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# Prove this 'simple' Special isosceles right triangle relationship.

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posted on Mar, 1 2012 @ 02:18 PM
One of my grandchildren came to me and asked me to prove a simple geometry question.

background:

A right triangle is any triangle with a 90 degree angle in it. Any right triangle can be solved using the Pythagorean theory easily. In one case, If you know 2 sides, you can easily compute the length of the other side by using the formula a^2 +b^2 = c^2. So a right triangle like a=3, b=4 and solving for the hypotenuse becomes 9+16 =25, take the square root of 25 and you have the answer, 5.

An isosceles triangle is any triangle with 2 sides the same.

A Special isosceles right triangle, the 45-45-90 , has two 45º angles and one right angle.
You can create 2 of them by taking a perfect square, say a 5 inch on a side square and divide it diagonally. You wind up with 2 Special isosceles right angle triangles, each with a 45-45-90 angle. 2 of the sides , a and b, will be length=5, so solving this for c would be: a =5*5 plus b=5*5 = 50, so c= the square root of 50 =7.07106781

The Problem:

So, given that, prove that there are no 45-45-90 triangles where a and b are integers, and the length of c calculates to be an integer.

I thought it would be simple, but it took some time. I have the answer, and will post it soon, but thought some of you might have fun with this.

edit on 1-3-2012 by charlyv because: correct a typo

posted on Mar, 1 2012 @ 02:28 PM

Originally posted by charlyv
So, given that, prove that there are no 45-45-90 triangles where a and b are integers, and the length of c calculates to be an integer.

The simple thought that comes to mind is that since a and b are the same length, I'm just going to call them both the same name, and this name and its length is "1".
The c length is therefore the square root of 2.

Since the square root of two is an irrational number then it really doesnt matter at all how much you scale up this triangle, because it will always be some multiple of an irrational number, which in itself is an irrational number.

Anyway, thats my thoughts after being awake for nearly 24 hours.

edit on 1-3-2012 by alfa1 because: (no reason given)

posted on Mar, 1 2012 @ 02:32 PM
I have to agree with alpha1. He just beat me to typing it out.

I am now very curious to see your "solution", because the preliminary math is showing it not to be possible.

Edit: Misread the OP's last sentence. I thought he said he found a solution. Alpha1 proves this quite nicely.
edit on 2012/3/1 by TLomon because: (no reason given)

posted on Mar, 1 2012 @ 02:50 PM

Originally posted by alfa1

Since the square root of two is an irrational number then it really doesnt matter at all how much you scale up this triangle, because it will always be some multiple of an irrational number, which in itself is an irrational number.

I just spent 30 minutes trying to figure out a way to concisely type that up....
Couldnt quite remember my HS trig well enough that it didnt sound like jibber jabber when i was done so I just closed the darn thread haha

Kudos!
edit on 1-3-2012 by youdidntseeme because: (no reason given)

posted on Mar, 1 2012 @ 03:14 PM
A pythagorean triple is a right triangle with all integer sides like a=3, b=4, c=5.

For a right triangle to be a pythagorean triple it must have the form: ka^2 + kb^2 = kc^2 with a,b,c = to existing pythagorean triple such as (3,4,5).

So another one would be: k=2; 2(3)^2 + 2(4)^2 = 2(5)^2 which is a=6, b=8, c=10.

You can see that a and b can never equal each other, so a 45,45,90 isosceles would not satisfy.

Also Euclid's formula: a = m^2 - n^2, b = 2mn, c = m^2 +n^2 where m > n
m^2 - n^2 cannot equal 2mn, thus a cannot equal b.
edit on 1-3-2012 by EnlightenIgnorance because: Euclid's

posted on Mar, 3 2012 @ 06:03 AM
Cool,
Alpha1 gets the prize. And very fast after the post.
There are no 45-45-90 triangles where a and b are integers, and the length of c calculates to be an integer.

Proof:
In a triangle like this, the length of the hypotenuse (c) always has the value a * SQRT(2). (or b * SQRT(2) as it does not matter since a always =b). Since the square root of 2 is an irrational number, anything multiplied by it is irrational as well, which can never produce an integer result.

Thanks,

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