Monday, January 29, 2007

Problems for 1-30-07

At the suggestions of readers, I'm reducing the number of daily problems to 3 or 4. I may also post problems on alternate days. Again, my main purposes in posting these questions is to supplement content, provide instructors with an additional resource of standardized test questions and encourage greater depth of understanding of important math concepts and strategies for problem-solving. Please view these questions in that light. I am very appreciative of your support, comments and suggestions for improving this process.


Totally_clueless said...

I almost got taken in by (1)

For (2), is it possible for the students' assumption of a rectangle with the auxiliary segment to actually hurt them in a multiple choice answer?

A neat (as decreed by me :-) generalization of (2) is to make the right vertical segment of length 7(or 8 or anything less than 6*sqrt(3)) (the two angles are still 60 degrees) and find the perimeter of the figure. I was surprised by the answer. The area seems to be more difficult in this case.

Jonathan said...

I liked 2. But 3 is very very slick. Could use it, time permitting, to let students try to 'break down the front door' and then generate some very nice discussion about symmetry, perhaps working backwards, etc. Handled correctly, it could be quite a memorable problem.

Dave Marain said...

i like that generalization! i want to play with it some more...
jonathan --
thanks for the encouragement! i needed that! this is h*** week for supervisors as the marking period ends, midyear evaluations of teachers are due, supplies budget is due, 3rd round of observations, covering an extra class for the next 3 weeks, dealing with an array of discipline issues and appeals for next year, trying not to fall behind in my AP class, and just a few more thousand issues going on at home... (yes, pls pls send me all the pity you have, then tell me your life is just as much fun!)...
my greatest ideas are far from original -- i see interesting problems in SAT books or math contests and i do what you do -- i ask 'what if' or how could i make an interesting variation (totally_clueless did just that); sometimes i just have an inspiration and one of these problems just happens; i've been writing these for 30+ years and writing for math contests for over 20 years (don't ask me to find my archives!); from sheer experience, i usually have a pretty good sense of how students will fare on these questions

well, at least the number of interesting math problems is of the order aleph-null or maybe 2 raised to that!
now, for homework boys and girls, for one million dollars, solve the Riemann Hypothesis -- explain it, you ask?? no way - it's a bonus!

Dave Marain said...

'Official' Answers"
1) (E)
2) Perimeter: 24 + 8*sqrt(3); A = 72 - 12*sqrt(3)
3) 6

your twist is definitely more difficult; if we want the intersections of the 60 degree rays from the endpoints of the vertical segments to 'meet', then I calculated that the longer segment must be less than or = to 6 + 4*sqrt(3) in which case a quadrilateral is formed with perimeter 24 + 12*sqrt(3); pls check my work for careless or logic errors

Eric Jablow said...

There is actually a fascinating historical discussion contained within (2). Forget about the perimeter and area problem; just consider the sides with length 6 and 12, and complete them to a quadrilateral.

In 1733, Giovanni Girolamo Saccheri S.J. published a book Euclides ab omni naevo vindicatus (Euclid Freed of Every Flaw) where he analyzed such a figure in an attempt to prove Euclid's parallel postulate from the other postulates. He considered three cases:

1. The remaining angles of the quadrilateral are right angles; this implies the parallel postulate.

2. The remaining angles are obtuse; he showed that this leads to a contradiction.

3. The remaining angles are acute; he derived results that were so non-intuitive that he felt them to be contradictions; these are in fact theorems of hyperbolic non-Euclidean geometry. He didn't have the language to realize what he could have been proving.

His book has been called the first work in non-Euclidean geometry, and these figures are called Saccheri quadrilaterals in his honor.

Anonymous said...

Hi Dave,

According to my calculations, if the length of the right vertical is l, with the left vertical being 6, and the horizontal being 12, then we need
4*sqrt(3)-6 <= l <= 4*sqrt(3)+6 for a concave pentagon to be formed.

The perimeter is 18+l+6*sqrt(3)

The area is (sqrt(3)/4)*(l-6)^2 + 6*sqrt(3)*(l-6)+12*sqrt(3)

It is not inconceivable that I have made mistakes in the above.


Dave Marain said...

i got your result for the upper bound; i din't check the lower bound yet...

cool stuff! i've read some about saccheri; this brings it all back -- thanks!