45 degrees optimum: geometric proof
In a previous post https://bbzippo.wordpress.com/2010/01/14/45-degrees-proof/ we saw that the 45 degree launch angle results in the maximum distance, because out of all rectangles having the same diagonal, the square has the maximum area. We used some basic algebra and the Pythagorean theorem to prove this fact. Now I’d like to present a purely geometric proof.
Again, let a and b be the sides of our rectangle and d be the diagonal. Let’s cut the rectangle along the diagonal, and take one half of it, which is a right triangle. Let’s arrange 4 copies of this triangle as shown in the picture. (You may recall this picture from one of the proofs of Pythagoras theorem). It is readily seen that the area of the triangle cannot be greater than 1/4 of the area of the square built on the diagonal, and that the maximum is achieved when the central square disappears, i.e. when a=b.
Now you don’t have any excuse for not teaching this to your third-graders.