MATH 306/506
pring 2013

Fifteenth and Last Group Work (April 22, 2013): We already knew that rotation, translation and reflection across a line are isometries, so I asked you to work in groups and look at compositions of two of those and see what isometry were you getting (see Section 6.1 In search of new isometries). On April 24 we discussed the outcome of your work.
  • Table 1: composition of two translations: first in the direction AB, second direction BC, is a third translation in the direction of AC. (The order doesn't matter here, outcome is the same.)
  • Table 2: composition of two rotations with the same center O and angles alpha and beta is a third rotation with center O and angle alpha+beta. (The order doesn't matter here, outcome is the same.)
  • Table 3: composition (M_m o M_l) of two reflections along parallel lines l and m is a translation along a direction AB where AB is perpendicular to both l and m, and AB is twice the length between l and m, and I forgot to mention the orientation, is the same as going from l to m. (Order matters, if we do (M_l o M_m) the segment is the same but orientation is reversed.) (see pages 293-295)
  • Table 4: composition (M_m o M_l) of two reflections along transversal lines l and m that meet at point O and with angle alpha from l to m is a rotation centered at O with twice the angle alpha. (Order matters, if we do (M_l o M_m) the rotation is about O but with angle twice the angle between m and l which is 180-alpha). (see pages 289-293).
  • Table 5: composition of a reflection across a line l first then a translation by AB parallel to l. This cannot be rotation or translation, because orientation is reversed (by the reflection). It is what is called a GLIDE-REFLECTION (see page 247). Order doesn't matter, and that was for table 6 to discover.
  • Table 6: composition of a translation by AB followed by reflection accross a line l parallel to AB is the same glide reflection found by Table 5.

    Fourteenth Group Work (April 17, 2013): We defined transformations in the plane (bijections in the plane or one-to-one and onto functions on the plane, they are always invertible). We defined geometrically translations by a directed segment AB, rotations about a point O by angle theta (counterclockwise) in between -180 and 180 and reflection across a line l (See Section 5.1 for the definitions). We described the inverse transformation for each one of them. We then defined when a transformation in the plane is an isometry (page 240).
  • Table 1 and 4: Show that translation by a directed segment AB is an isometry.
  • Tables 2 and 5: Show that rotation about a point O by angle theta is an isometry.
  • Tables 3 and 6: Show that reflection across a line l is an isometry.

    Thirteenth Group Work (April 15, 2013): Use geometric arguments to deduce trigonometric formulas. See Section 4.7 (Please I could not remember what I had assigned to tables 2, 3 and 4... I know the topics, can't remember the order... please confirm.)

  • Table 6: formula for the sine of the sum of two angles.
  • Table 5: formula for the cosine of the sum of two angles.
  • Table 4: law of the sines ? (see pages 222-223 and also Now Solve This 4.21)
  • Table 3: extended law of the sines? (see page 224, consider Now Solve This 4.21 for bonus points)
  • Table 2: law of the cosines? (see pages 224-225, consider Now Solve This 4.22 for bonus points).
  • Table 1: any of the above! Consider Item 5 (Brahmagupta's Theorem) in Now Solve This 4.22 for bonus points).

    Twelveth (spelling?) Group Work (April 10, 2013): Use similarity of triangles to show the following:

  • Table 1: If two chords AB and CD in a circle intersect at a point P inside the circle then show that AP.PB = CP.PD
  • Table 2: If two chords AB and CD in a circle intersect at a point P outside the circle then show that AP.PB = CP.PD
  • Table 3: Show that if two triangles are similar then their areas are proportional to r^2 where r is the proportion of their sides.
  • Table 4: Same as table 3 but for a "convex" quadrilateral. How about a convex pentagon, or n-agon?
  • Table 5: Let ABC be a right triangle at C. Let CD be the height dropped from C, and h=CD. Let a=AD and b=DB. Show that h= square root (ab).

    Assume a and b are positive numbers: The square root of ab is called the "geometric mean" of a and b. The "arithmetic mean" of a and b is (a+b)/2 The "harmonic mean" of a and b is the reciprocal of the arithmetic mean of the reciprocals of a and b.

  • Table 6: Show that the harmonic mean of a and b is less than or equal to the geometric mean of a and b, and the geometric mean of a and b is less than or equal to the arithmetic mean of a and b. (See now solve this 4.8 for an illuminating picture). You can prove these inequalities algebraically, but the geometric argument is very pretty.

    Eleventh Group Work (April 1, 2013): We took an axiomatic approach to area (see pages 118-119):

  • Axiom 1: each polygon can be assigned a unique positive number, its area.
  • Axiom 2: area of points and lines is zero.
  • Axiom 3: congruent polygons have equal area.
  • Axiom 4: Area is additive.
  • Axiom 5: Area of a square with sidelength a is a^2.
    With those 5 axioms each table was tasked to show:
  • Table 6: Area of a triangle with base b and height h is: bh/2
  • Table 5: Area of a rectangle with sides a and b is ab.
  • Table 4: Area of a parallelogram with base b and height h is bh.
  • Table 3: Area of a trapezoid with bases a and b and height h is h(a+b)/2.
  • Table 2: find area of a regular hexagon with side length a in terms of a.
  • Table 1: find area of a kite in terms of the length of its diagonals.
    Tables 5 and 6 discovered that it was simpler to derive formula for the rectangle from the axioms (Theorem 3.1 in the book, see picture 3.7). Then use that to get the area of a right triangle (half a rectangle), and use those to calculate the area of a general triangle (Theorem 3.3).
    One could also calculate the area of a parallelogram (Theorem 3.2) and then use that to find the area of a general triangle as half of the area of an appropriate parallelogram.
    For the Trapezoid we dropped some heigth and broke into a rectangle and two right triangles, and carefully calculated the total area in terms of the bases an the height (Theorem 3.4). We also broke into a parallelogram and a triangle and calculated those areas in terms of bases and height.
    For the regular hexagon, we observed its inscribed in a circle, and the 6 triangles created by joining the center to the vertices are all equilateral with side a. The area then is 6 times the area of the equilateral triangle. But we wanted the area of the triangle only in terms of the sidelength. It was necessary to calculate the height. For that Pythagorean theorem came handy... We observed that with the area axioms the proof that Kendall presented at the beginning of the semester worked (Proof 1 in the book page 131).
    Finally for the kite we know the diagonals are perpendicular to each other, can calculate the area of the two isosceles triangles in terms of those diagonal lengths, after a little algebra conclude that the area of a kite is the product of the diagonal's lengths divided by 2.

    Tenth Group Work (March 6, 2013): We worked on problems 25, 31 and 32 in Problem set 1.3 (p. 61-62). I want each one of you to write carefully the solutions of the three problems discussed today.

  • All tables worked first on problem 25. The conjecture was that the quadrilateral created was a square. Some tables showed first that pairs of opposite lines were parallel (hence the quadrilateral is a parallelogram), second they showed all four sides were congruent (hence this is a rhombus, in fact had you noticed this first you would conclude from properties of rhombus that it was a parallelogram), third you showed there was one right angle and hence all were right angles. It was very useful to show that the four rectangles in the picture are congruent right triangles, with hypothenuse the side of the big square. Let the lebghts of the legs be a and b (a Tables moved to problem 31, here you can reduce your problem to showing that one little angle equals x. To show that Table 5 drew some additional squares below the three given ones and two additional congruent lines that should help you get what we wanted...
  • Tables had not much time to think about problem 32. However we observed that if P is any of the vertices of the triangle, then x+y+z=h where h is the altitude of the given equilateral triangle. Same is true if P is the point of intersection of the altitudes=medians (here you must use that medians bisect each other on a 2:1 ratio). We then argued that if P is on the sides of the triangle, by drawing lines through P parallel to the other two sides, and noticing that two smaller equilateral triangles were constructed whose heights are the unknowns and the sum is still h. Finally for a generic point P inside the triangle, we drew through P three lines parallel to the sides of the triangles, noticed that these time three smaller equilateral triangles were created whose heights are the unknowns and add up to h. So for all points P x+y+z is equal to the height of the triangle, therefore the minimum value is....

    Ninth Group Work (March 4, 2013): The last 17 minutes we worked in groups on Now Solve this 1.13 (A proof that your conjecture on the Treasure Island Problem was correct). You are asked to use Figure 1.61 in page 56 to get the proof. In that picture M is the midpoint of the segment S_2S_1. I think all the tables understood that triangles T_1AS_1 and GT_1C are congruent and so are triangles S_2BT_2 and T_2CG... with that info we could show that N bisects T_2T_1. It remains to be shown that MN is congruent to T_2N (or to NT_1). I want this problem carefully written by Monday after Spring Break, I said one paper per table, but I actually want each one you to write down the whole argument.

    Eigth Group Work (Feb 25, 2013): From the book: Now Do This 1.7(page 36) Tables were tasked to write up for Wednesday the results of their work (please include the names of all the team-mates).

  • Tables 2 and 6 were in charge of part 1 (Proof of Theorem 1.17: given two lines and a transversal, if a pair of corresponding angles are congruent then the given lines are parallel). (I know Table 2 was checking the Exterior Angle Theorem, please include it in your write up).
  • Tables 4 and 5 were in charge of part 2 assuming known Part 1 (given a point P and a line l construct a line m parallel to P). After the explanation today about rhombus in part 3, please try to write up part 4 (construction without using lines perpendicular to a given line through a point).
  • Tables 1 and 3 where in charge of part 3: show that a rhombus (a quadrilateral with all for sides congruent) has opposite sides parallel. Show that the diagonals of the rhombus are perpendicular and they bisect each other. Also show that opposite angles are congruent. Can use Part 1, but cannot use that the sum of the angles of a triangle is 180.

    Seventh Group Work (Feb 18, 2013): Discuss your proofs for the Converse to Pons Asinorum (this was assigned as a homework, collected after the discussion). All together we had 5 different arguments:

  • 1) ultimate proof using ASA to conclude that triangles ABC and CBA were congruent. (here the congruent angles were at A and C)
  • 2) and 3) Assumed the opposite sides to the congruent angles were NOT congruent, hence one side was shorter than the other and we could fit it in the longer one (rulers axiom) introducing an auxiliary point D. 3) Linh's proof used SAS to conclude the larger triangle was congruent to the smaller one, but then we ended up concluding the measure of an angle was less than the measure of the angle, contradiction. 2) My argument used pons asinorum and the exterior angle theorem, to conclude the same... again contradiction. Therefore the opposite sides to the congruent angles must be congruent.
  • 4) Table 2 had observed that it sufficed to show that the third vertex must lie on the perpendicular bisector of the segment determined by the two congruent angles. If we assume is not, then again using Pons Asinorum and SAS we reached the same contradiction as in 2) and 3).
  • 5) Finally Dana's proof in table 5, drew the angle bisector to the third vertex (B), and used AAS to conclude the two triangles created were congruent, this implied BA congruent to BC. I noted that AAS is true but to prove it we need the fact that the sum of the angles of a triangle is 180, and that we will deduce from the Parallel Postulate. So is best to avoid using it for this homework. I also noted that we had not yet proven ASA. But we proved it today, so any argument based in ASA is now cleared.

    Sixth Group Work (Feb 13, 2013): We play a little bit with propositional calculus. Each table has to verify whether two given statements are equivalent or not, by calculating their truth tables and comparing them.

  • Table 1: p implies q, q implies p.
  • Table 2: p implies q, no q implies no p.
  • Table 3: p implies q, no p implies no q.
  • Table 4: p implies q, (p and no q) implies no p.
  • Table 5: p and q, (no p) or (no q)
  • Table 6: p or q, (no p) and (no q)
    We used this as a springboard to talk about converse of an implication, contrapositive, argument by contradiction, and de Morgan's Laws in logic.

    Fifth Group Work (Feb 11, 2013): From Problem Set 1 in the book:

  • Three Tables worked on Exercise 2: an argument is presented showing that the sum of the angles of a triangle is 180, the groups have to critizise the argument and discover its flaws.
  • Three Tables worked on Exercise 3: the book presents a proof of an statement, and the exercise presents a second argument (Jaimee's argument). The groups have to decide whether Jaimee was right or not, and highlight what they like about her proof, and what they don't (two tables were tasked to defend Jaimee the other to attack her work).

    Fourth Group Work (Jan 30, 2013): Think about how to define:

  • Table 1: right triangle.
  • Table 2: quadrilateral.
  • Table 3: square/rectangle/parallelogram.
  • Table 4: trapezoid.
  • Table 5: rhombus.
  • Table 6: kite.
    Here we realized the need for the parallelogram postulate to construct a parallelogram.

    Third Group Work (Jan 28, 2013): Define using the undefined objects (points. lines) and the incidence axioms, and what tables before yours have defined. For example, to define a triangle, Table 3 can use the definitions of segment and ray provided by Tables 1 and 2.

  • Table 1: segment.
  • Table 2: ray.
  • Table 3: angle.
  • Table 4: triangle.
  • Table 5: circle.
  • Table 6: interior of an angle.
    We realized the need to introduce other postulates such as Ruler Postulate to define segment, and the Plane-separation axiom to define the interior of an angle.

    Second Group Work (Jan 23, 2013): Each table (numbered from 1 to 6) will create an ornament S with N wires (where N is the Table number) and beads constructed using the following rules or axioms:

  • A1: every pair of wires has exactly one bead in common.
  • A2: Every bead is on exactly two wires.
  • A3: There are exactly N wires.
    I asked you to focus on two questions: How many beads are there in S? How many beads on each wire?

    First Group Work (week of Jan 15): Each group (defined by the people sitting in one round table) will read and discuss one of the following problems described in Chapter Zero then a spokesperson will describe the problem to the class and any additional insight the group gained on the problem (a solution or proof in general or for some particular configuration)

  • Table 1: 0.2 The Nine-Point Circle (pages 2-3)
  • Table 2: 0.3 Morley's Theorem (page 3)
  • Table 3: 0.4 The Hiker's Path (page 4)
  • Table 4: 0.5 The Shortest Highway (page 4)
  • Table 5: 0.6 Steiner's Minimun Distance Problem (pages 4-5)
  • Table 6: 0.7 The Pythagorean Theorem (pages 5-6)

    Return to: Department of Mathematics and Statistics, University of New Mexico

    Last updated: April 24th, 2013