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Pythagoras


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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.


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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.


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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.


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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 by Norman Albers
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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.


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


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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'.


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I guess I am alone in my stooopidity.

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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]

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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.


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


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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!

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