jeremiahsmith wrote:

Can't be done.

Let's assume that the lengths of the two equal sides are given by a, and the hypotenuse is b. Since this is a right triangle, by the Pythogorean theorem, a^2 + a^2 = b^2. Thus 2a^2 = b^2. If we take the square root of both sides, we get sqrt(2) * a = b, where sqrt is the square root function. Sqrt(2) is an irrational number, so if b is an integer, a can't be, and vice versa.

Should I point out the flaw here? My maths isn't up to scratch, but as I recall irrational numbers can exist, they simply can't be expressed with perfect accuracy. You've proven that a right-angled triangle cannot be an iscosceles triangle, which isn't true. sqrt(2) is a ratio, here. If you multiply a rational number by an irrational number, the result can be either. How on earth does a number being irrational preclude its use as a ratio? The ratio is (sqrt)2 : 1.

The most famous example is the irrational number Pi. You've proven that a circle

cannot exist. Pi is an irrational number, defined as the ratio between the diameter of a circle and its perimeter.

2Pi r = P

but Pi is irrational . . . follow?

Pi * D = P

The ratio would be

Pi : D = 1

But both the diameter and the perimeter of a circle exist. It looks like you're confusing the properties of irrational and *imaginary* numbers, to me.

Here's a working proof, however:

given that a^2 + b^2 = c^2, and a = b = c, we have

2a^2 = a^2

this statement is only true at a = 0, at which time the shape described is not a triangle. It's a point.

This of course assumes we're dealing with plane geometry . . . I wouldn't have a clue how to deal with anything else