Dear Students, Here is a collection of miscellaneous advice I printed up for previous editions of this course. It is lightly edited to fit our current needs. I hope it is useful to help you organize the course for the upcoming midterm, and beyond. 1. The Five Threads. Introduced in Week1 and reviewed F8. We have often referred, informally, to Euclid's geometry. We have studied Cartesian geometry in its contemporary form (vector geometry) in Chapters 1 and 3, and a small part of Chapter 2 of Tondeur's text. Cartesian geometry originally married algebra and geometry and thus enabled the discovery and development of the Calculus and all mathematics since. But vector algebra perfects Descarte's vision, and Todeur's book is an introduction of it. Our study of the role of perspective drawing in projective geometry is now complete, except of some amplification and elaboration along the way. We have not properly begun the fourth and fifth threads as yet. These will be covered after the midterm (weeks 10-15). 2. Drawing Equipment. You will need a proper compass (the kind you stick a pencil into is _NOT_ acceptable) and a 30-60-90 transparent plastic triangle. Pencils and erasors help a lot. 3. In Chapter 1, we, once again, skipped the topics involving complex numbers. But there were additions and amplifications you have in your notes. In particular the "teeter-totter" principle and barycentric coordinates. 4. Challenge problem (suitable for a project.) What is the 3-D analog to Ceva's Theorem? Note, partial solutions to challenge problems may be submitted at any time for inspection and comment. 5. Homework is posted on the calendar (on our website.) 6. Note that on homework as on tests, a minimal amount of "documentation" of your work is necessary. It must be possible to tell what your are doing. The reader is not someone that has access to your class notes nor to the book. I may subtract 1/2 point if a problem is not properly documented. That means, each problem should be stated. You may use shorthand notation, like "Let (ABC)" for "Let A,B,C be three distinct collinear points" but you must not leave out logical connectives such as "if" , "then", "and", etc. Think of writing an essay in complete sentence, someone else will read sympathetically, but who is under no obligation to guess what is going on in your mind. I will also subtract points for sloppy, hard to read, crowded writing, and poorly drawn figures. 7. Notation I have introduced which not only saves time but reveals logical symmetries. Assertions like "m=(ABC)" and "N=(hkl)" do NOT imply that point B lies between A and C on m, or that the cyclic order of the lines through N is hkl. But they DO imply that the three points are collinear etc, else their common line could not acquire a name. Indeed, I introduced yet another notation, (pQ), to mean that "line p goes through point Q" and "point Q lies on line p". With that in mind, notice that A(BC), B(CA), C(AB), and (ABC) all say the same thing. 8. The Teeter-Totter Principle as formulas. For (PQR) we have three equivalent expressions: pP+rR | Q-P r | for 3 nonzero numbers x,y,z Q = ----- | ---= - - | p+r | Q-R p | xP+yQ+zR=0 and x+y+z = 0 You should be able to derive each from either of the two others. Note that the left expression names the barycentric coordinates for Q relative to P and R. The middle expression is a ratio, and the right one is a criterion for collinearity. Consider just how many ratios there can be. Cycling on PQR produces 2 more. Then, there are reciprocals and negatives of each. In total, we deal with 12 ratios. The only relation between them that isn't obvious (negative, reciprocal) is Q-P P-Q P-Q P-R dz dz dy - --- = --- = --- --- -- = -- -- Q-R Q-R P-R Q-R dx dy dx which looks like it's true "by cancellation" but, just as for the chain-rule (on the right), there's something to be proved. 9. Finally, for both Ceva and Menelaus we have the same ingredients, a triangle, three additional point on each of its sides, and an expression consisting of the product of three ratios. The difference is that Ceva identifies the concurrency of the 3 cevians by the expression having the value -1, and Menelaus identifies the collinearity of the extra points by the expression being +1. Always read a triangle in cyclic (counterlockwise) order, and placing the extra point on each line in the middle thus (AC'B),(BA'C),(CA'B) Ceva: (AA')(BB')(CC') iff x= -1 C'-A A'-B B'-C Menelaus: (A'B'C') iff x= +1 x= ---- ---- ---- C'-B A'-C B'-A 10. Problem 22 has a double Menelaus figure, with two of the four possible ratios indicated. Note that the numbers on the segments are weights for the ratio, not lengths. To determine the other two ratios plug the known ratios into the Menelaus equation and solve. 11. Problem 23 is much harder that 24. But solving it, using the hint if you like, prepares you to tackle 24. Here is another hint: In the Ceva figure there are, for each cevian, two triangles which the cevian crosses (not at vertices). For instance, the cevian AA' crosses triangle BCB' thus (BA'C),(CAB'),(B'GB). Write out the Menelaus equation for this figure. Its first ratio is one you want to see again in the Ceva expression. The second one has the right letters (no G's) but in the wrong order. The third ratio has two G's in it. Note that its reciprocal comes from (BGB') which appears in one of the two possible Menelaus figures for which the cevian (CC') is the transversal. If you pick the right one, (consider the orientation) we get another Menelaus equation. Since both expressions equal 1, their product still equals 1 and terms begin to cancel (don't forget the relation in 2. above), leaving you with the Ceva equation. Consider the above paragraph as a recipe in an mathematical cookbook. Just follow it and the cake will come out of the oven. 12. Now you need only one hint for 24, namely the Menelaus figure (AC"B)(BB'D)(DA'A). Cycling, multiplying and applying the converse of Menelaus will yield the desired (A"B"C"). Try it! 13. Recall that Desargues' Theorem had three aspects not fully brought out by the text. For one, it is "obvious" from the viewpoint of perspective drawing. For another, it remains true when various points, even lines, become ideal. For a third, the statement has a logical symmetry, namely that if you exchange the words points and lines the theorem remains true, in the extended (aka Projective) plane. This is know as the Principle of Duality for Projective Geometry. 14. Pappus' Theorem. This concludes the formal parts of Chapter 1. There remains Pappus' theorem. The significance of Pappus' theorem, and its modern form, known as Pascal's theorem, is worthy of a separate chapter. Therefore we have not spent much time on Pappus' in class. Suffice it that you know the statement, also for cases some of its ideal points (parallels) become real points (concurrent lines). 15. Kepler's projective plane. Define the totality of lines parallel to a given line as an ideal point, or a point at infinity. In 3-space, think of it as a star, since for all practical purposes, it's infinitely far away, and all lines (in our world) pointing to it are parallel for all practical purposes. The totality of all infinite points of a plane is its line at infinity. So, if X is an ideal point, and Y is a real point, the XY is the line in the collection of parallel lines called X, which also goes through Y. Exercise: Check whether of not Desargues hold when the hypothesis, D=(AA')(BB')(CC'), has parallels in it. Case 1. D is ideal Case 2. A is ideal Case 3. AA' is ideal Can you think of other cases that aren't just relabeled, like the case that B' is ideal. 16. Some Questions and Answers. > On the recipe for creating a cube I have a few questions. On number 6, >it says to draw a corner at P, and a wall on PQ. Visually, a corner is where three edges meet. So, to draw a corner means to have three rays coming out of the same point in the picture plane. >Does drawing a corner at P, just mean a line perpendicular to the segment PQ? Yes/no. The real edges of the cube are mutually perpendicular. In the perspective drawing, we have two meanings for "perpendicular". There is "literally" perpendicular in the drawing, and there is the reality of the line being perpendicular to a plane, which you are depicting. Since the first meaning is used ONLY in the constructions, we should give it special term, say ppp for "picture-plane perpendicular". Then, when a recipe says to draw a "picture of a line through F perpendicular to a plane" you connect F to the zenith of the plane. There is no reason why this line should be ppp to the horizon line of that plane (generally, it won't be.) >Does drawing a wall on PQ, just mean to draw a line between PQ? No, a wall needs to suggest a region. Real walls don't look like lines except in very special views (edge-on). In general, they look like quadrilaterals, even when they are pictures of rectangles. > Then on number 8 it says to complete the cube. I'm not sure how to do this? >I have 3 vanishing points and D1 and D2 and the segment PQ (all of >the ingredients), but I'm not sure how to connect them all to get a cube. You have more. By step eight you have at least two diagnonal points on the two of the three horizons. You need those to measure off the walls to make the square instead of rectangular. In the Euclidean plane (reality!), given a rectangular sector (line with two perpendiculars rays on the same side). If you draw a 45 degree line (the diagonal direction) from either of the two corners, it will cut the opposite edge at a third corner of a square. By drawing one parallel to the base you finish the square. Do this into the third dimension, and you complete a cube. In perspective, we know how to draw parallels (lines ending at the same vanishing point). Once we can draw the diagonal direction from a corner, we get squares. Doing it on just two adjacent faces of the cube, we can complete the cube by just drawing lines. Once you have two square sides of the cube (in perspective, of course) which share an edge, you have all the information needed to complete the cube. Draw (perspective) parallels from each the four free corners of the the figure, and the edge opposite the common one, and you're done. 17. Some General Advice. For most of your experience with math in recent years you were taught pretty much by example. The instructor did a bunch of problems, and then you copied the method to do more problem of the same kind. It is possible to teach perspective drawing that way, but that wouldn't be mathematics. Math 303 is a senior level math course, and although the there examples, there are also principles from which you can derive the methods yourself. In the present section, these happen to be constructions of figures that mean something different that what they look like. The way to study these "recipes" is to first try them out in simple, Euclidean contexts to see what they mean, such as how to diagonalize a square. Then you translate this recipe into the perspective context. It will work, because you know that the perspective drawing is a true picture of reality. If two lines meett in a certain point in reality, their pictures must meet in the picture of the point. 18. Started lectures on Chapter 3. In particular why (G) The medians of a triangle are concurrent at the centroid. (D) The side bisectors are concurrent at the circumcenter. (H) The altitudes are concurrent at the orthocenter. any two of these three theorems implies the third. Background material (not in the book): Court's Proof of (G)and (D) implies (H). This also proves that the centroid, circumcenter and orthocenter are collinear on Euler's Line. Presented Tondeur's second proof (p.66) of this theorem 19. We need some properties (definition) of dilatations from Ch2 to the depth needed in this chapter, in particular, for Tondeur's first proof of Theorem (H), for the Euler-Line theorem, and for Feuerbach's nine-point circle construction. (incomplete, refer to your class notes for more.