1997-2006 Academic Years
10 March 1998 Hoi Huynh [Clemente HS]
She used graph paper to illustrate the Pythagorean Theorem, and showed a demonstration of it using the areas of the regions on the graph paper. She extended the resulting angle figure, rotated the figure, and applied ways of resolving the differences.
In the diagram, the area of the big square is (a + b)2, whereas the area of each of the four triangles of sides a and b is ab. The area of the inner square is c2, and
(a + b)2 = c2 + 4 ´½ab
a2 + b2 = c2.
Comment by Porter Johnson: The Pythagoreans were an ancient Greek cult who explored the mystical wonders of mathematics, and who were forbidden to reveal their mathematical discoveries upon punishment by death. Check out the websites http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Pythagoras.html and http://www-groups.dcs.st-and.ac.uk/~history/Diagrams/PythagorasTheorem.gif
16 March 1999: Al Tobecksen [Richard Voc HS]
He brought in some carpenter's squares and had us work with a way of finding the hypotenuse of a right triangle using a square (90deg) and a yard-meter stick. He passed out a table a, b, c. and had us use the scale on the right angles for the sides, and the meter- yard stick as the hypotenuse: and then use a calculator.
11 April 2000: Bill Shanks (Joliet Junior College, Music)
showed us how to generate Pythagorean numbers. He defined these as integers a, b, c that satisfy the relation
10 October 2000
Marilynn Stone (Lane Tech HS)
gave each of us a resealable sandwich bag containing these items:
She then challenged us to assemble the pieces together to form a rectangle. (Earl Zwicker was first to succeed - but he has had many years of practice doing Harald Jensen's Pythagorean puzzle http://mypages.iit.edu/~smile/ph9711.html to introduce the Phenomenological Approach to new SMILE teachers!) She then showed us how to prove the Pythagorean Theorem using the puzzle. She did this by projecting transparencies of the puzzle pieces so we could literally "see" the reasoning. Pretty! But then she showed us how to make a proof with just half the puzzle, using the large blue square and the red triangles at its sides to form an even larger square. Marilynn labeled each side of the blue square with a "c", and contiguous hypotenuse with a "c" also. Then the short sides of the 4 triangles were labeled "b", and the longer sides, "a". From this it was clear that the areas obeyed the following relation:2 green rectangles
3 blue squares
4 red triangles.
04 December 2001: Hoi Huynh (Clemente HS) Areas of Polygons
Hoi showed us a balanced, right procedure for calculating the area of any polygon, using the standard formula for the area of a trapezoid:
Area = [ (top length + bottom length) /2 ] ´ height = [ (a + b)/2 ] ´ h
The balance comes in because one must "balance" the length of
the top and the
bottom (a+b)/2, and the
right comes in because one must use the "right angle" or
perpendicular distance between the
sides; ie, the height h We can use the same principle for
For the parallelogram the top and bottom sides are equal, and we get
Area = a ´ h.
For the triangle the top side has length zero, and the bottom side has length a, so that the average is a/2, Thus, Area = a ´ h / 2.
By taking any polygon and cutting it into trapezoidal pieces, she showed us how to calculate its area.
Next Hoi showed us how to prove the Pythagorean Theorem using
formulas. She took a square of edge length (a + b), and divided
the edges into components of lengths a and b, as shown:
The total area of the square is (a + b) 2. Inside that square, there is a tilted smaller square of side c, with area c2. In addition, there are four congruent right triangles, each with sides a, b, and hypotenuse c. Thus, the area of the larger square may be written as the sum of the area of the inner square plus the areas of the four (identical) triangles. ...
Hoi told us that Mrs Pythagoras really invented the theorem while she was supposed to be knitting her husband's socks, and she let him pretend to discover it on his own. Perhaps this interesting story may not be completely accurate:
Pythagoras founded a philosophical and religious school in Croton (now Crotone, on the east of the heal of southern Italy) that had many followers. Pythagoras was the head of the society with an inner circle of followers known as mathematikoi. The mathematikoi lived permanently with the Society, had no personal possessions and were vegetarians. They were taught by Pythagoras himself and obeyed strict rules. The beliefs that Pythagoras held were :-Source: http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Pythagoras.html
Both men and women were permitted to become members of the Society, in fact several later women Pythagoreans became famous philosophers. The outer circle of the Society were known as the akousmatics and they lived in their own houses, only coming to the Society during the day. They were allowed their own possessions and were not required to be vegetarians.
- that at its deepest level, reality is mathematical in nature,
- that philosophy can be used for spiritual purification,
- that the soul can rise to union with the divine,
- that certain symbols have a mystical significance, and
- that all brothers of the order should observe strict loyalty and secrecy.
Very nice, Hoi!
25 March 2003: Hoi Huynh
[Mathematics Teacher] Maintaining Balance
Hoi demonstrated that, when a standing cylinder has better balance, it will have greater volume. In the previous example in making the lateral surface of a cylinder with an 8.5" ´ 11" sheet of transparency paper, Hoi pointed out that the shorter cylinder has better "balance" than the longer one. She said that this was connected to the fact that the top (circular) edge was closer to its geometrical center for the shorter cylinder than for the longer cylinder.
Hoi pointed out that, of all quadrilaterals that have a given perimeter p , the square has the greatest area, A = p2/16. For all shapes of a given perimeter p, the circle has the greatest area, A = p2/(4p). For three dimensional bodies of surface area A, the sphere has the greatest volume, V = [ A /(4p)]1.5 /3.
Hoi also mentioned that the formula for the Area of a Trapezoid of bases b1 and b2 and height h,
Thanks for the insights, Hoi!
08 March 2005: Bill Shanks [retired, Joliet New Lenox
Bill reminded us of the Pythagorean Theorem for a right triangle of sides (a, b, c = b + d), where c is the hypotenuse:
/| c = b + d / | b / | /___| a
|I||I + 1||J|
Note that the scale of subsequent triangles increases by a factor of about 6 each time. For discussion see the solution to Problem #152 on this website: http://www.dansmath.com/probofwk/probar16.html#anchor585356. For a general solution and a relation of these triangles to the Pell Equation, see http://wapedia.mobi/en/Pell_number.
Thanks for the ideas, Bill!