Monday, March 01, 2010

Cute "proof" of Pythagoras' theorem

[For reasons I have yet to fathom, the ASCIIMathML plugin doesn't want to work for this page...]

I'm reading The Mathematical Mechanic by Mark Levi, a tour of various mathematical results that can be obtained much more directly (albeit without proof) by appealing to our mechanical intuition.

Here's a beautiful proof from the book of Pythagoras' theorem, `a^2 + b^2 = c^2` where `a` and `b` are the opposite and adjacent sides of a right angled triangle and `c` is the hypotenuse.

Consider an ice skater of mass `m` on a perfectly smooth ice rink. The skater starts in the South West corner of the rink and pushes off against the South wall; the skater is now moving North with velocity `a` and hence has kinetic energy `ma^2/2`. Next the skater pushes off against the West wall and adds an Easterly component `b` to their velocity. The energy acquired from this second push is `mb^2/2`. Now, the skater's overall velocity is `c` and overall kinetic energy is `mc^2/2`. But this must be the sum of the energy acquired from each push, hence `ma^2/2 + mb^2/2 = mc^2/2`. Cancel the `m/2` terms and you have Pythagoras' theorem!

2 comments:

steve said...

Is momentum conserved? If it is then this is not a right angle, since the skater will still have a Northerly component after the push from the West wall. If the push from the wall stops the Northerly momentum, then where is the energy for this coming from in the equation, which is assuming kinetic (hence total) energy is conserved?

Rafe said...

Sorry, this is wayyyy late (I didn't realise anyone was reading this!)

The right angle comes from the components of the final velocity vector: a m/s North and b m/s East, with resulting velocity c m/s.

Momentum doesn't feature here because we're assuming the wall is immobile (i.e., the skater starts with zero KE, then adds some from his push off to the North, then adds some more from his second push off to the East).