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hey aserine

Discussion in 'Channel Zero' started by tue skinny, Feb 17, 2002.

  1. tue skinny

    tue skinny Elite Member

    Joined: Jul 3, 2001 Messages: 4,781 Likes Received: 0
    lets play hockey: and anyone else who wants to join

    street , ghetto style

    ur on my man !
     
  2. -Rage-

    -Rage- 12oz Loyalist

    Joined: Apr 12, 2001 Messages: 10,006 Likes Received: 27
  3. Pinche Hoto

    Pinche Hoto Guest

    Hey aserine why are you so fucking gay
     
  4. ASER1NE

    ASER1NE Veteran Member

    Joined: Oct 15, 2001 Messages: 7,577 Likes Received: 2
    i was speaking figuratively , considering you live many miles away , but if ever we meet its on .......TT would probably be down , among others

    hockey = best sport ever
     
  5. ASER1NE

    ASER1NE Veteran Member

    Joined: Oct 15, 2001 Messages: 7,577 Likes Received: 2
    Lick my sweaty ball sack .
     
  6. ASER1NE

    ASER1NE Veteran Member

    Joined: Oct 15, 2001 Messages: 7,577 Likes Received: 2
    learn something , its fun .

    -------------PYTHAGOREAN THEOREM----------------

    Let's build up squares on the sides of a right triangle. The Pythagoras' Theorem then claims that the sum of (areas of) the two small squares equals (the area of) the large one.
    In algebraic terms, a2+b2=c2 where c is the hypotenuse while a and b are the sides of the triangle.
    The theorem is of fundamental importance in the Euclidean Geometry where it serves as a basis for the definition of distance between two points. It's so basic and well known that, I believe, anyone who took geometry classes in high school couldn't fail to remember it long after other math notions got solidly forgotten.
    I plan to present several geometric proofs of the Pythagorean Theorem. An impetus for this page was provided by a remarkable Java applet written by Jim Morey. This constitutes the first proof on this page. There is nothing like learning while doing and, as an exercise in Java programming, I'll later offer an original Java applet. But, for now, let consider several plain HTML proofs.

    Remark
    The statement of the Theorem was discovered on a Babylonian tablet circa 1900-1600 B.C. Whether Pythagoras (c.560-c.480 B.C.) or someone else from his School was the first to discover its proof can't be claimed with any degree of credibility. Euclid's (c 300 B.C.) Elements furnish the first and, later, the standard reference in Geometry. Jim Morey's applet follows the Proposition I.47 (First Book, Proposition 47), mine VI.31. The Theorem is reversible which means that a triangle whose sides satisfy a2+b2=c2 is right angled. Euclid was the first (I.48) to mention and prove this fact.
    W.Dunham [Mathematical Universe] cites a book The Pythagorean Proposition by an early 20th century professor Elisha Scott Loomis. The book is a collection of 367 proofs of the Pythagorean Theorem and has been republished by NCTM in 1968.
    Pythagorean Theorem generalizes to spaces of higher dimensions. Some of the generalizations are far from obvious.
    The Theorem whose formulation leads to the notion of Euclidean distance and Euclidean and Hilbert spaces, plays an important role in Mathematics as a whole. I began collecting math facts whose proof may be based on the Pythagorean Theorem.
    Wherever all three sides of a right triangle are integers, their lengths form a Pythagorean triple (or Pythagorean numbers). There is a general formula for obtaining all such numbers.


    Proof #2
    We start with two squares with sides a and b, respectively, placed side by side. The total area of the two squares is a2+b2.
    The construction did not start with a triangle but now we draw two of them, both with sides a and b and hypotenuse c. Note that the segment common to the two squares has been removed. At this point we therefore have two triangles and a strange looking shape.
    As a last step, we rotate the triangles 90o, each around its top vertex. The right one is rotated clockwise whereas the left triangle is rotated counterclockwise. Obviously the resulting shape is a square with the side c and area c2.

    Proof #3
    Now we start with four copies of the same triangle. Three of these have been rotated 90o, 180o, and 270o, respectively. Each has area ab/2. Let's put them together without additional rotations so that they form a square with side c.

    The square has a square hole with the side (a-B). Summing up its area (a-B)2 and 2ab, the area of the four triangles (4·ab/2), we get

    c2 = (a-B)2+2ab = a2-2ab+b2+2ab = a2+b2

    Proof #4
    The fourth approach starts with the same four triangles, except that, this time, they combine to form a square with the side (a+B) and a hole with the side c. We can compute the area of the big square in two ways. Thus

    (a+B)2=4·ab/2+c2
    simplifying which we get the needed identity.

    Proof #5
    This proof, discovered by President J.A. Garfield in 1876 [Pappas], is a variation on the previous one. But this time we draw no squares at all. The key now is the formula for the area of a trapezoid - half sum of the bases times the altitude - (a+B)/2·(a+B). Looking at the picture another way, this also can be computed as the sum of areas of the three triangles - ab/2 + ab/2 + c·c/2. As before, simplifications yield a2+b2=c2.


    Proof #6
    We start with the original triangle, now denoted ABC, and need only one additional construct - the altitude AD. The triangles ABC, BDA and ADC are similar which leads to two ratios:

    AB/BC=BD/AB and AC/BC=DC/AC.
    Written another way these become
    AB·AB=BD·BC and AC·AC=DC·BC
    Summing up we get
    AB·AB+AC·AC=BD·BC+DC·BC=(BD+DC)·BC=BC·BC.

    Proof #7
    The next proof is taken verbatim from Euclid VI.31 in translation by Sir Thomas L. Heath. The great G. Polya analyzes it in his Induction and Analogy in Mathematics which is a recommended reading to students and teachers of Mathematics.

    In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle.
    Let ABC be a right-angled triangle having the angle BAC right; I say that the figure on BC is equal to the similar and similarly described figures on BA, AC.

    Let AD be drawn perpendicular. Then since, in the right-angled triangle ABC, AD has been drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC adjoining the perpendicular are similar both to the whole ABC and to one another [VI.8].

    And, since ABC is similar to ABD, therefore, as CB is to BA so is AB to BD [VI.Def.1].

    And, since three straight lines are proportional, as the first is to the third, so is the figure on the first to the similar and similarly described figure on the second [VI.19]. Therefore, as CB is to BD, so is the figure on CB to the similar and similarly described figure on BA.
    For the same reason also, as BC is to CD, so is the figure on BC to that on CA; so that, in addition, as BC is to BD, DC, so is the figure on BC to the similar and similarly described figures on BA, AC.

    But BC is equal to BD, DC; therefore the figure on BC is also equal to the similar and similarly described figures on BA, AC.

    Therefore etc. Q.E.D.

    Confession
    I got a real appreciation of this proof only after reading the book by Polya I mentioned above. I hope that a Java applet will help you get to the bottom of this remarkable proof. Note that the statement actually proven is much more general than the theorem as it's generally known.

    Proof #8

    Playing with the applet that demonstrates the Euclid's proof (#7), I have discovered another one which, although ugly, serves the purpose nonetheless.

    Thus starting with the triangle 1 we add three more in the way suggested in proof #7: similar and similarly described triangles 2, 3, and 4. Deriving a couple of ratios as was done in proof #6 we arrive at the side lengths as depicted on the diagram. Now, it's possible to look at the final shape in two ways:

    as a union of the rectangle (1+3+4) and the triangle 2, or
    as a union of the rectangle (1+2) and two triangles 3 and 4.
    Equating areas leads to


    ab/c · (a2+b2)/c + ab/2 = ab + (ab/c · a2/c + ab/c · b2/c)/2

    Simplifying we get

    ab/c · (a2+b2)/c/2 = ab/2, or (a2+b2)/c2 = 1

    Remark
    On a second look at the diagram, there is a simpler proof. Viz., look at the rectangle (1+3+4). Its long side is, on the one hand, plain c while, on the other, it's a2/c+b2/c and we again have the same identity.

    Proof #9

    Another proof stems from a rearrangement of rigid pieces, much like proof #2. It makes the algebraic part of proof #4 completely redundant. There is nothing much one can add to the two pictures.

    (My sincere thanks go to Monty Phister for the kind permission to use the graphics.)

    There is an interactive simulation to toy with.

    Proof #10
    This and the next 3 proofs came from [PWW].

    The triangles in Proof #3 may be rearranged in yet another way that makes the Pythagorean identity obvious.

    Proof #11
    Draw a circle with radius c and a right triangle with sides a and b as shown. In this situation, one may apply any of a few well known facts. For example, in the diagram three points F, G, H located on the circle form another right triangle with the altitude FK of length a. Its hypotenuse GH is split in the ratio (c+B)/(c-B). So, as in Proof #6, we get a2 = (c+B)(c-B) = c2 - b2.

    Proof #12
    This proof is a variation on #1, one of the original Euclid's proofs. In parts 1,2, and 3, the two small squares are sheared towards each other such that the total shaded area remains unchanged (and equal to a2+b2.) In part 3, the length of the vertical portion of the shaded area's border is exactly c because the two leftover triangles are copies of the original one. This means one may slide down the shaded area as in part 4. From here the Pythagorean Theorem follows easily.

    Proof #13
    In the diagram there is several similar triangles (abc, a'b'c', b'x, and a'c'.) We successively have

    y/b = b'/c, x/a = a'/c, cy + cx = aa' + bb'.

    And, finally, cc' = aa' + bb'. This is very much like Proof #6 but the result is more general.
     
  7. tue skinny

    tue skinny Elite Member

    Joined: Jul 3, 2001 Messages: 4,781 Likes Received: 0
    yah i talk to tt his always down for a game of street hockey.
     
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