Slickest proof of pythagorean theorem
lemm...@my-deja.com
theorem?
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Jim Heckman
>From: lemm...@my-deja.com
>Date: 1/12/01 5:47 PM Pacific Standard Time
>Message-id: <93oc49$hcp$1...@nnrp1.deja.com>
>
>What is the slickest (e.g. shortest, coolest) proof of the pythagorean
>theorem?
Draw a square, then inscribe another one inside, with each of the
vertices of the inner square touching a (different) side of the outer
one. Figure out the areas of the two squares, and of the four
congruent triangles in the corners of the outer square.
--
Jim Heckman
Zdislav V. Kovarik
:What is the slickest (e.g. shortest, coolest) proof of the
:pythagorean theorem?
:
A matter of taste, of course. There is a recipe for an
"animated" proof, where parts of the squares on the legs
slide, shear and rotate to become parts of the square
on the hypotenuse. Making a movie out of it is a simple
computer graphics project.
Consult
"Notices of the American Mathematical Society, November 2000
(Vol. 47, Number 10), page 1259 (article by Bill Casselman:
Pictures and Proofs).
(I wonder if anyone actually made that short movie.)
Have fun, ZVK(Slavek).
Dale Hall
I'm particularly fond of the proof wherein one drops the altitude from
the right angle to the hypotenuse, thus decomposing the whole triangle
into two triangles, each similar to the original, and for which the
hypotenuses are the two legs of the triangle. By appealing to a lemma
(area of similar figures is proportiional to the square of corresponding
sides), one has the theorem (once the proportionality constant is
removed).
Dale.
Will Self
then evaluate the whole notebook. You may want to divide it into
several separate cells. You will get a lot of pictures. Double click
on the cell markers at the right side to cause the several pictures to
collapse into one. Then choose Cell/Animate Selected Graphics, or just
press control-Y. The pictures should run in a movie. It's nicer if
you choose the "cycle" button at the bottom left. If this overloads
your computer, you can cut down on the number of frames generated.
Look for the line n=60 and change it to n=20. If you like this, I can
email you another one where the triangles rotate instead of translate.
I think this animated proof is nice, but even nicer is if you make it
up as pieces to a puzzle, like a child's puzzle where the pieces fit
into a preassigned area. You could make it out of picture matting, or
masonite if you have the tools. When cutting picture matting, use a
straightedge and a sharp Xacto or utility knife, and don't try to cut
all the way through in one pass. Work accurately and carefully and
make something you can hand down to your grandchildren.
Somewhere on the web I have seen a Java applet that shows the proof
involving dividing the square on the hypotenuse into two rectangles.
Try looking at www.jar.com or gamelan to try to find that applet.
Will Self
---------------------
The Mathematica Code:
a = 5; b = 12; c = Sqrt[a^2 + b^2];
blue = RGBColor[0, 0, 1]; red = RGBColor[1, 0, 0];
t0 = {blue, Polygon[{{0, 0}, {a, 0}, {0, b}}]};
t1[t_] := {blue,
Polygon[# - t{b, a}/c & /@ {{a + b, a}, {a + b, a + b}, {b, a +
b}}]}
t2[t_] := {blue, Polygon[# + {t, 0} & /@ {{0, b}, {b, a + b}, {0, a +
b}}]}
t3[t_] := {blue, Polygon[# + {0, t} & /@ {{a, 0}, {a + b, 0}, {a + b,
a}}]}
sq = {red, Polygon[{{0, 0}, {a + b, 0}, {a + b, a + b}, {0, a + b}}]}
Show[Graphics[{sq, t0, t1[c], t2[a], t3[b]}], AspectRatio -> Automatic]
n = 60; step = (a + b + c)/n; h = 7step; hh = 5step;
Do[Show[Which[
k <= 0, Graphics[{sq, t0, t1[0], t2[0], t3[0]}],
k <= c, Graphics[{sq, t0, t1[k ], t2[0], t3[0]}],
k <= c + hh, Graphics[{sq, t0, t1[c ], t2[0], t3[0]}],
k <= c + a + hh, Graphics[{sq, t0, t1[c], t2[k - c - hh], t3[0]}],
k <= c + a + 2hh, Graphics[{sq, t0, t1[c], t2[a], t3[0]}],
k <= c + a + b + 2hh,
Graphics[{sq, t0, t1[c], t2[a], t3[k - c - a - 2hh]}],
True, Graphics[{sq, t0, t1[c], t2[a], t3[b]}]],
AspectRatio -> Automatic],
{k, -h, c + a + b + 2hh + h, step}]
---------------------------------------------------
In article <93oeid$7...@mcmail.cis.McMaster.CA>,
dan...@my-deja.com
Look in
http://www.math.uu.nl/people/beukers/pythagoras/Pythagoras.html
Sincerely,
Dan
In article <93oeid$7...@mcmail.cis.McMaster.CA>,
kov...@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) wrote:
lemm...@my-deja.com
I know of this proof, and think it is cool. I was hoping someone would
reply with something like, "starting with the obvious identity
????????? (1)
merely multiply both sides of (1) by _____, and then subtract _ _ _.
The theorem is immediate."
I guess there isn't anything _that_ short.
Michael Hochster
: What is the slickest (e.g. shortest, coolest) proof of the pythagorean
: theorem?
Take a look at http://www.cut-the-knot.com/pythagoras/
for some candidates.
Mike
David Petry
lemm...@my-deja.com wrote
>What is the slickest (e.g. shortest, coolest) proof of the pythagorean
>theorem?
The slickest proof of the Pythagorean theorem is, of course,
the one I discovered myself.
It uses calculus!
You need to draw a picture to see it, but start with a right
triangle A-B-C with hypotenuse C. Then hold A constant, and
change B by a small amount, and draw a new C'. Then
drop a perpendicular line from the vertex opposite side
A to the new hypotenuse C'. If you have drawn the diagram
correctly, you will immediately see that
dC/dB = B/C, hence d(C^2) = d(B^2), hence C^2 = B^2 + K
for some constant K, and note that if B = 0, K = A^2
Will Self
That is outrageous, and good, and funny, all at the same time.
Will Self
Ray Vickson
Jim Heckman wrote:
A related question: it is true, isn't it, that if Pythagoras holds,
geometry is Euclidean? If so, how does the proof break down in a
non-Euclidean geometry? (I _think_ I know the answer, but am throwing out
the question for fun.)
>
>
> --
>
> Jim Heckman
--
R. G. Vickson
Department of Management Sciences
University of Waterloo
Waterloo, Ontario, CANADA
Jim Heckman
>Date: 1/15/01 2:49 PM Pacific Standard Time
>Message-id: <3A637E77...@engmail.uwaterloo.ca>
>
>Jim Heckman wrote:
>
>> >Subject: Slickest proof of pythagorean theorem
>> >From: lemm...@my-deja.com
>> >Date: 1/12/01 5:47 PM Pacific Standard Time
>> >Message-id: <93oc49$hcp$1...@nnrp1.deja.com>
>> >
>> >What is the slickest (e.g. shortest, coolest) proof of the pythagorean
>> >theorem?
>>
>> Draw a square, then inscribe another one inside, with each of the
>> vertices of the inner square touching a (different) side of the outer
>> one. Figure out the areas of the two squares, and of the four
>> congruent triangles in the corners of the outer square.
>
>A related question: it is true, isn't it, that if Pythagoras holds,
>geometry is Euclidean?
Assuming homogeneity and isotropy, yes.
>If so, how does the proof break down in a
>non-Euclidean geometry? (I _think_ I know the answer, but am throwing out
>the question for fun.)
You can't draw a (finite) square with 4 interior right angles in non-Euclidean
geometry. The sum of the angles must be >2pi for spherical, <2pi for
hyperbolic.
Also, areas of polygons are much more complicated functions of the lengths
of their sides in non-Euclidean geometry, generally involving spherical or
hyperbolic functions of these lengths as ratios of the "natural" length in such
spaces. (Of course in practice, one computes polygonal areas as the deviation
of the sum of the interior angles from that in flat space.)
--
Jim Heckman