Pythagoras

[Norman Albers]

What elements of carpenters' logic do you need to demonstrate the Pythogorean relation of any right triangle, that: [math]a^2+b^2=c^2[/math]?

[timo]

The assumption that the floor is flat and perpendicular to a plummet?

[Norman Albers]

Smart answer I had not thought of! We are talking planar geometry.

[StringJunky]

The 345 Rule. 3 lengths of wood, one each in a length ratio of 3:4:5 make a perfectly square corner when joined to form a triangle

[Norman Albers]

Yup, but prove this from a few geometric ideas.

[the tree]

What do you mean by Carpenter's Logic? I think this is something I don't know about.

[Norman Albers]

I mean what geometric principles do you need to prove this?

[Sisyphus]

Euclid takes 47 propositions to prove it from first principles, but those are real proofs. You can show it intuitively with a single picture, as shown at "proof #9" at this link:

 

http://www.cut-the-knot.org/pythagoras/index.shtml

[Norman Albers]

I added one line to the construction. I think I then use basically two principles to write a trigonometric algebraic proof. I do not use laws of sines and cosines, per se.

---

Merged post follows:

Consecutive posts merged

I have only started reading Penrose's monster book, The Road to Reality. He mentions Pythagorus in deep considerations of epistemology, how we know anything in mathematics in this case. I realized the problem made me feel stooopid for a couple of hours. I respect all such "simple exercises", especially when it is clear there is something to learn.

---

Merged post follows:

Consecutive posts merged

Maybe I use the word "trigonometry" wrongly, maybe I mean to say geometric algebraic proof. These are all the things I am sorting out in my own understanding.

---

Merged post follows:

Consecutive posts merged

The 345 Rule. 3 lengths of wood, one each in a length ratio of 3:4:5 make a perfectly square corner when joined to form a triangle

Isn't this an amazing quirk of numbers? I think so, and don't I recall hearing there are no other such integer squares? Edited December 27, 2009 by Norman Albers

[scalbers]

I recall the integer squares 5:12:13 and I see this can be constructed for every odd number:

 

http://www.friesian.com/pythag.htm

[Norman Albers]

COOL!!! I'd forgotten those. Isn't this sort of a Fermat's theorem? [sTERNLY, This is a bit off-topic because I don't really care about just these cases, but dammit I am the OP and am having fun.} OK now I see that the truth is the opposite, right, that there are always further tripletts? How cool. Thank you, my oh brother. When we are willing to feel stupid, it is possible we might learn something.

---

Merged post follows:

Consecutive posts merged

I composed a paperette called "Monty Pythagorus" and maybe scalbers can get it into my cache in a day or two.

---

Merged post follows:

Consecutive posts merged

So hurry on in with your answers, kids, to carve out your fifteen minutes of fame. HINT: The line I draw is inside the triangle. I need only 'carpenters' tools'.

---

Merged post follows:

Consecutive posts merged

I guess I am alone in my stooopidity.

Edited December 28, 2009 by Norman Albers
Consecutive posts merged.

[DJBruce]

You can do the proof by similar triangles. That can be done with just carpenters tools, and basic geometry. That proof will look like:

 

[math]\frac{b}{c} = \frac{AH}{b}[/math] and [math]\frac{a}{c} = \frac{HB}{a}[/math]

 

[math]b^{2}=(AH)©[/math] and [math]a^{2}=(HB)©[/math]

 

[math]a^{2}+b^{2}=©(AH + HB)=c^{2}[/math]

 

However, being a carpenter that knows my trigonometry I know I can use the pythagorean identity to prove the pythagorean theorem.

 

Given: [math] sin^{2}+cos^{2}=1[/math]

 

[math]\frac{opposite^{2}+ adjacent^{2}}{hypotenuse^{2}}=1[/math]

 

[math]opposite^{2}+ adjacent^{2}= hypotenuse^{2}[/math]

 

[math]a^{2}+b^{2}=c^{2}[/math]

[Norman Albers]

fifteen minutes of fame to DJBruce, HUZZAH, HUZZAH.. I am not here interested in the trigonometric stuff because I consider that derivative, later, understanding. I use similarity of triangles, as well as area of any triangle. First we see this as 1/2 ab, but then with the construction, it is also 1/2 c h. Further sober inspection tomorrow will hoppfuly confirm my inspcetions, and our h's.

---

Merged post follows:

Consecutive posts merged

Wow, it seems you use only the principles of similar triangles, and not the areas like I did. Good one.

---

Merged post follows:

Consecutive posts merged

DJBruce achieved a tighter solution than I did, so I think he deserves two FIFTEEN MINUTESES OF FAME.One does not need to appeal to the triangle area formula, if you write the two similarity relations as he does. Congrats! . . To hell with my paperette. Cheers!

Edited December 31, 2009 by Norman Albers
Consecutive posts merged.

[scalbers]

Wikipedia has a nice summary of various proofs, including the one above:

 

http://en.wikipedia.org/wiki/Pythagorean_theorem#Proofs

[Norman Albers]

I am fascinated to see the basic principles at work. I aspire to greater facility in 4-space. I should get my act together in 2-space, LOL.

[Norman Albers]

Thanks to timo for the headsup on wowies in the floor. This is what we get from GR, and old houses.

Edited January 3, 2010 by Norman Albers

[scalbers]

For all odd numbers one can construct a Pythagorian triangle.

 

Let A be odd.

Then B = (A^2 / 2) - 0.5

And C = (A^2 / 2) + 0.5

[phyti]

I was always impressed by the simplicity of this geometric demonstration seen in a text book years ago. The algebraic interpretation:

(a+b)^2 - 2ab = c^2.

[TheDrBraniac]

Here is a proof;

 

The rearrangement of the triangles is a visual proof.

---

Merged post follows:

Consecutive posts merged

Sorry the link didn't pop up;

 

here is the link;

 

http://en.wikipedia.org/wiki/File:Pythag.gif

 

Cheers!