Chapter 4 - Chapter 4 Isometries I Topics 1 Isometries as...

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Unformatted text preview: Chapter 4 Isometries I Topics : 1. Isometries as Product of Reflections 2. The Product of Two Reflections 3. Fixed Points and Involutions a1 a1 a1 a1 a1 a1 a65 a65 a65 a65 a65 a65 Copyright c circlecopyrt Claudiu C. Remsing, 2006. All rights reserved. 49 50 M2.1 - Transformation Geometry 4.1 Isometries as Product of Reflections A product of reflections is clearly an isometry. The converse is also true; that is, every isometry is a product of reflections . We prove now this fact in seven (small) steps. (Actually, we shall do better than that by showing the product has at most three factors, not necessarily distinct.) Looking at the fixed points of isometries turns out to be very rewarding in general. 4.1.1 Proposition. If an isometry fixes two distinct points on a line, then the isometry fixes that line pointwise. Proof : Knowing point P is on the line through distinct points A and B and knowing the nonzero distance AP , we do not know which of the two possible points is P . However, if we also know the distance BP , then P is uniquely determined. It follows that an isometry fixing both A and B must also fix the point P , since an isometry is a collineation that preserves distance. In other words, an isometry fixing distinct points A and B must fix every point on the line through A and B . a50 4.1.2 Proposition. If an isometry fixes three noncollinear points, then the isometry must be the identity. Proof : Suppose that an isometry fixes each of three noncollinear points A,B,C . Then the isometry must fix every point on triangle ABC as the isometry fixes every point on any one of the lines ←→ AB, ←→ BC, ←→ CA . Every point Q in the plane lies on a line that intersects triangle ABC in two distinct points. Hence the point Q is on a line containing two fixed points and, therefore, must also be fixed. So an isometry that fixes three noncollinear points must fix every point Q in the plane. a50 C.C. Remsing 51 4.1.3 Proposition. If α and β are isometries such that α ( P ) = β ( P ) , α ( Q ) = β ( Q ) , and α ( R ) = β ( R ) for three noncollinear points P,Q,R , then α = β . Proof : Multiplying each of the given equations by β- 1 on the left, we see that β- 1 α fixes each of the noncollinear points P,Q,R . Then β- 1 α = ι by Proposition 4.1.2 . Multiplying this last equation by β on the left, we have α = β . a50 4.1.4 Proposition. An isometry that fixes two points is a reflection or the identity. Proof : Suppose isometry α fixes distinct points P and Q on line L . We know two possibilities for α , namely ι and σ L . We shall show these are the only two possibilities by supposing α negationslash = ι and proving α = σ L . If α negationslash = ι , then there is a point R not fixed by α . So R is off L , and P,Q,R are three noncollinear points. Let R prime = α ( R ). So PR = PR prime and QR = QR prime , as α is an isometry. Therefore, L is the perpendicular bisector of RR prime as each of P and Q is in the locus of all points equidistant from...
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Chapter 4 - Chapter 4 Isometries I Topics 1 Isometries as...

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