Monthly Archives: October 2011

Learning to listen

Listen to and learn from your students; they hold the key to deeper understanding.

My precalculus class yesterday was exploring graphical behavior of rational functions.  We started with \frac {1}{(x-2)^3(x+3)^{6}} which the group easily handled with their understanding of even and odd vertical asymptotes (VAs) from the previous class.

The curve approaches 0^+ on the right, the VA at x=2 is odd so the rational function’s graph “passes through infinity” there, and “bounces off infinity” at x=-3.

We hadn’t explored what happened when rational functions had variable expressions in their numerators, and they had never seen holes in these curves, so I had no idea what they would do when I asked for the graphical behavior of \frac {x-2}{(x-2)^3(x+3)^{6}} at x=2.  As a student, I was taught to perform algebraic simplifications, but I thankfully remained silent.

My students were initially bothered by the \frac{0}{0}  form of the function at x=2, and various clusters were working toward different solutions when student NC declared, “at x=2, the denominator has the dominant exponent, so there still is a VA at that point.  It might be a different kind of VA, but it’s a VA.”  In an instant, everyone in the room understood what was happening.  Changing the degree of the numerator’s factor in later examples ultimately led to holes on and off the x-axis, but NC’s dominance argument yielded far deeper and lasting understanding than my plans for an algebraic approach ever would have.

OK, I admit that you can get to the same place by doing algebraic simplifications, but my deliberate silence allowed my students to develop their own subtly different understanding that exceeded what I had planned to offer.  After years of being the information dispenser in my classes, I’m learning the uncomfortable lesson that it’s often better to set them up and then shut up.

Three people and a monkey

Personally, I’ve always thought of educated Trial-and-Error as a valid problem-solving technique, or at least a problem-insight technique.  The challenge with this approach is the difficulty of trial-and-error to distinguish between situations with multiple solutions and those with only one.  I take every opportunity to remind my students of the dual goals of every problem solution:

  1. Show that your solution(s) is (are) correct, and
  2. Show that no other solution(s) exist(s).

Following is a variation of a fun problem from a member of the CPAM list-serve.

PARENTS:  This problem has been around for centuries in many different forms and more importantly, is a great problem-solving opportunity for elementary and middle school children.  Change the roles, genders, and animal to make the problem interesting for your children/students.

Three people and their pet monkey spend a day gathering bananas and go to sleep.  During the night, one wakes up, splits the bananas into three equal piles with one banana left over.  She gives the extra banana to the monkey, hides one pile for herself, combines the other two piles and goes back to sleep.
A little later, another person wakes and splits the remaining pile of bananas into three equal piles with one left over.  She gives the extra to the monkey, hides one pile for herself, combines the other two piles and goes back to sleep.
The third person wakes later and repeats the process.
In the morning, the remaining pile is divided evenly among the three peo
ple with nothing left for the monkey. 

What is the smallest number of bananas that could have been in the original pile?

Don’t read any further if you want to solve this problem for yourself.

Because the problem asks for the smallest size of the original pile, there is only one answer and the fundamental weakness of a trial-and-error solution has been eliminated.

The recent CPAM post asked:  “Please help me to solve the attached problem in the way I could explain it to grade 6-7 students. I came up with very complicated equations. Is there any other method to solve it?”

From experience, I knew that I could approach this with lots of equations involving fractions, but I could instantly hear the groans of so many middle school students if this approach was the first attempted.  That’s when I tried another great problem-solving approach:  Work Backwards!

Because the problem wanted the smallest possible pile, I wondered if there could have been 1 banana each in the divided piles the next morning.  If so, there would have been 3 bananas in the re-combined pile, an impossible situation when you remember that this pile was the result of combining the two piles that remained after the third person had split the piles and hid her “share.”

Because the sum of three odd numbers is always odd, the evenly divided piles the next morning must have an even number of bananas each to avoid the impossible situation from that of the paragraph above.

So what if there were 2 bananas each in the divided piles the next morning?

  • That would give 6 bananas in the re-combined pile after the 3rd person went back to sleep.
  • The 6 bananas would have been in piles of 3 & 3 after the 3rd person’s pile of 3 was hidden and the monkey had 1, giving 10 (3+3+3+1) bananas in the re-combined pile after the 2nd person went back to sleep.
  • Those 10 bananas would have been in piles of 5 & 5 after the 2nd person’s pile of 5 was hidden and the monkey had 1, giving 16 (5+5+5+1) bananas in the re-combined pile after the 1st person went back to sleep.
  • Those 16 bananas would have been in piles of 8 & 8 after the 1st person’s pile of 8 was hidden and the monkey had 1, giving 25 (8+8+8+1) bananas in the original pile.

The answer must be 25. 

GENERALIZATION:  The original problem suggests that there are lots of solutions, so algebra may be the best way from here.  The approach learned from the trial-and-error approach can guide the global solution.

  • If there are x bananas in each final pile the next morning, then there were 3x after the 3rd went to sleep.
  • That means there were \frac{3x}{2} in each of the 3rd person’s piles for a total of 3*\frac{3x}{2}+1=\frac{9x+2}{2} bananas after the 2nd went to sleep.
  • So there were \frac{(9x+2)/2}{2}=\frac{9x+2}{4} in each of the 2nd person’s piles for a total of 3*\frac{9x+2}{4}+1=\frac{27x+10}{4} bananas after the 1st went to sleep.
  • Finally, there were \frac{(27x+10)/4}{2}=\frac{27x+10}{8} in each of the 1st person’s piles for a total of 3*\frac{27x+10}{8}+1=\frac{81x+38}{8} original bananas.

Notice that x=2 gives \frac{81*2+38}{8}=\frac{200}{8}=25, confirming the trial-and-error solution from earlier.

CONCLUSIONS:

  • Because the answer is an integer, 81x+38 must be a multiple of 8 which can only happen for even values of x, confirming the earlier hypothesis.
  • 81x+38=(80x+32)+(x+6)=8(10x+4)+(x+6), therefore the only solutions, x, to this problem are those which make x+6 a multiple of 8.  The smallest positive value for which this is true is x=2, again confirming the trial-and-error solution.  All such final pile values of x form an arithmetic sequence:  2, 10, 18, 26, ...  which correspond to initial piles of size 25, 106, 187, 268, ....

In my opinion, the initial trial-and-error solution is attainable by any elementary school student who knows how to add, but the logic to get there might need to be scaffolded for younger students.  The arithmetic behind the generalization can be expected of middle school students, but the inclusion of a variable on top of the trial-and-error procedure would push it out of reach of most students who haven’t had a pre-algebra course.  There are many high school students who eventually could understand the factoring arguments in the conclusion, but that logic, again, seems to be absent from most curricula.

This is a good problem that can be approached again and again by students as their mathematical sophistication matures.  I’d love to hear how others tackle it or specific results of student attempts.

Nested Isosceles Triangles

Check out this fun little problem from @daveinstpaul.

I’m sure there is a much more elegant solution, but given my technology interests, I thought this would be a cool way to incorporate CAS.  Based on Dave’s restrictions, the following angle measures apply.

There are many ways to write equations from this setup, many of which are identical forms of the same information.  One way to keep from writing dependent equations is to use combinations.

From each small single triangle, you get 2A+Z=180, 2X+Y=180, and 2W+V=180.

From the top two triangles, one relationship is  A+Y=180. Because the large triangle is isosceles, the bottom two triangles give X+V=W.  There’s no convenient way to combine the top and bottom triangle.

Combining all three triangles, one relationship is X+W+Z=180.

That’s a system of 6 equations in 6 variables which Wolfram Alpha solves to give A=45 degrees.

I’d love to see a non-algebraic approach.