July 2nd, 2017, 07:29 PM  #1 
Banned Camp Joined: Jul 2010 Posts: 118 Thanks: 0  Special triangles
A new method that produces right triangles , with angles close 45 deg' Select a number close to ( 1 + root of 2 ) for example a = 2.4141 b = 0.5 ( a^2  1 ) = 2.41394 c = b +1 = 3.41394 a^2 + b^2 = c^2 
July 2nd, 2017, 07:57 PM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,408 Thanks: 480 Math Focus: Yet to find out. 
No it doesn't. And what does "Select a number close to (1 + root of 2)" have to do with anything? Why not just say, select a number a and b and the result is another arbitrary constant.

July 2nd, 2017, 08:17 PM  #3 
Banned Camp Joined: Jul 2010 Posts: 118 Thanks: 0  It's so simple
I choose a = 2.41 and that's all. b = 0.5 ( a^2  1) = 2.40405 c = b + 1 = 3.40405 a^2 + b^2 = c^2 
July 2nd, 2017, 08:24 PM  #4  
Senior Member Joined: May 2016 From: USA Posts: 823 Thanks: 335  Quote:
I went to Columbia in the 60's, when you could drink in New York at 18. I heard far more cogent discussions in the freshman dorm after the bars closed.  
July 2nd, 2017, 08:24 PM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,900 Thanks: 716 
We don't need your method. Quit being silly... 
July 2nd, 2017, 08:38 PM  #6  
Banned Camp Joined: Jul 2010 Posts: 118 Thanks: 0  The perfect triangle Quote:
b = 0.5 ( a^2  1 ) = 1 + root of 2 c = b + 1 = 2 + root of 2 a^2 + b^2 = c^2  
July 2nd, 2017, 09:04 PM  #7  
Senior Member Joined: May 2016 From: USA Posts: 823 Thanks: 335  Quote:
$b = 0.5(a^2  1) = 0.5(2\sqrt{2} + 3  1) = \sqrt{2} + 1 \implies \\ b^2 = 2\sqrt{2} + 3.$ $c = b + 1 = \sqrt{2} + 2 \implies c^2 = 4\sqrt{2} + 6.$ $a^2 + b^2 = 2\sqrt{2} + 3 + 2\sqrt{2} + 3 = 4\sqrt{2} + 6.$ $a^2 + b^2 = c^2.$ And we deduce from this what exactly?  
July 2nd, 2017, 10:16 PM  #8  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,031 Thanks: 2342 Math Focus: Mainly analysis and algebra  Quote:
We also have methods for generating every single rightangled triangle with integer sides. Do you really think the world cares about a rightangled triangle with angles "close" to $45^\circ$? If you are going to use irrational numbers, you really ought to get comfortable with them.  
July 2nd, 2017, 10:36 PM  #9 
Banned Camp Joined: Jul 2010 Posts: 118 Thanks: 0  The number a (that selected,between zero and infinite)
Determines all the triangle data. b = 0.5 ( a^2  1 ) c = 0.5 ( a^2  1 ) + 1 circumference = a^2 + a aria = 0.25 ( a^3  a) height = ( a^3  a) : (a^2 +1) angle tg against a = 2a : ( a^2  1) half angle tg against a = 1 : a thanks 
July 2nd, 2017, 11:26 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 18,145 Thanks: 1418 
If b = (a²  1)/2, a² = 2b + 1, and so a² + b² = b² + 2b + 1 = (b + 1)² = c², satisfying Pythagoras. This isn't new. As a = 1 + √2 satisfies (a²  1)/2 = a, choosing a to be "close to" 1 + √2 ensures that b is also "close to" 1 + √2, and so the acute angles of the triangle are close to 45°. In particular, a = 2.4141 implies b = 2.413939405 and c = 3.413939405. To ensure the triangle has an angle of exactly 90°, the values of b and c shouldn't be rounded to fewer decimal places. 

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