INSTRUCTIONAL STRATEGIES SERIES: Teaching for Meaning - More Than Just A Geometry/Algebra Problem

HAPPY THANKSGIVING!

 

Alright, you're teaching about the rule for slopes of perpendicular lines in Algebra or Geometry. 

Here are some of the instructional strategies or approaches you may have used...

(1) State the theorem without explanation followed by 3-4 demo examples of how it's used
(2) Motivate the theorem using the lines y = (3/4)x and y = (-4/3)x, choosing the points (4,3) and (-3,4) to demonstrate why these lines are perpendicular
(3) A more abstract approach using the following diagram

NOTE: Q(-b,a) is the point on line M in quadrant II. The label is too far from the dot!

FROM THE GIVEN INFORMATION IN THE DIAGRAM PROVE THAT ∠QOP IS A RIGHT ANGLE, THAT IS, LINES L AND M ARE PERPENDICULAR.

Comments

(a) If your group was advanced, would you omit the perpendiculars QR and PS?

(b) Would you draw the diagram to scale to prevent confusion for most students?

(c) Would you even consider Option (3) with a regular or weaker group of students? Would Option 2 be more than enough to get at the main idea?

(d) To more strongly suggest the use of slopes and/or similar triangles, would it be better to use the points (4,3) and (-6,8) on the lines? I personally would prefer this (and not give the equations of the lines). What do you imagine most students would do with this problem a few weeks (or even days!) later? Would they make the connection to slopes immediately if they had moved on to another unit or if this appeared on an assessment?

(e) Would some students need more than one example to suggest a generalization? Exactly what questions would you ask to promote a generalization?

(f) What have you done with this topic and/or how would you modify the above ideas??? The floor is open..

By the way, do you believe it is likely or unlikely that some version of this problem might appear on a standardized test like ADP's Algebra 2 End of Course Test or the SATs?

The Return of the WarmUp Challenges!

Just when you thought that MathNotations is on permanent hiatus or in hibernation, here are a couple of WarmUps/Problems of the Day/Test Prep/Challenges/// to consider for your students. 

Actually, I'm embarking on a new venture - an online tutoring website with live audio and video for OneOnOne math tutoring for Grades 6-14 (through Calculus II). In addition, I'm also working on setting up a small group (5-10 students) online SAT or ACT Course grouped by ability (a 600-800 SAT group, a 450-600 group, etc.).  If you're interested in getting more information about these before the official launch just contact me at dmarain at gmail dot com.


Update: Answers/comments are at the bottom...

1.   NOTE: ANGLE B IS A RIGHT ANGLE IN DIAGRAM BELOW - THANKS TO JONATHAN FOR CATCHING THAT OVERSIGHT!

2.   If 10^-1000 - 10^-997 is written as a decimal, answer the following:


(a) How many decimal places are there, i.e., how many digits to the right of the decimal point?
(b) One can show that the decimal digits end in a string of 9's. How many 9's?
(c) How many zeros are to the right of the decimal point and to the left of the string of 9's?

Notes:
(1) If we write the negative exponent expressions as rational numbers, this is perfectly appropriate for middle schoolers and, in fact, I think they need more of these experiences!
(2) The "Make It Simpler - Look for a Pattern" Strategy should be second nature to our youngsters, but when they see questions like these on the SATs, how many of our students really think of it!
(3) The fact that some calculators return a value of zero for the expression in the problem is a teachable moment - seize it!!
(4) See below for an algebraic approach.



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ANSWERS


1. 9√3


2. (a) 1000   (b) 3   (c) 997


An Algebraic Approach to #2:
First, students need to be familiar with the basic pattern:
10^-1 = 1/10 = .1 Note that there is one decimal digit.

10^-2 = 1/10² = 1/100 = .01  Note that there are two decimal places, etc.


10^-1000 - 10^-997 = 1/10¹⁰⁰⁰ - 1/10⁹⁹⁷
Using 10¹⁰⁰⁰ as the common denominator, we obtain
1/10¹⁰⁰⁰ - 10³/10¹⁰⁰⁰ =
-999/10¹⁰⁰⁰ from which the results follow (with some additional reasoning)...

Note: I could have worked directly with the exponent form by factoring out 10^-1000 but I chose rational form for the younger student.

THE OPEN-ENDED CONTEST PROBLEM AND SOLUTIONS

As promised, here is the open-ended, rubric-based, holistically scored, performance-assessed, student-constructed first problem from MathNotation's Third Contest:

1. A primitive Pythagorean triple is defined as an ordered triple of positive integers (a,b,c) in which a² + b² = c² and the greatest common factor (divisor) of a, b and c is 1. If (a,b,c) form such a triple, explain why c cannot be an even integer.

Comments
(a) The content here is number theory. Is some of this covered in your district's middle school curriculum or beyond? More importantly, at what point do students begin to formulate and write valid mathematical arguments?

(b) The immediate reaction of most students was that this seemed like a fairly simple problem. However, only a couple of teams scored any points. Perhaps the challenge here was the construction of a deductive argument, although as you will see below, there is one challenging part.

(c) There were two successful approaches used by the teams. Both involved indirect reasoning. Do your students begin to do these in middle school or are "proofs" first introduced in geometry?

(d) I allowed students to assume without proof the following:

(i) The general rules of parity of the sum of two integers
(ii) The square of a positive integer has the same parity as the integer

(e) Interestingly, none of the teams considered an algebraic approach to the one challenging case, i.e., demonstrating that the sum of the squares of two odd integers is not divisible by 4.

If a and b are odd, they can be represented as
a = 2m+1 and b = 2n+1, where m and n are integers.
Then a² + b² = (2m+1)² + (2n+1)² =
(4m² + 4m + 1) + (4n² + 4n + 1) =
4(m² + n²) + 4(m + n) + 2, which leaves a remainder of 2 when divided by 4.
BUT, if c is even, say c = 2k, then c² = 4k², which is divisible by 4.

(f) The two best solutions came from our first and second place teams, Chiles HS in FL and Hanover Park Middle School in CA. Both used the ideas of congruence modulo 4.

Here is the indirect method used by Chiles:

Let's assume that c can be an even integer. We'll prove by contradiction. An even integer can be summed in two ways:
1. with two even integers or
2. two odd integers
If it is the latter case, then looking at the residuals of modulo 4, the two odd integers summed will be equal to 2, but this is not the case as 2 is not a modulo of 4 residue. If it is the former case, then it does not satisfy the problem as then a, b, and c have common factor of 2. Therefore c must be an odd integer. Q.E.D.


Here is the indirect method used by Hanover Park:

Suppose, for the sake of contradiction, that there is a PPT (primitive Pythagorean Triple) s.t. c is even. Then c² ≡ 0 (mod 4).

We break this into cases based on the parity of a,b.

Case I: Both a and b are even; gcd(a,b,c) ≥ 2 because a,b,c are even, a contradiction.

Case 2: One of a and b is even. Then, a² + b² ≡ 0 + 1 ≡ 1
not ≡ 0 (mod 4), a contradiction.

Case 3: Both of a, b are odd. Then a² + b² ≡ 1 + 1 ≡ 2
not ≡ 0 (mod 4), a contradiction.
We have covered all cases for a, b with no valid cases. Thus, in a PPT, c cannot be even.

Both of these arguments represent a more sophisticated understanding of mathematics and the methods of proof. Clearly, these students are quite advanced and exceptional, however, I feel many middle school teachers begin early on to encourage their students to explain their thought processes both orally and in writing. Am I right? I would like to hear your thoughts on this...

RESULTS OF THIRD MATHNOTATIONS CONTEST and OTHER NEWS...

FINALLY -- THE RESULTS ARE IN!!

I apologize for the delay in getting these results out. The participating schools have all been notified.
NOTE: If any participating school did not receive an email from me, the advisor should email me. Also, if I misspelled anyone's name pls let me know and I'll correct it immediately!


INITIAL COMMENTS ON CONTEST, ETC...

• MEAN SCORE: 5.6 PTS OUT OF 12
• TOPICS INCLUDED Number Theory, Geometric Sequences, Function Notation, Geometry, Discrete Math, Quadratic Functions, and Absolute Value Inequalities (advanced level)
• Twenty schools registered from around the world, but only about half were able to actually give the contest.
  
• I will post the open-ended number theory problem later on but I didn't want to take away from recognizing the efforts of these outstanding students and their dedicated advisors.
• The next contest will be announced in a few weeks. Sign up early!
• After the 5th contest, you will be able to purchase all contests and solutions via download.
  

THIS WAS A CHALLENGING CONTEST, PARTICULARLY FOR YOUNGER STUDENTS, ALTHOUGH, AS YOU CAN SEE BELOW, THEY HELD THEIR OWN!! CONGRATULATIONS TO ALL PARTICIPANTS FOR A JOB WELL DONE!

FIRST PLACE - 12 OUT OF 12 POINTS!

CHILES HIGH SCHOOL
TALLAHASSEE, FL

Marshall Jiang - 11th
William Dunn - 12th
Wayne Zhao - 9th
Andrew Young - 11th
Jack Findley - 12th
Danielo Hoekman - 11th

Advisor, Steve Friedlander

SECOND PLACE - 11 OUT OF 12 PTS

HARVEST PARK MIDDLE SCHOOL
PLEASANTON, CA

Eugene Chen - 8th
Jerry Li - 8th
Brian Shimanuki - 8th
Christine Xu - 8th
Jeffrey Zhang - 8th
Ian Zhou - 8th

Advisor, Randall S. Lomas


THIRD PLACE - 9 OUT OF 12 PTS

CANADIAN ACADEMY - PINK PANDA TEAM
KOBE, JAPAN

Kevin Chen - 11th
Sean Qiao - 11th
Alice Fujita - 11th
Cathy Xu - 11th
Steven Jang - 11th
Sooyeon Chung - 10th

Advisor, Ms. Elizabeth Durkin


FOURTH PLACE - 7 OUT OF 12 PTS

CANADIAN ACADEMY - BLACK SWAN TEAM
KOBE, JAPAN

Hyun Song - 11th
Max Mottin - 11th
Ron Lee - 10th
Kyoko Yumura - 10th
Selim Lee - 10th
Evangel Jung - 10th

Advisor, Ms. Elizabeth Durkin


FIFTH PLACE - 4 OUT OF 12 POINTS

MEMORIAL MIDDLE SCHOOL - TEAM DAVID
FAIR LAWN, NJ

David Bates - 8th
Isaiah Chen - 8th
Kajan Jani - 8th
Thomas Koike - 8th
Priya Mehta - 8th
Joseph Nooger - 8th

Advisor, Ms. Karen Kasyan


SIXTH PLACE TIE

WALLINGTON JR/SR HS - SENIOR TEAM
WALLINGTON , NJ

Nicole Bacza - 12th
Tomasz Hajduk - 12th
Martyna Jezewska - 12th
Thomas Minieri - 12th
Urszula Nieznelska - 12th
Damian Niedzielski - 12th

Advisor, Stephanie Regetz


FAIR LAWN HS - TEAMS A & B
FAIR LAWN, NJ

Team A
Egor Buharin - 12th
Kelly Cunningham - 12th
Elizabeth Manzi - 12th
Gurteg Singh - 12th
Daniel Auld - 12th
Richard Gaugler - 12th

Team B
David Rosenfeld - 12th
Gil Rozensher - 12th
Roger Blumin - 9th
Mike Park - 9th
Jason Bandutia - 9th
Alexander Lankianov - 9th

Advisor, Victoria Velasco


SEVENTH PLACE TIE

WALLINGTON JR/SN HS
WALLINGTON, NJ

Junior Team
Konrad Plewa - 11th
Matthew Kmetz - 11th
Eman Elhadad - 11th
Patrick Sudol - 10th
Marek Kwasnica - 10th
Anna Jezewska - 10th

Advisor, Stephanie Regetz

MEMORIAL MIDDLE SCHOOL - TEAM SIMRAN
FAIR LAWN, NJ

Simran Arjani - 8th
Aramis Bermudez - 8th
Allan Chen - 8th
Kateryna Kaplun - 8th
Harsh Patel - 8th

Advisor, Ms. Karen Kasyan