[Cozi101]

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?

 

[IbanezroadstarII]

Working.....

 

[IbanezroadstarII]

sorry I dont think i can help.

 

[mesa_boogie_man]

Perhaps this link will help?

http://www.tsm-resources.com/alists/trip.html

 

[Big Job]

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]

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

51 squared is 2601. so the pattern is 1300 ans 1301.

 

[ZJH]

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.

 

[Cozi101]

Thanks Guys that really helped.

 

[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!

 

[Drew]

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]

(by the way, i have a feeling the fact the first two are merely one integer apart is purely coincidental)

 

[AirGuitarHero]

You have all the triads in mesa_boogie_man's link:

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

 

[VanWyck]

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...

 

[ZJH]

You have all the triads in mesa_boogie_man's link:

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

Voila, you have your first pythagorean triad. ;p

Just repeat the process and you'll have the rest in no time. ;p

 

[Drew]

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

 

[VanWyck]

Yeah, I couldn't stop trying to figure it out! Very addictive... I'm on call at the hospital tonite, and between trips to the ED I sat for almost an hour scratching furiously at some paper. Of course there was much pointing and giggling by the other residents.

Any more algebra? It's 2am and I still have six hours to kill...

 

[Cozi101]

Lol Sorry that's all my maths homework done. I posted it the night before the assignment was due and then checked this morning and finished it off with mesa's link and big job's explanation, Thanks guys.

 

[Devo]

I'm on call at the hospital tonite ....It's 2am and I still have six hours to kill...

perhaps you should choose your words more carefully!

 

[Big Job]

Your very welcome for the help! I'm glad I could be of assistance.

 

[The Euphor]

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

I'm sorry Drew. It was not meant to be serious. The numbers are logical though, but not in any way related to anything Pythagoras.