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...

Another Sample Contest Problem - Counting...

There is still time to register for the upcoming MathNotations Third Online Math Team Contest, which should be administered on one of the days from Mon October 12th through Fri October 16th in a 45-minute time period.

Registration could not be easier this time around. Just email me at dmarain "at" "gamil dot com" and include your full name, title, name and full address of your school (indicate if Middle or Secondary School).

Be sure to include THIRD MATHNOTATIONS ONLINE CONTEST in the subject/title of the email. I will accept registrations up to Fri October 9th (exceptions can always be made!).

BASIC RULES
* Your school can field up to two teams with from two to six members on each. (A team of one requires special approval).
* Schools can be from anywhere on our planet and we encourage homeschooling teams as well.
* The contest includes topics from 2nd year algebra (including sequences, series), geometry, number theory and middle school math. I did not include any advanced math topics this time around, so don't worry about trig or logs.
* Questions may be multi-part and at least one is open-ended requiring careful justification (see example below).
* Few teams are expected to be able to finish all questions in the time allotted. Teams generally need to divide up the labor in order to have the best chance of completing the test.
* Calculators are permitted (no restrictions) but no computer mathematical software like Mathematica can be used.
* Computers can be used (no internet access) to type solutions in Microsoft Word. Answers and solutions can also be written by hand and scanned (preferred). A pdf file is also fine.


Ok, here's another sample contest problem, this time a "counting" question that is equally appropriate for middle schoolers and high schoolers:

How many 4-digit positive integers have distinct digits and the property that the product of their thousands' and hundreds' digits equals the product of their tens' and units' digits?

Comments
The math background here may be middle school but the reading comprehension level and specific knowledge of math terminology is quite high. This more than counting strategies is often an impediment. If this were an SAT-type question, an example would be given of such a number to give access to students who cannot decipher the problem, thereby testing the math more than the verbal side. On most contests, however, anything is fair game!

Beyond understanding what the question is asking, I believe there are some worthwhile counting strategies and combinatorial thinking involved here. Enjoy it!

Click More to see the result I came up with (although you may find an error and want to correct it!)




My Unofficial Answer: 40
(Please feel free to challenge that in your comments!!_

...Read more

A Practice PSAT/SAT Quiz with Strategies!!

UPDATE #2: Answers to the quiz are now provided at the bottom. If you disagree with any answers or would like clarification, don't hesitate to post a comment or send an email to dmarain "at gmail dot com".

UPDATE: No comments from my faithful readers yet -- I suspect they are giving students a chance to try these! I will post answers on Friday 9-25. However, students or any readers who would like to check their answers against mine need only email me at dmarain "at" gmail "dot" com and I will let them know how they did!


With the SAT/PSAT coming in a few weeks, I thought it would be helpful to your students to try a challenging "quiz". Most of these questions represent the high end level of difficulty and some are intentionally above the level of these tests. Then again, difficulty is very subjective. A student taking Honors Precalculus would have a very different perspective from the student starting Algebra 2!

Also, these questions can also be used to prepare for some math contests such as the THIRD MATHNOTATIONS FREE ONLINE MATH CONTEST! Yes, another shameless plug, but time is running out for your registration...

A Few Reminders For Students

(1) Do not worry about the time these take although I would suggest about 30 minutes. The idea is to try these, then correct mistakes and/or learn methods/strategies. It's what you do after this quiz that will be of most benefit!

(2) I added strategies and comments after the quiz. I suggest trying as many as you can without looking at these. Then go back, read the comments and re-try some. I will not provide answers yet!

(3) Don't forget these problems are copyrighted and cannot be reproduced for commercial use. See the Creative Commons License in the sidebar. Thank you...


PRACTICE PSAT/SAT QUIZ
1. If n is an even positive integer, how many digits of 100²ⁿ - 100^2n-2 will be equal to 9 when the expression is expanded?

(A) 2 (B) 4 (C) 8 (E) 2n (E) 2n - 4


2. The sides of a triangle have lengths a, b and c. Let S represent (a+b+c)/2. Which of the following could be true?

I. S is less than c
II. S > c
III. S = c

(A) I only (B) II only (C) I and II only (D) I and III only (E) I, II and III


3. The mean, median and mode of 3 numbers are x, x+1 and x+1 respectively. Which of the following represents the least of the 3 numbers?

(A) x (B) x - 1 (C) x - 2 (D) x-3 (E) 2x - 2


4. (10/√5)⁵⁰⁰ (1/(2√5))⁵⁰⁰ = _________


5. A point P(x,y) lies on the graph of the equation x²y² = 64. If x and y are both integers, how many such points are there?

(A) 4 (B) 8 (C) 16 (D) 32 (E 64


6. Each side of a parallelogram is increased by 50% while the shape is preserved. By what percent is the area of the parallelogram increased? __________


7.


AB is parallel to CD , AB = 3, CD = 5, AD = BC = 4. If segments AD and BC are extended to form a triangle ABE (not shown), what would be the length of AE?
Ans_________

Figure not drawn to scale



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STRATEGIES/COMMENTS

1. Most students learn to substitute numbers for n here although it can be done algebraically by factoring. However, the real issue here is figuring out what the question is asking. Reading interpretation - ugh!!

2. When you are not given any information about what type of triangle it is, just choose a few special cases and draw a conclusion. O course, if one recalls a key inequality theorem from geometry, this problem can be done in short order.

3. If you don't feel comfortable setting this up algebraically (preferred method), PLUG IN A VALUE FOR x...

4. Your calculator may not be able to handle the exponent so skills are needed. The large exponent also suggests a Make it Simpler strategy. This is a "Grid-In" question so if one is guessing remember that most answers are simple whole numbers! Finally, if one knows their basic exponent rules and basic radical simplification, none of the above strategies are needed!

5. Possibilities should be listed carefully. It is possible to count these efficiently by recognizing the effect of reversals and signs. Easy to get this one wrong if not careful.

6. For those who do not remember or want to apply a key geometry concept about ratios in similar figures, there are a couple of essential test-taking strategies which all students should be aware of of:
(a) Consider a special case of a parallelogram
(b) choose particular values for the sides.
In the end, even strong students often make a different error, however. That darn ol' percent increase idea!

7. Should you skip this if you have no idea how to start? Absolutely not! Draw a complete diagram and even if you don't recognize the similar triangles, make an educated guess! It's a grid-in and there's no penalty for guessing. Further, answers tend to be positivc integers!!

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ANSWERS

1. B

2. B

3. C

4. 1

5. C

6. 125

7. 6

...Read more

A Morning Warmup for Middle and High Schoolers - No Calculators Please!

How many integers from -1001 ro 1001 inclusive are not equal to the cube of an integer?

Hint: This could be a real 'Thriller'!

Click Read more for comments...


Comments
1) Do you think daily exposure to these kinds of problems as early as 7th grade will improve student thinking, careful attention to details (reading!) and ultimately performance on assessments? I think you can guess my answer!

2) I've published many similar questions on my blog but I couldn't resist this tribute to MJ.

3) I strongly believe we must occasionally remove the calculator to force their thinking. The stronger student recognizes immediately that 1000 and -1000 are perfect cubes and that one does not need to count the cubes but rather the integers which are being cubed (aka, their cube roots). The student with less number sense and weaker basics will feel lost at first but eventually their minds will develop as well if challenged regularly.

4) I added some complications to this fairly common 'counting' problem, similar to many SAT problems. This type of question is also typical of 8th grade math contests. Where do you think the common errors would occur assuming the student has some idea of how to approach this? Is understanding the language the primary barrier or not?

5) Let me know if you use this in September to set the tone for the year!

...Read more

An Equation Which May Be More 'Complex' Than It Appears!

Maybe I should rename this blog to Saturday 'Morning' Post. After all, no one reads that either anymore!

As the school year comes to a close (and I'm assuming it's already over for some), here's an innocent-looking equation which might be worth discussing with your advanced algebra/precalculus students now or next year. I might have considered saving this for our next online math contest but it's complex nature makes it more suitable for discussion in the classroom than on a test. Have you seen exercises like this in your Algebra or Precalculus texts? Do students often delve beneath the surface of these? It's kind of like a black box. We often feel we simply cannot reveal too much of the mystery here or we will not finish required content. Well, you know my philosophy of 'less is more' and I don't even live in Westport, CT. (Ok, that's a post for another day!).

SOLVE (by at least two different methods):

2a^-3/2 - a^-1/2 - a^1/2 = 0

Preliminary Comments/Questions/Issues

• Is the term solve ambiguous here, i.e., should we always specify the domain to be over the reals or over the complex numbers or is that understood in the context of the problems? I'm guessing that most advanced algebra students learn that the domain of the variable or solve instructions may impact on the result, but, that is precisely one of the objectives of this problem.
  
• Should students immediately change all fractional exponents to radical form? OR use the gcf approach (which requires strong skill)?
  
• It's not hard to guess that 1 is a solution but is it the only solution? Can we make a case for -2 being the other solution? The graph doesn't reveal this and surely, -2 doesn't make sense or does it....
  
• Is there ambiguity in raising a negative real number to a fractional exponent (never mind raising i to the i)? Why? Isn't there a principal value for such an expression? How is it defined? This problem raises fundamental and sophisticated issues about numbers which can be taken as far as one chooses to go Just how complex can complex numbers get?
• What is the role of the graphing calculator here? Mathematica? Wolfram Alpha? In addition to verifying solutions or determining answers, can these tools also be useful in clarifying ideas or raising new questions?
  
• Students (and the rest of us) are now capable of quickly filling in the gaps in their knowledge base by visiting Wolfram's MathWorld or Wikipedia for more background. Should this impact on how we present material? Typically, in the pre-web days teachers would avoid opening up a can of worms like complex solutions here, but, with your more capable groups, the sky's the limit now IMO...
  

Two Geometry Problems To Sharpen The Mind - Never Too Late In the Year For That!

Well, the June SATs have arrived today so these problems come too late for that, but these kinds of questions can be used to review basic ideas while strengthening thinking skills. Both questions below are appropriate for both middle and secondary students, although the second requires knowledge of a fundamental geometry principle regarding the sides of triangles.

There are other important principles embedded in these problems as well. In the end, I believe that students need to be exposed to many of these "contest-type" challenges to improve reading skill, learn how to pay attention to detail and think clearly. As a separate issue, performing well under testing conditions requires extensive training. You may not feel this is an important objective for math teaching in the classroom, but testing is a reality for the student...

These questions may appear fairly straightforward at first but be careful! I believe the second is more challenging than the first. These are not so different from the "gotcha" problem on our latest online contest.

1) The dimensions of a rectangle are odd integers and its perimeter is 100. How many different values are possible for its area?

2) The perimeter of an isosceles triangle is 96 and the lengths of its sides are even integers. How many noncongruent triangles satisfy these conditions?

For my "unofficial" answers, click on Read more...


Unofficial Answers (no solutions):
1) 13
2) 11
Feel free to challenge these answers or express agreement!

Comments

Which of the following do you believe would cause the most difficulty for students?

• The wording/terminology (e.g., noncongruent); general reading comprehension issues
  
• The sheer number of details (e.g., odd vs. even, perimeter vs. area, integer values)
• A precise counting/listing strategy vs. an abstract or commonsense approach
• The "square is also a rectangle", "equilateral is also isosceles" traps
  
• The issue of different areas for #1
  
• The triangle inequality for #2
• Other concerns?
  

...Read more

Classic Exponent Challenge for SATs, Algebra 2, Math Contests...

Don't forget to register for the upcoming 2nd MathNotations Free Online Contest for secondary students. Click here for more info.



The first 4 terms of a sequence are 2, 6, 18, and 54.
Each term after that is three times the preceding term.
If the sum of the 49th, 50th and 51st terms of this sequence is expressed as k⋅3⁴⁹, then k = ?

Click Read more to see the answer, solution, discussion...


Answer: 26/3

Suggested Solution
The first three terms can be written as
2(3⁰), 2(3¹) and 2(3²). (***)
In general, the nth term is 2(3^n-1).
The sum of the 3 desired terms would then be 2(3⁴⁸) + 2(3⁴⁹) +2(3⁵⁰). Factoring out 3⁴⁹, we obtain 3⁴⁹(2/3 + 2 + 6) = (26/3)(3⁴⁹), so k = 26/3.

Comments
(1) Too hard for SATs? Similar (but slightly easier) problems have appeared on the test.
(2) Could students use the "Make it simpler strategy" here to reduce the problem to the sum of just the first three terms? But this is the essence of geometric sequences (or exponential functions):
From (***) above, this sum would be
2(3⁰) + 2(3¹) + 2(3²) = 2 + 6 + 18 = 3(2/3 + 2 +6) = 3(26/3). The coefficient 26/3 would be the same for any three consecutive terms! Is this concept/technique worth developing?

...Read more

Updates and A Challenge Problem From Our First Contest

Updates...

• There are some 'new' math blogs (in some cases, new to me!) that I wish to bring to your attention. One of these was just started by one of my former students who is still in high school. She made excellent contributions to MathNotations last year. I'll post a link shortly. Look for creative thinking, challenging problems and an engaging writing style!
• Look for the Math Problem of the Day in the right sidebar. This is a 'gadget' made available by Blogger. Nice problems which are accessible to advanced middle schoolers and secondary students. They change every day so try them and check the solution link which is provided.
• There's a wonderful fraction calculator out there on the Web to which I will post a link and publish an article in a few days. If you haven't seen it, it will blow your mind! Imagine seeing the decimal expansion of any rational number to any desired number of places (within limits) instantly and that's not all it does!
  
• Yes, Pi Day is coming so I will post links to previous articles I've published and other excellent resources out there.
• Some of you know there's an excellent free resource of Singapore Math assessments for primary grades and more. You can download these and use them for your students. they make wonderful Problems of the Day and discussion points for your next department meeting! I'll post a link in a few days. Wait till you see the level of thinking and the content in the Grade 3 assessment, for example.
  

The following was question 5 from our contest. I'll leave it up for you to try. Feel free to comment or solve. This question proved to be of moderate difficulty for the teams. One has to be very careful about adhering to all the conditions regarding points P, Q and V. Have fun with it!

Problem 5 (2 pt question)

The graphs of y = 2x+3 and y = -x² + bx + c intersect in 2 distinct points P and Q, where P is on the y-axis. Let V denote the vertex of the graph of the parabola.

(a) Determine all values of b for which the points Q and V coincide.
(b) Determine all values of b for which Q and V are distinct and the slope of line QV equals 3.

A Presidential Contest Digit Problem: Fours are Wild!

With MathNotation's First Math Contest less than two weeks away (look here for details), I wanted to provide another sample contest question (multi-part). By the way, we now have several middle schools, high schools and even homeschool teams registered from all overv the country! It takes only a few minutes to register and there's still time!
--------------------------------------------------------------------------------------------------------
For President Obama, the number four has special significance. The most obvious is that he's the 44th president. You can ask your students to think of several other connections between our new president and the number four. But for now, we will focus on 44...

(a) Since 44 = 2^5 + 2^3 + 2^2, 44 equals 101100 in base 2 (binary representation).
Let S be the set of all base-10 numbers (positive integers) whose binary representation consists of six digits, exactly three of which are 1's. Find the sum of these base-10 numbers to reveal part of the mystery behind the title of this post!
Note that the leftmost binary digit must be "1".

Comment: This is a fairly straightforward 'counting' problem accessible to middle schoolers as well as older students. One could simply make a list of the numbers and add them. However, there's a more systematic way to count the 'combinations' and a "different" way of adding here that may help you solve the next problem. Can you find it?

(b) Consider the set of all base-10 numbers (positive integers) whose binary representation consists of ten digits, exactly three of which are 1's. Show that the sum of these base-10 numbers can be written 44(2^9) - 4 - 4. "Fours are wild!"

Note: This seems like a tedious generalization of Part I, but, again, if you find the right way to count and add it won't take long!

BTW, if you're wondering how I came to find all these 4's, well, it might have been serendipity. After all, serendipity has 11 letters and 11 is a factor of 44 and... (Twilight Zone music playing in the background...). Also, if you're wondering what my outside sources for these kinds of problems are, do you really think there's anyone else out there whose mind could be this warped!

A Preview of the Contest: Probability Investigation with Replacement

Don't forget to email me if you want your students to participate in the first MathNotations online math contest on Tue Feb 3rd. There is still time! Look here for info.


There may not be a probability question on the first contest but the following gives you a flavor of the type of multi-part question I'm talking about -- an investigation in more depth.

You will find many variations of the following problem in texts. From experience we know that the student needs to have numerous experiences with these. How do many students do on this topic when the exam question is slightly different from the ones reviewed in class!

THE PROBLEM STATEMENT
Five cards are numbered 1 through 5 (different number on each card). Typical scenario, right?
George chooses cards randomly one at a time. After he selects a card, he marks a dot on the card, then puts it back (replacement!) in the pile of 5 cards, reshuffles them and draws the next card and so on. The game continues until he selects one of the "marked" cards.

INSTRUCTIONAL STRATEGY
Before a technical analysis of this experiment (sample space, random variable, specific probabilities, expected value), I would typically ask students a broad intuitive question or ask them to suggest questions one might ask about this "game".

Intuitively, I might ask:
In the long run, how many draws would you "expect" it to take for the game to end?

With five cards, what do you think most students would guess? Draw three? Draw four? I think asking this initial question is crucial. In most cases, we want the mathematical result to be reasonable and to roughly agree with our intuition (not always of course, there are paradoxes in math which are counterintuitive!).

THE INVESTIGATION

Part I
What is the probability that George chooses a "marked" card on his second draw for the first time? On the 3rd draw for the first time? 4th draw? 5th draw? 6th draw?
Another way to ask these are: What is the probability that the game "ends" after 2 draws? 3 draws, etc.

Part II
"On average", how many cards would George need to draw to get one of the marked cards for the first time?

Note: In more technical language we are asking for the expected number of draws before the game ends?

Normally, I don't publish answers to these questions but, in this case I will give partial results. Please check for accuracy.

The probability the game ends after 3 draws is 8/25 or 32%.
The expected value for the number of draws for the game to end is approximately 3.51. What does this mean!

An Algebra 2 Math Contest Problem and a Warmup Too!

Not the most challenging contest problem but something to give to your students to develop logical careful thinking and some "basic skills." There's a slight 'twist' but nothing that will faze our math experts out there.

SOLVE

(x² - 6x + 9)^{(x2 - 4x + 3)} = 1

1,2,2,3,3,3,4,4,4,4,... What is the 2008th term? SAT-type Questions vs. Math Contest Problems

Don't forget to submit the name of our Mystery Mathematician. Contest ends around 4-15-08. Thus far, only one correct submission (emailed of course!).

The problem in the title is the math contest version. Knowing the formula for the nth triangular number would be helpful (so might a calculator). Do all middle school and hs students become familiar with triangular and other figurate numbers? Should they? My vote: Yes!


The SAT -type would be:

What is the 56th term of the sequence 1,2,2,3,3,3,4,4,4,4,..., in which each positive integer N occurs N times?

Comment: This is considerably easier than the contest problem as one could do it by listing with or without a calculator. It also may reveal your strong number sense students who will see the idea fairly rapidly. Try it as a warm-up in class!

Another SAT-type (probability, counting) to help students prepare for the May Exam:

Let S be the set of all 3-digit positive integers whose middle digit is zero. If a number is chosen at random from S, what is the probability that the sum of its digits is even?

Note: I wrote this question to demonstrate basic principles of probability and counting. Although one could use the Multiplication Principle (aka, Fundamental Principle of Counting), students should also be encouraged to make an organized list and, by grouping, see why the answer is 1/2.

In the long integer 36912151821...9999, what is the 1107th digit?

The title question would be a medium level math contest problem.

The following could be the SAT version:

369121518...99
The integer above is formed from consecutive positive integer multiples of 3 from 3 to 99 inclusive. What would be the 50th digit in this number?

A student could theoretically take the time to list all the digits up to the 50th, however, this would be more time consuming than using logic.

Possible approach for SAT-type:

369 uses 3 digits.
This leaves 50-3 = 47 more digits.
Starting with 121518... there are two digits for each multiple of 3, so we divide 47 by 2, producing 23 with remainder 1. We now need to determine the 23rd multiple of 3 beginning with 12:

12 + (23-1)3 = 78. [At what point should students know this formula for arithmetic sequences?]

The remainder of 1 means we need to look at the first digit of the multiple of 3 that comes after 78: 81
So the answer would be 8, but, of course, I could have erred in my logic or calculation, so pls check that!

One could now assign the title problem for extra credit or as an extension.

Comment: How many readers believe that middle schoolers on up tend to rely on their calculator to find remainders. This has been discussed previously on this blog and a couple of algorithms were given for computing the remainder from the decimal result given by the calculator.

Monthly Math League Challenge: List all ordered triples of positive integers (x,y,z) whose product is 4 times their sum and x>y>z.

The problem in the title is another wonderful challenge for our readers or for students. Questions like these are powerful tools to develop student reasoning and problem-solving prowess. Once again I have received permission from the directors of the Math League to publish this question on MathNotations. This was the last question from the first contest this year.

Please cite the question in the title as:
Copyright Mathematics Leagues Inc 2007. May not be reproduced without permission of the copyright holder.

You can learn more about Math League Contests at the Math League website.

Now you know that I'm not going to simply copy a problem and just leave it at that!
How can we make this more of an enrichment experience for our algebra students who may not quite be ready for the contest level. Can you guess? Scroll down...


Well, what makes the contest problem particularly formidable is the use of three variables. In fact, two numbers is already a difficult problem for most! So we use the "Let's Make It Simpler" strategy:

Find all ordered pairs (x,y) of positive integers whose product is four times their sum and x>y.

I'll begin some analysis, starting with our basic equation:
xy = 4(x+y).

Before one starts the traditional solving for y in terms of x ritual (then let's go to the graphing calculator), we need to remind our students that this is a positive integer problem, which allows for a somewhat different kind of approach.
I also frequently suggest to students to consider the case that x = y even though the restrictions do not allow this.
If x=y, then x² = 8x leading to x=8, y = 8 (or ____?). Even though this is not allowed, it could suggest other solutions. Those who enjoy graphical solutions will also appreciate that this solution is one of the two points of intersection of the graph of our basic equation with the line y = x. I'll leave it to our readers to find the other.

Note that from xy = 4x + 4y we can see that neither x nor y can equal 4. For example, if x=4, we'd obtain 4y = 16+4y, which is impossible.
Similarly, neither x nor y could be less than 4. This is more readily proved algebraically. I'll omit the details.

Note: Another important tool for students in solving these kinds of problems is to consider symmetry. The basic equation is symmetrical in x and y. Symmetry can be very useful.

Ok, while most students are using guess-test methods (they would call it 'plugging in'), we will solve for y in terms of x (skipping a few steps):
y = 4x/(x-4).
There's a well-known algebraic device students need to see here when finding integer solutions, which is equivalent to long division. Rewrite the previous equation as:
y = (4x-16)/(x-4) + 16/(x-4). I'll let you guess why I subtracted 16 in the first fraction then added it back in the second. This is equivalent to:
y = 4 + 16/(x-4).

Now we use the integer condition (and the fact that x > 4 and x > y) to find solutions:
x-4 has to be a factor of 16 which implies that x-4 could equal 16, 8, 4, 2, or 1. In fact we can show that only 16 and 8 are possible. I'll leave the rest to you...

Classic AMC Contest Square Dissection Problem and more...

Recognize this diagram from a famous math contest problem which I first saw many years ago on an old AHSME contest (now known as AMC)? We'll start with this one and then modify it, creating variations on the basic theme. Finally, we will ask our readers/students to generalize the result algebraically.


In the diagram at the left, nothing is labeled, so we will describe it verbally and hope it will make sense.
We start with a square and dissect it by drawing 4 segments, each connecting a vertex to a midpoint of a side.

THE CLASSIC
Explain why the area of the shaded region is one-fifth of the area of the original square.

Notes/Comments:
(1) This is a wonderful exercise to develop spatial reasoning and to demonstrate a visual approach to a geometry problem when dimensions are not given. Of course, one could use an algebraic or numerical approach if one chooses.
(2) Students who 'see' the jigsaw puzzle approach of rearranging the pieces rarely consider what assumptions are being made. To make the problem even more meaningful, the instructor could ask why the shaded region is, in fact, a square.
(3) Simpler versions of this often appear on the SATs.

VARIATION #1

This time, both diagonals are drawn. The additional two segments join the midpoint of the bottom side to the midpoints of two other sides.

(a) The red shaded region (does it have to be a square?) is not one-fifth of the original square. What fractional part is it?
(b) The total shaded area is what part of the original square?

VARIATION #2

This time the smaller segments divide the sides into a 1:2 ratio. The figure is not drawn to scale. The 3 segments on the base are supposed to be equal!

(a) The blue shaded region (is it a square?) is now what fractional part of the original square?
(b) The total shaded area is now what part of the original square?



THE GENERALIZATION OF VARIATION 2
Use the diagram from Variation 2. Assume the original square has a side length of 1 unit. If the smaller segments divide the sides of the square into an x:(1-x) ratio, do parts (a) and (b) again, expressing your results in terms of x. What restrictions on x make sense here? Make sure your expressions agree with the results above.

Another Math League Challenge - Medians of a Triangle...

As promised...

In a triangle, two sides and the median to the third side have respective lengths 5, 13, and x. List all possible integer values of x.

“Copyright Mathematics Leagues Inc 2007. May not be reproduced without permission of the copyright holder.”


Comments:

• There is an elegant solution/proof here! I found a far more complicated coordinate argument.
  
• Some students might make an educated guess based on insight but the real challenge is to explain one's conjecture!
  
• Challenge yourselves with this - if you see how to do it, suggest an approach but don't give it all away for at least a few hours!
• We generally think of these as only appropriate for our best and brightest but there's no harm in throwing it out to all of our geometry students- amazing things happen when we open up these kinds of problems to everyone!
  

An Exceptional Math Contest Problem

Update: The solution I emailed to the director of the Math League is at the bottom after you scroll down a ways. Note how this ties into the difference of squares problem I posted the other day!

A very good friend of mine who runs one of the most successful math league contests in the US has given me permission to reprint the question below with the following attribution:
“Copyright Mathematics Leagues Inc 2007. May not be reproduced without permission of the copyright holder.”

This question was the last question on a recent contest. It certainly made me think! I felt the need to be rigorous about justifying one of the steps and that was the reason for the
M² - N² = 12 post from the other day. There are at least 3 methods that have been found thus far, according to my friend. Teachers often send in their own solutions and some even get included with the official solutions. My feeling is that questions like these are vital for our students intending to pursue higher mathematics, not just for those who happen to participate in contests. I also believe that variations on great questions like this will have an eternal life...

THE QUESTION

There's exactly one real number a for which ax² + (a+3)x + (a-3) = 0 has two positive integer solutions for x. What are these values of x?

Have fun with this!
For more information about the Math League, visit here.

Scroll down to see the solution I sent to the directors of the Math League...











First of all, I'll state the answer I obtained before giving the details:

A real number a which produces two positive integer solutions is a = -3/7 which does lead to the solutions x = 4 and x = 2. Proving that this is the only real value of a is far more challenging and I suspect that there's a much easier solution than the one I found. Anyway, here's my approach:

In order to simplify the original equation, divide both sides by a to make the leading coefficient equal to 1:

x^2 + (1+(3/a))x + (1 - (3/a)) = 0
To simplify further, let b = 3/a:
x^2 + (1+b)x + (1-b) = 0
From a well-known rule about the roots of a quadratic equation, the sum of the roots in this case equals -1-b and the product of the roots is 1-b. Since the roots must be positive integers, it follows that b is an integer less than -1. Remember for later:
b must be negative!

From the quadratic formula, we obtain the roots to be:
x = ((-1-b) +- SQRT(b^2+6b-3))/2.
In order for the roots to be integers, the expression b^2+6b-3 must be a perfect square.
I will rewrite this expression by completing the square:
(b+3)^2 - 12 and this must equal a perfect square , call it N^2.
Let b+3 = M, then we must have M^2 - 12 = N^2 or
M^2 - N^2 = 12. It's easy to guess a difference of perfect squares equal to 12, namely 16-4, but I will now prove this is the only solution:

Factoring, we have (M+N)(M-N) = 12 where M,N are integers.
The only pairs of factors to consider are: 12,1; 6,2; 4,3 as well as their negatives.
In order for M,N to be integers, the factors of 12 must BOTH be EVEN so the only possibilities are 6 and 2 or -6 and -2..
If M+N = 6, M-N = 2, then, by adding, M=4 and N=2. From b+3 = M, we would have b = 1, but this doesn't work since b must be negative (see above!)
Therefore M+N = -6 and M-N = -2. From this we have M=-4 and N=-2.
Then b = M-3 = -4 - 3 = -7.
b =-7 leads to the quadratic equation: x^2 + (1+(-7))x = (1-(-7)) = 0 or
x^2 - 6x + 8 = 0. Factoring we obtain the roots to be x = 4 and x = 2.

Since b = 3/a, it follows that -7 = 3/a or a = -3/7.

Some Algebraic Reasoning (Math Contest/SAT-Type) to Whet Your Appetite

While we're waiting for Carnival of Mathematics #11 and Part II of Recursive Sequences, here are a few algebra problems to store away for your students...
Using math contest problems (AMC, MathCounts, etc.) in the classroom can be very helpful in raising the level of expectation for many students. One thing experienced teachers know is that students will only learn how think at higher levels when we provide these kinds of challenges. Math educators are always seeking more examples of this nature. Look no further than released math contests, samples of which can be found online. Use these problems as models which can be revised to match the skill and conceptual level of your group. This is not easy but it's worth the effort. Assuming that more basic level math students would have little or no chance at these (or will become too frustrated) can be a self-fulfilling prophecy. Those of you who have been following this blog since its inception 6 months ago, know that I challenged just such a group of 9th graders this year. Deciding what prerequisite skills were needed and how to develop some of the challenges incrementally required considerable thought and planning but their engagement in the activities and a feeling of accomplishment on their part made it worthwhile.

Questions 1 and 2 below are appropriate for 1st or 2nd year algebra students. The objective of these conceptually-based problems is to develop algebra sense rather than provide mechanical practice with algorithms.

1. For how many values of x is
(x-3)(x-4)(x-5)(x-6) = (3-x)(4-x)(5-x)(6-x)?
(A) 0 (B) 2 (C) 3 (D) 4 (E) more than 4

2. For how many values of x is
(x-3)(x-4)(x-5)(x-6)(x-7) = (3-x)(4-x)(5-x)(6-x)(7-x)?
(A) 0 (B) 3 (C) 4 (D) 5 (E) more than 5

3. What is the greatest integer value of N, less than one million, for which
√(1+√N) is a positive integer?

Note: Is estimation worthwhile here as a starting point? How would some students use a calculator to 'solve' this? Watch them! Do algebra textbooks provide methods and practice for solving positive integer problems? Are there special methods one needs for these?