MATH 243 Winter 2008 Geometry II: Transformation Geometry Solutions to Problem Set 1 Completion Date: Monday January 21, 2008

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1 MTH 4 Winter 008 Geometry II: Transformation Geometry Solutions to Problem Set 1 ompletion Date: Monday January 1, 008 Department of Mathematical Statistical Sciences University of lberta Question 1. Let u v be nonzero vectors, both parallel to a line l. (a Show that u + v is parallel to l. (b Show that ku is parallel to l for each k R, k 0. Solution: (a Let be an arbitrary point in the Euclidean point space E, apply the vector u in the translation space V of E to the point to get the point such that = u. Next apply the vector v to the point to get the point such that = v, since u v are parallel to l, then the directed line segments (, (, are also parallel to l, therefore have the same support line, thus, the points,, are collinear. From the definition of vectors in V, this means that u + v = is parallel to l also. (b ssume that k > 0, let an arbitrary point in the Euclidean point space E, apply the vector u in the translation space V of E to the point to get the point such that = u. Let be the point on the support line of the directed line segment (, such that = k. From the definition of scalar multiplication In V, we have = k u, the directed line segment (, is parallel to l, that is, k u = is parallel to l also. Question. Let u v be nonzero vectors, both parallel to a plane Π. (a Show that u + v is parallel to Π. (b Show that ku is parallel to Π for each k R, k 0. Solution: (a Let ( e 1, e be a planar direction for the plane Π, then we can write for some scalars a, b, c, d, so that u = a e 1 + b e v = c e 1 + d e u + v is also parallel to the plane Π. u + v = (a + c e 1 + (b + d e, (b Similarly, if k 0, then k u = ka e 1 + kb e, so that k u is parallel to the plane Π.

2 Question. Given an arbitrary point O, let,,, be the midpoints of the sides,, of, show that O + O + O = O + O + O. Solution: In the figure O we have similarly therefore O = 1 O + = 1 O + ( 1 O O = ( O + O, O = 1 ( O + O O = 1 ( O + O, O + O + O = O + O + O. Question 4. Given, let G G be their centroids, respectively. Show that GG = 1 ( + +. Solution: If we introduce the point O E, then GG = OG 1 ( OG = O + O + O 1 = 1 ( O O + 1 ( O O + 1 ( O + O + O ( O O = 1 ( + +. Question 5. Show using vectors that the segment joining the midpoints of the nonparallel sides of a trapezoid is parallel to either base congruent to half the sum of the bases. Solution: In the figure E F D we have EF = E + + F EF = ED + D + F so that EF = + D, since E + ED = 0 F + F = 0, therefore 1 ( EF = + D.

3 Question 6. If the point P lies in the plane of, but is distinct from the vertices of the triangle, the parallelograms P, P, P are completed, show that the segments [ ], [ ], [ ] bisect each other. Solution: In the figure we have completed the parallelograms P P. P Let Q R be the midpoints of the segements, respectively (not shown. then P Q = P + 1 = P + 1 ( P P = 1 P + 1 ( P + = 1 P + 1 ( P + P, P Q = 1 ( P + P + P. lso, = 1 ( P + P P = 1 P + 1 ( P + P R = 1 P + = 1 P + 1 ( P + P, P R = 1 ( P + P + P. Therefore Q = R, the line segments bisect each other. Similarly, the line segments bisect each other.

4 Question 7. Suppose that the points,, lie on the sides [], [], [] of, respectively, = =. (a Show that the centroids of coincide. (b If the parallelograms are completed, show that are parallel to the median of through. Solution: In the figure let,, be points on the sides [], [], [] of, respectively, such that where k > 0 (if k = 0, then the triangles coincide. = = = k, (a We know from Question 4 that if G G are the centroids of, respectively, then GG = 1 ( + +, in order to show that the centroids coincide, we only need to show that Note that since + + = 0. = k, then = k, so that = + = k + 1 k = + = + lso, note that since = k, then = k, so that = + = k + 1 k = + = + k. (1 k + 1 k. ( k + 1

5 Finally, note that since = k, then = k, so that = + = k + 1 k dding (1, (, (, we have + + = since + + = 0. = + = + k. ( k + 1 ( k ( + + = 0 k + 1 (b In the figure below we have completed the parallelograms, let M be the midpoint of the side. We want to show that are both parallel to M. M Let t =, so that = t, then = + = (, that is, = 1, = = lso, t = =, so that = t, = + = ( so that Therefore, Subtracting (4 from (5 we have = 1. t. (4 = + = + = 1 +. (5 = = 1 ( + = M, is parallel to the median through. Similarly, is parallel to the median through.

6 Question 8. If 1,,, 4, 5, 6 are the midpoints of consecutive sides of a hexagon, show that have the same centroid. Solution: Given the hexagon DEF in the figure, let 1 = midpoint of, = midpoint of, = midpoint of D, 4 = midpoint of DE, 5 = midpoint of EF, 6 = midpoint of F, let O be an arbitrary point in the Euclidean point space E. 1 6 F 5 O E 4 D Let G odd be the centroid of 1 5 G even be the centroid of 4 6, then OG odd = 1 ( O1 + O + O 5 ( = 1 ( O + O + 1 ( O + OD + 1 OE + OF ( = 1 ( O + OF + 1 ( O + O + 1 OD + OE = 1 ( O6 + O + O 4 = OGeven. Question 9. If P, Q, R, S are the midpoints of the edges,, D, D, respectively, of a tetrahedron D, show that P R QS bisect each other at the centroid of the tetrahedron. Solution: In the figure, let O be an arbitrary point in the Euclidean point space E, let G be the centroid of the tetrahedron. P S O Q R D

7 From the definition of the centroid of the tetrahedron, we have OG = 1 ( O + O + O + OD 4 ( = 1 ( O + O + 1 O + OD = 1 G is the midpoint of the segment P R. ( OP + OR Similarly, OG = 1 ( O + O + O + OD 4 ( = 1 ( O + O + 1 O + OD = 1 ( OQ + OS G is the midpoint of the segment QS. Question 10. Let D D be two parallelograms, not necessarily coplanar, show that the midpoints I, J, K, L, of [ ], [ ], [ ], [DD ], respectively, are the vertices of a parallelogram. Solution: In the figure, D D are parallelograms, O is an arbitrary point in the Euclidean point space E. I D L D J K O Note first that = D, = D, D =, D =. Now IJ = OJ OI = 1 ( O + O 1 ( O + O = 1 ( O O + 1 ( O O = 1 ( +.

8 lso, LK = OK OL = 1 ( O + O 1 ( OD + OD = 1 ( O OD + 1 ( O OD = 1 ( D + D. Since = D = D, then IJ = LK, so that IJ is parallel to LK IJ = LK. Similarly, IL = OL OI = 1 ( OD + OD 1 ( O + O = 1 ( OD O + 1 ( OD O = 1 ( D + D, JK = OK OJ = 1 ( O + O 1 ( O + O = 1 ( O O + 1 ( O O = 1 ( +. Since D = D =, then IL = JK, so that IL is parallel to JK IL = JK. Therefore, IJKL is a parallelogram. Note that the above did not depend on the way the figure was drawn. The figure was drawn to make the process more transparent. Question 11. Show that the angle bisectors of can be used to construct a new triangle if only if is equilateral. Solution: learly, if is equilateral, then the angle bisectors are the perpendicular bisectors of the sides, all have the same length. Therefore the angle bisectors can be used to construct a new triangle, in fact, an equilateral triangle. onversely, let the angle bisectors of the angles at,, hit the opposite sides at the points,,, respectively, as in the figure, let a =, b =, c =.

9 Since the internal angle bisectors divide the opposite side (internally in the ratio of the adjacent sides, then we have = = c b, = = c a, = = b a, if the angle bisectors,, can be used to form a new triangle, then we must have + + = 0. Since then since = + = + = + = + = + = + c b + c a a + c b, a + b c + a + b = 0, b + c a + c a + b + + = 0, ( this implies that Subtracting c ( from (, we have b + c ( a a + c c ( + b + c c + a + b = 0. ( b + c a + c a + b b a + b since the set of vectors {, } is linearly independent, then a a + c c b + c = 0 b a + b that is, c = ab b = ac, so that b c = ac ab = a(b c. c = 0, b + c c b + c = 0, If b c, this implies that b + c = a, that is, a + b + c = 0, which is impossible. Therefore b = c, then ab = c = b implies a = b, so that is equilateral.

10 Question 1. Let X 1, X,..., X n be n points on a circle, let G be their centroid. Denote by Y 1, Y,..., Y n the second points of intersection of the lines X 1 G, X G,..., X n G with the circle, respectively. (a Show that X 1 G GY 1 + X G GY + + X ng GY n = n. (b Show that the set of points P inside the circle that satisfy X 1 P P Y 1 + X P P Y + + X np P Y n = n is the circle with diameter OG, where O is the center of the circle. Solution: (a Let r be the radius of the circle let O be its center. From the Power of a Point, we know that if P is any point inside, then for any two chords XY intersecting at the point P, we have P P = XP P Y. X P Y O In the figure we apply this result to a chord XY the diameter that both pass through P, then XP P Y = (r + OP (r OP = r OP, therefore X k P P Y k = X k P 1 = X k P P Y k r OP Since each of the points X k are on the circle, then we have X k P = X k P = OP OX k X k P. = OP OP OX k + OX k = OP + r OP OX k, where OP OX k is the Euclidean inner product of the two vectors OP OX k. Therefore, X k P = n(op + r n OP OX k, by definition of the centroid G, we have OX k = n OG, (

11 so that ( X k P = n OP + r OP OG, ( Setting P = G, we have so that X k G = n(r OG, X k G 1 = GY k r OG X k G = n(r OG r OG = n. (b The argument above which lead to ( is valid for any point P inside the circle, clearly the equation X 1 P + X P + + X np = n P Y 1 P Y P Y n is equivalent to which is equivalent to OP = OP OG, ( OP P G = OP OG OP = OP OG OP = 0, that is, that the vectors OP P G are perpendicular, this condition defines the circle with diameter OG.