Prove that if you draw a right triangle and then draw a circle with its center on the hypotenuse's midpoint such that it intersects at least 1 vertex, it will in fact intersect all three.
Consider any right triangle ABC, where B is the right angle and the center of the circle lies at the midpoint of AC.
Rotate a copy of the triangle 180 degrees about this midpoint and you now have a rectangle with its diagonals crossing at the center of the circle. Obviously the distance from this point to any of the four corners of the rectangle is equal and therefore if any one of the corners is on the circle, they all must be.
I know this problem is old, and this isn't a fancy mathematical proof, but it just seemed such an easy, intuitive solution I had to add it to the others. ;)
Posted by tomarken
on 2006-03-02 12:39:54