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)
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Last updated: April 24th, 2013