Tuesday, August 26, 2008

Starting the Year off in Precalculus/Algebra 2 - A Little Review?

Don't forget our August-September MathAnagram. No responses yet but this mathematician deserves our recognition. 'Relatively' speaking, this mathematician was truly unique and, perhaps, the last of a dying breed.

PRINCE? NAH! E-ROI!

Sometimes I found that a coordinate problem was a good way of reviewing the geometry and algebra needed to refresh memories after a 2-month layoff. Here's a fairly straightforward one that admits many different approaches and might be used to set the tone. Encourage students, working in groups, to find at least THREE different methods. This will extend the thinking of those who "solve" it rapidly and sit there complacently. This problem is a bit more appropriate for the students who completed Algebra 2 although Geometry and Algebra 1 methods are possible.
To reiterate: The problem itself is not particularly challenging. The purpose here is to provide review of several ideas, methods, theorems and strategies.

Given points A(0,0) and B(12,0). Determine the coordinates of all points C(x,y) such that ∠ACB is a right angle and ΔACB has area 18.

Eric Jablow said...

I wonder how many students would reason that ACB is a semicircle on diameter AB.

Dave Marain said...

Eric,
That's the method I personally prefer. If the students do not think of it, it is up to us to guide them. Unfortunately, I do not think many will see this. But isn't one of the purposes of a good problem to deepen insight...

There are so many ways to look at this question:
Equation of circle;
Product of slopes = -1;
30-60-90 triangle (sides are 6,3,3√3);
Pythagorean;
Distance formula
and, of course, all the methods students would think of that we might not!

My guess is that the distance formula or Pythagorean method would be used by a majority of students. If anyone uses this in class, let us know!

Kate said...

Leery of springing this on Algebra 2 students early in the year even though they should know enough to make a decent attempt at it. I'd be interested to see what they do with it.

We found 2 straightforward ways that would both be a stretch for our regular-paced 11th graders - on the circle with diameter AB, there has to be four locations for C such that the height of the right triangle (with base AB) is 3. Then use the mean proportional x/3 = 3/(12-x) to find the distance along the x-axis from the origin to the altitude. This gives coordinates for C of (6+-3rad3, +-3).

Another way was to solve the system: a^2 + b^2 = 144 and ab = 36, where a and b are the legs of the right triangle in question. This gives a quartic equation that has to be solved by completing the square or quadratic formula. The completing the square answer looked different +-sqrt(sqrt(3888)+72) but was equivalent. Then you would still have to use pythagoras to find the x-coordinates of C.

Sean said...

Eric,
Not many would probably notice that, you're right there. However, it's NOT a semicircle. It's ALMOST a semicircle. We cannot allow C=A or C=B. So it's actually a semicircle, minus two points, (0,0) and (0,12).

Sean said...

oops! Omit (12,0)...not (0,12)....got my x's and y's switched.

Dave Marain said...

Kate--
I really like using the altitude on hypotenuse theorem. You also recognized that there are 4 solutions that result from the quadratic-quadratic system. I'm confused though. Did you actually use this with your students or was this a discussion with other teachers?
At any rate, this problem is taking on a life of its own. Feels like we could use this to review a good part of geometry and algebra 2 and I haven't even mentioned a parametric solution! Once students recognize the circle they should be able to "see" the 30-60-90 triangle!

Sean -- Nice catch on those endpoints! Do you like the problem?

Dave

Joshua Zucker said...

Hi Kate,
You wrote:

Another way was to solve the system: a^2 + b^2 = 144 and ab = 36,

And I agree, but there are easier ways to solve it!

For instance, add 2ab to the first equation, (a+b)^2 = 216, a+b = 6*sqrt(6) (maybe with a - sign),
then subtract 2ab, (a-b)^2 = 72, a-b = 6*sqrt(2) (again maybe a - sign), then add those two equations,
2a = 6*(sqrt(6) + sqrt(2)),
a = 3*(sqrt(6) + sqrt(2)).

I don't know how much that helps, since I'm too lazy to do the algebra of finishing the problem from there, but I like looking for ways to use those perfect squares! It seems to give the answer in a simpler form, anyway.

Dave Marain said...

Nice Joshua! That method should be required for all mathletes!

Kate--
Even if students do not recognize that C is on a semicircle, there's another less well-known theorem that is usually helpful for these kinds of questions:
The midpoint of the hypotenuse of a right triangle is equidistant from the 3 vertices, i.e., the length of the median on the hypotenuse is half the length of the hypotenuse.

Using this theorem, students should recognize that the length of the segment from C to the midpoint, M, of AB is also 6. This produces the 30-60-90 triangle I've been referring to with sides of length 3,6,3√3. The rest follows...

Sean said...

It's an ok problem, using lots of different properties and concepts. I don't personally have an Algebra 2 class, but I would imagine that this would be a great question to get them started for the year.

Kate said...

Dave - I haven't used it with a class yet, school starts next week. We were just playing with it in a district faculty PD lecture thing. The speaker was actually pretty good, I just got distracted by math again.

Also Dave - I like the median solution! These types of lessons are challenging and it's always nice to know different ways the students might go with it.

Joshua - That is a more elegant way to solve that system. I wouldn't have thought of it. Thanks!

Kate said...

Follow up - tried it with precalculus on the first day of school. Most of them went with the pythagorean theorem equation plus the area equation, but very few were able to solve. Nobody had ever heard of completing the square. Nobody could quickly write the proportion involving the altitude without re-drawing similar triangles oriented the same way. Nobody thought about writing expressions for slope. All in all disappointing...this group seems to want to cling to rote procedures and avoid creative problem solving. I have my work cut out for me.

Dave Marain said...

Kate--
Thanks for trying it and reporting back. Do not despair! Your experience matched most of mine and and many others I'm sure. Changing how youngsters look at these problems and overcoming the inclination to use mechanical procedures is always our challenge as students move up to precalculus and higher courses.

I do think that changes in the algebra curriculum and instruction will facilitate this. This is why I am so encouraged by more standardization of the algebra curriculum as we are now seeing with the ADP project. Of course curriculum does not necessarily lead to changing student mindset about problem-solving but it's a start. Also, better and more uniform algebra assessments can help.

I suspect that by the end of the year your students will see mathematics in a new light -- the light you will shine on them...