Transformation Geometry Solutions & Comments

0. See class notes! A transformation is a one-to-one mapping of the points of the plane to new points of the same plane. An isometry is a transformation which preserves distances, and, thus, figures– that is, the image of any figure is a congruent figure. Isometries are also called "rigid motions". There are four types of isometries of the plane, described here. In each case, P & P’ refer to a point P and its transformed image P’ (and PP’ is the segment joining P to P’). A translation of the plane is an isometry in which the directed line segment (vector) joining each point to its image (P to P') is constant; i.e. every point moves the same distance, same direction. A rotation of the plane is an isometry which fixes one point, O, (the center of rotation); all other P move in such a way that P and P' are equidistant from O, and POP' define angles congruent to the given angle of rotation. That is: points rotate on arcs of circles centered on O; the arcs all subtend central angles with the same measure. A reflection of the plane through the line l is an isometry in which l is the perpendicular bisector of every segment PP'. The points on l are fixed (do not move). That is: except for the points on l, all points move on paths perpendicular to l, to the opposite side of l. P and P’ are equidistant from l, and figures are thus orientation-reversed, or “flipped over”, because the entire plane is flipped over. A glide-reflection can be thought of as a combination-move: reflection followed by translation along a vector parallel to the line of reflection (or vice versa!). All points move, any figure is reversed. 1. See exercises using multiple reflections on the TG-1 through TG-5 worksheets: C1, M5, C5 2. Translations leave a figure “facing the same direction”. All other transformations can change the direction a figure is facing (although this may n ot be obv ious in specia l circumstan ces, such a s a 180° rota tion of a sym metric figure, e.g. a rectan gle ). A figure that is “facing NW” will remain “facing NW” after any number of translations of the plane. The figure never gets to “turn”, as it does in a rotation. As for reflections, we have seen how two reflections can cause a rotation, and just one reflection results in the figure “facing the opposite direction”. Q: Which of the four images is facing the C C (2) (4) same direction as the original, shaded C (3) C [ A: only the 2nd. The transformations are (1) (1) Reflection (2) Translation (3) Rotation C (4) Glide-reflection ] 3. Only Reflections and Glide-reflections change the clockwise sense of a figure. That is because these are the only transformations that “flip” the plane over, and “flipping” is required to reverse this orientation or sense of the figure. The effects on a figure, of translating and rotating the plane can be visualized by sliding and turning a figure, and neither of these results in a “flipped over” figure. 4. Translation fig. 1 Reflection fig. A fig. B Glide-Reflection fig. C Rotation fig. D fig. E fig. F Figure E is the result of a dilatation (actually, “contraction”), a type of size transformation, not an isometry. Figure F is not even similar to Figure 1. We do not study any such transformations (as F) in this course. 5. V 6. 6a. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A“ Symmetries of a square– Four lines: . . . . . . . . . . . . . . . . . . l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A'' A' . . . . . . . . . . . . . . . . . . P . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Translate the figure A (gray) by vector V, label that A'. Then rotate A' 90° about P ; label that A''. Then reflect A'' through line l . Label the final result A “ . A and Four rotations: B D C B C C D D A A B A D B A C B D C 90° 180° 270° 360° 6b. There are five line symmetries of the regular pentagon, one from each vertex through the midpoint of the opposite side. There are also five rotational symmetries of the regular pentagon, the smallest is one-fifth of a rotation, 72°; the other four are, of course, 144°, 216°, 288° and 360°. _ 6c. The regular hexagon has six line, and six rotational, symmetries. However, the figure given has no line symmetries and only 180° and 360° rotational symmetries, because of the “flags” that have been added. _ 6d. About the individual letters in “S Y M M E T R I C” : The letter S, as printed, is like the letters N and Z– although at first glance S may appear to have some line symmetry, it has none. What we are noticing is the 180° rotational symmetry that all these letters have. Of course there is also the 360° rotation. The letter I has two line symmetries (it can be flipped on the horizontal and vertical axes and land back on top of its original footprint). This also has 180° and 360° rotational symmetries. The letter R has no line nor rotational symmetry. Y and M and T each have one line symmetry, about their vertical axes. No rotational symmetry. E and C both have one line symmetry, about their horizontal axes. No rotational symmetry. 7. See class notes titled “Tessellation of the plane”. 8. a. See class notes titled “Tessellation of the plane”. This contained an illustration of one such semi-regular tessellation; the code for this is 4-3-4-3-3. In class, we found another such tessellation, using the green and orange pattern blocks; the code for that one is 4-4-3-3-3 (which some abbreviate as “4233) . b. No semi-regular tessellation of the plane using regular pentagons and equilateral triangles (EqTs) is possible. A vertex (where 3 or more edges meet) must be surrounded by polygonal regions whose interior angles total 360°. The interior angles of a regular pentagon measure 108°, and those of the EqTs, 60°. No combination of 108° and 60° add up to exactly 360°. (One pentagon doesn’t work: 108° + 4@ 60° = 348° & 108° + 5@ 60° = 408°. Two pentagons w on’t work: 2 @108° +2@ 60° = 336° & 2@108° + 4@ 60° = 396 °. Three pentagons w on’t either: 3@108° +1@ 60° = 3 84°. ) 9. The plane may be tessellated by ANY quadrilateral. This is because the sum of the interior angles of the quadrilateral is 360°. Joining four quadrilaterals with their four different vertices put together, but in a way that places matching sides together creates a guaranteed tessellation pattern (illustration below). The plane may be tessellated by ANY triangle (rotate a triangle 180° around the midpoint of any side, and see... that you have created a parallelogram. Since every quadrilateral tiles the plane, this stack of two triangles does too. 2 a. This pentagon with three right angles and an axis of symmetry tessellates the plane. You should be able to easily find a pattern that covers the plane with non-overlapping copies of this figure. ... Just like b & e. Every quadrilateral tessellates the plane; see above! the example above right. c. A triangle tessellates the plane. See above! 135° 135° 90° d. The regular octagon alone cannot tessellate the plane. We have seen previously that its 135° interior angles are so large, that joining two of these figures edge-to-edge covers 2@135° = 270°, leaving only a 90° gap, which is insufficient to insert a third regular octagon. (Of course w e have also se en that tha t 90° gap is jus t right for insertion of a square with a matching side length, to form a semi-regular tessellation. But the questio n here asks for the octagon to tessellate the plane alone.) 10. There are MANY ways to make each figure given have a line symmetry, but NO rotational symmetry. Check yours using tracing paper. FYI: Both a and b can be accomplished with the addition of just one square (in the right location!). Part c needs no additional squares, as it already satisfies the requirements. But if we read the question as requiring the addition of a square, placing one in the middle of the top row works. Some additional comments on # 2 and #4: The transformations are (1) Reflection (2) Translation (3) Rotation (4) Glide-reflection 2. After reflection, before the glide, the C C (2) (4) ¹ image is here. C (3) C Reflection (1) line for (1) C Transla tion ü vector ¹Line of reflection for the glide reflection The rotation here is a 180° rotation. The center of a 180° rotation is always easy to find, because it is halfway between a point and its image. Other rotations require more effort to locate the center. 4. fig. A 4. fig. 1 fig. B Translation fig. C Reflection fig. D Glide-Reflection fig. E Rotation Dilation/ (Contraction) Notice that figure A faces the same direction as the original figure (fig. 1). Also notice all po ints fig. A Showing the vector of translation. 4. fig. 1 4. fig. 1 move the same distance and direction. Notice that all points move on parallel paths, but different distances.... Showing the line of reflection. fig. B Showing the vector of translation, and the line of reflection. Notice points move on non-parallel paths. fig. C 4. ... and orientation is reversed (clockw ise-> cou nter...) fig. 1 Finding the center of rotation is trivial in the case of 180° rotation. C angle of rotation is 180° Notice that points move on non-parallel paths, but orientation is unchanged... what was clockwise, remains clockwise. 4. fig. D fig. 1 fig. E The scale factor is about b . Figure E can be obtained from fig. 1 by a contraction of the plane (dilation in reverse), then translating. Don’t stress over the “contraction” terminology. By the way, when we place figures 1 & E in a “perspective” arrangement, fig 1 is about 3/2 as far away from the center as fig E. So the dilation factor from fig. E to fig. 1 is 3/2 (or 1.5 if you prefer decimal form). Thus the contraction factor from fig. 1 to fig. E is the opposite ratio, 2/3.