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**Milos****Member**- Registered: 2005-05-06
- Posts: 44

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.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

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 ....

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**Milos****Member**- Registered: 2005-05-06
- Posts: 44

Yes, the idea is good, you are on the right way. Now we need proofs.:)

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**Roraborealis****Member**- Registered: 2005-03-17
- Posts: 1,594

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?

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**Milos****Member**- Registered: 2005-05-06
- Posts: 44

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.

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**Mr T****Member**- Registered: 2005-03-30
- Posts: 1,012

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

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**Mr T****Member**- Registered: 2005-03-30
- Posts: 1,012

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

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

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**NIH****Member**- Registered: 2005-06-14
- Posts: 33

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.

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