Suggested problems - solutions

Transcription

1 Suggested problems - solutions Examples and models Material for this section references College Geometry: A Discovery Approach, 2/e, David C. Kay, Addison Wesley, In particular, see section 2.2, pp The problems are all from section 2.2. Problems: 1, 2, 4, 5, 6, 7, 8, 9 (tied to example 2) #1, #2 See the linked pages for these and similar puzzles. One thing that can suggest where the problem is is to calculate the areas of the individual pieces. Solution: Look at #1. For a triangle, A = 2 1bh, and for a trapezoid, A = 2 1 (B + b)h. Now, the triangles in the first picture appear to have base b = 8 and height h = 3. Total area is 2( 1 2 )(8)(3) = 24 The trapezoids have long base B = 5, short base b = 3, and height h = 5. Total area is 2( 2 1 )(5 + 3)(5) = 40. And the total total area in that picture (assuming that those are triangles and trapezoids) is 64, not 65. A contradiction. Since our only assumption is that those are triangles and trapezoids... that must mean that they aren t. Look very carefully at the slopes of the lines. Here s an enlarged picture of the top triangle: It s a bit more obvious at this magnification.

2 #4 Show that this model violates the axioms in Example 5: Members: Committees: John, Dave, Robert, Mary, Kathy, Jane Executive Committee: Mary, Robert Steering Committee: John, Robert Nomination Committee: Jane, Kathy, Dave Solution: This model violates the second axiom, that every member is on exactly one committee. Here, Robert is on two. #5 Find a third distinctively different model. Solution: You can t, and still satisfy all three of the axioms in example 5. That s what categorical means. If you ve got something significantly different... it s violating an axiom somewhere. #6 Find two different models for the following set of axioms: Axiom 1: Axiom 2: Axiom 3: Every line is a set of at least two points. Each two lines intersect in a unique point. There are precisely three lines. Solution: The main thing to note when doing this is that although there s a specified number of lines, there s no restriction on how many points are in the model, or the maximum number of points that can be on a line. So Model 1: Points: S = {1, 2, 3, 4} Lines: l 1 = {1, 2}, l 2 = {2, 3}, l 3 = {1, 3, 4} Model 2: Points: S = {1, 2, 3, 4, 5, 6} Lines: l 1 = {1, 2, 5}, l 2 = {2, 3, 6}, l 3 = {1, 3, 4}

3 #7 Show that the diagram constitutes a model for the Euclidean Plane for the axioms given in Example 1, Section 2.1. Notice that two of the points aren t marked (at intersections) and should be. Axiom 1: Each line is a set of four points. Axiom 2: Each point is contained by precisely two lines. Axiom 3: Two distinct lines which intersect do so in exactly one point. Solution: You can tell at a glance at the diagram that all the axioms are satisfied - which is the great thing about diagrams if they re accurate. To show it rigorously, you need to do some labeling and listing: Points: S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Lines: l 1 = {1, 6, 4, 10}, l 2 = {1, 7, 8, 3}, l 3 = {5, 6, 7, 2}, l 4 = {5, 10, 9, 3}, l 5 = {2, 8, 9, 4} Note: sets are unordered, and the ordering of the points in each line implies nothing about the ordering of the points in space, or which points are between which. (They happen to match the lines because it s easier to match them to the diagram that way, but l 1, for example, could just as well have been written as {1,4, 6,10}. Axiom 1 is immediately observed to be true: each line is a set of four points. To verify Axiom 2, list points and the lines that contain them: Point Contained by Point Contained by 1 l 1, l 2 6 l 1, l 3 2 l 3, l 5 7 l 2, l 3 3 l 2, l 4 8 l 2, l 5 4 l 1, l 5 9 l 4, l 5 5 l 3, l 4 10 l 1, l 4 Axiom 3 is verified by listing all (10) possible intersections of lines... which conveniently works out to be the same data as above in a different form: l 1 l 2 = {1} l 2 l 4 = {3} l 1 l 3 = {6} l 2 l 5 = {8} l 1 l 4 = {10} l 3 l 4 = {5} l 1 l 5 = {4} l 3 l 5 = {2} l 2 l 3 = {7} l 4 l 5 = {9} The model cannot be categorical because we already know there is a distinctly different model which satisfies the axioms - the model given in Example 1. There is no way we can have an isomorphism (bijection) between the 10 point model and an infinite point model.

4 #8 Show the sum of the measures of the angles of any convex quadrilateral is 360. Solution: This isn t the best of problems, because you don t really have a precise definition of diagonal yet. Or between, for that matter. We re used to working with convex quadrilaterals anyway, where the diagonals are the things that go through the middle from vertex to opposite vertex, but the whole point here is to contrast this with the non-convex quadrilateral in Example 1, Figure 2.6. So let s get a decent working definition of diagonal: in a quadrilateral ABCD, where the notation indicates that the vertices proceed in a clockwise order from A to B to C to D, the diagonals are the segments AC and BD. Non-convex quadrilateral. Convex quadrilateral. AC between AB and AD allows you to assert that m BAD = m BAC + m DAC and similarly that m BCD = m BCA + m DCA BD between BA and BC allows you to assert that m ABC = m ABD + m CBD and similarly that m ADC = m ADB + m CDB Diagonals divide figure into triangles. One of the diagonals does not lie between adjacent sides. Because of this, we cannot assert all the angle sum relationships that we can in the convex case. In this particular picture, for example. m ABC m ABD + m CBD We need the idea of betweenness for line segments to guarantee the vertex angles are cut by the diagonals. Still no formal definition of betweenness, but the intuitive one displayed in these diagrams is essentially correct. Diagonals do not divide figure into triangles. Oh, and the actual proof? That is now simply the proof given in Example 1, p 62, so I won t repeat it here. Kay s comment is that the proof does not use the fact that the figure is a parallelogram - that s half right. The proof does however implicitly use the fact that the figure is a convex quadrilateral, the instant it asserts the figure is divided into two triangles. The flaw is that it doesn t say it explicitly. That would be the only adjustment needed - assert that your convex quadrilateral has (by the definition of convex quadrilateral) the property that those angle sums hold up, and proceed with the proof.

5 And with that statement explicitly there, we know it can t be generalized to the non-convex case.

6 #9 If you do all the calculations correctly and draw exactly according to scale, you should get this: I ve also gone and calculated the lengths of the sides. Now, go look at example 2 in this section - see the difference in the diagram? OC is to the right of OB, and no matter how far you pull C closer to D, it s still not going to switch to the other side. There s a little bit of an optical illusion going on in example 2 - you can bet that those right angles aren t right. Notice that the proof is fine up to the congruence OAE = OBC (they really are congruent, by construction, and the measurements bear that out). It s after that - the claims that the angle sums hold when they don t. So we can safely say that not all obtuse angles are right angles after all. Which is a relief. Now, for the calculations that went into that diagram: A at (0, 0) and E at (0, 10) were labeled to give rectangle ABCD. To verify that BC = BD (which is clearly 10), the distance formula d = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 was applied: BC = (26 20) 2 + (8 0) 2 = = 10

7 The midpoint formula ( x1 + x 2 (x mid, y mid ) =, y ) 1 + y was used to verify that L was the midpoint of E(0, 10) and C(26, 8): Then, the slope formula ( L(x mid, y mid ) =, ) = (13, 9) 2 2 m = y 2 y 1 x 2 x 1 was applied to get the slope of EC (where E at (0, 10), C at (26, 8): m EC = = 1 13 The slope of a perpendicular to that line would be the negative reciprocal: m = = 13 So the equation of the perpendicular through L would be the equation of a line with slope = 13 through the point (13, 9). Use the point-slope formula: y y 1 = m(x x 1 ) y 9 = 13(x 13) y = 13x 160 The vertical line at point M simply has the equation x = 10 (recall x = x 1 : vertical, y = y 1 : horizontal). To solve y = 13x 160 and x = 10 simultaneously, all you need to do is evaluate: y = 13(10) 160 = 30 The point of intersection is (10, 30). Placing point O(10, 30) on the diagram gives an accurate picture. Isn t that a great problem? Everything you ever learned about coordinate geometry in Algebra II all rolled into one thing!