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