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#1 2005-05-07 21:30:48
- Milos
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Prove this
Prove that if the sides of a triangle are prime numbers its surface can not be whole number.
Try to work this out - it is interesting and not too difficult.
I will give you the possible solution in few days.
#2 2005-05-07 21:57:09
- MathsIsFun
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Re: Prove this
Thank you Milos, something for us to think about ...
I might be able to get this STARTED, because there is a formula for working out the area of a triangle from its three side, it is called "Heron's Forumula", after a Greek mathematican called Heron who lived about 2000 years ago.
The formula is:
Area of Triangle = sqrt( s(s-a)(s-b)(s-c) )
Where a, b and c are the lengths of the three sides, and s (a+b+c)/2 (ie s=half the perimeter of the triangle)
But I haven't figured what to do next ....
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#3 2005-05-08 01:34:18
- Milos
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Re: Prove this
Yes, the idea is good, you are on the right way. Now we need proofs.
#4 2005-05-08 18:40:24
- Roraborealis
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Re: Prove this
What if all the sides are 2cm (presuming it's an equilateral) then we do 2x2 divided by two and the answer is two? Two is a whole number.
School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?
#5 2005-05-09 04:03:13
- Milos
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Re: Prove this
If the triangle is equilateral then his angles are
180:3=60, and formula for surface you mentioned is for right-angled triangle.
If a=b=c=2 than P=(2xh)/2 . You can calculate h. cos30=h/a ; h=sqr3. So P=sqr3 - as you see this is not a whole number. This also could be the first proof.
But you should try to do it using Heron's formula which administrator mentioned.
#6 2005-05-11 07:59:17
- Mr T
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Re: Prove this
ye but we're stupid
I come back stronger than a powered-up Pac-Man
I bought a large popcorn @ the cinema the other day, it was pretty big...some might even say it was "large
Fatboy Slim is a Legend #7 2005-05-31 21:09:02
- Mr T
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Re: Prove this
why has no-one replied
I come back stronger than a powered-up Pac-Man
I bought a large popcorn @ the cinema the other day, it was pretty big...some might even say it was "large
Fatboy Slim is a Legend #8 2005-06-25 02:21:40
- mathsyperson
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Re: Prove this
Substitute in s = (a+b+c)/2:
A = sqrt( ((a+b+c)/2)((a+b+c)/2-a)((a+b+c)/2-b)((a+b+c)/2-c) )
Yuck.
Simplifying: A= sqrt( (a+b+c)(-a+b+c)(a-b+c)(a+b-c)/16)
Take out the denominator: A=sqrt ((a+b+c)(-a+b+c)(a-b+c)(a+b-c)/4
I think I've taken the wrong route, but anyway...
Multiply out the brackets: A= sqrt((-a^4+2a^2*b^2+2a^2*c^2+2b^2*c^2-b^4-c^4))/4
Quite a few things cancelled out there, or it would be even more horrible!
To prove that A isn't a whole number we need to prove that the horrible thing inside the brackets isn't 4 times a square number.
Anyone?
Why did the vector cross the road?
It wanted to be normal.
#9 2005-06-26 02:57:52
- NIH
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Re: Prove this
You're almost there, mathsyperson!
It's better to leave the expression as A^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c)/16. If a, b, c are all odd then the numerator is the product of four odd numbers, and hence is itself odd. So A^2 is not an integer, and neither is A. The same argument applies if two of a, b, c are equal to 2 and the remaining side length is odd.
Milos has already shown that if a = b = c = 2, then A = sqrt(3). If a = 2 and b, c are odd, then, to form a triangle, we must have b = c, with b > 1. (Try drawing a triangle with sides 2, 9, 11!) So we have an isosceles triangle with base 2 and height sqrt(b^2 - 1). b^2 - 1 is not a perfect square for b > 1, so the height is irrational, and hence also the area.
So we've proved a stronger result, a triangle with all sides lengths equal to either 2 or an odd number does not have integral area.
2 + 2 = 5, for large values of 2.