Monday, February 5, 2007

Problems 2-6-07

I'm sure most visited today hoping I would forget about my 'cause' and go back to writing problems. Well, I haven't given up the fight for what I believe, but I also will try to keep up my usual routine, although perhaps not as often.

Question #1 below is a challenge for Algebra 2 students who want to test their factoring skills with fractional exponents. This type of manipulation is rare these days in most textbooks, probably because it is seen as 'old-school'. Do our top math-science students still need exposure to these or have these questions been rendered obsolete by 'Symbolic Algebra Manipulators'?
Question #2 is a hard version of a standard SAT problem, comparing two series. Many students think only in terms of summing each series separately.
Question #3 is fairly easy using trig, but I'm encouraging strategies using only the Pythagorean Theorem and one other key idea (thinking about areas when you see a perpendicular!).


15 comments:

mahtmom said...

Ooh, I like #2 a lot. And cute hint.

Answer: 5050

Reasoning:
By subtracting the terms pairwise, we see that E-D = (2^2-1^2) + (4^2-3^2) + (6^2-5^2) + ... + (100^2-99^2)

Factor each pair using difference of squares:

E-D = (2+1)(2-1) + (4+3)(4-3) + (6+5)(6-5) + ... + (100+99)(100-99)

The second half of each product is always going to be one (since they are all of the form (n+1 - n)) so we can drop them all.

Then E-D = (2+1)+(4+3)+(6+5)+...+(100+99)

removing parentheses and rearranging we see that
E-D = 1+2+3+4+5+6+...+99+100
using Gauss' trick (adding pairwise from both ends) we see that E-D = 101 * 50 = 5050

I love teaching Gauss' trick to my students. It comes in handy when doing handshake problems, so they get to practice with it. I get to see kids for many years in a row, teaching all the different levels in our small school, and this trick is one that usually does stick after being shown it a couple of time.

Dave Marain said...

Answers and solutions will be posted later on Tue after more time is given for others to try these...

dear mathmom--
i'm glad you appreciate the Gauss hint! even apparently unmotivated students perk up when they see that; actually, i believe what little karl did (he was a close personal friend of mine!) was to write the series backwards, line it up with the original series and add, producing a sum of 101 one hundred times, then divided by 2!!

#1 appears to be mindless symbol-pushing but try it anyway; there's a satisfaction our students (me too) derive from working through some complicated manipulation and watching how a simple expression results; besides, how does one become proficient with these if one doesn't do them!

#3 falls easily if we use the law of cosines to find the cos of the 'middle' angle to be 4/5, thus one can form a 3-4-5 triangle or use coordinates to determine the tan of the difference of 2 angles from slopes, but using only 'synthetic' methods from geometry requires thinking a different way which i laways encourage (and find so difficult for myself).

Finally, thanks for the support with the National Math Panel. i didn't expect Fox News to be coming to my house or, for that matter, Secretary Spellings! I do expect that everyone will see the nedd for classroom educators to be on the Panel since that's far less controversial than suppoting a scary national math curriculum with Big Brother watching over us. In truth, committees like these rarely include teachers, a demeaning of our profession. That's the real tragedy here...

jonathan said...

For #3, start with sides 2n. The half sides are n.

AE = AF = n(sqr(5))

Areas:
[ABE] = n^2
[ADF] = n^2
[FEC] = (1/2)n^2
so [AEF] (not drawn) = (3/2)n^2

Look more closely at AEF. AF is the base, EG is that altitude.
(AF)(EG)/2 = (3/2)n^2
n(sqr(5))(EG)/2 = (3/2)n^2
(sqr(5))(EG)/2 = (3/2)n
EG = (3/sqr(5))n


AEG is a right triangle
AG^2 + EG^2 = AE^2
AG^2 + [(3/sqr(5))n]^2 = [n(sqr(5))]^2
AG^2 + 9n^2/5 = 5n^2
AG^2 =16n^2/5
AG = 4n/sqr(5)

Hm, could have done better. If we had made the area of the square = 20, the sides are 2sqr(5), the half sides are sqr(5), the areas are 5, 5, 5/2, and 15/2. AE is 5. EG is 3. AG is 4.

It would be lousy to start with "Assume WLOG the area of the square is 20." Then everything falls out like magic and the kids are dazzled and learn nothing.

Totally_clueless said...

For (1), the denominator is a perfect cube, while the numerator has four factors, one of which cancels with one term of the denominator. Were you expecting something more here?

It would have taken me a long time to get (3) without the area hint.

For problems like (2), do you encourage the students to write the general term (in this case (2n)^2 - (2n-1)^2), or write specific terms and generalize? It is probably a good skill to write the general term.

TC

mathmom said...

Dave,

I've actually heard that the Gauss story might be apocryphal. I have heard it both ways -- as you said, writing it backward, or the way I said, folding it in half.

Here's a kind of interesting page on someone's attempt to verify the story:
Gauss's Day of Reckoning

I'll try to look at the other problems if I have time tonight.

Dave Marain said...

jonathan--
nice analysis of the geometry; if some students will benefit from particular values that make the calculator work better, i don't discourage it. This is particularly true when dealing with ratios as in similar triangles.

tc--
if i'm working with a stronger group, i would encourage a general expression for hte differences of squares, however, getting most students to use the pairing strategy, recognize the difference of squares, then recognize the sum of consecutive integers is more than enough!!

There's nothing fancy about #1 -- it's a pure skill problem that i believe is still part of the training needed to excel in math and science but. of course, this is considered heresy by some!

mathmom said...

Oh, dear, I don't want to tell you how long I stared at #3 thinking it impossible because, with leg lengths of 1 and 2, triangles ABE and AFD must be 1:2:sqrt(3) or 30-60-90 triangles, which would make AEG also a 30-60-90 triangle which is NOT similar to a 3-4-5 triangle. How many days ago was it that I said I could always remember that the 2 had to be on the hypotenuse since it was the longest side? ;-)

OK, now that I'm over that delusion, I'll actually try the problem the right way...

mrc said...

mathmom -- I'm glad I'm not the only one that got caught up in that thought process! It's very educational to see how we can all get tricked, even when we think we know what we're doing. I puzzled over it last night and came back today to see if anyone had a hint for us. Thanks, Jonathan!

mathmom said...

I did finally get #3 on my own, using reasoning very similar to Jonathan's, except that I let the length of the side of the square be 2, for convenience. Since we are only talking about ratios, you can let it be whatever you want. sqrt(20) would have been an odd thing to start with, though, at least before solving the problem once with other numbers. :)

I need to brush up on factoring polynomials before I can do #1. I'd start with "let x=a^1/5 and y=b^1/5" to get those in more manageable looking forms, but then, like I said, I need to brush up on the actual factoring.

mathmom said...

Can someone help me with the factoring of the denominator of #1. I am not seeing the perfect cube there. I get different coefficients if it were a perfect cube.

Dave Marain said...

It's interesting that #1 is actually pretty annoying. Readers are talking around it but not cutting through it. I don't see it as involving cubes at all!
In fact it's all about gcf's, differences of squares, and factoring by grouping! The fractional exponents were inserted to disguise some of this. Here are some hints:
1) In the numerator, start by taking out the gcf of
b^(1/5). The resulting factor is of the form x^4 - y^4 so it can be factored as a diff of squares and then one of the factors can be factored again.
2) The terms in the denominator can be grouped in pairs. E.g., factor out a^(2/5) from the first 2 terms, then 9b^(2/5) from the next 2. You should see it from there.
As I said earlier, there still seems to be a need for these kinds of manipulations for some of our students and why not prsent it as a challenge to others (make it worth 2 bonus points and watch them try it!) There is both an art (recognizing forms) and a skill (executing the forms) to this. I am certainly not recommending it as a significant unit in algebra however.
Another aspect to all of this is thinking about math problems as 'puzzles' as Jonathan calls his challenges. What distinguishes a difficult problem from a 'puzzle?' I think he uses the terms synonymously. Mathematicians by definition (!) only work on hard problems, i.e., those for which there is no 'obvious' solution. Problems for which there is no roadmap -- one has to create a solution path by experimenting with different approaches as one might go through an intricate maze (weak analogy). Calling it a puzzle seems to make it more engaging though. What child doesn't love to try a puzzle, but call it a math problem and ugh...
Could we argue that math is the art of solving 'puzzles' then, at least puzzles involving number, space or the other abstract objects we study?? I think there is more subtlety here than meets hte eye, it's more than just a 'name'. I guess I'll leave it to jonathan to clarify this, but I'd be interested in other points of view.

Totally_clueless said...

On (1), Dave got me again. I got fooled into thinking the denominator was a cube based on the coefficients.

Setting a^(1/5) and 3* b^(1/5) simplifies the problem and gets the nice elegant answer in a simple fashion.

mathmom said...

For #1 I finally got

b^1/5(a^1/5-3b^1/5)

Hope that's right ;-)

Here's what I did:
let x=a^1/5 and y=b^1/5

Numerator becomes x^4y-81y^5
Factor out y, then use difference of squares to get
y(x^4-81y^4)
=y(x^2+9y^2)(x^2-9y^2)
one more difference of squares in the final term gives:
=y(x^2+9y^2)(x+3y)(x-3y)

Denominator becomes:
x^3+3x^2y+9xy^2+27y^3
factoring in pairs as Dave suggested:
x^2(x+3y) + 9y^2(x+3y)
= (x^2+9y^2)(x+3y)

Cancelling out common terms in numerator and denominator leaves:

y(x-3y)

subbing the a's and b's back in:

b^1/5(a^1/5-3b^1/5)

Dave Marain said...

great job, mathmom!
i knew i wouldn't have to type the solution or answer myself!
well, was rhe effort worthwhile? did it bring back your facoring skills? is it just meaningless symbol-pushing which our top-performing students don't need any more?

mathmom said...

Yes, Dave, it was worth the effort. Not because I need the skills it brought back in my life, but because it's a kind of "puzzle" and solving it is rewarding.

Generally, I'd say any "non-routine" problem can be considered a puzzle. And yes, I think the name can help make it seem more appealing to some.

I suspect our top students now have calculators that can do the symbol pushing. So the skill may not be so important in a practical sense, but like I said, the skill of solving the "puzzle" of it is a transferrable skill, in my opinion, and worth the effort.