(A + B) = A·A + A·B + B·A + B·B = |A|2 + 2 A·B + |B|2 = |A|2 + |B|2

which is the conclusion of the Pythagorean Theorem.

However, in order to justify this formal calculation, we must define the concept of "vector" and "vector addition," in terms of known geometric objects, and provide an explicit formula for the scalar product. Two approaches commonly appear in the textbooks:

1. Define a vector in the plane as an ordered pair of numbers. Then vector addition may readily be defined coordinate-wise: (u, v) + (w, x) = (u + w, v + x). The length of a vector is defined by the usual Cartesian distance formula - and now we are in trouble: as we have just seen, the Cartesian distance formula is just a restatement of the Pythagorean Theorem, which we are hoping to prove. A scalar product may be introduced according to the familiar formula: (u, v)·(w, x) = uw + vx. With this formula, the commutative and distributive laws are readily verified by simple algebra. The formula relating the length of a vector to the scalar product follows directly. If we identify an ordered pair with the line-segment from the origin to the corresponding point on the Cartesian plane, then the Orthogonality criterion is a mere paraphrase of the negative-reciprocal rule for slopes mentioned above. Apparently, our "vector algebra" has simply restated the Cartesian calculation in more elegant notation.

2. Define a vector geometrically as a "directed segment" in the plane, and stipulate that the length of a vector is the same as its ordinary length as an (undirected) segment. It will be convenient to regard a vector as "fixed" in place at its initial point. Addition of vectors may be defined by the "triangle" law: to add A + B, make a copy of the vector B (same length, same orientation) so the initial point of the copy of vector B coincides with the terminal point of vector A; then define A + B as the vector with the same initial point as vector A, and the same terminal point as the copy of vector B. In this geometric context, the scalar product can be defined by setting U·V = |U||V|cos(θ), where θ is the angle between the orientations of U and V. Of course, if we take this approach, we must assume enough of the underlying structure of geometry to make sense of the notion of cos(θ). This includes the realization that all right triangles in which one of the acute angles measures θ are similar. Of course, once we notice this fact, we are but a line or two away from some of the standard proofs of the Pythagorean Theorem [These can be essentially algebraic in nature (see proofs #6 and #22 on the Pythagorean Theorem page) or rely on the observation that the ratio of the areas of similar figures is the square of the ratio of the lengths of the corresponding features]. In this context, the length formula, commutative law, and orthogonality formula are readily verified. The distributive law can be reduced to a simple construction.

The scalar product may also be defined in terms of the "projection operator" - the projection of U on V is the length of the segment from the origin of V determined by the foot of the perpendicular from the endpoint of U to V. Then U·V = |V| (projection of U on V). This avoids explicit mention of the cosine, but ultimately gains us little: in order to verify the Commutative Law for the scalar product, one must again invoke the similarity properties of right triangles.

Finally, what are we to make of the "mechanical proof," which obtains the Pythagorean Theorem from the observation that if a triangle contains a uniform "pressure", there is no net torque about any of the vertices. On its face, this is a strikingly "extra-geometric" assumption. On closer examination, a geometrical underpinning emerges:

First, we must examine the concept of "torque" itself. Select an arbitrary origin. Apply a force F at the terminal point of a radius vector r drawn from the origin. The usual definition of torque is the product of the length of the radius vector and the component of the force perpendicular to the radius vector. If we allow ourselves explicit trigonometry, we may express this in the form: torque = |r||F|sin(θ), where θ is the angle between the force vector F and the radius vector r. In general, the torque will depend on the choice of origin. Often, it is helpful to compute the torque in an alternative way: instead of multiplying the length of the radius vector by the component of the force perpendicular to the radius vector |F|sin(θ), multiply the magnitude of the force by the component of the radius vector perpendicular to the force |r|sin(θ). If we define the torque geometrically, a calculation using similar triangles verifies that the two products are equal. [Whether we determine these components geometrically or trigonometrically, an understanding of similar triangles is clearly required here.]

The mechanical proof of the Pythagorean Theorem was based on the observation that the torques on the sides of the gas-filled triangular box sum to zero. This was justified by appeal to Newton's laws and the statistical mechanics of gases.

In fact, we may dispense with the gas altogether. Instead, imagine a triangle (right-angled or otherwise) at rest, but free to move and to rotate - say floating over the surface of an air-hockey table. Since the triangle forms a closed figure, if we take the sides of the triangle as vectors laid head to tail, we conclude that the three vectors which represent the sides sum to zero.

Now simultaneously apply, at the mid-point of each side of the triangle, a force, proportional to the length of the side, directed outward along the perpendicular bisector of the side. Since each of the force vectors along the perpendicular bisectors is proportional to the corresponding side-vector, subject to a 90° rotation, it follows that the three force vectors likewise sum to zero. This guarantees that the total (linear) momentum of the box is conserved - that is, the box (to be specific, the center of gravity of the box) does not accelerate. Since the box was initially at rest, it remains at rest (Newton's First Law of Motion). Does the box begin to rotate in place?

It can be shown that when the center of gravity of an object is at rest, as here, the angular momentum is independent of the choice of origin [see Goldstein, p. 7]. In this case, it is easy to see that if an object is acted upon by three planar forces whose lines of action meet at a single point, then the net torque is zero: choose the common intersection as the origin for computing the torques! [This is Tokieda's Lemma, p 699.] But the perpendicular bisectors of a triangle are concurrent (that is, they meet at a single point). Thus we conclude that the triangle will not rotate.

Filling a triangular box with gas under pressure is merely a simple way to realize forces, effectively acting along the perpendicular bisectors of the sides, proportional to the lengths of the sides.

From here, the derivation of the Pythagorean Theorem may proceed as described previously. Of course, an argument by similar triangles is apparently still required.

In summary, these "extra-geometrical" proofs of the Pythagorean Theorem are apparently closer in spirit to the traditional proofs by the methods of High-School geometry than they at first appear. Nevertheless, they direct our attention to the flatness of the plane, and to the central role of the Pythagorean Theorem in the fabric of geometrical thinking.

Scott E. Brodie

10/21/98

### References

1. H. Goldstein, Classical Mechanics, Addison-Wesley, Reading, MA, 1950 [This was for many years the standard reference for Newtonian Physics in the US]

2. T. F. Tokieda, Mechanical Ideas in Geometry, American Mathematical Monthly, 105:697-702, October, 1998. 