# sols

## Transcript

1 Selected Solutions 3.3.1 Show that any three points not on a line lie on a unique circle. P, Q, R . Let be the line equidistant from P Let the points be called L . (As proved earlier, it is the perpendicular bisector of the segment and Q Q P , and it’s defined by the property that for all points on connecting and Q .) Now let M this line, their distance to P is equal to their distance to R . and Q be the line equidistant from is perpendicular to P Q and Suppose they were parallel. Then since L M QR P Q is parallel to QR , which is perpendicular to , it would follow that all lie on the same line, which is not true by P, Q, R would imply that and M are not parallel. assumption. We can conclude that L be the point of intersection of L and S Therefore they intersect. Let . Then d ( S, P ) = d M S, Q ) because S ∈ L and d ( S, Q ) = d ( S, R ) since ( S M . This means the distance to all three points is the same, so a circle ∈ S centered at of that radius goes through all three points. The circle is unique because any other circle would also have to be centered at a point equidistant from all three of P, Q, R , and S is the only such point because M can only intersect in one point. L and 2 2 +2 y − 1 = 0 3.4.3 x The line and y x from the equations found by eliminating the of the circles should have some geometric meaning. What is it? Anything satisfying the equations of both circles satisfies the sum and difference of the equations as well. (Because if A = B C = D , then and A + C = B + D and A − C = B − D ; this is actually one of Euclid’s common notions.) The equation x y − 1 = 0 is obtained by subtracting the two + 2 equations of the circles, so it must be satisfied by any point on both circles. The circles intersect in two points (as shown in the previous exercises), so the equation must be the line determined by (that is, going through) those two points of intersection. 1 Reflect the plane in the x -axis, and then in the line = 3.6.2 y . Show that the 2 ( x, y ) to ( x, y + 1) , so it is the translation t . resulting isometry sends , 1 0 x f which sends ( x, y ) The reflection of the plane in the -axis is the flip 1 is the flip } to ( L = { y = x, − y ). The reflection of the plane in the line 2 f ). To check that this is correct, note that which sends ( x, y ) to ( x, 1 − y L 1 = this map fixes every point on the line y and sends (0 0) to (0 , 1), which , 2 L agrees with what should happen when flipping over the line . To check an isometry, it suffices to check three non-collinear points, so that shows that this is the right map f . L Putting them together, we get f , ◦ f ( x, y ) = f ) ( x, − y ) = ( x, 1 − ( − y )) = ( x, 1 + y L L as desired. This map doesn’t fix any points, so it’s not a rotation, and it maps every line of the form x = k (that is, every vertical line) to itself, so it must be a translation. It sends (0 , 0) to (0 , 1), so it is the translation by the vector v = (0 , 1). 1

2 2 3.6.7 Deduce from Exercise 3.6.6 that any combination of three reflections is a glide reflection. (The previous exercise showed that any combination of three reflections, ′′ ′ ′ ′′ ′ ′′ ′ f f f ◦ f ◦ , can be rewritten as f ◦ ◦ f , where M M ‖ L .) and N ⊥ L L M N M N This one was done in lecture. You showed on homework that reflecting in two parallel lines one after the other produces a translation in a direction perpendicular to the lines. ′ ′ ′′ So f ◦ f N is a translation in the direction . But then the combined L M ′ (recall that the one effect of the three reflections is to first flip over N on the right is done first, since this is composition of functions) and then translate in the direction of that line. This is precisely what it means to be a glide reflection: flip and then slide in the direction of the line of reflection. 3.7.4 State a simple test for telling whether f is a translation, rotation, or glide f ( A ) , f ( B reflection from the positions of , and f ( C ) . ) In this case, the three points are not on a line, so the positions A, B, C f determine what isometry it is. A simple test: draw of their images under , f ( , f ( B ) ) ( C ). Check if it has the same the triangle whose vertices are A f by checking if going orientation as the original triangle with vertices A, B, C A to B to C is clockwise or counterclockwise in the first triangle, and from checking similarly for the image triangle. If the orientations are different, it must be a glide reflection (translations and rotations preserve orientation). A Now draw vectors pointing from f ( A ), from B to f ( B ), and from C to to ( C ). If those are all the same vector (same magnitude, same direction), f f then is a translation (by that vector!). If not, it’s a rotation, since these are the only three possibilities. State and prove a vector Desargues theorem. 4.2.2 (Recall that the Desargues theorem says that if two triangles are “in per- spective” from a point — with their vertices on three distinct lines through the point — then if two corresponding sides are parallel, the third sides are too. Look at the diagram on page 13 for a picture of this configuration.) Assume without loss of generality that the perspective point is the ori- gin (because translating a point to the origin does not affect whether the relevant lines are parallel). Suppose the first triangle has points , s , t and the second has points r = , w . Then being in perspective means that u = a r , v , b s , and w = c t . u v r − s = d We’re supposing that u − v ) (one set of lines is parallel) and ( s t = e ( v − w ) (the next set is parallel). We need to show that the last − r − t is a multiple of u − w . set of lines is parallel, namely that Adding the two equations from the previous paragraph, we get ( ? ) r − t = d u − d v + e v − e w . Now let’s turn our attention to and c . We have a, b, r − s = d u − d v = da r − db s , s − t = e v − e w = eb s − ec t . Rearranging, . − da ) r + ( db − 1) s = 0 (1 (1 − eb ) s + ( ec − 1) t = 0 ,

3 3 But s , t are linearly independent since they all point in different directions r , 1 1 = = from the origin. We can conclude that and b = c = a b , which e d d b = c and means = e . But now, plugging back in to equation ( ? ), we = a r − t = d u − d v + d v − d w = d ( u − w ), as desired. finally have 2 Give an example of two rotations of whose product is a translation. R 7.5.1 How can you tell if an isometry is a translation? Well, an isometry is determined by what it does to a triangle. A translation must preserve slopes of lines, so the image of a triangle has its new sides parallel to the original sides of the triangle. (Note a rotation does not do this, and a reflection/glide reflection does not do this. Only a translation does.) about one point and then So we want to do rotation by some angle α about a different point. (Notice: if you do two rotation by some angle β rotations about the same point, their composition is still a rotation!) The first rotation changes the slopes of the sides of the triangle, so we need the second rotation to change them back by what they used to be! So an example of two rotations whose composition is a translation: First do α , and then do rotation about (1 , 0) by − α . rotation about the origin by α π/ 2, for instance, then the composition sends (0 , 0) to (1 , 1), so it is = If v = (1 , 1). translation by Show that the distance from the center to any vertex of a four-dimensional 7.7.2 cube is equal to the length of its side. 1 1 1 1 The vertices of the hypercube are ± . The sidelengths ± ± i ± k j 2 2 2 2 are 1 because if you change any one of the signs, you get a point one unit away: ) ) ( (( ) √ 1 1 1 1 1 1 1 1 2 2 2 2 , , , , , , . + 0 d + 0 , , + 0 = = 1 1 − 2 2 2 2 2 2 2 2 In the previous problem, you showed they were all distance 1 from the origin by calculating the distance: √ √ 1 1 1 1 2 2 2 2 + (0 / + (0 ± 1 / 2) ± + (0 ± 1 / 2) 1 = 2) / 2) + 1 + ± + (0 = 1 . 4 4 4 4 So for the unit hypercube centered at the origin, the sidelengths are equal to the distance of each vertex to the center. Now you can get any hypercube from this one by a combination of translations and dilations, but neither of those change the fact that sidelength is equal to distance of each vertex to the center!

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