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Mechanical Proof of the Pythagorean TheoremI've been asked several times whether a generalization of the Pythagorean Theorem to Euclidean spaces (vector spaces with scalar product) provides a proof of the "regular" Pythagorean Theorem. The answer is Yes it does but it does not belong here. As a matter of fact, proof depends on a set of selected axioms. The theory of vector spaces is derived from the set of axioms altogether different from Euclid's. For example, the Fifth Postulate is easily proven from the analytic form of the equation of a straight line. The whole of Euclidean geometry can be derived in the framework of vector spaces and analytic geometry. And as M.M.Postnikov wrote in his Lectures in Geometry, there might be a good reason to do so:
Of course, no one actually follows through with basing Euclidean Geometry on the notion of vector, not in a high school course. (Postnikov's was a freshmen course at the Moscow State University intended for math majors. Euclidean Geometry was only mentioned on a few rare occasions.) The reasons are many of which not the least are the intuitive character of Euclidean Geometry and the inherent charm of many of its theorems and proofs. This said, one can't deny the simple elegance of the analytical proof of the Pythagorean Theorem. The fact is that various sets of axioms are used in various branches of mathematics but also in other sciences as well. Newton's laws that form a foundation of (non-relativistic and non-quantum) mechanics are nothing but a special set of axioms (the first law being a consequence of the second one.) Other common laws (theorems of mechanics) are derived from those axioms. For example, for a body whose motion is restricted to rotation around a fixed axis, one obtains the principle of conservation of angular momentum a particular case of which can be stated in the following manner:
I'll define the torque for a plane body and the axis of rotation passing through point O perpendicular to the plane. Let a force F apply at point P of the body. The force F can be split into two components. One acting along the line OP and another perpendicular to that line. Regardless of how big is the former, the only component that may cause the body to rotate is the one perpendicular to the line OP. Then the corresponding torque is defined as OP·PQ. If, to start with, F is perpendicular to OP then the torque is equal to ±OP·|F|, where the vertical bars indicate the magnitude of the force. The sign is taken to be "+" if the induced rotation is counterclockwise, and "-" otherwise. 
Let in This is another consequence of Newton's laws and additional assumptions that are made in statistical mechanics (mechanics of gases) that gas inside the box pushes against a face of the box with the force proportional to the area of the face. To horizontal faces being equal and parallel, the upward and downward forces cancel each other. For the sides, the forces may be regarded as acting on the centers of the sides perpendicular to the surface. The three magnitudes are h·AB, h·BC, and h·AC. 
Here is the view from the top. Forces that apply to sides AB and BC try to revolve the box counterclockwise. The force acting on side AC tries to revolve it in the clockwise direction. In mechanics of gases this is accepted as a fact (since, statistically speaking, any other behavior has negligible probability) that gas will not cause the box to change its position. The box can be considered to be at rest although there are forces acting on its sides. We may apply the principle of conservation of angular momentum. 
Torques of the forces acting on sides that contain BC and AC are easy to compute. One is BC/2·h·BC and the other is AC/2·h·AC. The torque at the side AB equals CM·PM, where PM is perpendicular to CM and M is the center of BC. From similar triangles CBM and FPM, CM/BM = MF/PM. Therefore CM·PM = BM·MF = AB/2·h·AB. Getting rid of the common factor h/2, we obtain the Pythagorean Theorem:Do we have another proof of the famous theorem? Yes we do. However, the proof is based on axioms not usually included into Euclidean Geometry. The underlying axiom - Newton's second law - is not even geometric. What we found is, in a sense, that Newton's second law implies Pythagorean Theorem which, as we already know, is equivalent to the Fifth Postulate. In other words, Newton's laws may only hold in the world where Euclidean Geometry (as opposed to non-Euclidean geometries where the Fifth Postulate does not hold) reigns. Gaining this insight is a sufficient justification for learning a proof that does not belong in a Geometry course. Reference
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ABC angle B be right. Consider the thin box (right prism of height h) with lid and bottom equal to 
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