High School Mathematics-Physics SMILE Meeting
1997-2006 Academic Years
Pythagorean Theorem

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 ;
or
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

a2 + b2 = c2.
An example most of us know is 3, 4, 5. These are very useful, especially to physics teachers making up exam problems involving vectors where you want the answers to work out as integers. Bill asked us to give him a 2 digit odd number, so one of us named "17." Bill wrote it on the board, and then he wrote
172 + 1442 = 1452.
How did he do that in his head?!! He did it again with "14" (left as an exercise to the reader!), but then he explained. For the sides of a right triangle write: side a, side b, and hypotenuse c = b+2. Then
a2 + b2 = (b+2)2.
A little algebra results in b = (a/2)22 - 1. Let a = 2x, and let x be any integer you like, say 4. Then, a = 8, b = 15, and c = b + 2 = 17. So
82 + 152 = 172  !
It really works! Now make a table with x taking on various integer values, 3, 4, 5, 6,... and you thus generate a whole series of Pythagorean numbers! See the website http://www.cut-the-knot.com/pythagoras/ Thanks, Bill!

10 October 2000 Marilynn Stone (Lane Tech HS)
gave each of us a resealable sandwich bag containing these items:

2 green rectangles
3 blue squares
4 red triangles.
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 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:
(a + b)2 = c2 + 4 ´ (ab/2)
{Just draw the picture!) One needs then only to complete the algebra to show that a2 + b2 = c2. Very nice, Marilynn! She then told us she had generated the transparencies by printing from her computer printer onto a transparency sheet. Neato! She gave us a handout page for with directions for her students to actually measure a, b, c on four different right triangles and check out the Pythagorean Theorem numerically for each one.
Good reinforcement.

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:

TRAPEZPOID

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 other figures

PARALLELOGRAM

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 she pointed out that the area of a circle of radius r [such as a pie] is equal to pie ´ r2, or p r2. [PJ comment:  You could slice the pie into ever smaller equal-sized pieces, and turn the pieces alternately "up" and "down", to make a figure that becomes a simple trapezoid in the limiting case. Half of the pie edge is on top, and half is on the bottom, so that the "height" of the trapezoidal figure is r, the circle radius.  Thus, by the trapezoidal formula, in the limiting case the area is   [ (p r+ p r) /2] ´ r, or p r2].

Next Hoi showed us how to prove the Pythagorean Theorem using area formulas.  She took a square of edge length (a + b), and divided each of 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. ...

(a + b) 2  =  c2 + 4 [a  ´ b / 2]
Writing out both sides of this expression, we get
a2 + 2a ´ b +b2 =  c2 + 2a  ´ b

or
a2 + 2a ´ b +  b2 =  c2 + 2a  ´ b

with the result
a2 + b2  =  c2  
which is the Pythagorean Theorem.

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 [2]:-
  1. that at its deepest level, reality is mathematical in nature,
  2. that philosophy can be used for spiritual purification,
  3. that the soul can rise to union with the divine,
  4. that certain symbols have a mystical significance, and
  5. that all brothers of the order should observe strict loyalty and secrecy.
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.
Source: http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Pythagoras.html

Very nice, Hoi!

25 March 2003: Hoi Huynh [Mathematics Teacher]      Maintaining Balance in Mathematics
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,

Area = (b1 + b2) /2 ´ h
yields the areas of a rectangle (b1 = b2) and a triangle {b2 =0) as special cases.

Thanks for the insights, Hoi!

08 March 2005: Bill Shanks [retired, Joliet New Lenox environs]              Geometry
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
c2 = a2 + b2
(b + d)2 = a2 + b2
2 b d + d2 = a2 
2 b d = a2 - d2
b =  (a2 - d2 ) / (2d)
Bill pointed out, when a and b are chosen so that b is an integer, we obtain a right triangle with integer sides.  Bill found that Interesting Right Triangles were obtained for the cases d = 1, 2, 8, 9, 18, 25, 49, 50, ... .  These numbers are all of the form  d = 2p nq, where n is odd, and p is either zero or an odd number. For example, with  d = 1, we get b = (a2 -1) / 2, and a must be an odd number for b to be an integer.  We thus obtain these right triangles with integer sides:
a b c=b+1
3 4 5
5 12 13
7 24 25
9 40 41
11 60 61
13 84 85
25 312 313
35 612 613
999   499000   499001
For d =2, we obtain  b = (a/2)2 -1, so that a must be an even number.  We thus obtain the table
a b c=b+2
4 3 5
6 8 10
8 15 17
10 24 26
12 35 37
100 2499 2501
1000   249999   250001
For d = 8, we obtain b = (a/4)2 -4, so that a must be divisible by 4. We obtain these triangles:
a b c=b+8
12 5 13
16 12 20
20 21 29
24 32 40
28 45 53
36 77 85
888   49280  48288
For d = 9, we obtain b = [ (a/3)2 - 9 ] / 2, so that a must be odd and divisible by 3. We obtain these triangles:
a b c=b+9
15 8 17
21 20 41
27 36 45
33 56 65
39 80 89
111 680 6893
999   55440   55449
Bill finds this useful in making up exam problems, when he wants all the sides of the right triangle to be integers. Fred Schaal mentioned Nearly Isosceles Integer Right Triangles, for which the the sides of the triangle are  (I, I+1, J), where I and J are integers.  Here are the first six examples of this infinite set:
I I + 1 J
3 4 5
20 21 29
119 120 169
696 697 985
  4059    4060     5741  
  23660    23661     33461  

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 the following PDF file:  http://hometown.aol.com/jpr2718/pell.pdf.

Thanks for the ideas, Bill!