Sunday, April 3, 2011

Book Review: Rediscovering Mathematics, by Shai Simonson










  1. How about this:
    Breaking the figure down into mutually exclusive regions, there are 3 unique shapes:
    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
    z = 1+pi/3-sqrt(3)

  2. 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.

  3. Draw 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

    The 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.

  4. I borrowed the "bulging square" problem, but then couldn't remember where I got it from! Hao pointed me back here.

    My 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.


  5. 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.

    I'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.


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