 Pythagoras Theorem NEED HELP - Jemsite
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Old 05-29-2005, 11:02 AM Thread Starter

Join Date: Jan 2005
Location: Melbourne, Australia
Posts: 260 Pythagoras Theorem NEED HELP

Hey I've been on this question for so long on my maths assignment and I just can't get it and since everyone that buys jems must be smart I decided to give it one last go on the forum.

You need to continue the pattern:

3 4 5
5 12 13
7 24 25
11 __ __
13 __ __
15 __ __
17 __ __
19 __ __

Lol pretty sad that I'm asking you to do assignments for me but I'm really stuck, I also need to explain how I got it so I need that because my teacher will ask me how I got it and I won't know what to say. Thanks to math geniuses

P.S. What pathagorean triad starts with 51?
Cozi101 is offline

Old 05-29-2005, 02:35 PM

Join Date: Mar 2004
Posts: 405
Re: Pythagoras Theorem NEED HELP

Working.....
Old 05-29-2005, 02:43 PM

Join Date: Mar 2004
Posts: 405
Re: Pythagoras Theorem NEED HELP

sorry I dont think i can help.
Old 05-29-2005, 02:58 PM

Join Date: Feb 2005
Location: San Antonio, TX
Posts: 76
Re: Pythagoras Theorem NEED HELP

http://www.tsm-resources.com/alists/trip.html
mesa_boogie_man is offline
Old 05-29-2005, 03:15 PM

Join Date: Jun 2001
Location: South West England
Posts: 385
Re: Pythagoras Theorem NEED HELP

This could be crap but here is what I have figured out...

for the first like 3 squared is 9 and the four and the five add up to nine. 5 squared is 25 the 12 and the 13 add up to 25. And this pattern continues.

So forlike the 11 do 11 squared = 121. Then you have to get two numbers that are ONE DIGIT appart for the answer. So this would be 60 and 61.

The one for 13 would be 84 and 85.

This appears to work, but it could be utter tripe.

Big Job
Big Job is offline
Old 05-29-2005, 03:18 PM

Join Date: Jun 2001
Location: South West England
Posts: 385
Re: Pythagoras Theorem NEED HELP

And if the same thing applies for the triad then...

51 squared is 2601. so the pattern is 1300 ans 1301.
Big Job is offline
Old 05-29-2005, 03:20 PM

Join Date: Apr 2005
Location: Snoresville, CA
Posts: 148
Re: Pythagoras Theorem NEED HELP

Big Job has a thing goin' there, but it wouldn't be mathematically correct. It's a good strategy, but that's basically assuming that the shortest side is the hypotenuse of the triangle. His method works, but if you notice the first two, the first number is the "a," the second is "b," and the third is "c" in the Pythagorean Theorem (a²+b²=c²). Keep that in mind. I hope I helped. ZJH is offline
Old 05-29-2005, 06:21 PM Thread Starter

Join Date: Jan 2005
Location: Melbourne, Australia
Posts: 260
Re: Pythagoras Theorem NEED HELP

Thanks Guys that really helped.
Cozi101 is offline
Old 05-29-2005, 06:24 PM

Join Date: Jul 2001
Location: Trondheim, Norway
Posts: 3,614
Re: Pythagoras Theorem NEED HELP

3 4 5
5 12 13
7 24 25
11 48 49
13 70 71
15 96 97
17 126 127
19 169 170

I even calculated everything without any help! The Euphor is offline
Old 05-29-2005, 07:07 PM

Join Date: Nov 2001
Location: Somerville, MA
Posts: 6,200
Re: Pythagoras Theorem NEED HELP

Quote:
Originally Posted by The Euphor
3 4 5
5 12 13
7 24 25
11 48 49
13 70 71
15 96 97
17 126 127
19 169 170

I even calculated everything without any help! doesn't quite work - this is basically a series of values for the pythagorean theorem, a²+b²=c². 17 squared is 289, and 126 squared is 15876. Summed, these give us 16165, which definitely does NOT equal 127 squared, or 16129. Close, but not quite.

I almost posted something this morning, but honestly, it's been so long since I've done this stuff, that I have no idea how to solve it, lol. the upshot, I believe, is since you know a, you need to solve for b in terms of c, and then replace all instances of b with whatever it's equivalent in c is. From there, solve for C and then use the pythagorean theorem to give you your value for b.

I'll take a crack at it while cooking dinner, but my algebra is VERY rusty, lol.

-D
Drew is offline
Old 05-29-2005, 07:08 PM

Join Date: Nov 2001
Location: Somerville, MA
Posts: 6,200
Re: Pythagoras Theorem NEED HELP

(by the way, i have a feeling the fact the first two are merely one integer apart is purely coincidental)
Drew is offline
Old 05-29-2005, 09:36 PM

Join Date: Jul 2004
Location: China
Posts: 186
Re: Pythagoras Theorem NEED HELP

3 4 5
5 12 13
7 24 25
11 - 60 - 61
13 - 84 - 85
15 - 112 - 113
17 - 144 - 145
19 - 180 - 181

Big Job got it right too. Anyway, the numbers are a2 = b2 + c2 and the difference between b and c is just 1, so c=b+1. That gives you a2 = b2 + (b+1)2, therefore a2 = 2b2 + 2b + 1. You have "a", so just substitute and solve for b (like Drew said) What you are doing here is finding the Pythagorean triads in which c = b+1 and which correspond to the given 'a'

Now, you can be a smartass and give both solutions for each equation (the positive and the negative integers) Geometrically it makes no sense, but it is algebraically correct I wonder the point of this exercise. It doesn't exactly teach you much, does it? Oh yes, I know: find the solution on the internet, then work backwards and find the reason why LOL

Last edited by AirGuitarHero; 05-29-2005 at 09:43 PM.
AirGuitarHero is offline
Old 05-29-2005, 10:04 PM

Join Date: Nov 2002
Location: New Jersey
Posts: 412
Re: Pythagoras Theorem NEED HELP

I think I got it.
Solve the equation A^2 + B^2 = C^2 where C = B + 1 (the second and third integers in each example are one apart) so that A^2 + B^2 = (B+1)^2
then A^2 + B^2 = (B+1)(B+1)
then A^2 + B^2 = B^2 + 2B + 1
then A^2 = 2B + 1
At this point we can solve for any given "A"
If A=51 then 2601 = 2B + 1 or b=1300, and then c=1301

Now having a migraine...
I had been a math major before going in to medicine. Medicine is easier...
VanWyck is offline
Old 05-29-2005, 11:19 PM

Join Date: Apr 2005
Location: Snoresville, CA
Posts: 148
Re: Pythagoras Theorem NEED HELP

Quote:
Originally Posted by AirGuitarHero

3 4 5
5 12 13
7 24 25
11 - 60 - 61
13 - 84 - 85
15 - 112 - 113
17 - 144 - 145
19 - 180 - 181

Big Job got it right too. Anyway, the numbers are a2 = b2 + c2 and the difference between b and c is just 1, so c=b+1. That gives you a2 = b2 + (b+1)2, therefore a2 = 2b2 + 2b + 1. You have "a", so just substitute and solve for b (like Drew said) What you are doing here is finding the Pythagorean triads in which c = b+1 and which correspond to the given 'a'

Now, you can be a smartass and give both solutions for each equation (the positive and the negative integers) Geometrically it makes no sense, but it is algebraically correct I wonder the point of this exercise. It doesn't exactly teach you much, does it? Oh yes, I know: find the solution on the internet, then work backwards and find the reason why LOL
Actually, your algebra's kinda off. ;p

a² does not equal b²+c². Your derivation would actually be "a²=c²-b²"

And since the pattern indicates that c=b+1, you could easily solve for the other two integers given the value of a. Deriving a formula using that given value...

a²=(b+1)²-b²
=b²+2b+1-b²
=2b+1

Thus, if you used the a=3 triad...

a²=2b+1
3²=2b+1
9=2b+1
8=2b
b=4

Substituting that into c=b+1...

c=4+1=5

Just repeat the process and you'll have the rest in no time. ;p
ZJH is offline
Old 05-30-2005, 01:12 AM

Join Date: Nov 2001
Location: Somerville, MA
Posts: 6,200
Re: Pythagoras Theorem NEED HELP

Well, airguitar's algebra might be off, but at least the numbers check out this time - go team. depressingly, I actually thought that was kinda fun, lol
Drew is offline

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