Math 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 4 Solutions

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1 Math 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 04 Homework 4 Solutions Exercises from 1.7 of the notes: 7.1, 7.6, 7.7, 7.10, 7.14, 7.16, 7.17 Exercise 7.1 (Isometries form a group). Show that the identity map id : X X is always an isometry, that the composition of two isometries is an isometry, and that the inverse of an isometry is an isometry. Solution. As d(x, y) = d(id(x), id(y)) for any x, y X, the bijection id is an isometry. If f : X X and g : X X are isometries, then we have d(x, y) = d(g(x), g(y)) = d(f(g(x)), f(g(y))) = d(f g(x), f g(y)). As f g is a bijection if f and g are, the map f g is an isometry. Finally, for any x, y X we have f(f 1 (x)) = x and f(f 1 (y)) = y, so that d(f 1 (x), f 1 (y)) = d(f(f 1 (x)), f(f 1 (y))) = d(x, y). Thus the bijection f 1 is also an isometry. Exercise 7.6. Given f : X X, let Fix(f) = {x X f(x) = x}. Show that Fix(f g f 1 ) = f(fix(g)), whenever f, g are both bijections X X. Solution. Suppose that x Fix(f g f 1 ), so that x = f g f 1 (x). Applying f 1 to both sides of this equality, we find f 1 (x) = g(f 1 (x)). Thus f 1 (x) Fix(g), and applying f to the set Fix(g) we find that x Fix(g). Thus Fix(f g f 1 ) Fix(g). On the other hand, if x f(fix(g)) then x = f(y) for some y Fix(g). In that case, y = f 1 (x) and g(y) = y, so that g(f 1 (x)) = f 1 (x). Applying f to both sides of this equation, we find that x = f g f 1 (x) so that x Fix(f g f 1 ). Thus Fix(g) Fix(f g f 1 ). Exercise 7.7 (Isometries send lines to lines). Let x, y R n. By the triangle inequality, a point z lies on the line through x and y if and 1

2 only if either d(x, y) + d(y, z) = d(x, z), d(x, z) + d(z, y) = d(x, y), or d(z, x) + d(x, y) = d(z, y). Use this to show that if if l is a line in R n and f : R n R n is an isometry, then f(l) is also a line. Solution. If the point z is on the line through x and y, either z is between x and y (in which case d(x, z)+d(z, y) = d(x, y)), x is between z and y (in which case d(z, x)+d(x, y) = d(z, y)), or y is between z and x (in which case d(x, y) + d(y, z) = d(x, z)). On the other hand, if one of these equalities holds, then by Corollary 1.6 (the triangle inequality) x, y, and z must be collinear. Suppose now that f : R n R n is an isometry and l is a line in R n, and pick a pair of points x, y on l. We claim that f(l) is the unique line through f(x) and f(y). Choose a third point f(z) f(l), so that z l. One of the three equalities in the statement of the problem is satisfied by x, y and z, say without loss of generality Since f is an isometry, we have d(x, y) + d(y, z) = d(x, z). d(f(x), f(y)) + d(f(y), f(z)) = d(f(x), f(z)), so that the work above implies that f(x), f(y), and f(z) are collinear. In particular, each point of f(l) is contained in the line through f(x) and f(y). Call this line L. Evidently, we have l = {x + t(y x) t R} and L = {f(x) + t(f(y) f(x)) t R}. Suppose that the point x + t(y x) l is sent by f to the point f(x) + s(f(y) f(x)). In this case, we show below that s must be equal to t: Since f is an isometry, we have d(x + t(y x), x) = d(f(x) + s(f(y) f(x)), f(x)), which implies that t d(x, y) = s d(f(x), f(y)). We have d(x, y) = d(f(x), f(y)), so that s = t. Similarly, we have d(x + t(y x), y) = d(f(x) + s(f(y) f(x)), f(x)), which implies that t 1 d(x, y) = s 1 d(f(x), f(y)). Once again, f is an isometry so that d(x, y) = d(f(x), f(y)) and t 1 = s 1. Suppose that s = t. Then t 1 = s 1 = ( t) 1 = t + 1.

3 We have either t 1 = t + 1 (an impossibility) or t 1 = t 1, in which case t = 0. Then s = 0 as well. Thus s = t. Thus we have shown that f(x + t(y x)) = f(x) + t(f(y) f(x)), and the image f(l) is the entire line L. 3 Exercise Show that reflections are isometries. Solution. As on p. 3 of the notes (or discussed in class), the reflection R l through the line l = {p + tv t R} can be described by R l (x) = T p R {tv t R} T p (x). This implies that R l is given by the composition of translations and the reflection through a line through the origin, so in order to check that R l is an isometry, by Exercise 7.1 above it suffices to check that the reflection through a line through the origin is an isometry. As discussed in class, when l passes through the origin it can be written as {tv t R} for a non-zero vector v. Now Rl is given by For x, y R, we have R l (x) = proj v (x) x. d(r l (x), R l (y)) = proj v (x) x (proj v (y) y) = (proj v (x) proj v (y)) x + y = proj v (x y) (x y) = d(r l (x y), 0). In order to see that d(r l (x), R l (y)) = d(x, y), by the above it suffices to see that d(r l (x y), 0) = d(x y, 0). But this can be seen independent of x and y: For any p R, by Exercise 1.4 0, p, and proj v (p) form a right triangle with base lengths proj v (p) and proj v (p) p. But then 0, p, and R l (p) form a right triangle with the same nonhypotenuse lengths (since l is the perpendicular bisector of the segment p R l (p)), so that the hypotenuses of these triangles are equal, i.e. d(r l (p), 0) = d(p, 0). Exercise Show that if l and l intersect at a point p, the composition R l R l is the rotation O p,θ, where θ is twice the angle from l to l.

4 4 Solution. By Corollary 7.1, if a pair of isometries agree on three distinct points, then they are equal. Evidently, we have R l R l (p) = p = O p,θ (p). Choose a point x l. We claim that R l R l and O p,θ agree at x. Indeed, R l (x) = x, so that R l R l (x) = R l (x). The triangles (p, proj l (x), x) and (p, proj l (x), R l (x)) are right triangles with equal length legs, so they are congruent. We conclude that the angles x p proj l (x) and R l (x) p proj l (x) are congruent, and the points x and R l (x) are equidistant from p. This implies that R l (x) is O p,θ, where θ is twice the measure of the angle x p proj l (x). The angle x p proj l (x) is the angle from l to l, so we are done. Exercise Show that any composition T v R l is a glide reflection, unless v is perpendicular to l, in which case the composition is a reflection. Solution. Let l = {p + tw t R}. By the formula at the top of p. 3, we have R l (x) = x proj w (x p), so that T v R l (x) = v + x proj w (x p). Let l = {p + 1v + tw t R} and let v = proj w (v). Then we have: (x p 1 ) v T v R l (x) = v + x proj w ( 1 v ) = proj w (v) + x proj w (x p) proj w = proj w (v) + proj w (v) + x proj w (x p) = v + x proj w (x p) = T v R l (x). Note that in the first line we ve used the description of the composition T v R l obtained above, in the second and third lines we ve used the linearity of the projection map proj w, and in the fourth line we ve used the fact that v = proj w (v) + proj w (v) for any v, w R. This last fact can be checked easily with a bit of linear algebra: The linear map proj w + proj w : R R takes w to w and w to w. The vectors {w, w } form a basis for R, and a linear map that fixes basis elements is the identity map. Note that if the translation vector v = proj w (v) is nonzero then it is parallel to w, which is the direction of the line l. Thus T v R l is a

5 5 glide reflection unless v = 0. In this last case, T v so that T v R l = R l, a reflection. is the identity map, Exercise If p / l, show that O p,θ R l is a glide reflection. Solution. By Exercise 7.14, the rotation O p,θ can be written as the composition R l1 R l, where l 1 and l are any pair of lines intersecting at p, and so that the angle from l to l 1 is θ/. We make the choice of the pair l 1 and l (intersecting at p with angle θ/) so that l is perpendicular to l, say at the intersection point p. Note that O p,θ R l = R l1 R l R l. Because l and l intersect orthogonally at p, by Exercise 7.14 the composition R l R l is the rotation O p,π. Using Exercise 7.14 once more, we can write O p,π as the composition of reflections R l R l, for any pair of lines l and l intersecting orthogonally at p. We make such a choice for l and l so that l is orthogonal to l 1. Note that we now have O p,θ R l = R l1 R l R l, where l 1 and l are both orthogonal to l. Since l 1 and l have a common orthogonal, they are parallel. This implies that R l1 R l is a translation in the direction commonly perpendicular to l 1 and l, i.e. in a direction parallel to l. Thus R l1 R l R l is a glide reflection.

6 6 Problem A6. Given three points p, q, and r in R n, the barycenter of p, q, and r (or, sometimes, the barycenter of the triangle (p, q, r) formed by p, q and r) is the point given by the convex combination p 3 + q 3 + r 3. Consider the points p = (0, 0), q = (a, 0), r = (b, c) in R, and let x denote the barycenter of p, q and r. See the figure below. r = (b, c) y x p = (0, 0) q = (a, 0) (a) Find the coordinates of x. (b) Find the area of the triangle (p, q, x). (c) Find the area of (p, r, x). Hint: Use the labelled point y, the projection of x to the side of (p, q, r) between p and r. (d) Prove that, for any triangle (p, q, r), the barycenter of (p, q, r) divides it into three triangles, each of one-third the area of (p, q, r). Hint: You may assume that isometries of R take barycenters to barycenters and preserve area. Solution (a). We have x = p 3 + q 3 + r 3 = (0, 0) + (a/3, 0) + (b/3, c/3) = 1 (a + b, c). 3 Solution (b). When the side pq is chosen as the base of the triangle (p, q, x), the altitude is given by the distance from x to the x-axis, i.e. the absolute value of its y-coordinate. This implies that we have Area( (p, q, x)) = 1 a c = ac 3 6.

7 7 Solution (c). When the side pr is chosen as the base of the triangle (p, r, x), the altitude is given by the distance from x to the line through p and r. By Exercise 1.4, this distance is given by the length of the vector x proj r p (x). We have x projr p (x) = x projr (x) = x x r r r r 1 = 1 3 (a + b, c) (a + b, c) (b, c) 3 (b, c) (b, c) (b, c) = 1 3 (a + b, c) ab + b + c (b, c) b + c = 1 ( ) ab 3 (a + b, c) b + c + 1 (b, c) = 1 3 (a ab b + c, abc ) c + c = a 3 (1 b b + c, bc ) b + c = a ( c 3 b + c, bc ) b + c ac = 3(b + c ) (c, b) = ac 3 b + c. Now the area of (p, r, x) is given by Area( (p, r, x)) = 1 b + c ac 3 b + c = ac 6. Solution (d). Given any triangle (p, q, r), we may apply the isometry T p translating p to the origin, and follow that by a rotation taking the point T p (q) to a point on the x-axis. The result is a triangle of the form given above; one vertex is at the origin and a second vertex is on the x-axis. Because isometries take barycenters to barycenters and preserve area, if we know that the three triangles (x, q, r), (p, x, r) and (p, q, x) have equal area in the setting given in the statement of Problem A6, we will know that these triangles have the same area for any triangle (p, q, r). Now parts (b) and (c) show that

8 8 Area( (p, q, x)) = Area( (p, x, r)). It is easy to see that Area( (p, q, r)) = 1 ac a c =. Because x is inside (p, q, r) (as the triangle is convex and x is a convex combination of the vertices) we have Area( (x, q, r)) = Area( (p, q, r)) Area( (p, q, x)) Area( (p, x, r)) Thus we have as desired. = ac ac 6 ac 6 = ac 6. Area( (p, q, x)) = Area( (p, x, r)) = Area( (x, q, r))