Pythagoras (Middle School)

Greeks apparently got a lot of their geometric insight playing with tiles (their version of blocks).  For example, one ancient demonstration of Pythagoras' proof was to take any right triangle and make 4 copies of it.

These 4 triangles can be arranged in a big square shape in the two ways shown below. 

Clearly, the sides of each big square are of equal length (a + b) and so must have equal area.  Also, taken together, the 4 triangles take up the same area in each big square.  So if we subtract out the triangles, the area of colored square "C" must equal the combined areas of the colored squares "A" and "B".  That is to say “area of square C” = “area of square A” + “area of square B”.  But "C" is a square with sides of length "c" and "B" is a square with sides of length "b" and "A" is a square with sides of length "a", so c² = a² + b². 

The key thing to note here is that the "insight" to solving this problem is not to think about two column geometric proofs, but to understand the implications of the patterns you see when you play with tiles.

Oxen, Beware!

Pythagoras sacrificed an ox when he discovered this.  I have to admit it's a pretty neat trick, but I don't think I'd sacrifice an ox over it.

Acknowledgements:

This problem was (very briefly) described here.

Update:

This solution is is also described on page 154 of Martin Gardner's Sixth book of Mathematical Diversions.  He mentions that the proof may predate Pythagoras himself, in that the figures appear in the Chou Pei, a Chinese manuscript that goes back to the Han dynasty (202 BC to 220 AD).