When my review copy arrived yesterday I was disappointed to see the hard cover. I suspected that would mean a higher price than most people would find comfortable. The damage is even worse than I expected - $65 (or $53 for MAA members). I think it's a shame to limit the distribution of such good material with a cover price like that. [I had considered approaching the MAA about publishing my book, but perhaps they don't know how to produce books at affordable prices.] Note added five years later: Now it's available used for just over $30. Much better!
The only other objection I have to this lovely book is that solutions immediately follow the problems posed, so it's hard to resist the temptation to peek. I just now managed to resist temptation, and sat with the problem below (from page 175) until I got it. I don't teach geometry, so my geometric intuitions aren't well-honed. It took me a while, and I reverted to algebraic reasoning for parts of it.
Given the square ABCD, with side length 1 and circular arcs centered at each vertex, find the area of the region at the center - without using calculus.
I broke it down into shapes I could handle, and got an answer something like his but not quite. I haven't yet found the discrepancy. (My process was very different from his.)
Simonson includes a number of problems I haven't seen before, which is quite a feat after all the grazing I've done online in the past few years. And the problems are at lots of levels, so there is much to chew on whatever your mathematical sophistication.
I like his perspective on how math should be taught and learned. Here he is on memorization (page 169):
Every mathematical idea has a story. To remember the idea, just recall the story. In mathematics, the stories are proofs and the endings are theorems. The more you turn a proof into a story, the easier it is to remember the ending. Can you tell me what you did last summer? Of course you can. Did you memorize that? Surely not; there is a context and one thought leads to another. Of course it can get a little tedious recalling a story a hundred times just to get to the ending, so sooner or later one just knows the ending. This is the kind of memorizing that a student should do with mathematics.An example of this comes earlier. On page 131 he writes, "In the 1960's, it was popular in the U.S. for the middle school math curriculum to include a square root algorithm. The logic of the algorithm ... is clever but cumbersome, unintuitive, and inaccessible to students and most teachers." He compares this method to a Babylonian method: To find the square root of x, guess (r1), divide x by your guess (x/r1), for a better estimate average your guess with this quotient (r2=(r1+x/r1)/2), repeat. Simple, elegant, transparent.
I didn't learn an algorithm for finding the square root in my U.S. math classes, but I did learn one during my junior year, while attending high school in Brazil (as an exchange student). I've wished I could remember how it worked, but never could dredge it up. I don't think it was anything this straightforward.
Most students will not forget the Babylonian method because it makes sense. It is a story. It can be remembered because it can be reconstructed. There is context, purpose, and structure. (page 132)
I'll be sharing quite a few nuggets from this book with my students. If you can afford it, I highly recommend getting this book.
How about this:
ReplyDeleteBreaking the figure down into mutually exclusive regions, there are 3 unique shapes:
CDF, CEF, EFGH
Let the areas of those shapes be x, y, and z, respectively.
[1] 2x + y = 1-pi/4 (quarter circle)
[2] 4x + 4y + z = 1 (total square)
[3] x + 2y + z = pi/3-sqrt(3)/4 (using the fact that triangle ADE is equilateral)
add [1] + [3]:
3x + 3y + z = 1+pi/12-sqrt(3)/4
multiply by 4:
12x + 12y + 4z = 4 + pi/3 - sqrt(3)
multiply [2] by 3;
12x + 12y + 3z = 3
subtract:
z = 1+pi/3-sqrt(3)
Very nice! It took me a bit to see [3], but I've got it now. You agree with the book, so I must be wrong. I'll write mine out neatly, so I can find my mistake. Mine is more complicated. I divided the figure up with vertical and horizontal lines through the middle.
ReplyDeleteDraw a square EFGH. This divides the region of interest into the square EFGH and four segments of a circle each of radius 1 and included angle theta, congruent to segment EDH
ReplyDeleteThe triangle AED is equilateral, so the altitude of E altE to AD is sqrt(3)/2. Similarly, the altitude of G altG to BC is sqrt(3)/2.
Therefore, the diameter of the square
EG = 1 - (1-altE) - (1-altG)
= 1 - 1 + altE - 1 + altG
= altE + altG - 1
= √(3)/2 + √(3)/2 - 1
= √(3) - 1.
The area of
EFGH = (EG)^2/2
= (√(3) - 1)^2/2
= (√(3)^2 - 2√(3) + 1)/2
= (3 - 2√(3) + 1)/2
= (4 - 2√(3))/2
= 2-√(3).
The angle ADE is τ/6, the angle CDH is τ/6, and the angle ADC is τ/4. So the angle
EDH = ADC - (ADC - ADE) - (ADC-CDH)
= ADE + CDH - ADC
= τ/3 - τ/4
= τ/12.
The area of a segment is R^2(θ - sin θ)/2. In this case, R=1, θ = τ/12, and sin θ = 0.5. So the area of segment EDH = (τ/12 - 0.5)/2.
So the total
area = EFGH + 4EDH
= 2-√(3) + 4(τ/12 - 1/2)/2
= 2-√(3) + τ/6 - 1
= 1 + τ/6 -√(3).
Which is the same answer Hao got.
I learned a manual method of doing square roots in school in the early 80's. I can't remember if it was taught, or if I just learned it. The method superficially resembled long division, in that you would find successive digits by doing a trial multiplication, subtraction, and "bringing down" the next group of digits. It works, but can be slow, and requires a small amount of "guessing" for each digit. You don't have to decide in advance how many digits you go.
The Babylonian method doesn't require any guessing, but I don't know how tolerant it is of partial calculations (can you increase your precision as you go?). It is also equivalent to Newton's Method.
I borrowed the "bulging square" problem, but then couldn't remember where I got it from! Hao pointed me back here.
ReplyDeleteMy friend started differently. He determined that EH is π/6 worth of arc, and found the area of that sector. Not so interesting yet? Then he subtracted the area of ΔDEH. See that little piece left? (is that called a lune?) There are four of them, plus the square.
The π/6, by the way, comes from recognizing the equilateral triangles.
Me? I got stuck and quit.
Jonathan
Ahh, Hao, I didn't recognize you over at JD's, as Mr. H. I have finally discovered your blog and added it to my list.
ReplyDeleteI'm loving Pat's extension (from JD's blog): "... given eight unit spheres centered on the vertices of a unit cube… find the volume of intersection… And in case you wondered, I just made that one up, although I can’t believe I would be the first…"
I haven't begun to think about it, I think I need to let it stew in my brain for a while.