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Pythagorean Soliloquyby Total Gadha - Wednesday, 31 December 2008, 02:07 PM

Years ago, my Italian language teacher and my classmates learning Italian used to question me how I couldpick up the language so fast. “Perche sono un insegnante di matematica,” was my standard answer. Thatwould leave them baffled. What was the relation between language and mathematics? Little do my classmatesor many other students, especially our CAT aspirants, understand that language and mathematics are the twosides of the same coin; both consist of symbols connected through logic. And more often than not, even thelogic they follow is the same. The symbols for both were gradually developed. The mathematician Euler wasresponsible for our common, modern-day use of many famous mathematical notations—for example, f(x) for afunction, e for the base of natural logs, i for the square root of –1, Π for pi, Σ for summation. Shakespeare iscredited with coining nearly 1700 words in the English language. These are our symbols. The logic was alsodeveloped gradually; the grammar, mathematical proofs, critical reasoning, logical reasoning, etc. form theparts of logic that bind those symbols. Therefore, if you are a logical creature, chances are that you would notface difficulty understanding either of these subjects, provided you are comfortable with the symbols- English

vocabulary or mathematical notations, whatever the case maybe.

The progression of thoughts of a writer is not markedly different from that of a mathematician. While reading a good passage one canautomatically guess the next paragraph. In the same manner, a mathematician proceeds from one thought to another in a logical manner,removing obstacles in his path to discovery and finding answers to the problems one after another.

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Re: Pythagorean Soliloquyby Gowri Nandana - Wednesday, 31 December 2008, 02:22 PM

Awesome article........

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Re: Pythagorean Soliloquyby Gowri Nandana - Wednesday, 31 December 2008, 02:35 PM

Happy New year to the super TG team.

May God gift u with 48 hours a day to manage ur wonderful site

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Re: Pythagorean Soliloquyby Dagny Taggart - Thursday, 1 January 2009, 05:28 PM

A very Hppy New Year to you too, Gowri.

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Re: Pythagorean Soliloquyby Gul Gul - Thursday, 1 January 2009, 07:19 PM

One of the best ....!

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Re: Pythagorean Soliloquyby sandeep somavarapu - Thursday, 1 January 2009, 08:55 PM

Sir ji...What an explantion sir ji.....------------one concept can change your thought processI wish you and TG family a very happy New year.

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Re: Pythagorean Soliloquyby rajat shukla - Friday, 2 January 2009, 08:28 PM

happy new year...to all of u..

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Re: Pythagorean Soliloquyby rashi agarwal - Friday, 2 January 2009, 11:59 PM

hello TG sir, A very happy New Year to you...Great article......

sir, i hav a little confusion.To calculate the no. of right triangles with a given length N as one of its leg,

N= 2a x pb x qc x .................

The formula is [(2a -1)(2b+1)(2c+1)....-1 ]/ 2

But sir if we want to calculate with 25 as the length of the leg. then a =0. then whole term will be -ve. this method will not be used for the oddnumbers..please sir correct me if I am wrong.

rashi

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Re: Pythagorean Soliloquyby Total Gadha - Saturday, 3 January 2009, 02:18 AM

Hi Rashi,

Don't consider the term 2a - 1 if N is an odd number, i.e. no power of 2 is present.

Total Gadha

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Re: Pythagorean Soliloquyby Dipanjan Biswas - Sunday, 4 January 2009, 11:48 PM

Thank you sir a lot for such a nice article..It really fosters neural network of thoughts..Simultaneously it covers up number system...Thanks onceagain..Happy New Year to all of TG familyTG rockzzz

Dipanjan

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Re: Pythagorean Soliloquyby Tuhin Banerjee - Monday, 5 January 2009, 06:24 PM

Hi TG,

Really a awesome article.

Thanks a zillion mam.

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Re: Pythagorean Soliloquyby gadha abc - Monday, 5 January 2009, 06:49 PM

Each sentence speaks the intelligence..Happy new TG sir and Mam.

Sir, can u please little elaborate how that formula u derive N+5C5 for the dice problem.

Regards

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Re: Pythagorean Soliloquyby himanshu mishra - Sunday, 11 January 2009, 10:49 PM

hi,tg

This article is superb...........all the comments are useless for this nugget of wisdom.

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Re: Pythagorean Soliloquyby Sharadha Kuntumalla - Monday, 26 January 2009, 12:57 AM

Hello TG Sir,

I want to copy the article in a word document so that I can print it, highlight important points and keep revising it. But the content is not gettingaligned properly on the word document. I won't find this difficulty when I am copying verbal articles/exercises. Can you please help me with aprintable version of these wonderful articles.

Shadh

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Re: Pythagorean Soliloquyby Total Gadha - Monday, 26 January 2009, 11:09 AM

Hi Sharadha,

Many of these articles are images. Right click on them, 'copy' them and paste them in paint and then save them as images. Then you can printthem out.

Total Gadha

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Re: Pythagorean Soliloquyby sumit jamwal - Thursday, 29 January 2009, 12:50 AM

hi sir,as you stated in the articleif i take 25 As hypotenuse

then there are

25=5^2

so ((2*2+1)-1)/2 = 2bt i cud find only

5^2=3^2 +4^2

confused

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Re: Pythagorean Soliloquyby Total Gadha - Thursday, 29 January 2009, 10:25 AM

Hi Sumit,

You are taking 5 or 25 as hypotenuse? If you are taking 25, then you need to take 625 = 252.

Total Gadha

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Re: Pythagorean Soliloquyby Anurag Rai - Sunday, 1 February 2009, 02:06 PM

Nice and Brilliant

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Pythagorean Soliloquyby Abhishek Bansal - Monday, 2 February 2009, 02:17 AM

As i was attending a class on Geometry. I was told that there there is one triplet such as (20,21,29). I started searching and wanted to find away to get to the number of triplets if i am given one of the sides.

A side can be the longest side (i.e. hypotenuse) or base/perpendicular.





























i will say its by sheer coincidence, that someone told me about this site today when i was taking the class - and in that class i took the case ofone of the sides being 5^11.

i knew a way to find when 5^11 is not the longest side or any number for that matter when it is not the longest side..

if we are looking for integral solutions then (a-b)*(a+b) should be even or odd...

so 11 triangles are possible.

I wrote a code in Matlab (just to ensure that i am not the one doing the calculation) - i took 5^(n) ... n starting from 1,2,3,..... 12,13...

at n = 15 it started showing some errors related to memory.

I really need to check the code, and the results.. because at n =12 it gives me 16 triangles which are possible when 5^12 is the hypotenuseand 5^13 gives me 60 triangles.

what i saw till 5^11.. the answer will be the power itself.

and 3^n and 7^n won't give me any triangles.

and 31^n and 37^n.. will start giving me triangles for some value of n > 2,3

so can u check the validity.

I can be wrong. i just have my code and the results and those i will post tomorrow.

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Re: Pythagorean Soliloquyby Abhishek Bansal - Monday, 2 February 2009, 02:42 AM

It is quite possible that the number suddenly increases because of the precision errors.

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Re: Pythagorean Soliloquyby Harish Bansal - Wednesday, 4 February 2009, 12:22 AM

Hi TG,

I tried to derive the same formula that u have written at the top by using the same example of 60^2. I got a different result.

My result is: {(a-1)(b+1)(c+1)(d+1)....-1}/2

Please clarify

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Re: Pythagorean Soliloquyby sumit jamwal - Wednesday, 4 February 2009, 08:44 AM

Thanks TG ..my mistake

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Re: Pythagorean Soliloquyby Total Gadha - Wednesday, 4 February 2009, 09:25 AM

Hi Harish,

Maybe if you can explain how you derived the formula, I can point out the mistake.

Total Gadha

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Re: Pythagorean Soliloquyby alsadra @TG - Monday, 16 February 2009, 04:01 PM

sir i 've been reading this article for more than 90 minutes. had to imagine a lot. can u please explain me how u calculated the no of ways of

expressing 65 as a sum of two squares towards the end of the article.

the article is very informative and very dense in information. thanks sir...

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Re: Pythagorean Soliloquyby srihari sankararaman - Thursday, 19 February 2009, 05:49 AM

Hello Sir,

Sheer brilliance... I was not able to prove the last theorem. Could you tell me how you came to this conclusion.(n+5)c5?

Once again,Kudos to TG.

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Re: Pythagorean Soliloquyby syed haque - Wednesday, 25 February 2009, 05:53 PM

Hi TG,

Can you please explain why you have divided the number of factors by 2?

As two different pairs can give different values of a and b and will result in two right triangles.Sir please explain me this.

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Re: Pythagorean Soliloquyby amrit jajodia - Sunday, 1 March 2009, 12:28 AM

Hi,Just one word for the article ..... Awesome!!!!!

I am not clear about the no of distinct outcome formula.Just for the case if N = 2 then no of instinct outcome possible is nos from 2 to 12. that is 11 outcomes but with the formula i get 21. Can youplease clarify on this.

And moreover can you please put up the answer for those three questions also.

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Re: Pythagorean Soliloquyby Abhishek Bansal - Sunday, 1 March 2009, 10:46 AM

you are talking about the sum... when you see 2 to 12....

see the total number of outcomes when you throw two dice will be 36 if the dice are not identical to each other....

1,2 is different from 2,1

but when you throw two identical dice together....

1,2 is same as 2,1

there is no difference as you won't be able to distinguish...

so those kind of cases will be 21...

1,1 2,2 3,3 4,4 5,5 6,6 if you take this cases out of the sample space.. there are 30 more cases left... now these 30 cases have (p,q) and (q,p)where p is not equal to q.

so these cases are counted twice....

30/2 + 6 = 15 +6 = 21cases

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Re: Pythagorean Soliloquyby yogesh bansal - Tuesday, 23 June 2009, 02:47 PM

hi...answer of first 2 questions out of last 3 questions...

1.) 48

2.) 27

Am i correct???

Tg plz let me knw..Show parent | Reply

Re: Pythagorean Soliloquyby Mohit Bhambri - Friday, 4 September 2009, 10:06 PM

























1) 48since we have to get 10 distinct [(2a-1)(2b+1)...-1]/2 =10 which gives (2a-1)(2b+1)...= 21let it be 3 and 7 taking 7 for powers of 2 and 3 for powers of 3 we get 2^4*3 =48 ..pls confirm

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2) I too think it is 2^7. For a number to be hypotenuse it must have a prime no of form 4n+1 in its factors. So if we start with a number of form5x(..) so we have to get remaining 5 triangles from side as 5x(..) which means[(2a-1)(2b+1)(2(1)+1)-1 ]/2 =5 cause we already let 5 be a factor thats y the 2(1)+1*if there is a power of2 which gives 3x something =11 not possible hence having a single 4n+1 factor is ruled outsimilarly having 2 4n+1 factors is ruled out i.e 13, 5 at the same time therefore only possibility to have 6 triangles is with it as SIDE ONLY nothypotenuse.

by logic of 1) we get N= 2^7Correct me if im wrong

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There is some problem in framing question 3 as it doesnt specify 3 traingles having integral sides of equal areas.

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Re: Pythagorean Soliloquyby srinivasan ravi - Friday, 11 September 2009, 12:20 AM

hi mohit,can u explain the 2nd question more clearly..thanks..

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Re: Pythagorean Soliloquyby Raju Singh - Monday, 28 September 2009, 02:34 AM

@ Mohit

Could we hav some simple approach for sol of 2 Question asked:Smallest number for 6 distinct right triangle for integer sides:

[{(2a-1)(2b+1)(2c+1)}...-1]/2=6(2a-1)(2b+1)(2c+1)=2x6+1(2a-1)(2b+1)(2c+1)=13

As RHS is 13 which is prime number so any of LHS term must have 13 as one of the number.

For number to be minimum only (2a-1) must be equal to 13 and rest all terms as 1Therefore,(2a-1)=13, (2b+1) =1 and (2c+1)=1 etc..hence a=7 and b=c=...=0And thereby 2^7 as the smallest number for rt triangles with integer side lengths.Mohit what u hv elaborated I'm not getting plz clarify it.

Thanks

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Re: Pythagorean Soliloquyby Deepak Kumar - Saturday, 9 October 2010, 12:31 PM

Hi TG Sir,

Really a great artical. I liked the way you have written the artical, logically coming to the next step, the way you told in the preface of the artical.

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Re: Pythagorean Soliloquyby abhishek abhi - Saturday, 1 October 2011, 08:46 PM

sir in the first problem ,,to find out odd pairs 3*3 is done ,,sir could you please explain me how to find odd pairs in those factor pairs

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Re: Pythagorean Soliloquyby Just Gadha - Wednesday, 5 October 2011, 10:58 AM

How do u get the total no. of solutions of the equation a1+a2+a3+a4+a5=Nas N+5C5

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Re: Pythagorean Soliloquyby Kamal Joshi - Friday, 7 October 2011, 05:40 AM

Sir,

Please explain, how did you derive (2^2+1^2)(2^2+3^2)(2^2+1^2)(2^2+3^2) = 63^2+16^2 = 33^2+56^2???

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Re: Pythagorean Soliloquyby Kamal Joshi - Friday, 7 October 2011, 04:22 PM

Ok got it, sir!!! You derived different combinations of 65^2 by (a^2+b^2)(c^2+d^2)= (ac+bd)^2+(ad-bc)^2 = (ac-bd)^2+(ad+bc)^2. Correct?

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Re: Pythagorean Soliloquyby prabhat taneja - Wednesday, 17 October 2012, 05:38 PM

one word.. SUPERB.. I liked the article overall, but the footnote made my day.. Thanks to Mr Mahajan for the unique solution and TG for such

fabulous a post.

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