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Unformatted text preview: Math 348 (2006): Assignment #3: Solutions 1. Let be a circle passing through O . Let A be the point on diametrically opposite from O . Let A be the inverse of A . Now consider any point P on other than O and A . Since OPA is an angle inscribed in a semicircle, we know that it is a right angle. Let P be the inverse of P . Then OP OP = k 2 = OA OA , and hence we have OP OA = OA OP from which it follows by SAS for similarity that 4 OPA 4 OA P . Since the corre sponding angles are congruent, this means OA P = 90 . Since P was arbitrary, we have shown that the inverse of all the points (other than O ) on the circle is the line ` through A which is perpendicular to line OA . 2. Suppose is a circle passing through O and C . Then we know that the inverse of with respect to inversion in will be a line not through O , and since passes through C , the line must pass through C , the inverse of C . Since C is the centre of the circle , and...
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This note was uploaded on 02/20/2011 for the course MATH 348 taught by Professor Karigiannis during the Summer '06 term at McGill.
 Summer '06
 Karigiannis
 Math, Geometry

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