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Unformatted text preview: Math 348 (2006): Assignment #2: Solutions 1. Let ` and ` be a pair of parallel lines. Suppose the line m intersects ` and ` at points P and P , respectively. Since ` k ` , the alternate interior angles are congruent. Now let f be an isometry. We know it takes lines to lines, and preserves angles. Therefore f ( ` ), f ( ` ), and f ( m ) are all lines. Also f ( m ) intersects f ( ` ) and f ( ` ) at points f ( P ) and f ( P ), and the alternate interior angles are the same as they were before, (since f preserves angles), so they are still congruent. Therefore by the converse of the alternate interior angle theorem, f ( ` ) k f ( ` ). 2. Let r ` be reflection across a line ` . Clearly the line ` is invariant under r ` . Suppose m is another line. If m k ` , then m is not invariant, since m = r ` ( m ) will be on the opposite side of ` from m . Therefore we can suppose that m intersects ` at P , and let Q be a point on m different from P , and let R be a point on ` different from P . Then PQR = P Q R . If m is to be an invariant line, then Q must be on m also. Therefore PQR must be congruent to its supplement, so it is a right angle and m is perpendicular to ` . For a circle to be invariant under r ` , its centre O must be a fixed point, and hence must be on ` . Its easy to see that a circle centred on ` is invariant under r ` , using congruent triangles. 3. Let R O, be rotation about O by an angle . Clearly any circle centred at O is invariant under R O, , since the distance to O is preserved. Any circle centred at a point different from O cannot be invariant, since the centre would not be fixed. Consider a line ` not through O . Drop a perpendicular from O to ` at P . Then ` is tangent to the circle centred at O with radius OP . Since is less than 360 , the line ` is sent to ` , a line tangent to the same circle but now at the point P , different from P , so ` is not invariant. Its easy to see that lines though O are not invariant under R O, unless = 180 , in which case all...
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