NAME:Mathematics 133, Winter 2009, Examination 2Answer Key11. [15 points] Suppose that the points A, B and C have coordinates (1, 1), (0, 0)and (2, 1) respectively. Using barycentric coordinates, show that (4, 3) lies in the interiorof6ABC.SOLUTION.To find the barycentric coordinates of (4, 3) we first need to solve the equation(4, 3) − B = u(A − B) + w(C − B)for u and w. Since B = 0, this equation is just(4, 3) = u(1, 1) + w(2, 1)and if we solve for u and w we obtain w = 1 and u = 2. The third barycentric coordinate isgiven by v = 1 − u − w = 1 − 3 = −2, but we really do not need this because the conditionsu, w > 0 already imply that (4, 3) lies in the interior of6ABC.22. [20 points] Suppose that we are given6ABC and6DEF such that |6ABC| =|6DEF |, and let G ∈ Int6ABC and H ∈ Int6DEF be such that |6ABG| = |6DEH|.Prove that |6CBG| = |6F EH|.SOLUTION.The Additivity Postulate implies the equations|6ABC| = |6ABG| + |6CBG| , |6DEF | = |6DEH| + |6HEF |and we can rewrite these in the following forms:|6ABC| − |6ABG| = |6CBG| , |6DEF | − |6DEH| = |6HEF |By our assumptions the left hand sides of both equations are equal, and therefore the righthand sides must also be equal.33. [20 points] Suppose that we are given a triangle ∆ ABC and D ∈ (BC) such thatd(B, A) = d(B, D). If |6ABC| = 50◦and |6ACB| = 30◦, find |6ADC|. [Hint: Drawinga picture is probably worthwhile.]SOLUTION. By the Supplement Postulate we know that |6ADC|+|6ADB| = 180◦,so if we can find |6ADB then we can find |6ADC|. Since ∆ADB is isosceles with legs[BD] and [BA], it follows that180◦= |6ABC| + 2 |6ADB| = 50◦+ 2 |6ADB|so that |6ADB| = 65◦. If we substitute this into the original equation, we find that|6ADC| = 180◦− 65◦= 115◦.44. [15 points] Given an ordered list four coplanar points A, B, C, D such thatno three are collinear, state the conditions for these points (in the given order) to be thevertices of a convex quadrilateral.SOLUTION.If (A, B, C, D) is the ordered list of four points, then the conditions are that(i) A and B lie on the same side of CD,(ii) B and C lie on the same side of DA,(iii) C and D lie on the same side of AB,(iv) D and A lie on the same side of BC.55. [10 points] Let L be the line in R3which contains the points (1, 2, 3) and (4, 5, 6),and let M be the line through (2, 1, 3) which is parallel to L. Find a second point on M.SOLUTION.The line L may be written as (1, 2, 3) + W , where W is spanned by the difference(4, 5, 6) − (1, 2, 3) = (3, 3, 3). Therefore W is the set of all vectors of the form (t, t, t)for some real number t. It follows that the line M may be written as (2, 1, 3) + W , andconsequently the second point can be any point of the form (2+ t, 1 + t, 3 + t), where t 6= 0.66. [20 points] Suppose that we are given ∆ABC such that d(A, B) = 8 andd(B, C) = 17, and we know that d(A, C) is an integer which is a perfect square. Usethe Triangle Inequality to show that d(A, C) must be equal to 16.SOLUTION.The Triangle Inequality implies that17 = d(B, C) < d(B, A) + d(A, C) < 8 + d(A, C)so that d(A, C) > 9 and alsod(A, C) < d(A, B) + d(B, C) = 8 + 17 = 25so that d(A, C) < 25. Sine the only perfect square between 9 and 25 is 16, it follows thatd(A, C) =