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m348-assignment3sol

# m348-assignment3sol - Math 348(2006 Assignment#3 Solutions...

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Math 348 (2006): Assignment #3: Solutions 1. Let δ be a circle passing through O . Let A be the point on δ diametrically opposite from O . Let A be the inverse of A . Now consider any point P on δ other than O and A . Since OPA is an angle inscribed in a semi-circle, 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 OPA 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 δ is the line through A which is perpendicular to line ←→ OA . 2. Suppose β is a circle passing through O and C . Then we know that the inverse β of β with respect to inversion in γ will be a line not through O , and since β passes through C , the line β must pass through C , the inverse of C . Since C is the centre of the circle δ , and β is a line through C , β is perpendicular to δ . Since inversion preserves angles,

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m348-assignment3sol - Math 348(2006 Assignment#3 Solutions...

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