SOLUTIONS TO EXERCISES FORMATHEMATICS 133 — Part 3Fall 2006NOTE ON ILLUSTRATIONS. Drawings for several of the solutions in this file are availablein the file (s)http://math.ucr.edu/∼res/math133/math133solutions3figures.∗where the extension ∗ is one of doc, ps, or pdf.II . Linear algebra and Euclidean geometryII.1 : Measurement axioms1. First of all, f is 1–1. Define kXso that X = A + kX(B − A). Then f(X) = f(Y )implies kXd(A, B) = kYd(A, B), and since d(A, B) is positive this means kX= kY. Also, if r isan arbitrary real number and k = r/d(A, B), then it follows that f maps X = A + k(B − A) to r.Therefore f is onto. Finally, to verify the statement on distances, note that the distance from Xto Y is equal to|X − Y | =[A + kX(B − A)] − [A + kY(B − A)]=[kX(B − A)] − [kY(B − A)]=(kX− kY)(B − A)= | (kX− kY) | · | (B − A) | = | (kX− kY) | · d(A, B) =| (kXd(A, B) − kY)d(A, B) | = | f(X) − f(Y ) |which is the identity to be shown.2. Follow the hint, and let h = gof−1. By construction, h is a 1–1 onto map from the realnumbers to themselves such that |u − v| = |h(u) − h(v)| for all u and v. If k(t) = h(t) − h(0), thenelementary algebra shows that we also have |u − v| = |k(u) − k(v)| but also k(0) = 0. Thereforewe have |k(t)| = |k(t) − k(0)| = |t − 0| = |t| for all t. In particular, this means that for each t wehave k(t) = εt· t, where εt= ±1. We claim that εtis the same for all t 6= 0; for t = 0 the value ofε does not matter. But suppose that we had k(u) = u and k(v) = −v, where u, v 6= 0. Then wecould not have |u − v| = |k(u) − k(v)|; If u and v have the same sign, then the right hand side isgreater than the left, and if they have opposite signs, then the right hand side is less than the left(why?). Therefore k(t) = ε · t where ε = ±1, and hence also h(t) = k(t) + h(0) = ±t + h(0), whichis the form required in the exercise.3. Write things out using barycentric coordinates. We have X − A = (2, −6), whileB − A = (−5, −9) and C − A = (4, −9). Thus we need to solve(2, −6) = y(−5, −9) + z(4, −9) = (−5y + 4z, −9y − 9z)and if we do so we obtain z =1281and y =681; using x + y + z = 1 we also obtain x =79. Thus allthree barycentric coordinates are positive and the point lie in the interior.1To work the second part, we have X − A = (10, k − 10), and we need to consider the system(10, k − 10) = y(−5, −9) + z(4, −9) = (−5y + 4z, −9y − 9z)and determine those values of k for which z > 0 and x = 1 − y − z > 0. Solving for the barycentriccoordinates, we findz =150 − 5k81, y =50 − 4k81, x =9k − 10981.The point will lie in the interior if and only if x and z are positive, which is the same as sayingthat the numerators 150 − 5k and 9k − 109 should both be positive. This happens if and only if1219< k < 30.4. In this case we need to work the first part of the problem when X is either (30, 200) or(75, 135), so that X − A is either (23, 190) or (68, 125). The barycentric coordinate z in both casesare negative and thererore neither point lies in the interior of the angle.5. We now have X − A = (−1, 0), while B − A = (3, −6) and C − A = (−1, 20). Thus weneed to solve(−1, 0) = y(3, −6) + z(−1, −20) = (3y − z, −6y − 20z)and if we do so we obtain z = −19so that the point cannot lie in the interior of the angle.To work the second part, we have X − A = (21, k − 8), and we need to consider the system(21, k − 8) = (3y − z, −6y − 20z)and determine those values of k for which z > 0 and x = 1 − y − z > 0. Solving for the barycentriccoordinates, we findz =3k + 11854, y =k + 41254, x =50k − 47654.The point will lie in the interior if and only if x and z are positive, which is the same as saying thatthe numerators for x and z should both be positive. The inequality 50k − 476 > 0 implies k > 0and hence we see that z is positive whenever x is positive, so that the condition for the point to liein the interior of the angle is simply k > 238/25.6. The original file had a misprint; the open segment should be (AC). Suppose thatX ∈ (AC), so that A ∗ X ∗ C is true. By theorems on plane separation, this implies that A and Xlie on the same side of BC, and C and X lie on the same side of AB. But these are the two criteriafor a point to lie in the interior of6ABC, and therefore we know that X lies in the interior of thisangle.7. It looks as if the open ray (DE meets the triangle in exactly one point. See theillustration in the file of figures.8. The segment (AC) should be corrected to (AX). With this correction, proceed asfollows: If Y lies on (AX), then A ∗ Y ∗ X is true, so that X and Y lie on the same sides of ABand AC. However, B ∗ X ∗ C implies that X and B lie on the same side of AC and X and C lieon the same side of AB. Therefore we also know that Y and B lie on the same side of AC and Yand C lie on the same side of AB. All that remains is to show that Y and A lie on the same sideof BC. But this follows because A ∗ Y ∗ X and X ∈ BC.29. By the Protractor Postulat there is a point E on the side of BC opposite A such that|6EBC| = |6ABC|, and by a consequence of the Ruler Postulate there is a point D ∈ (AE suchthat d(D, B) = d(A, B). By construction the distance equation holds, and the angle measurementequation holds because [AD = [AE. Finally D is on the side of BC opposite A because D ∈ (BE,while E and A lie on opposite sides and all points of (BE lie on a single side of BC.10. The interior of the triangle is contained in the interior of6BAC, so by the CrossbarTheorem we know that (AD meets (BC) in some point E. It will suffice to show that we have theorder relationship A ∗D ∗E (take A = X and E = Y …