Manifold Mirrors: The Crossing Paths of the Arts and by Felipe Cucker

By Felipe Cucker

Amazon: http://www.amazon.com/Manifold-Mirrors-Crossing-Paths-Mathematics/dp/0521728762

Most artistic endeavors, even if illustrative, musical or literary, are created topic to a suite of constraints. in lots of (but now not all) instances, those constraints have a mathematical nature, for instance, the geometric alterations governing the canons of J. S. Bach, a few of the projection structures utilized in classical portray, the catalog of symmetries present in Islamic paintings, or the principles referring to poetic constitution. This interesting e-book describes geometric frameworks underlying this constraint-based construction. the writer presents either a improvement in geometry and an outline of the way those frameworks healthy the inventive approach inside numerous paintings practices. He in addition discusses the perceptual results derived from the presence of specific geometric features. The e-book all started existence as a liberal arts direction and it truly is definitely compatible as a textbook. notwithstanding, somebody attracted to the facility and ubiquity of arithmetic will get pleasure from this revealing perception into the connection among arithmetic and the humanities.

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Extra info for Manifold Mirrors: The Crossing Paths of the Arts and Mathematics

Example text

Therefore, the two angles at m need to be the same, and since their sum is 180◦ each of them is 90◦ . 5 one shows that ϕ is a reflection with axis . We now assume that P = P. Then, since ϕ is an isometry, dist(A, P) = dist(A, P ) = dist(A, P ) and the points P, P , P are in the circle of centre A and radius dist(A, P). 2 If you feel that P cannot coincide with P since P ∈ , you are right. But, as it happens, we do not need to prove this fact. 37 2 Motions on the plane Again, since dist(A, P) = dist(A, P ) and dist(P, P ) = dist(P , P ), it follows that the triangles APP and AP P are equal.

Y y x x 21 1 Space and geometry The equation y = mx therefore describes this line in the same sense as above: any solution of this equation is (the name of) a point in this line and, conversely, any point on the line is a solution of the equation. This includes the point (0, 0). We finally consider arbitrary lines (non-vertical, since we already have equations for these lines).

This includes the point (0, 0). We finally consider arbitrary lines (non-vertical, since we already have equations for these lines). Let be one such line. Since it is not parallel to the y-axis it intersects this axis at a point (0, q). Consider the line parallel to and passing through the origin. As we have just seen, the equation of is y = mx for some real number m. For any point P in , consider the line through P parallel to the y-axis and let P be the intersection of this line with . e. opposite sides are parallel).

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