Sunday, January 24, 2016

Division M2 session on Jan 24, 2016

Instructor:Mr. Ahmed Hefny
Attendance:Ayan, Adan, Omar, Baseer
Handout:Combinatorics Problems I
Problems Solved in Class:2,9 and 11
Concepts:Counting, Factorials, Permutations and Combinations
Student Difficulties:Although the terms factorials, and combinations seemed to ring a bell. There was difficulty solving more general counting problems. I believe the general material may be new to the students.
Homework:Attempt rest of problems in the handout, For questions that ask for probabilities, solve for "In how many ways could X happen" instead of "what is the probability of X happening".
Notes:Most students had difficulty with problem 1 in the last geometry homework.
Winner:
Main ideas:* The number of outcomes of a multistep decision sequence is the product of the number of possibilities for each decision.
Example 1:
Number of ways to order n elements = n * (n-1) * (n-2) * ... * 1 = n!

* If there are different alternatives to decide (part of) the outcome, the number of possibilities of these alternatives should be added.
* When confused, draw a tree of decisions and then go bottom-up: each leaf represents a single outcome. The number of outcomes from a on-leaf node is that sum of outcomes from its children. Usually, you won;t need to draw the tree down to the leaves.
* One strategy to find the number of possible values of part of the outcome you care about (e.g. 4 out of 10 numbers) is to use the following factorization:
number of possible full outcomes = number of possible values of the part we care about * number of ways to decide the values of the part we don't care about.
The quantity in red is the one we want to compute, the other two are usually easy to compute.

Example 2 - number of ways to select r elements from n elements where order matters (nPr):
number of ways to order n elements = nPr * number of ways to order the remaining n-r elements
--> n! = nPr * (n-r)! --> nPr = n! / (n-r)!

Example 2 - number of ways to select r elements from n elements where order does not matter (nCr):number of ways to select r elements from n elements where order matters = nCr * number of ways to order the selected r elements
--> nPr = nCr * r! --> nCr = nPr / r! = n!(n-r)! / r!

Division M1 session on Jan 24, 2016

Instructor:Mr. Zia Hydari
Attendance:AbdulRafay, Adeel, Bilal, Ramy, Sohaib, Areej, Abdul Qadir
Handout:Angles (definitions, review problems, challenge problems)
Problems Solved in Class:2.28, 2.29, 2.30, 2.31
Concepts:Axioms, Euclid's Axioms, Angles, Parallel Lines
Student Difficulties:Students who did not do homework had difficulties which slowed down the class as I had to review concepts.
Homework:Complete Khan Academy missions: LinesAngles, Shapes

Solve (or at least attempt) Problems 2.32 to 2.46 (p. 46–47) in the handout. Bonus points for finishing challenge problems (2.47–2.57). 
Notes:Most student did not complete the Khan Academy mission on Lines and Angles before class. Those who did, found the material easy.
Winner:Areej (highest class score for solving problems on the board)

Monday, January 18, 2016

Division M2 session on Jan 17, 2016

Instructor:Ahmed Hefny
Attendance:
Handout:Level 5 Geometry Problems (All homework except problem 5). Please download the handout by clicking here.
Problems Solved in Class:Problem 5
Concepts:Pythagorean theorem, congruent triangles, similar triangles
Student Difficulties:Square roots, solving geometry problems by constructing intermediate shapes.
Notes:Congruent Triangles
Two triangles are congruent if they have equal side lengths and angle measures. 
We do not need to test ALL lengths and measures; there are minimal tests for congruency.  

Usually the solution strategy goes as follows: 
  • There is missing piece of information about triangle A (e.g. missing side length) 
  • The information we have about A is sufficient to establish congruency with another triangle B using one of the minimal tests. 
  • Once congruency is established, we determine missing information about A from what we know about B. 

Minimal tests for congruency: 
  • Three sides 
  • Two sides and the angle between them 
  • Two angles and the side between them 
  • If the two triangles are right. Any two sides suffice (due to Pythagorean theorem) 

How do we know these tests are sufficient ? 
The information used in the tests uniquely identify all lengths and angles of a triangle. 

Example 1: 
Knowing the lengths of all sides uniquely determines the shape of a triangle. Figure 1 shows how to draw a triangle knowing the lengths of the three sides using a ruler and a compass. 

Example 2: 
Knowing the lengths of two sides and the measurement of an angle other than the one between them does not uniquely determine the shape of a triangle. 

Similar triangles: 
  • The ratios of side lengths are the same for all corresponding sides. 
  • Corresponding angles have equal measures. 
Each property implies the other one. 

If two shapes are similar with side length ratio r, the are ratio is r*r and the volume ratio is r*r*r. 

Important right triangles: 

Sunday, January 17, 2016

Division E Session on Jan 17, 2016

Instructor:Ms. Rudina Morina
Attendance:
Handout:A combination of more challenging problems not covered in the past weeks
Problems Solved in Class:5/7 problems given in the handout 
Concepts:The problems covered a number of topics, but the emphasis was on problem solving, especially considering the different approaches to arrive at the same solution. 
Student Difficulties:
Notes:Students were divided into groups of two and for each problem they were given ~5 min to work with their partners. Then a volunteer student solved the problem in the board followed by the teacher's explanations. This gave us a possibility to consider at least two different approaches to a problem: that of the student and the teacher. 
Assigned Work: None

Division M1 session on Jan 17, 2016

Instructor:Mr. Zia Hydari
Attendance: ARS, AM, AP, MS, AK, RM, SH
Handout:Angles (definitions, review problems, challenge problems
Problems Solved in Class:2.27, 2.47, and 2.48
Concepts:Axioms, Euclid's Axioms, Angles, Parallel Lines
Student Difficulties:The students found the Challenge Problems too difficult so we reverted to the Review Problems.
Homework: Read and understand the definitions on the handout.

If you do not understand the definitions in the handout, complete the following missions on Khan Academy: Lines and Angles.

Solve (or at least attempt) Problems 2.26 to 2.35 (p. 46) in the handout. We will give points to the students who can solve these problems on the board during the next class.

For Fun:
See video on Golden Ratio by clicking here
See video on number sets by clicking here
Notes:Student were not ready for challenge problems

Sunday, January 10, 2016

Division M session on Jan 10, 2016



Instructor:Mr. Ahmed Hefny
Attendance:11 Students
Handout:
Problems Solved in Class:
Concepts:
Student Difficulties:
Assigned Work: None.
Notes:
  1. Both Division M classes were combined into one.
  2. Meriem, Omar, and Ayan won prizes for their Geometry HW submission.


Whiteboard Images (please click on the image to enlarge):




















Division E session on Jan 10, 2016



Instructor: Mr. Zia Hydari
Attendance: 16 Students
Handout: pp. 33--34
Problems Solved in Class: All problems on page 33 only
Concepts: Averages
Student Difficulties:
Assigned Work: None.
Notes: Both Division E classes were combined into one.

Friday, January 8, 2016

Geometry Problem Set for Dec 20, 2015 and Jan 3, 2016

Download Geometry Problem Set from the following link:
Geometry Problem Set

Geometry Mission on Khan Academy
The Geometry Mission on Khan Academy is a systematic way to learn the background material for the Geometry Problem Set.

Especially Useful Content

Triangle similarity
AA similarity
Right triangles and trigonometry
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  1. This problem is harder than the other Level 2 problems.
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Geometry Level 3 Helpful Videos
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Geometry Level 4 Helpful Videos