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# Equilteral triangle proof/whatever? - can't find any online

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1. Is an equilateral triangle, with a height and edges that are all integers possible?

Take a right handed triangle (half an equilateral triangle) - if:
hypotenuse = 2x
width = x
and height of original equilateral = h

would it be possible for both x and h to be an integer? All the numbers i've tried haven't work but i've only tested a few 2 digit numbers

(*This isn't really specific to study help, but none of the other topics really seemed to fit.)
Is an equilateral triangle, with a height and edges that are all integers possible?

Take a right handed triangle (half an equilateral triangle) - if:
hypotenuse = 2x
width = x
and height of original equilateral = h

would it be possible for both x and h to be an integer? All the numbers i've tried haven't work but i've only tested a few 2 digit numbers

(*This isn't really specific to study help, but none of the other topics really seemed to fit.)
What level is this for? As in, what methods of proof have you been taught?
3. Consider an equilateral triangle with length, x. Find the height in terms of x and the answer should be obvious.
4. (Original post by offhegoes)
What level is this for? As in, what methods of proof have you been taught?
Just for fun, bc i'm a nerd, and I'm at GCSE at the moment - have been taught pythagoras + basic trigonometry stuff
5. Hint: If x=1, what's h? What does that tell you about h for other (integer) values of x?
Just for fun, bc i'm a nerd, and I'm at GCSE at the moment - have been taught pythagoras + basic trigonometry stuff
Can you see that, calling half the length of each side 'x', that the height will be "root 3 times x".

Root 3 multiplied by an integer will never be an integer.

So x and h can never both be integers.
7. h = 2x sin 60

h = rt3 x

This does mean that for both of these to be rational would be an impossibility.
8. (Original post by offhegoes)
What level is this for? As in, what methods of proof have you been taught?
Use Pythagoras' theorem with h and x as adjacent and opposite.

Assume h and x are integers.

You'll end up with h = +- xroot3

EDITED: Which implies either x is not an integer or h is not an integer.

There is a contradiction, as you originally assumed both were integers.

Therefore impossible
9. (Original post by Armpits)
Use Pythagoras' theorem with h and x as adjacent and opposite.

Assume h and x are integers.

You'll end up with h = +- xroot3

Which implies x is an integer but h isn't.

There is a contradiction, as you originally assumed h was an integer.

Therefore impossible
Oops yes forgot to square root the x sqaured!

But yes, that's pretty much as I said. Since OP hasn't studied methods of proof yes I wasn't really bothering with formal methods.
10. (Original post by Armpits)

Which implies x is an integer but h isn't.

There is a contradiction, as you originally assumed h was an integer.

Therefore impossible
That doesn't actually imply x is an integer but h isn't. All it does imply that one of x and h aren't integers.
11. What's more it shows you can't even have an equilateral triangle where the sides and height are both rational.
12. (Original post by Zacken)
That doesn't actually imply x is an integer but h isn't. All it does imply that one of x and h aren't integers.
Ah yes, my mistake.
13. thanks everyone for the responses
thanks everyone for the responses
It's a good question.

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